Electrochemistry

Electrochemistry:

Electrochemistry deals with the relationship between electrical energy and chemical energy and the interconversion of these two forms of energy. Chemical reactions can produce electrical energy. Also, certain nonspontaneous chemical reactions can be driven in the forward direction by applying an electric current.

Both these processes involve electrochemical reactions, i.e., chemical reactions using or generating electricity. The batteries we use in cars, toys, mobile phones, etc., are all examples of electrochemical cells. Methods like electroplating, employed to prevent corrosion, and electrorefining of metals involve electrochemical reactions.

Metallic and Electrolytic Conductance

Substances that allow the passage of electric current are called conductors. Metals are good conductors of electricity. Examples include copper, aluminum, and silver.

Apart from metals, there are other substances that allow the passage of electric current through them when they are in the molten state or in an aqueous solution. These substances are called electrolytes and the species responsible for the flow of current in electrolytes are the ions produced by the dissociation of these substances. This is in contrast to metallic conduction, in which the flow of current is due to the movement of free electrons.

The conduction of electricity by ions in solution is called ionic conduction. The conductivity of a solution depends upon various factors the

1. Nature of the electrolyte
2. The size of the ions in the solution,
3. The nature of the solvent and its viscosity
4. The temperature and concentration of the electrolyte.

On the other hand, metallic conduction occurs through electrons and is therefore also known as electronic conduction.

It depends on:

1. The nature (density) and structure of the metal
2. The number of valence electrons per atom
3. Temperature.

Summarises the important characteristics of metallic and electrolytic conduction.

A comparison of metallic and electrolytic conduction:

Substances like sucrose, which do not conduct electricity either in their molten state or in their aqueous solutions, are called nonelectrolytes.

Conductance and Conductivity

Electrolytic conduction can be studied by applying a potential difference between two electrodes dipped in the solution of a sample. The ions of the sample which is an electrolyte move across the electrodes and result in the flow of electric current through the solution.

Under this condition, the electrolyte obeys Ohm’s law, according to which the potential difference between the electrodes, V, is equal to the product of the current I flowing through the solution and the resistance offered, R.

V=IR.

The reciprocal of resistance is called conductance, G, of the electrolytic solution.

\(G=\frac{1}{R}\)

As resistance is expressed in Ω, conductance can be expressed in Ω. The SI unit of conductance is Siemens, S.

\(1 S=1 \Omega^{-1}\)

Specific resistance and specific conductance

Consider a uniform bar of a conductor of length 1 cm and cross-sectional area A cm². Imagine that the cross-section is rectangular and that the bar is divided into cubes of side 1 cm.

The resistance of the bar is then equal to that of the I layers, in series with one another. Each layer is equivalent to A cubes, each cube of length 1 cm, whose resistances are in parallel. If p is the specific resistance (resistance of 1 cm cube) of the conductor, the resistance of one layer may be worked out from

\(\frac{1}{r}=\frac{1}{\rho}+\frac{1}{\rho}+\) ………. (A times)

Or \(\frac{1}{r}=A\left(\frac{1}{\rho}\right)\)

Or \(r=\frac{\rho}{A}\)

If R is the resistance of the whole bar, equivalent to the resistance of I layers, then

\(R=l r=l \cdot \frac{\rho}{A}=\rho \cdot \frac{l}{A} \Omega\)

This equation is valid for all types of conductors including electrolytic solutions. In this equation, p is called resistivity (or specific resistance).

Substances can be classified as conductors, semiconductors, or insulators based on their resistivities. If I is expressed in cm and A in cm², the unit of p will be 2 cm. The SI unit is ohm m (22 m). The reciprocal of p is called specific conductance, or conductivity, denoted by K.

\(\kappa=\frac{1}{\rho}\)

Resistivities of some materials:

Equation 3.2 can now be written as

\(R=\frac{1}{\kappa} \cdot \frac{l}{A}\)

The quantity \(\frac{l}{A}\) is called cell constant and is denoted by G*.

\(G^*=\frac{l}{A}\)

Equation 3.3 can now be written as

\(R=\frac{G^*}{\kappa}\)

or, \(G^*=R \kappa\)

Taking the reciprocal of Equation 3.2, we get

\(\frac{1}{R}=\frac{1}{\rho} \cdot \frac{A}{l}\)

or, \(G=\kappa \cdot \frac{A}{l}\)

From Equation 3.4 it may be seen that the conductance G of a solution decreases with 1 and increases with A.

The SI unit of K may be obtained as follows.

\(\mathrm{\kappa}=G \cdot \frac{l}{A}=\frac{1}{R} \cdot \frac{l}{A}=\frac{1}{\Omega} \cdot \frac{\mathrm{m}}{\mathrm{m}^2}=\Omega^{-1} \mathrm{~m}^{-1}=\mathrm{S} \mathrm{m}^{-1}\)

Lists the conductivity values of some electrolytes:

Conductivity may be defined as the conductance of a solution held between electrodes of a unit area of cross-section (1 cm² or 1 m²) and unit distance apart (1 cm or 1 m). It may be noted that

\(1 \mathrm{~S} \mathrm{~cm}^{-1}=1 \frac{\mathrm{S}}{\mathrm{cm}}=1 \frac{\mathrm{S}}{10^{-2} \mathrm{~m}}=100 \mathrm{~S} \mathrm{~m}^{-1}\)

Measurement Of The Conductivity Of Ionic Solutions

You must have studied in your physics lessons that the unknown resistance of a solid conductor can be measured using a Wheatstone bridge. However the resistance of an ionic solution cannot be measured by using a Wheatstone bridge because a solution cannot be connected to the bridge like a metallic wire.

Therefore, a specially designed conductivity cell is used to measure the electrical resistance of a solution of known concentration. Unlike in the Wheatstone bridge, where d.c. is used, in a conductivity cell, alternating current is used to minimize redox reactions.

Electrode reactions are prevented in an a.c. field since the electrodes change polarities very fast under the operating frequency. The reaction which occurs in the first half of the cycle is reversed in the second half.

The conductivity cell (in which the electrolyte is taken) consists of two platinum electrodes coated with platinum black. The electrodes have a cross-sectional area A and are a distance / apart.

In the conductivity cell, two known fixed resistances (R_x and R₂) and a variable resistance R₃ are arranged.

Tire current may flow along ABC and ADC as long as the potentials at B and D are different. The galvanometer will show a deflection. Once the two are balanced, no current flows and the galvanometer shows no deflection (null deflection). The resistance is adjusted to obtain null deflection on AB and at this point,

\(I_1 X=I_2 R_1 \text { and } I_1 R_3=I_2 R_2\)

where X is the resistance of the electrolyte solution in the conductivity cell.

Hence \(\frac{R_3}{X}=\frac{R_2}{R_1}\)

\(X=R_3 \cdot \frac{R_1}{R_2}\)

Instruments based on the above setup are called conductivity bridges. Nowadays instruments which directly give the conductivity value are also available.

They are called conductometers. There are many types of conductivity cells, varying in size and shape, but they can broadly be divided into two categories the solution is poured into the cell, which acts as a vessel, and another dip-type cell in which the cell is dipped inside a beaker containing the electrolyte solution.

It is customary to determine the cell constant of the cell being used before measuring the conductance of the solution under study. For this purpose, the conductance of a solution of a known concentration of potassium chloride is measured.

For example, let us say the conductance of a 0.1 M KCl solution is observed to be 0.023 S. The specific conductance of this solution is noted from the literature and the cell constant is then calculated using the equation.

\(\frac{l}{A}=\frac{\kappa}{G}\)

The cell constant for the cell in the above example, therefore, is found to be

\(\frac{l}{A}=\frac{\kappa}{G}=\frac{0.01167}{0.023}=0.5 \mathrm{~cm}^{-1}\)

Gives the conductivity values of some standard solutions of potassium chloride.

Specific conductance (conductivity) of KCl solutions at 20°C for various concentrations:

Molar conductivity

If the solution of one electrolyte is more concentrated than that of another, for a fixed volume of a solution held between the electrodes of a conductivity cell, the former will have more ions, and hence its conductivity may be higher.

To compare the conductances of different electrolytes, it was necessary to define another quantity, namely molar conductivity A. It is defined as the electrolytic conductivity к divided by concentration c.

Molar conductivity,

\(\Lambda_m=\frac{\kappa}{c}\)

If the cell constant is known, the conductivity of any solution can be measured by filling the conductivity cell with the solution and measuring the resistance. Conductivity depends on the nature of the material.

The SI unit of A is S m² mol-1. If x is expressed in S m1, concentration should be expressed in mol m3. If c is expressed in mol L-1 (M), the formula becomes

\(\Lambda_m=\frac{\kappa\left(\mathrm{S} \mathrm{m}^{-1}\right)}{c\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)}\) \(\frac{\kappa}{c} \cdot \frac{\left(\mathrm{S} \mathrm{m}^{-1}\right)(\mathrm{L})}{(\mathrm{mol})}\)

\(\frac{\kappa\left(\mathrm{S} \mathrm{m}^{-1}\right)\left(\mathrm{dm}^3\right)}{c(\mathrm{~mol})}\)  [… 1L = 1 dm3]

\(\Lambda_m=\frac{10^{-3} \kappa}{c} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}\)

The CGS unit of A is S cm2 mol1. The two types of units are related to each other as

\(1 \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}=10^4 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

And \(1 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}=10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}\)

Example 1. The conductance of a solution of acid as read from a conductometer is 5×10 S. If the conductance of a 0.1 M KCl solution using the same cell and conductometer is found to be 1.07 x 10³ S, calculate the conductivity of the acid solution. The conductivity of a 0.1 M KCl solution is 0.001289 S cm¹.
Solution:

Given

The conductance of a solution of acid as read from a conductometer is 5×10 S. If the conductance of a 0.1 M KCl solution using the same cell and conductometer is found to be 1.07 x 10³ S

Whenever conductance has to be measured, first the conductance of a standard solution of KCl is determined and the cell constant is calculated. Then the conductance of the sample solution is determined and multiplied by the cell constant to obtain its conductivity.

\(\text { Cell constant } G^*=\frac{\text { conductivity } \mathrm{\kappa}}{\text { conductance } G}=\frac{0.001289}{1.07 \times 10^{-3}}=1.2 \mathrm{~cm}^{-1}\)

Now conductivity of acid solution = K=5×10 -3 x 1.2=6×10 -3 S cm -1.

Example 2. The conductance of a 0.10 mol L-1 solution of NaCl was found to be 6.13×1032^-1. Calculate its molar conductivity if the cell constant is 1.5 cm¹.
Solution:

Given

The conductance of a 0.10 mol L-1 solution of NaCl was found to be 6.13×1032^-1.

For calculating molar conductivity, the conductivity has to be calculated.

K = \(G \cdot G^*=4 \times 0.005=0.02 \mathrm{~S} \mathrm{~m}^{-1}\)

\(9.195 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1} \text { or } \mathrm{Scm}^{-1}\)

Molar conductivity,

\(\Lambda_m=\frac{\kappa \times 1000}{c(\text { in M })}=\frac{9.195 \times 10^{-3}}{0.1} \times 1000=0.0919 \times 10^3 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Example The conductance of a 0.1 M solution of an electrolyte of formula AX2 is 4@@@-1. Calculate the molar conductivity of this solution if the cell constant is 0.005 m -1.

Solution Conductivity K = \(G \cdot G^*=4 \times 0.005=0.02 \mathrm{~S} \mathrm{~m}^{-1}\)

Molar conductivity \(\)

But 10 dm = 1 m ⇒ 1dm = 0.1 m

∴ \(\Lambda_m=\frac{\kappa}{c}=\frac{0.02 \mathrm{~S} \mathrm{~m}^{-1}}{0.1 \mathrm{~mol} \mathrm{dm}^{-3}}\)

\(\frac{0.02 \times(0.1)^3}{0.1} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}=0.002 \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}\)

Example The conductance of 0.01 M HCl was found to be 3.296×103 using a cell in which the distance between the electrodes was 1 cm and the area of the cross-section of electrodes was 0.80 cm2. Calculate its molar conductivity.

Solution Cell constant \(\mathrm{G}^*=\frac{1}{0.80} \mathrm{~cm}^{-1}=1.25 \mathrm{~cm}^{-1}=1.25 \times \frac{1}{\mathrm{~cm}}=\frac{1.25}{10^{-2} \mathrm{~m}}\)

= 125 m^-1

Conductivity K=G x G* = \(3.296 \times 10^{-3} \times 125 \mathrm{~S} \mathrm{~m}^{-1}=0.412 \mathrm{~S} \mathrm{~m}^{-1}\)

Molar conductivity = \(\frac{10^{-3} \times \kappa}{c\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)}=0.0412 \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}\)

Variation Of Conductivity And Molar Conductivity With Concentration

Conductivity always decreases with a decrease in concentration for all kinds of electrolytes (both strong and weak). This can be understood from the relation between conductance and conductivity.

When both the distance I and the area of cross-section a are unity, the conductance measured is the specific conductance or conductivity. We are basically measuring the conductance of one unit volume of solution to obtain the specific conductance.

When the concentration decreases, i.e., the solution becomes dilute, the same unit volume of solution contains fewer ions and hence the conductivity decreases. In contrast to conductivity, molar conductivity increases on dilution. We can relate the molar conductivity to the volume of the solution as follows.

\(\Lambda_m=\frac{\kappa}{c}\)

If c is expressed in mol m^-3

\(\Lambda_m=\kappa \times\left(\frac{1}{c}\right)=\kappa V\)

where V is the volume of a solution containing 1 mol of the electrolyte (c moles contained in 1 m³; 1 mol contained in = \(\frac{1}{c} \mathrm{~m}^3\))

Am is directly related to the volume containing 1 mol of the substance. In a dilute solution, the volume containing 1 mol of the electrolyte will be larger and hence from Equation 3.6, A will be larger. As the concentration increases, the molar conductivity decreases.

We can therefore define molar conductivity as the conductance of a volume of a solution containing 1 mol of a dissolved substance when placed between two parallel electrodes which are a unit distance apart and are large enough to contain between them the whole solution.

Strong And Weak Electrolytes

Between 1860 and 1880, Friedrich Kohlrausch and his coworkers performed a series of measurements of molar conductivities of aqueous solutions with varying concentrations of electrolytes. The molar conductivity was found to vary with the concentration of the electrolyte.

This variation is due to the fact that the number of ions per unit volume of the solution is not proportional to the molar concentration of the electrolyte. For example, in case of a weak acid, doubling the concentration of the acid does not double the number of ions. In addition, since ions interact with each other, the conductivity of a solution is not proportional to the number of ions present.

Based on the dependence of molar conductivity on concentration, electrolytes are divided into two classes-strong and weak. The characteristic of a strong electrolyte is that its molar conductivity decreases only slightly as its concentration is increased.

On the other hand, the molar conductivity of a weak electrolyte is appreciable at concentrations close to zero but falls sharply as the concentration increases. The classification of electrolytes also depends on the solvent employed. For example, lithium chloride is a strong electrolyte in water but a weak electrolyte in propanone.

Strong Electrolytes

Strong electrolytes are completely ionized in solution. Examples include ionic solids such as NaCl and KCl, and strong acids. The number of ions in a solution of a strong electrolyte is proportional to the concentration of the electrolyte added.

Kohlrausch, from a series of experiments, showed that at low concentrations the molar conductivities of strong electrolytes vary linearly with the square roots of their concentrations.

\(\Lambda_m=\Lambda_m^0-A \sqrt{c}\)

This is called Kohlrauseh’s law. The constant \(\Lambda_m^0\), is the limiting molar conductivity, the molar conductivity in the limit of zero concentration. This means that it is the molar conductivity of the solution when its concentration approaches zero. In other words, it refers to a solution that is so dilute that there are very few ions and they are far apart and do not interact with one another.

\(\Lambda=\Lambda_m^0 \text { as } c \rightarrow 0\)

Factor A is a constant and depends more on the stoichiometry of the electrolyte (i.e., whether it is M_A, M_2A, etc.) than on its specific identity. For example, NaCl and KCl are a particular type of electrolytes, i.e., 1-1, having the same value for A. Similarly CaCl₂ and MgSO₄ are respectively of types 2-1 and 2-2 respectively.

The value of \(\Lambda_m^0\), can be obtained by plotting \(\Lambda_m \operatorname{vs} \sqrt{c}\) and then extrapolating the graph to √c = 0. It can be seen that the plot of \(\Lambda_m \operatorname{vs} \sqrt{c}\) is nearly linear for KCl, a strong electrolyte, but \(\Lambda_m\) decreases slightly as the concentration increases.

The number of ions per unit volume of the solution increases as the concentration increases and an increase in Am is expected. However, there is a considerable change in interionic interaction (between K⁺ and Cl^–) as the concentration changes. The attraction between the ions increases with an increase in concentration.

The ability of the ions to move independently decreases, as a result of which the conductivity decreases (but only slightly) as the concentration increases. As the solution is diluted, the concentration decreases, the ions are far apart, the attraction decreases, and the conductivity increases.

When the concentration becomes very low, the interionic interactions are the least and the molar conductivity approaches the maximum limiting value A°m, a constant. In other words, the limiting molar conductivity is the molar conductivity in the limit of such low concentration that ions are no longer able to interact with one another.

The tin value is characteristic of every strong electrolyte. It is also called the molar conductivity at infinite dilution. The ions are very well separated at infinite dilution or in the limit of zero concentration and therefore behave independently of one another.

Kohlrausch gave the law of independent migration of ions. The law states that the limiting molar conductivity of an electrolyte can be expressed as the sum of the individual contributions of the anion and cation of the electrolyte.

If the limiting molar conductivity of cations is denoted by X°+ and that of anions by λ°_, then according to Kohlrausch’s law of independent migration of ions,

\(\Lambda_m^0=v_{+} \lambda_{+}^0+v_{-} \lambda_{-}^0\)

where v+ and v_ are tire numbers of cations and anions respectively per formula unit of the electrolyte. For example, for KC1 and \(v_{+}=v_{-}=1 ; \text { for } \mathrm{MgCl}_2, v_{+}=1, v_{-}=2\)

Weak Electrolytes

Weak electrolytes are not fully ionized in the solution. They include weak Bronsted acids and bases such as CH₃COOH and NH₃. The following equilibrium exists in solutions of weak electrolytes.

\(\mathrm{HA}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})\)

At low concentrations, the equilibrium shifts towards the right, thus increasing the number of ions in solution due to which the conductivity increases. Therefore, it is obvious that the conductivity depends on the degree of ionization, a, of the weak electrolyte. Therefore, if the molar concentration of the acid HA is c then the concentrations of HA, H₂O+, and A- at equilibrium will be

\([\mathrm{HA}]=(1-\alpha) c,\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\alpha c \text { and }\left[\mathrm{A}^{-}\right]=\alpha c \text {. }\)

The acidity constant or the acid dissociation constant is given by

\(k_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)

or, \(k_a=\frac{\alpha^2 c}{1-\alpha}\)

At infinite dilution (c→0), the electrolyte is fully ionized and the molar conductivity is A°nr According to Arrhenius, at infinite dilution, only a fraction a is actually present as ions in the solution, and the measured molar conductivity is given by

\(\Lambda_m=\alpha \Lambda_m^0\)

or, \(\alpha=\frac{\Lambda_m}{\Lambda_m^0}\)

Substituting this value of an in Equation (3.8), we get

\(k_a=\frac{\left(\frac{\Lambda_m}{\Lambda_m^0}\right)^2 c}{1-\left(\frac{\Lambda_m}{\Lambda_m^0}\right)}=\frac{c \Lambda_m^2}{\Lambda_m^0\left(\Lambda_m^0-\Lambda_m\right)}\)

Using Kohlrausch’s law of independent migration of ions \(\ Lambda_m^0te] for any electrolyte can be calculated from the X° of individual ions (Equation 3.7). Both the degree of ionization and dissociation constant for a weak electrolyte can be determined if [latex]\Lambda_m \text { and } \Lambda_m^0\) are known at a given concentration c.

Example Calculate the molar conductivity at infinite dilution (\(\Lambda_m^0\)) of acetic acid given that \(\Lambda_m^0\) of HC1, NaCl, and CH₃COONa are 426,126 and 91 Ω^-1 cm² mol^-1respectively.

Solution \(\Lambda_{m, \mathrm{HAC}}^0=\Lambda_{m, \mathrm{NaAc}}^0+\Lambda_{m, \mathrm{HCl}}^0-\Lambda_{m, \mathrm{NaCl}}^0\left(\text { Ac stands for acetate, } \mathrm{CH}_3 \mathrm{COO}^{-}\right)\)

∴ \(\Lambda_{m, \mathrm{HAc}}^0=91+426-126=391 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

If the units are to be converted to S m² mol^-1, since 1 cm = 10-2 m, 1 cm2 = 10^-2 x 10^-2 m², multiply by 10-4 m²

∴ \(\Lambda_{m, \mathrm{HAc}}^0=391 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}\)

= 0.0391 Sm² mol^-1

Example If the molar conductivity of 0.001 M acetic acid is 0.00492 S m² mol^-1, and its\(\Lambda_m^0\) is 0.03907 S m² mol⁴, calculate the degree of dissociation, &&&. Also, find the equilibrium constant for the dissociation of this acid.

Solution \(\alpha=\frac{\Lambda_m}{\Lambda_m^0}=\frac{0.00492}{0.03907}=0.1259\)

\(\mathrm{HAc} \rightleftharpoons \mathrm{H}^{+}+\mathrm{Ac}^{-}\) \(k=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{Ac}^{-}\right]}{[\mathrm{HAc}]}=\frac{(c \alpha)^2}{c(1-\alpha)}=\frac{c \alpha^2}{1-\alpha}\)

Thus the dissociation constant of acetic acid (HAc) is

\(k=\frac{0.001 \times(0.1259)^2}{1-0.13}=1.94 \times 10^{-5}\)

Electrochemical Cells

An electrochemical cell is a setup in which a spontaneous chemical reaction occurs to produce electrical energy or, a nonspontaneous chemical reaction is carried out by using an external source of are called galvanic cells or voltaic cells, and the latter, electrolytic cells All electrochem electrodes dipping into either one or two electrolyte solutions.

Current can pass. It may be made of a metal such as copper, silver zinc, etc. or electric Electrodes come in a variety of shapes-they may be wireSPplates or rodsd.

AnddectrolyteTs In The electrode and its surrounding electrolyte solution are contained in electrodes may or may not share the same compartment.

Compartment The two are made of zinc and copper. These electrodes are dipped in zinc sulfate and copper sulfate The electrodes are respectively, taken in different compartments. To complete the electrical circuit the solution (electrolytes) salt bridge. The following redox reaction occurs in the cell.

\(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\)

The chemical energy of this reaction is converted to electrical energy, which can be used to light a bulb, rim a motor, and so on.

An electrolytic cell. A battery supplies electrical energy to carry out a nonspontaneous reaction.

Cell diagram

Any electrochemical cell can be represented by a cell diagram that shows the oxidized as well as the reduced forms of the electroactive substance and any other species that may be involved in the chemical reaction.

The electrodes, denoted by their atomic symbols, are indicated at the two ends of the diagram. Any insoluble substances or gases are denoted inside, next only to the electrodes.

The soluble species are represented in the middle of the diagram. The states of aggregation (solid, liquid, or gas) are also specified within brackets. In an abbreviated diagram, some of the above information is ignored and only the main components are given.

In a cell diagram, a phase boundary is indicated by a solid vertical bar; a junction between two miscible liquid phases is represented by a single dashed vertical bar.

A double dashed vertical bar indicates a junction between two miscible liquid phases at which the junction potential has been eliminated using a salt bridge. Commas separate different soluble species in the same phase. It is assumed that the right-hand electrode is the cathode and the left-hand electrode is the anode.

Some examples of cell diagrams are given below.

1. \(\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(a=0.35): \mathrm{Cu}^{2+}(a=0.49)\right| \mathrm{Cu}(\mathrm{s})\) (complete diagram)(a denotes the activity of the ions)

\(\mathrm{Zn}\left|\mathrm{Zn}^{2+}: \mathrm{Cu}^{2+}\right| \mathrm{Cu} \text { (abbreviated diagram) }\)

This cell is called the Daniell cell.

2. \(\mathrm{Pt}\left|\mathrm{H}_2(\mathrm{~g})\right| \mathrm{HCl}(m)|\mathrm{AgCl}(S)| \mathrm{Ag}(\mathrm{s}) \mid \mathrm{Pt}\)

3. \(\mathrm{Pt}\left|\mathrm{H}_2(\mathrm{~g}), \mathrm{H}^{+}: \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}\right| \mathrm{Pt}\)

4. \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}\left(c_1\right) \vdots \mathrm{Zn}^{2+}\left(c_2\right)\right| \mathrm{Zn}\)

(m denotes molality, c₁ and c₂ denotes the cons=centrations.)

Galvanic (Voltaic) Cell

A galvanic cell is an electrochemical cell that produces electricity from the spontaneous chemical reactions occurring inside it. Consider a cell made up of a zinc rod dipped in ZnSO₄ solution and a copper rod dipped in

CuSO₄ solution. This is a Daniell cell—a type of galvanic cell. To avoid direct mixing of the electrolyte solutions, these are taken in separate compartments, and connection between them is established through a salt bridge.

A salt bridge is made of a glass tube containing an inert electrolyte solution (whose ions do not react with the ions of the main electrolytes or the electrodes) such as KCl, KNOa, or NH₄NO₃, kept in agar-agar, a geL The two portions of the cell are called half-cells or redox couples.

In one half-cell, the metal from an electrode dissolves in the solution leaving behind electrons, making the electrode negatively charged and simultaneously, in the other half-cell, the metal ions from the solution deposit on the other metal electrode, making this electrode positively charged. The actual reactions occurring at the two electrodes (half-cell reactions) are as follows.

Zn electrode (anode) Oxidation occurs at this electrode.

\(\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}\)

Cu electrode (cathode) Reduction occurs at this electrode.

\(\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\)

The overall reaction is obtained by adding the above two equations.

\(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\)

We can see that Zn undergoes oxidation while the Cu²⁺ ions undergo reduction. Thus the Zn electrode is negatively charged and the copper electrode is positively charged. The electrode which is negatively charged is called the anode and that which is positively charged is called the cathode. In the solution, the cations Zn^2- and Cu²⁺ move towards the cathode while the anions SO⁴⁺ move towards the anode.

Since the two electrodes are oppositely charged, a potential difference occurs between them and electrons flow from the anode to the cathode. The direction of current is opposite to that of electron flow.

The conventional method is to write the reduction reactions for both electrodes (even though in reality oxidation may be occurring at one electrode) and obtain the overall reaction by subtracting the reduction reaction at the anode (left-hand electrode) from that occurring at the cathode (right-hand electrode). The half-reactions for the reduction of copper by zinc are as follows.

If the two solutions are not separated from each other, the Cu²⁺ ions react directly with the zinc bar.

\(\mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Zn}^{2+}(\mathrm{aq})\)

In such a situation, no useful electrical work is obtained.

The cell potential of a galvanic cell:

A galvanic cell converts the chemical energy liberated during a chemical reaction to electrical energy. It produces a driving force through the chemical reaction that pushes electrons through an external circuit. The work done by a given transfer of electrons depends on the cell potential.

The cell potential or emf of the cell is the potential difference between the two electrodes. It may also be called cell voltage. The unit of cell potential is the volt (V).

Under balanced reversible conditions, a cell reaction can proceed in either direction. Here eversibility refers to the change in direction of the movement of ions by reversing the polarity of the electrodes. In a cell diagram, the cell reaction is written in such a manner that electrons are shown to be accepted from the external circuit by the electrode on the right and given up by the electrode on the left.

Consider the following example In which ft cell In represents an

\(\mathrm{Pt}\left|\mathrm{H}_2(1 \mathrm{bar}) \& \mathrm{HCl}(1 \mathrm{M})\right| \mathrm{AgCl} \mid \mathrm{Ag}\)

The reactions that occur in the electrodes of the cell are as follows,

Conversely, if the above cell diagram is shown as ns

\(\left.\mathrm{Ag}|\mathrm{AgCl}| \mathrm{HCl}(1 \mathrm{M}) \mid \mathrm{H}_2 \text { (1 bar }\right) \mid \mathrm{Pt}\)

then the reactions that occur at the two electrodes get reversed,

As you can see the emf of the reaction represented by Equation 3,9 Is positive and hence, by convention, the reaction will proceed from left to right. However, the emf of the reaction shown in Equation 3,10 is negative, indicating that the reverse of this reaction is spontaneous,

Measurement of the cell potential of a galvanic cell and that of a Daniell cell To measure the cell potential of a galvanic ceil, a potentiometer is placed in the circuit. In a potentiometer, a steady current from a battery flows through a resistor, A sliding contact is used to vary the potential difference across the slide wire connected to an electrochemical cell, If the applied potential difference (V) is less than the emf of the galvanic cell (Ecell) the cell discharges electricity spontaneously. When the two are exactly the same (V = Ea4l), then no current flows through the cell and this is the equilibrium potential difference (emf or Ece]|) we are interested in.

In practice, the two substances (electrodes and electrolytes) forming the cell are kept in their standard states (all the solutes are at 1 M and the gases are at 1 atm).

The cell potential measured under the standard state conditions is called the standard cell potential, Efcn. The superscript is used to indicate that all reactants and products are in their standard states.

When 1 M solutions of ZnSC)₄ and CuSO₄ are used for the redox reaction in the Daniell cell, the is found to be 1.1 V at 25°C. A galvanic cell has a positive emf if the cell reaction is spontaneous, i.e„ electrons are accepted by the right-hand electrode (cathode) and released by the left-hand electrode (anode). Therefore, Zn reduces copper ions to Cti(s) according to Equation 3.11 for which EceU is positive.

\(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Zn}^{2+}(\mathrm{aq})\)

But, for the reverse reaction

\(\mathrm{Cu}(\mathrm{s})+\mathrm{Zn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})\)

the emf is negative, i.e., \(E_{\text {cell }}^\theta=-1.1 \mathrm{~V}\) Tire reverse reaction is made possible by applying an external opposite potential to the galvanic cell. As long as the external voltage is less than 1.1V (\(E_{\text {cell }}\), of the Daniell cell) the exception proceeds spontaneously; zinc gets oxidized and Cu²⁺ ions get reduced deporting copper at the cathode.

Current flows from copper to zinc. When the external potential becomes equal to Ewll, neither do electrons flow through the cell nor does any reaction take place. When EwU is exceeded, the reaction proceeds in the reverse direction# electrons flow from copper to zinc as copper gets oxidized and Zn²⁺ ions get reduced, deputing zinc at the zinc electrode, which now acts as the cathode. The cell now functions as an electrolytic cell

Half-Cell Potential Of A Daniell Cell

Each compartment consisting of an electrode dipped in the corresponding electrolyte solution is called a half-veil As you already know, the electrode may be positively or negatively charged with respect to the solution, Hence, a potential difference develops between the electrode and the electrolyte, called electrode potential bach half-cell in a cell can therefore be assigned a half-cell potential which is nothing hut the electrode potential The ha I heel I potential could be either oxidation potential when oxidation occurs or reduction potential when reduction occurs, The sum of half-cell potentials gives the cell potential For the Daniel! cell formed of copper and zinc electrodes operating under standard state conditions, the anode or oxidation potential is 0.763 V and the cathode or reduction potential is 0.337 V at 25°C. Therefore, the cell potential \(\left(E_{\mathrm{cell}}^{\Theta}\right) \text { is } 1.1 \mathrm{~V}\)

It is, however, important to note that by convention, the half-reactions are written as reduction actions and the corresponding half-cell potentials, which are both reduction potentials, are considered fir calculating E-cell

When the species are in their standard states, the half-cell reduction potential is called standard electrode potential. The overall emf of the cell, is then obtained by subtracting the electrode potential of the anode (left-hand electrode) from that of the cathode (right-hand electrode) or simply by the relation

\(E_{\text {cell }}^{\ominus}=E_{\text {right }}^{\ominus}-E_{\text {left }}^{\ominus} \quad \text { (both } E_{\text {right }}^{\ominus} \text { and } E_{\text {left }}^{\ominus} \text { are reduction potentials) }\)

For the above Daniell cell,

\(E_{\text {cell }}^{\Theta}=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\Theta}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^\theta\)

\(E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\Theta} \text { (for the reaction } \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn} \text { ) }\) is equal to the negative of that of the reverse reation,

i.e., \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}[/lattex] whose potential is represented as [latex]E_{\mathrm{Zn} / \mathrm{Zn}^{2+}}^\theta(0.763 \mathrm{~V})\)

Hence \(E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\Theta}=-0.763 \mathrm{~V}\)

Therefore,

\(E_{\mathrm{cell}}^{\Theta}=0.337-(-0.763) \mathrm{V}\)

or, \(E_{\text {cell }}^{\Theta}=1.1 \mathrm{~V}\)

The standard electrode potential of a galvanic cell

Although we can measure the overall cell potential of a galvanic cell, there In no satisfactory method to measure the individual actual potentials of the two electrodes. However, a standard potential can be assigned to one electrode and the other electrode can be assigned a relative value of potentials based on this.

The most commonly used reference electrode is the standard hydrogen electrode (SHE), which is assigned a potential of zero volts at all temperatures. The potentials of all other electrodes are reported relative to that of the SHE.

The standard hydrogen electrode is a gas-ion electrode. In a hydrogen electrode, hydrogen gas is bubbled through a solution of hydrogen ions in which a platinum foil coaled with very finely divided platinum black is dipping partially. Here platinum metal acts as an inert electrode; it does not participate in the reaction—it only provides its surface for the redox reaction. The net ionic equation for reduction in the electrode is

\(2 \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g}) \div 2 \mathrm{H}_2 \mathrm{O}(\mathrm{I})\)

The potential of a gas electrode changes with the pressure of die gas and with the concentration of the other components. If a hydrogen electrode operates at 298 K with hydrogen gas at a pressure of 1 bar bubbling around a platinum foil immersed in a 1M solution of hydronium ions, it is called a standard hydrogen electrode.

Measurement of the standard electrode potential of a galvanic cell:

In order to measure the potential of an electrode relative to the SHE, a galvanic cell is constructed with the standard hydrogen electrode and electrode of interest. The cell potential is the potential of the other half-cell, i.e., the electrode whose potential is to be measured.

For example, the potential of a copper electrode can be determined by coupling it with a SHE. The cell diagram is \(\mathrm{Pt}(\mathrm{s})\left|\mathrm{H}_2(1 \mathrm{~atm})\right| \mathrm{H}^{+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M}) \mid \mathrm{Cu}(\mathrm{s}) .\). The SHE acts as the anode and the copper electrode as the cathode. The overall redox reaction and the two half-reactions in the cell are

\(\mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{H}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Cu}(\mathrm{s})+2 \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})\)

Cathode: \(\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\)

Anode: \(\mathrm{H}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+2 \mathrm{e}^{-}\)

The observed potential for the cell is 0.337 V at 25°C.

\(E_{\text {cell }}^{\Theta}=E_{\text {cathode }}^{\Theta}-E_{\text {anode }}^{\Theta}\) \(0.337=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\Theta}-E_{\mathrm{H}^{+} / \mathrm{H}_2}^{\Theta}\) \(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\Theta}-0\)

Thus, \(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}=0.337 \mathrm{~V}\)

The standard electrode potential of copper, \(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{\prime}} \text {, is } 0.337 \mathrm{~V}\) the subscript Cu²⁺/Cu indicating the reaction \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}\) If we wish to determine the standard electrode potential (the reduction potential) of a zinc electrode, the SHE acts as the cathode and the zinc electrode acts as the anode.

The cell diagram is

\(\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right| \mathrm{H}_2(1 \mathrm{~atm}) \mid \mathrm{Pt}(\mathrm{s})\)

The two half-reactions and the overall reaction are as follows.

The \(E_{\text {cell }}^{\ominus}\) was found to be 0.76 V.

\(E_{\text {cell }}^{\Theta}=E_{\text {cathode }}^{\Theta}-E_{\text {anode }}^{\Theta}\) \(0.76=E_{\mathrm{H}^{+} / \mathrm{H}_2}^{\Theta}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\Theta}\) \(0.76=0-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\ominus} \Rightarrow E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\Theta}=-0.76 \mathrm{~V}\)

Thus tire standard reduction potential of zinc, \(E_{\mathrm{Zn}^{2+} / \mathrm{Zn}^{\prime}} \text { is }-0.76 \mathrm{~V}\)

Lists the values of standard electrode potentials (reduction potentials) for some half-cells.

The positive value of standard electrode potential in the example of copper indicates that Cu²⁺ ions get reduced more easily than the H⁺ ions. The negative value of standard electrode potential in case of zinc indicates that hydrogen ions can oxidize zinc easily.

The standard electrode potentials are useful in finding out the £Te’ for a combination of different half-cells. It is important to know the following facts about the standard electrode potential values,

1. The \(E^{\Theta}\) values are applicable to the half-cell reactions as read in the forward direction.

2. The half-cell reactions are reversible. A particular electrode can act as anode or cathode, depending on the electrode that is coupled to it,

3. The values of reduction potentials also indicate the tendency of a reaction to occur. The higher the value, the greater is the tendency for reduction, Those electrodes which have a positive electrode potential can be more easily reduced than those having a negative electrode potential. A positive electrode potential indicates a greater tendency to reduce than hydrogen ions.

4. The more positive the value of \(E^{\Theta}\), the greater is the tendency for the substance to be reduced. A look at Table 3.5 shows that P2 has the greatest tendency to get reduced or, in other words, it is the strongest oxidizing agent, Lithium figures at the bottom of the table, indicating that it has the lowest ability to get reduced or the greatest tendency to get oxidized—thus, it is the strongest reducing agent. Thus we see that the relative tendency of metals to get oxidized can be judged from the values of the standard electrode potentials of the corresponding half-cells. The electrochemical series is an arrangement of metal/metal ion electrode potentials in order to decrease the tendency to lose electrons.

5. If any two electrodes are chosen to form a cell, then the electrode that figures first in the table as we move from top to bottom acts ns the cathode and reduction occurs at this electrode. Conversely, the electrode that figures Inter in the table acts as the anode and oxidation occurs at that electrode.

6. Changing the stoichiometric coefficients of a half-cell reaction does not affect the value of \(E^{\Theta}\) as electrode potential is an intensive property.

7. The sign of \(E^{\Theta}\) changes when the reaction is reversed.

Example Calculate the emf of a cell consisting of two half-cells \(\mathrm{Ni}^{2+} / \mathrm{Ni} \text { and } \mathrm{Cd}^{2+} / \mathrm{Cd}\). Given \(E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^\theta=-0.25 \mathrm{~V}\) concentrations of Nl2+ and Cd2+ are 0.2 M and 0.3 M respectively.

Solution As the reduction potential of the cell Cd₂l/Cd is more negative, it has a tendency to undergo oxidation, the two electrode reactions are

\(E_{\mathrm{cell}}^{\mathrm{O}}=E_{\mathrm{kH}}^0-E_{\mathrm{l}, \mathrm{HI}}^{\mathrm{H}}=-0.25+0.4=0.15 \mathrm{~V}\) \(E_{\text {cell }}=E_{\text {cell }}^{\Theta}+\frac{R T}{n F} \ln \frac{\left[\mathrm{Cd}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]}=0.15+\frac{8.314 \times 298}{2 \times 96500} \ln \left[\frac{(0.2)}{(0.3)}\right]\)

=0.145 V

Types Of Electrodes

The term ‘electrode’ is used in two different ways—one to refer to n wire or some conductor that delivers electrons through the external circuit and the other to refer to a complete half-cell. Let us touch upon the following types of electrodes to illustrate this point.

Gas-ion electrodes A gas-ion electrode consists of an inert collector of electrons such as platinum or graphite in contact with a gas ami a soluble ion, The hydrogen electrode is one example,

\(\mathrm{Pt}^{\mathrm{t}}\left|\mathrm{H}_2(\mathrm{~g})\right| \mathrm{H}^{+}\)

We can also have a chlorine electrode, represented by

\(\mathrm{Cl}_2\left|\mathrm{Cl}^{-}\right| \text {graphite }\)

Metal-metal ion electrodes A metal strip or rod dipped in a solution containing the metal ions constitutes a metal-metal ion electrode. The Zn²⁺ /Zn and Cu²⁺ /Cu electrodes of the Daniell cell are examples. Metal-insoluble salt-anion electrode Such an electrode consists of a bar of metal immersed in a solution containing a solid insoluble salt of the metal and anions of the salt.

The silver-silver chloride electrode, \(\mathrm{Cl}^{-}|\mathrm{AgCl}(\mathrm{s})| \mathrm{Ag}(\mathrm{s})\), and the calomel electrode, \(\mathrm{Cl}^{-}\left|\mathrm{Hg}_2 \mathrm{Cl}_2\right| \mathrm{Hg}\) are two examples.

Oxidation-reduction electrodes Although electrodes generally involve either oxidation or reduction in their operation, there are certain electrodes called oxidation-reduction electrodes. Such electrodes have an inert metal collector, usually platinum dipping in a solution that contains two soluble species in different states of oxidation. An example is the ferric-ferrous ion electrode.

Electrical Work And Free Energy

The chemical reaction in a galvanic cell proceeds spontaneously in the forward direction obeying laws of chemical equilibrium. Electrical work can be obtained using the galvanic cell and the total amount of energy available from a particular cell depends on the cell’s potential and the number of electrons involved. Let us now see how E^u and AGe are related. The unit of potential difference is the volt.

The work done to move a charge across a potential difference is given by the product of the charge and the potential difference. The unit of electrical work is the joule. Thus, one volt (V) is the potential difference required to impart one joule 0) of energy to a charge of one coulomb (C). In other words, one joule of energy is available when a charge of one coulomb passes between electrodes having a unit potential difference.

1J = 1V x 1C

In this context, it is important to recall that in a galvanic cell, the electrons flow in the direction determined by the potential difference between the two electrodes. When the two electrodes acquire the same potential, there is no passage of current. More work is available when more charge is moved between the electrodes.

The quantity of charge flowing through an electrical circuit connected to a galvanic cell equals the product of the number of moles of electrons (n) flowing through the circuit and the Faraday constant. The Faraday constant is the electrical charge contained in 1 mol of electrons. Experiments have shown that 1 faraday is equivalent to 96,487 coulombs.

Thus,

\(1 F=96,487 \frac{\mathrm{C}}{\mathrm{mol} \mathrm{e}^{-}} \text {or } 96,487 \frac{\mathrm{J}}{\mathrm{V} \mathrm{mol} \mathrm{e}^{-}} \text {and } C=n F\)

The electrical work (Wcle) done on the circuit is equal to the product of this charge and the electromotive force pushing it. When the work is expressed as the work done by the cell, a negative sign is introduced.

\(W_{\text {ele }}=-C E_{\text {cell }}=-n F E_{\text {cell }}\)

where n is the number of electrons transferred, F is the Faraday constant and Ecell is the cell potential. A negative cell potential means that energy is consumed while a positive potential means that work is done by the cell. The maximum amount of electrical work that can be done by the system is equal to the maximum amount of useful work. Hence,

\(W_{\max }=W_{\text {ele }}=-n F E_{\text {cell }}\)

A cell can do no work when the potential is zero (or the reaction is at equilibrium).

When the current produced by the cell is used to do work, the energy of the system (the cell) is reduced and the sign of work done is negative but the cell potential is positive. Thus the magnitude of the cell potential and the maximum amount of work Wmax available from the cell have opposite signs.

As you already know, the Gibbs free energy for a system (AG) is equal to the maximum possible useful work done by the system, Equation 3.12 can be further modified as

\(\Delta G=W_{\max }=-n F E_{\text {cell }}\)

where ΔG is the Gibbs energy change for the reaction,

The free energy of the system (here cell), therefore, decreases (i.e., has a net negative value) when the system does some useful work. The actual amount of work that a cell can do is, in fact, less than the calculated maximum amount of work because some energy is converted into heat owing to the resistance in the circuit.

When the concentrations of the reacting species are unity, the corresponding free energy change, and cell potential values are standard state values.

\(\Delta G^{\ominus}=-n F E_{\text {cell }}^{\ominus}\)

Thus, cell potential can be used to calculate \(\Delta G \text { or } \Delta G^{\ominus}\) for a process.

Example 1. Calculate the standard free energy change for the reaction occurring in the following cell.

\(\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(\mathbf{1} \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathbf{1} \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\)

\(\text { Given } E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^\theta=-0.76 \mathrm{~V} ; E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^\theta=0.34 \mathrm{~V} ; F=96500 \mathrm{C} \mathrm{mol}^{-1}\)
Solution:

Since \(\Delta G^{\ominus}=n F E_{\text {cell }}^{\Theta}\)

the number of electrons involved in the reaction (n) = \(2 \text { and } E_{\text {cell }}^{\Theta}=\underset{\substack{\text { cathode } \\(\mathrm{RHE})}}{E_{\text {and }}^{\Theta}}-\underset{\text { (LHE) }}{E_{\text {anode }}}=0.34-(-0.76)\)

\(E_{\text {cell }}^{\ominus}=0.34+0.76=1.10 \mathrm{~V}\)

∴\(\Delta G^{\ominus}=-2 \times 96500 \times 1.1=-212300 \mathrm{~J}=-212.3 \mathrm{~kJ} \quad(1000 \mathrm{~J}=1 \mathrm{~kJ})\)

Example 2. Calculate the maximum possible electrical work that can be obtained from the following cell under standard conditions of temperature and pressure.

\(\mathrm{Zn} / \mathrm{Zn}^{2+}(\mathrm{aq}) \| \mathrm{Sn}^{2+}(\mathrm{aq}) \mid \mathrm{Sn}\)

Given \(E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\Theta}=-0.76 \mathrm{~V} \text { and } E_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\Theta}=-0.14 \mathrm{~V}\)
Solution:

Electrical work done, \(W_{\text {ele }}=\Delta G^\theta\) under standard conditions and \(\Delta G^{\Theta}=-n F E_{\text {cell }}^{\Theta}\)

Also \(E_{\mathrm{cell}}^{\Theta}=E_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\Theta}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\Theta}=-0.14-(-0.76)\)

\(E_{\text {cell }}^{\ominus}=0.62 \mathrm{~V} ; n=2\) (two electrons are involved in the cell reaction).

∴ \(\Delta G^{\ominus}=W_{\text {ele }}=-2 \times 96500 \times 0.62=-1,19,660 \mathrm{~J}\)

Thus, 119.7 kJ of work can be obtained from this cell.

It is important to note that since ΔG is an extensive thermodynamic property and the value depends on n, it changes with the change of stoichiometric coefficients. For example, the values of ΔG for the reactions

\(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\)

and \(2 \mathrm{Zn}(\mathrm{s})+2 \mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Cu}(\mathrm{s})\)

\(\text { are }-2 F E_{\text {cell }} \text { and }-4 F E_{\text {cell }} \text { respectively. }\)

Effect Of Concentration On Cell Potential (Nernst Equation)

We may not always measure the potential of a cell under standard conditions. What happens when the concentrations of the species involved are not unity? The cell potential of the cell changes with the change in concentrations of the species

Let us consider a galvanic cell reaction of the general type

\(a \mathrm{~A}+b \mathrm{~B} \rightleftharpoons c \mathrm{C}+d \mathrm{D}\)

The reaction quotient Q is given by

\(Q=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{~A}]^a[\mathrm{~B}]^b}\)

where the square brackets indicate the concentrations of the corresponding species and all quantities are raised to the power of the respective stoichiometric coefficients. Note that products appear in the numerator and reactants in the denominator.

The free energy change AG for the reaction is given by

\(\Delta G=\Delta G^{\Theta}+R T \ln \left[\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{~A}]^a[\mathrm{~B}]^b}\right]\)

Substituting \(\Delta G \text { and } \Delta G^{\ominus} \text { in terms of } E_{\text {cell }} \text { and } E_{\text {cell, }}^{\ominus} \text { we get }\)

\(-n F E_{\text {cell }}=-n F E_{\text {cell }}^\theta+R T \ln Q\).

\(E_{\text {cell }}=E_{\text {cell }}^{\Theta}-\frac{R T}{n F} \ln Q\).

Here n is the number of moles of electrons exchanged in the cell reaction. Equation (3-13) is known as the Nemst equation after the German chemist and physicist W H Nemst, who was awarded the Nobel Prize in Chemistry in 1920.

By converting the natural logarithm to the base 10 and at a temperature of 298 K, and substituting for the values of R and F, the above equation can be written as

\(E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.05916}{n} \log Q\)

For a half-cell reaction

\(\mathrm{M}^{n+}(\mathrm{aq})+n \mathrm{e}^{-} \rightarrow \mathrm{M}(\mathrm{s})\)

the electrode potential at any concentration measured by connecting it to an SHE can be represented by

\(E_{\mathrm{M}^{n+} / \mathrm{M}}=E_{\mathrm{M}^{n+} / \mathrm{M}}^{\Theta}-\frac{R T}{n F} \ln \frac{[\mathrm{M}]}{\left[\mathrm{M}^{n+}\right]}\)

Since M is a solid, its concentration is taken as unity and the above equation then becomes

\(E_{\mathrm{M}^{n+} / \mathrm{M}}=E_{\mathrm{M}^{n+} / \mathrm{M}}^{\Theta}-\frac{R T}{n F} \ln \frac{1}{\left[\mathrm{M}^{n+}\right]}\)

Consider the example of the Daniell cell; the cell potential is given by

\(E_{\text {cell }}=E_{\text {cell }}^{\Theta}-\frac{0.05916}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right][\mathrm{Cu}]}{[\mathrm{Zn}]\left[\mathrm{Cu}^{2+}\right]}\)

Since the activities of solid copper and zinc are unity,

\(E_{\text {cell }}=E_{\text {cell }}^\theta-\frac{0.05916}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)

or \(E_{\text {cell }}=E_{\text {cell }}^{\Theta}+\frac{0.05916}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Zn}^{2+}\right]}\)

As you can see, the value of the cell potential depends on the concentrations of both Cu²⁺ and Zn²⁺. It increases with an increase in the concentration of Cu²⁺ and decreases with an increase in the concentration of Zn²⁺.

In case of cells with electrodes of some type dipping in same type of elect roly tie solutions hut wllli different concentrations (known as consent ml ion cells), (lie standard potentials of the electrodes cancel each other and the cell potential is given in terms of the concentration of electrolytic solutions, for example, In the concentration cell,
\(\mathrm{Cu}\left|\mathrm{CuSO}_4\left(c_1\right) \| \mathrm{CuSO}_4\left(c_2\right)\right| \mathrm{Cu}\) the cell potential is given by where q and c2 are the concentrations of the two solutions.

\(E_{\text {cell }}=\left(\frac{R T}{2 F}\right) \ln \left(\frac{c_2}{\epsilon_1}\right)\)

where c₁ and c₂ are the concentrations of the two solutions.

Cell Potential And Equilibrium Constant

The \(\Delta G^\theta\) of a reaction is related to the equilibrium constant K as

\(\Delta G^\theta=-R T \ln K\)

but \(\Delta G^\theta=-n F E_{\text {cell }}^\theta\)

Hence,

\(-n F E_{\mathrm{cell}}^{\Theta} \equiv-R T \ln K\) \(E_{\text {cell }}^\theta=\frac{R T}{n F} \ln K\)

If the standard potential of a cell is known then the equilibrium constant of a reaction taking place in that ceil can be obtained from Equation (3.15). On substituting the values of R and P and conversion to common logarithm at 298 K, we get

\(E_{\text {oell }}^{\Theta}=\frac{0.05916}{n} \log K\)

In an electrochemical cell, the voltage as read on a voltmeter drops gradually and becomes zero at some point of time. This is because as the reaction in the cell proceeds, the concentrations of reactants keep decreasing and those of products increase till equilibrium is attained, when there is no change in the concentrations of either. At that time, cell potential is zero.

Example 1. Calculate the end of a cell consisting of two half-cells \(\mathrm{Al}^{3+} / \mathrm{Al} \text { and } \mathrm{Ag}^{+} / \mathrm{Ag}\). Given that \(E_{\mathrm{Al}^{3 *} / \mathrm{Al}}^6=-1.66 \mathrm{~V}\) and \(E_{\mathrm{Ag}^{+} / \mathrm{Ag}^{\Theta}}^{\Theta}=0.8 \mathrm{~V} \text { at } 298 \mathrm{~K}\)
Solution:

The standard reduction potential of the half-cell reactions are

\(E_{\text {cell }}^{\Theta}=E_{\mathrm{RHE}}^{\Theta}-E_{\mathrm{LHE}}^{\Theta}=0.8-(-1.66)=0.8+1.66=2.46 \mathrm{~V}\) \(E_{\text {cell }}=E_{\text {cell }}^{\Theta}+\frac{R T}{n F} \ln \left[\frac{\left[\mathrm{Al}^{3+}\right]}{\left[\mathrm{Ag}^{+}\right]^3}\right]=2.46+\frac{8.314 \times 298}{3 \times 96500} \ln \frac{0.2}{(0.01)^3}\)

= 2.46 + 0.104

= 2.564 V.

Example 2. A cell is made up of two electrodes \(\mathrm{Cr}^{2+} / \mathrm{Cr} \text { and } \mathrm{Co}^{2+} / \mathrm{Co}\). The standard reduction potentials are \(E_{\mathrm{Cr}^{2+} / \mathrm{Cr}}^{\Theta}=-0.91 \mathrm{~V} ; E_{\mathrm{Cu}^{2+} / \mathrm{Co}}^{\Theta}=-0.28\) at 298 K. If the cell is 0.659 V and the concentration of Cr²⁺ 0.1 M, find the concentration of CO²⁺.
Solution:

\(E_{\mathrm{cell}}^{\Theta}=E_{\mathrm{Co}^{2+} / \mathrm{Co}_0}^{\Theta}-E_{\mathrm{Cr}^{2+} / \mathrm{Cr}}^{\Theta}=-0.28+0.91=0.63 \mathrm{~V}\) \(E_{\text {cell }}=E_{\text {cell }}^{\Theta}+\frac{R T}{2 F} \ln \frac{\left[\mathrm{Cr}^{2+}\right]}{\left[\mathrm{Co}^{2+}\right]}\) \(\left(E_{\text {cell }}-E_{\text {cell }}^{\Theta}\right) \times \frac{2 F}{R T}=\ln \frac{\left[\mathrm{Cr}^{2+}\right]}{\left[\mathrm{Co}^{2+}\right]}\) \(\frac{(0.659-0.63) \times 2 \times 96500}{2.303 \times 8.314 \times 298}=\log \frac{\left[\mathrm{Cr}^{2+}\right]}{\left[\mathrm{Co}^{2+}\right]}\) \(\frac{\left[\mathrm{Cr}^{2+}\right]}{\left[\mathrm{Co}^{2+}\right]}=9.6 \Rightarrow \frac{0.1}{9.6}=\left[\mathrm{Co}^{2+}\right] \Rightarrow\left[\mathrm{Co}^{2+}\right]\)

= 0.01 M.

Example 3. For a reaction, K = 1.8x 10 7 at 300 K. What is the value of \(\Delta G^{\ominus}\) at this temperature?
Solution:

\(\Delta G^\theta=-R T \ln K=-8.314 \times 300 \times \ln 1.8 \times 10^7\) \(-41667.8 \mathrm{~J} \mathrm{~mol}^{-1}=-41.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Example 4. Calculate the equilibrium constant of the reaction

\(\mathrm{Ni}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Ni}^{2+}(\mathrm{aq})\)

\(E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\Theta}=-0.25 \mathrm{~V}, E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\Theta}=0.34 \mathrm{~V}\);

\(R=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, F=96,500 \mathrm{C} \mathrm{mol}^{-1}\)
Solution:

\(\Delta G^\theta=-R T \ln K\) \(\text { Also } \Delta G^{\ominus}=-n F E_{\text {cell }}^{\ominus}\)

and \(E_{\mathrm{cell}}^{\ominus}=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\Theta}-E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\ominus}=0.34-(-0.25)=0.59 \mathrm{~V} \text { and } n=2\)

∴ \(\Delta G^{\ominus}=-2 \times 96500 \times 0.59=-113,870 \mathrm{~J} \mathrm{~mol}^{-1}\)

\(\ln K=-\frac{\Delta G^\theta}{R T}=\frac{113,870}{8.314 \times 298}=45.96\)

2.303 log K = 45.96

log K = 19.96

K = antilog (19.96)

= 9.12 x 10 19.

Electrolysis and electrolytic cells

When electrodes (metallic conductors) are dipped in a solution of an electrolyte and a sufficient potential difference, of the order of several volts, is applied across the electrodes, chemical reactions are observed at the electrodes. This process is known as electrolysis.

Electrolysis is observed in an electrolytic cell (or an electrochemical cell) in which a nonspontaneous reaction is driven by an external source of current. When electrolysis occurs in an electrolytic cell the electrode that is charged positively by the applied potential resulting in a deficit of electrons is called the anode and that chareed negatively with an excess of electrons is called the cathode.

In this context, it is also important to distinguish between inert electrodes and reacting electrodes. Inert electrodes, usually platinum wires, serve only to transfer electrons to and from the solution. Reacting electrodes chemically enter into an electrode reaction. Metals except platinum are examples of reacting electrodes and contribute metal ions to the solution.

The two major electrode reactions that occur in electrolysis are oxidation and reduction. At the anode, loss of electrons (oxidation) occurs whereas at the cathode electrons are introduced (by the external circuit) and reduction occurs.

Electrolysis was studied extensively by Michael Faraday in 1820. He observed that the amount of charge passed through an electrolyte was quantitatively related to the amount of products formed at the electrodes.

These quantities axe conveniently related by introducing the faradav unit of charge with the symbol Fr defined as the charge of 3 mol, or an Avogadro number, of electrons. We have

1F = NA x (electronic charge)

= \(6.023 \times 10^{23} \times 1.602 \times 10^{-19}=96,485 \mathrm{C} \mathrm{mol}^{-1}\)

Based on his experimental findings, Faraday gave two laws of electrolysis, which are as follows.

First law:

The amount of a chemical substance liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the cell.

Second law:

The amounts of different substances produced by a given quantity of electricity (passing through the cell) are proportional to the molar masses of the substances.

The laws are applicable to molten electrolytes as well as to solutions of electrolytes and are independent of temperature, pressure or nature of the solvent. One important application of electrolysis is in the extraction of metals such as lithium, sodium, magnesium, aluminum and calcium.

Electrolysis of molten sodium chloride:

The commercial electrolysis of molten sodium chloride is carried out in a Down’s cell. This cell consists of an airtight vessel containing molten sodium chloride, two inert electrodes and a porous separator screen that permits diffusion of ions through the cell but prevents intermixing of the reactants and products. At the cathode, sodium ions get reduced to sodium metal At the anode the chloride ions get oxidized to chlorine gas. The two half-reactions are as follows.

Cathode: \(\mathrm{Na}^{+}(\mathrm{l})+\mathrm{e}^{-} \rightarrow \mathrm{Na}(\mathrm{s})\)

Anode: \(2 \mathrm{Cl}^{-}(\mathrm{l}) \rightarrow \mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{e}^{-}\)

To balance the electrons, by multiplying Equation 3.17 by 2, we get

\(2 \mathrm{Na}^{+}(\mathrm{l})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Na}(\mathrm{s})\)

The net reaction is obtained by adding Equations 3.18 and 3.19.

\(2 \mathrm{Na}^{+}(\mathrm{l})+2 \mathrm{Cl}^{-}(\mathrm{l}) \rightarrow 2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_2(\mathrm{~g})\)

For this process, a voltage of about 5 V is applied with a current of thousands of amperes. Li, Mg, and Ca are obtained by the electrolysis of their respective chlorides—they are produced at the cathode.

Quantitative aspects To reduce one sodium ion, one electron is required. Therefore, to reduce one mole of sodium ions (Na~), one mole of electrons is required, and one mole of sodium ions gives rise to one gram atom, i.e., 23 of sodium metal. Thus, on passing one mole of electrons (one faraday of electricity), 23 g of sodium metal is produced. At the same time, if one faraday of electricity is removed from the anode, 35.4 g of chloride ions (CT) is discharged, Le., 0.5 mol of Cl2 gas is produced according to the reaction

\(\mathrm{Cl}^{-}(\mathrm{l}) \rightarrow \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g})+\mathrm{e}^{-}\)

The charge carried by n moles of electrons is

Q = nF.

If the oxidation reaction is

\(2 \mathrm{Cl}^{-}(\mathrm{l}) \rightarrow \mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{e}^{-}\)

then the total charge carried by the electrons is Q = 2F. Therefore, on passing two faradays of electricity, one mole of Cl₂(g) is produced. The charge passed through the cell during electrolysis is equal to the product of current (1) in amperes and time (t) in seconds.

Q = It.

Example 1. Calculate the amount of product formed at the two electrodes when a current of 0334 A is passed through a cell constructed for the electrolysis of calcium chloride for 2 h.
Given: Atomic mass of Ca = 40.08 and that of Cl = 35.4.
Solution:

The only species that we get from CaCl₂ are Ca²⁺ and Cl ions. Hence it is easy to guess that Cl ions will get oxidised while Ca²⁺ ions will get reduced.

Amount of charge passed through the cell,

\(Q=I t=0.334 \mathrm{~A} \times 2 \mathrm{~h} \times \frac{3600 \mathrm{~s}}{1 \mathrm{~h}} \times \frac{1 \mathrm{C}}{1 \mathrm{~A} \cdot \mathrm{s}}=2404.8 \mathrm{C}\)

Since 1 mol of e- = 96,487 C and 2 mol of electrons is required to reduce 1 mol of Ca²⁺ ions, the mass of Ca formed at the cathode is calculated as follows.

Amount of Ca = \(2404.8 \mathrm{C} \times \frac{1 \mathrm{~mol} \mathrm{e}^{-}}{96,487 \mathrm{C}} \times \frac{1 \mathrm{~mol} \mathrm{Ca}}{2 \mathrm{~mol} \mathrm{e}^{-}} \times \frac{40.08 \mathrm{~g} \mathrm{Ca}}{1 \mathrm{~mol} \mathrm{Ca}}=0.5 \mathrm{~g}\)

For the anodic reaction, 2 mol of electrons is required to oxidize Cl ions or to produce 1 mol of Cl₂ gas.

Amount of Cl₂ (in grams) = \(2404.8 \mathrm{C} \times \frac{1 \mathrm{~mol} \mathrm{e}^{-}}{96,487 \mathrm{C}} \times \frac{1 \mathrm{~mol} \mathrm{Cl}_2}{2 \mathrm{~mol} \mathrm{e}^{-}} \times \frac{2 \times 35.4}{1 \mathrm{~mol} \mathrm{Cl}_2}=\mathbf{0 . 8 8 2}\)

Example 2. How much charge is required to be passed through a cell containing a Cd²⁺/ Cd electrode so that 0.5 mol of Cd is deposited at the electrode?
Solution:

The reaction occurring in the cell is

\(\mathrm{Cd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(\mathrm{s})\)

1 mol of Cd (s) requires 2 mol of electrons.

∴ 0.5 mol Cd (s) requires 1 mol of electrons.

Charge on 1 mol of electrons = \(1.6023 \times 10^{-19} \mathrm{C} \times 6.023 \times 10^{23} \text { electron } \mathrm{mol}^{-1}=96,485 \mathrm{C}\)

Hence 96,485 C of charge should be passed for the deposition of 0.5 mol of Cd.

Example 3. How much charge is required to be passed through a cell containing a Cu²⁺/Cu electrode so that 25 g of Cu (atomic weight = 63.5) gets deposited at the electrode?
Solution:

\(\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\)

1 mol of Cu²⁺ requires 2 mol of electrons (v charge of 1 mol of electrons = 96,485 C)

= 2x 96485 C ≅ 2 x 96500 C

= 193, 000 C.

For 63.5 g (1 mol) 193,000 C of charge is required.

∴ For 25 g of Cu \(\) of charge is required.

= 75,984 C of charge is required.

Example 4. Find the mass of Al deposited according to the reaction \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\), given that a current of 4 A is passed for 50 min. Atomic mass of AI = 27.
Solution:

Since \(\frac{193,000}{63.5} \times 25 \mathrm{C}\) = y, total charge passed = 4x50x60As = 12,000 C.

3 mol of electrons or 3 x 96,500 C of charge deposits 1 mol of Al.

∴12,000 C of charge will deposit \(\frac{12,000}{3 \times 96,500}\) = 0.0414 mol of  AI = 0.0414 x 27 g of AI = 1.117 g of AI.

Example 5. Find the amount of Na deposited at an electrode due to the reaction \(\mathrm{Na}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Na}\) when passed through molten NaCl for 80 min (atomic mass of Na = 23).
Solution:

According to the reaction, 1 mol of electrons is required for depositing 1 mol of Na.

Charge on 1 mol of electrons = 96,500 C.

Charge passed = 2 x 80 x 60 = 9600 C.

∴ amount of Na deposited = \(\frac{9600}{96,500} \times 1 \approx 0.1 \mathrm{~mol}\) = 0.1 x 23 = 2.3 g.

Example 6. How much time is required for the deposition of 15.875 g of Cu according to the electrode reaction \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}\) when 4 A of current is passed through the cell? The atomic mass of Cu = 63.5.
Solution:

According to the electrode reaction, for 1 mol of Cu, 2 mol of electrons is required, i.e., for 63.5 g of Cu, 2 x 96,500 C of charge is required.

For 15.875 g of Cu, \(\frac{2 \times 96,500}{63.5}\) x 15.875 = 48,250 C of charge is required.

Charge = current x time.

∴ time required for the deposition of 15.875 g of Cu

\(=\frac{48,250}{4}=12,062.5 \mathrm{~s}=\frac{12,062.5}{3600} \text { hours }\)

= 3.35 hours.

Products Of Electrolysis

As the amount of the substance or products liberated at the electrode depends on the amount of electricity passed [ through the cell, the products of electrolysis depend on the state of the substance (i.e., molten or aqueous) and the nature of the electrodes being used in the electrolysis. For instance, the electrolysis of molten sodium chloride yields sodium metal and chlorine gas. But what will happen if an aqueous solution of sodium chloride is used? f Let us discuss this case along with some more examples.

Electrolysis of aqueous sodium chloride:

In tire electrolysis of an aqueous solution of sodium chloride, more than one reaction is possible at each electrode. At the cathode, the sodium ion as well as water can be reduced.

\(\mathrm{Na}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Na} \quad E_{\text {red }}^{\Theta}=-2.714 \mathrm{~V}\) \(2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2+2 \mathrm{OH}^{-} \quad E_{\mathrm{red}}^{\ominus}=-0.828 \mathrm{~V}\)

Tire species with a higher reduction potential can get more easily reduced. Between sodium ions and water, water has a larger reduction potential and hence the reaction involving water is the one favored at the cathode.

At the anode, the chloride ion as well as water can be oxidised. In order to optimise the production of chlorine, n concentrated solution of sodium chloride is used.

The reactions occurring at the two electrodes and the overall reaction are as follows.

Because the sodium ions remain unchanged, more and more sodium hydroxide is formed in the solution as electrolysis proceeds. Hence the electrolysis of an aqueous solution of sodium chloride can be used for the commercial production of H₂ gas, Cl₂ gas and NaOH. This is possible by using currents of 10,000 – 60,000 A at about 3.8 V.

Electrolysis of aqueous HCI:

When an aqueous solution of hydrochloric acid is electrolyzed, hydronium ions are reduced to hydrogen gas at the cathode.

\(2 \mathrm{H}_3 \mathrm{O}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2+2 \mathrm{H}_2 \mathrm{O}\)

At the anode, two oxidation reactions are possible.

\(2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_2+2 \mathrm{e}^{-} \quad E_{\mathrm{ov}}^{\Theta}=-1.36 \mathrm{~V}\) \(6 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{O}_2+4 \mathrm{H}_3 \mathrm{O}^{+}+4 \mathrm{e}^{-} \quad E_{\mathrm{ox}}^{\Theta}=-1.23 \mathrm{~V}\)

The oxidation potentials of these two reactions are comparable and both can occur with equal ease. However, which one of them occurs or whether both of them occur depends upon the concentration of chloride ions in the solution.

In a very dilute solution of HCl, H₂O is electrolysed and oxygen is released at the anode.

In concentrated solutions of HCl, chloride ions are reduced to form chlorine gas at the anode (Equation 3.20). The reactions are as follows.

A voltage of 1.36 V (or more) has to be applied for the above reaction to occur at standard state conditions. At moderate concentrations of HCl, both the reactions—the electrolysis of water and the oxidation of chloride ions—occur.

Electrolytic deposition of metals:

Chrome plating, bronzing, and refining of copper, all employ electrolysis.

In bronzing, a metallic strip of copper is used as the anode. Copper is oxidized to form copper ions.

\(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}\)

The cathode may be any conducting material or a nonconducting material dusted with graphite powder to make it conduct. The copper ions formed are reduced at the cathode and copper is deposited there.

\(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}\)

Thus, the material used as the cathode is bronzed.

If copper is used as the cathode, refined copper is obtained. Copper is purified electrochemically to improve its conductivity. The process is called electrorefining. Impure copper is made of the anode whereas the cathode is pure copper. During electrolysis, the less active metals such as gold and silver, which do not get as easily oxidized as copper settle down as ‘anode mud’, at the bottom of the cell. Impurities

such as iron and zinc, which are more easily oxidized than copper, get oxidized at the anode and go into the solution as ions. When the potential difference between the electrodes is carefully controlled, copper is oxidized to copper ions at the anode and Fe²⁺ and Zn²⁺ are not reduced at the cathode. Only Cu²⁺ ions are reduced to form metallic copper, and deposit at the cathode (which can also be a metal other than copper). Thus pure copper is obtained at the cathode.

Corrosion

On exposure to air and water, the surfaces of many metallic objects, in particular those made of iron, rust. Rusting is nothing but corrosion. The presence of an electrolyte accelerates corrosion, which is an electrochemical process. When iron is in contact with even a droplet of water, it is oxidized and releases electrons. These electrons, on coming into contact with air, reduce oxygen forming the hydroxide ion. Thus two half-cells, whose reactions are given below, are created.

Anode: \(\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}\)

Cathode: \(\mathrm{O}_2+2 \mathrm{H}_2 \mathrm{U}+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}\)

The Fe2+ and OH- ions thus formed diffuse together to form insoluble iron (2) hydroxide.

\(\mathrm{Fe}^{2+}(\mathrm{aq})+2\left(\mathrm{OH}^{-}\right)(\mathrm{aq}) \rightarrow \mathrm{Fe}(\mathrm{OH})_2(\mathrm{~s})\)

Ferrous hydroxide is rapidly oxidized by oxygen to rust with an approximate composition of Fe₂O₃. H₂O.

Corrosion may cause severe damage to buildings, ships, bridges and all objects made of metals, especially iron. Therefore prevention of corrosion is very important to avoid accidents such as the collapse of a bridge or the nonfunctioning of a part in an engine. The simplest of the methods is to prevent the metallic surface from coming in contact with air by coating the surface with paint or other chemicals.

Corrosion can also be prevented by coating iron with grease or asphalt or with ceramic enamel as used in sinks, refrigerators, etc. A tough coating of Fe₃O₄ is obtained by exposing iron to superheated steam. We can use alloys of iron such as stainless steel, which is corrosion-resistant.

Metals such as chromium, tin or zinc afford a more durable surface coating for iron than paint. The steel used in automobiles, for example, is coated with zinc by dipping it into a bath of molten zinc. This process is known as galvanization. Zinc is more easily oxidized than iron and whenever oxidation occurs, zinc is oxidized rather than iron.

\(\mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}) \quad E^{\Theta}=0.45 \mathrm{~V}\) \(\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(\mathrm{s}) \quad E^{\ominus}=-0.76 \mathrm{~V}\)

Any oxidation, if at all, of iron is reversed immediately as Zn can reduce Fe₂ to Fe^2-.

Instead of covering the entire surface of a metal with another metal, a simple electrical contact with a second metal can save the first one from corroding. This is called cathodic protection.

Cathodic protection is an important application of electrochemistry used for protecting iron or steel. It is used in underground pipelines that are in contact with soil. The iron present as a component is connected by a wire to a more active metal such as zinc, aluminum, or magnesium. It then becomes a cathode where oxygen is reduced rather than an anode where iron is oxidized. The other metal acts as a sacrificial anode, which corrodes instead of iron.

An electric current flows between the two metals because of the difference in their activities; this results in the corrosion of the more active metal and thus iron remains protected. Taking the example of magnesium as the more active metal, the reactions are as follows.

The active metal is slowly consumed and must be replaced periodically. This method of protection is used for large steel structures such as pipelines, storage tanks, bridges, and ships.

Batteries

Batteries are cells connected together so as to make maximum use of the reaction occurring in the cell and hence deliver a large current. Cells are of two types—primary and secondary.

Primary cells

Cells or batteries that cannot be recharged or become dead after use over a period of time are called primary cells or batteries. Examples include the Daniell cell and batteries used for torches, toys, etc.

A dry cell is a commonly used primary cell. Dry-cell batteries are used to power flashlights, radios, and other portable electronic devices. These are called ‘dry’ cells as the electrolyte inside them is a solid, in fact a paste. Since this type of cell was invented by the French engineer Georges Leclanche, it is also called a Leclanche cell.

The cell consists of a zinc container serving as an anode mid a graphite rod as the cathode. The graphite rod is surrounded by a moist mixture of ammonium chloride, manganese dioxide, zinc chloride and carbon. A porous paper liner serves as a separator of the two compartments and also as a salt bridge. When the cell is put to use, the zinc anode is oxidised forming zinc ions and electrons which flow through the external circuit to the cathode and reduce manganese dioxide at the cathode.

The reactions are:

anode: \(\mathrm{Zn}+2 \mathrm{NH}_4 \mathrm{Cl} \rightarrow 2 \mathrm{Zn}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \text {or } \mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}\)

cathode: \(2 \mathrm{MnO}_2+2 \mathrm{e}^{-}+2 \mathrm{H}^{+} \rightarrow \mathrm{Mn}_2 \mathrm{O}_3+\mathrm{H}_2 \mathrm{O}\)

A cell commonly used in watches and also in pacemakers is the mercury cell. It consists of a zinc anode and a cathode made of steel in contact with mercury(II) oxide (HgO) in an alkaline medium of KOH and Zn(OH)^–,. Zinc is oxidized at the anode and HgO is reduced at the cathode. The anodic and cathodic reactions are as follows.

The small buttom-shaped batteries used in camars, calculators and watches are alkaline cells. Apart from these, dry cells have been replaced by longer lasting alkaline cells. An alkaline cell also consists of zinc and manages dioxide but the electrolyte contains potassium hydroxide instead of ammonium cloride. Under alkaline conditions, the anodic half reaction is

\(\mathrm{Zn}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Zn}(\mathrm{OH})_2(\mathrm{~s})+2 \mathrm{e}^{-}\)

while at the cathode MnOz is reduced to solid Mn₂O₃

\(2 \mathrm{MnO}_2(\mathrm{~s})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}_2 \mathrm{O}_3(\mathrm{~s})+2 \mathrm{OH}^{-}(\mathrm{aq})\)

Alkaline cells have a greater shelf life (as Zn in the cell oxidizes slowly) and also last longer than dry cells as they have greater quantities of reactants than dry cells. Lists some commonly used batteries.

Some commonly used batteries and their applications:

Secondary cells

Secondary cells are galvanic cells whose reactions can be reversed by the application of an external electric potential in a direction opposite to that of the discharge. This recharges the cell. The lead storage battery and nickel-cadmium (NICAD) cells are secondary cells.

The lead-storage or lead-acid battery is a dead one used in an automobile. The electrodes consist of lead-alloy grids; one set of grids is filled with lead(TV) oxide and the other with spongy lead metal The electrolyte used is dilute sulphuric add. When the battery delivers current, lead is oxidized to lead ions, these combine with the sulfate ions of sulphuric add and coat the lead electrode with insoluble lead sulfate.

The electrons released by the anode reaction flow through the external circuit to the cathode and reduce lead(IV) oxide to lead(II) ions, also forming water. These lead(2) ions also combine with sulfate ions forming lead sulfate and coating the cathode. The two half-reactions are

Anode: \(\mathrm{Pb}(\mathrm{s})+\mathrm{SO}_4^{2-}(\mathrm{aq}) \rightarrow \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{e}^{-}[/laetx]

Cathode:

[latex]\mathrm{PbO}_2(\mathrm{~s})+4 \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{PbSO}_4(\mathrm{~s})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

When used in a car, an alternator (power supply) recharges the battery by producing an external potential that pushes electrons in the reverse direction through the cell. The cell becomes an electrolytic cell during recharging. During recharging, the reactions at the cathode and anode are as follows.

Anode:

\(\mathrm{PbSO}_4(\mathrm{~s})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{PbO}_2(\mathrm{~s})+4 \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{SO}_4^{2-}(\mathrm{aq})+2 \mathrm{e}^{-}\)

Cathode:

\(\mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{SO}_4^{2-}(\mathrm{aq})\)

The net reversible cell reaction in the lead storage battery is

\(\mathrm{Pb}+\mathrm{PbO}_2+4 \mathrm{H}_3 \mathrm{O}^{+}+2 \mathrm{SO}_4^{2-} \underset{\text { recharge }}{\stackrel{\text { discharge }}{\rightleftharpoons}} 2 \mathrm{PbSO}_4+6 \mathrm{H}_2 \mathrm{O}\)

NICAD batteries contain nickel and cadmium electrodes. Cadmium acts as the anode and a hydroxide solution forms the electrolyte. NiO2 is reduced at the cathode.

The two half-reactions are:

anode:

\(\mathrm{Cd}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cd}(\mathrm{OH})_2(\mathrm{~s})+2 \mathrm{e}^{-}[/laetx]

cathode:

[latex]\mathrm{NiO}_2(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}(\mathrm{OH})_2(\mathrm{~s})+2 \mathrm{OH}^{-}\)

The net reaction is

\(\mathrm{Cd}+\mathrm{NiO}_2+2 \mathrm{H}_2 \mathrm{O} \underset{\text { recharge }}{\stackrel{\text { discharge }}{\rightleftharpoons}} \mathrm{Ni}(\mathrm{OH})_2+\mathrm{Cd}(\mathrm{OH})_2\)

A solid coating of Cd(OH)₂ forms around the anode and a solid coating of Ni(OH)₂ forms around the cathode when the battery is discharged; these are converted back to the starting materials on recharging.

Earlier, Ni-Cd batteries were used in laptops but improvements were made and Ni-metal hydride (NiMH) batteries were developed. Here the cathode is NiO(OH) as in Ni-Cd batteries but the anode is made of a transition metal alloy such as LaNi₅. The two electrodes are separated by a porous spacer containing an aqueous solution of KOH.

The reactions are

Cathode:

\(\mathrm{NiO}(\mathrm{OH})(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{I})+\mathrm{e}^{-} \rightarrow \mathrm{Ni}(\mathrm{OH})_2(\mathrm{~s})+\mathrm{OH}^{-}(\mathrm{aq})\)

Anode:

\(\mathrm{MH}(\mathrm{s})+\mathrm{OH}^{-} \rightarrow \mathrm{M}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{e}^{-}\)

Lithium-Ion Batteries

High power demands by the late 1990s in high-speed processors, large video displays, and modem laptop computers led to the manufacture of lithium-ion batteries. These are based not on redox chemistry but on the concentration-driven migration of lithium ions. The anode is made of pure graphite and lithium ions are stored in it. The cathode is highly porous, made of transition-metal oxides such as MnOz, and forms a stable complex with lithium ions.

There is a high concentration of Li+ at the anode and a low concentration at the cathode. The migration of Li+ from the anode to the cathode (during discharge) is accompanied by that of electrons in an external circuit from the cathode to the anode. The electrolytes used in these batteries are solutions of lithium salts in non-aqueous solvents such as tetrahydrofuran, ethylene carbonate or propylene carbonate. Aqueous solutions cannot be used as electrolytes, because electrodes, particularly anodes, react with oxygen and water.

The electrode reactions are

Anode: \(\mathrm{Li}(\mathrm{s}) \rightarrow \mathrm{Li}^{+}+\mathrm{e}^{-}\)

Cathode: \(\mathrm{MnO}_2(\mathrm{~s})+\mathrm{Li}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{LiMnO}_2(\mathrm{~s})\)

Fuel cells

In fuel cells, which are again galvanic cells, the electrode materials in the form of gases are supplied continuously to produce electricity. Such cells are used in space shuttles. The anode is a porous electrode with a catalyst such as finely divided platinum or palladium on its surface. Hydrogen gas is diffused through the anode.

The cathode is a porous electrode impregnated with cobalt oxide, platinum or silver as a catalyst. Oxygen gas is diffused through this electrode. The two electrodes are separated by a concentrated solution of sodium hydroxides or potassium hydroxide acting as an electrolyte. The two half-reactions are

Anode: \(2 \mathrm{H}_2+4 \mathrm{OH}^{-} \rightarrow 4 \mathrm{H}_2 \mathrm{O}+4 \mathrm{e}^{-} \quad E^{\ominus}=-0.828 \mathrm{~V}\)

Cathode: \(\mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-} \quad E^{\ominus}=0.401 \mathrm{~V}\)

The net cell reaction therefore leads to the formation of water as steam.

\(2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}\)

Methane or other hydrocarbons can also be used instead of hydrogen. Hydroxide ions formed at the cathode migrate through the electrolyte to the anode where they combine with hydrogen to form water.

When acidic electrolytes have used the reactions at the anode and cathode are anode:

Anode: \(\mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \quad E^\theta=0.00 \mathrm{~V}\)

Cathode: \(\mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \quad E^{\Theta}=1.229 \mathrm{~V}\)

Fuel cells are promising alternative energy sources for automobiles.

Most of the fuel cells used in vehicles have a proton exchange membrane (PEM) between the two halves of the cell. Gas diffusers bring H₂ and O₂ to opposite sides of the membrane. At the anode side, H₂ is oxidized to H+ ions. These ions migrate through die membrane to the cathodic side where they react with oxygen to produce water.

Another large-scale application of fuel cells is in power generation. Such cells have a lot of advantages over conventional thermal plants generating electricity. Their efficiency is 60 per cent more than that of thermal plants. In a thermal plant the liberated free energy of a chemical fuel (on burning) is used to first convert water into steam which is then used to drive a turbine. The turbine coupled with a generator produces electricity. A thermal plant causes pollution due to the burning of fossil fuels. On the other hand, fuel cells emit only water vapour—they do not pollute the air.

Electrochemistry Multiple-Choice Questions

Question 1. The flow of current in an electrolyte is due to the movement of

1. Electrons
2. Protons
3. Ions
4. None Of These

Answer: 3. Ions

Question 2. Metallic conductors conduct electricity through the movement of

1. Ions
2. Metal Atoms
3. Free Electrons
4. None Of These

Answer: 3. Free Electrons

Question 3. Which of the following substances does not conduct

1. Crystalline NaCl
2. Graphite
3. CuSO, solution
4. NaCl crystals with a defect

Answer: 1. Crystalline NaCl

Question 4. Conductance (G) is the reciprocal of electricity.

1. Specific Resistance
2. Current
3. Resistance
4. Concentration

Answer: 3. Resistance

Question 5. The specific conductance (K) or conductivity of any conducting material is defined as the

1. Reciprocal Of Current
2. Reciprocal Of Specific Resistance
3. Reciprocal Of Resistance
4. Product Of Specific Resistance And Current

Answer: 2. Reciprocal Of Specific Resistance

Question 6. The unit of specific conductance (x) is

1. Ωcm
2. Ω^-1 cm
3. -1 cm^-1
4. Ω cm^-1

Answer: 3. -1 cm^-1

Question 7. The cell constant of an electrochemical cell is given by

1. \(\frac{k}{R}\)
2. RK
3. R2k
4. \(\frac{R}{k}\)

Answer: 2. RK

Question 8. Calculate the molar conductivity of a 0.1 M H2SO, solution in S m2 mol^-1 given that its specific conductance is 26 x 102 cm^-1.

1. 26×10^-1
2. 26×10^-4
3. 52×10^-4
4. 52×10-1

Answer: 2. 26×10^-4

Question 9. On increasing the dilution, the specific conductance

1. Increases
2. Decreases
3. Remains Constant
4. None Of These

Answer: 2. Decreases

Question 10. Strong electrolytes are those which

1. Dissolve Readily In Water
2. Conduct Electricity
3. Dissociate Into Ions Even At High Concentration
4. Dissociate Into Ions At High Dilution

Answer: 3. Dissociate Into Ions Even At High Concentration

Question 11. Substances whose aqueous solutions are good conductors of electricity are called

1. Strong Electrolytes
2. Weak Electrolytes
3. Nonelectrolytes
4. Catalysts

Answer: 1. Strong Electrolytes

Question 12. The conductivity of an electrolyte depends on the

1. Molecular Mass Of The Electrolyte
2. The Boiling Point Of The Solvent
3. Volume Of The Solvent
4. Degree Of Ionization

Answer: 4. Degree Of Ionization

Question 13. In a galvanic cell

1. Chemical Energy Is Converted Into Electrical Energy
2. Electrical Energy Is Converted Into Chemical Energy
3. The Cathode Is The Negative Terminal And The Anode Is The Positive Terminal
4. None Of The Above

Answer: 1. Chemical Energy Is Converted Into Electrical Energy

Question 14. A salt bridge may contain

1. A Saturated Solution Of Kcl And Agar-Agar
2. A Saturated Solution Of Kno3 And Agar-Agar
3. A Saturated Solution Of Nh4no3 And Agar-Agar
4. All Of These

Answer: 4. All Of These

Question 15. The emf of a Daniell cell (given below) at 298 K is E.

\(\mathrm{Zn}\left|\mathrm{ZnSO}_4(0.01 \mathrm{M}) \| \mathrm{CuSO}_4(1.0 \mathrm{M})\right| \mathrm{Cu}\)

When the concentration of ZnSO, is 1.0 M and that of CuSO4 is 0.01 M, the emf changes to E2. What is the relationship between E and E2?

1. E₁ >E₂
2. E₁ < E₂
3. E₁ =E₂
4. E₂ = 0 ≠ E

Answer: 1. E₁ >E₂

Question 16. The Nearest equation

\(E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{R T}{n F} \ln Q\)

indicates that the equilibrium constant kc will be equal to Q when

1. \(E=E^{\ominus}\)
2. \(\frac{R T}{n F}=1\)
3. E=0
4. \(E^{\Theta}=1\)

Answer: 3. E=0

Question 17. The value of the reaction quotient Q for the cell

\(\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(0.01 \mathrm{M}) \| \mathrm{Ag}^{+}(1.25 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})\) is

1. 156
2. 125
3. 1.25×10^-2
4. 6.4 x 10^-3

Answer: 4. 6.4×10^-3

Question 18. For the cell reaction

\(\mathrm{Cu}^{2+}\left(c_1, \mathrm{aq}\right)+\mathrm{Zn}(\mathrm{s}) \rightleftharpoons \mathrm{Zn}^{2+}\left(c_2, \mathrm{aq}\right)+\mathrm{Cu}(\mathrm{s})\)

the change in free energy, AG, at a given temperature is a function of

1. In c1
2. \(\ln \left(\frac{c_2}{c_1}\right)\)
3. In c2
4. In (c1+c2)

Answer: 2. \(\ln \left(\frac{c_2}{c_1}\right)\)

Question 19. The standard reduction potentials of three metals A, B and C are +0.5 V, -3.0 V and -1.2 V respectively. Th arrangement of the metals, in order of their reducing power, is

1. B>C>A
2. A > B >C
3. C>B>A
4. A>C>B

Answer: 1. B>C>A

Question 20. For a cell reaction involving a two-electron change, the standard emf of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be

1. 1.0×10^-10
2. 29.5×10^-2
3. 10.0
4. 1.0 x 10¹⁰

Answer: 4. 1.0×10¹⁰

Question 21. Given \(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}=0.337 \mathrm{~V} \text { and } E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}^{\Theta}=0.153 \mathrm{~V} \text {, calculate } E_{\mathrm{Cu}^{+} / \mathrm{Cu}^{\circ}}^{\Theta}\)

1. 0.184 V
2. 0.827 V
3. 0.521 V
4. 0.490 V

Answer: 1. 0.184 V

Question 22. A cell reaction is spontaneous when

1. \(E_{\text {red }}^{\ominus} \text { is negative }\)
2. \(E_{\text {red }}^{\ominus} \text { is positive }\)
3. \(\Delta G^{\ominus} \text { is negative }\)
4. \(\Delta G^{\ominus} \text { is positive }\)

Answer: 3. \(\Delta G^{\ominus} \text { is negative }\)

Question 23. The electrode potential for the following half-cell reactions are

\(\mathrm{Zn} \rightleftharpoons \mathrm{Zn}^{2+}+2 \mathrm{e} ; \quad E^{\ominus}=+0.76 \mathrm{~V}\) \(\mathrm{Fe} \rightleftharpoons \mathrm{Fe}^{2+}+2 \mathrm{e} ; \quad E^{\ominus}=+0.44 \mathrm{~V}\)

The emf for the cell reaction

\(\mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}\)

1. +0.32 V
2. +1.20 V
3. -0.32 V
4. -1.20 V

Answer: 1. +0.32 V

Question 24. 9650 C of charge is passed through a solution of AgNO₃. The mass of silver deposited on the cathode is (Atomic mass of Ag is 108)

1. 1.08 g
2. 10.8 g
3. 21.6 g
4. 100.0 g

Answer: 2. 10.8 g

Question 25. Three faradays of electricity are passed through molten Al₂O₃, an aqueous solution of CuSO, and molten NaCl taken in a separated electrolytic cell. The amount of Al, Cu, and Na deposited at the cathodes will be in the molar ratio of

1. 1:2:3
2. 3:2:1
3. 1:15:3
4. 1.5: 2:3

Answer: 3. 1:15:3

Question 26. Which of the following aqueous solutions remains neutral after electrolysis?

1. CuSO₄
2. AgNO₃
3. K2SO₄
4. NaCl

Answer: 3. K₂SO₄

Question 27. How many faradays are required to reduce one mole of \(\mathrm{MnO}_4^{-} \text {to } \mathrm{Mn}^{2+}\)

1. 1
2. 2
3. 3
4. 5

Answer: 4. 5

Question 28. Aluminum oxide may be electrolyzed at 1000°C to = 96,500 coulombs). The cathode reaction is Al³⁺+ 3e and would require furnished aluminum metal (at. mass = 27 amu, 1 faraday → AL To prepare 5.12 kg of aluminum metal by this method

1. 5.49 × 10 C of electricity
2. 5.49 x 10 C of electricity
3. 1.83×10⁷ C of electricity
4. 5.49×10¹ C of electricity

Answer: 1. 5.49 × 10⁷ C of electricity

Question 29. Copper containing zinc as an impurity is refined by electrolysis. The cathode and anode used are

Answer: 3.

Cathode: Pure copper

Anode: Impure copper

Question 30. Galvanized iron is

1. Iron Coated With Tin
2. Tin Coated With Iron
3. Zinc Coated With Iron
4. Iron Coated With Zinc

Answer: 4. Iron Coated With Zinc

Question 31. Rusting of iron is

1. A Decomposition Process
2. A Photochemical Process
3. An Electrochemical Process
4. A Reduction Process

Answer: 3. An Electrochemical Process

Question 32. In the rusting of iron, which of the following cell reactions occurs at the cathode?

1. Fe²⁺/Fe
2. O₂/H₂O
3. Fe³⁺/Fe²⁺
4. Fe/Fe3+

Answer: 3. Fe³⁺/Fe²⁺

Question 33. In a lead storage battery

1. Pb is oxidized to PbSO, at the anode
2. PbO₂ is reduced to PbSO, at the cathode
3. both electrodes are immersed in the same aqueous solution of H₂SO₄
4. All The Above Are True

Answer: 4. All The Above Are True

Question 34. Which of the following statements is correct in the context of a battery?

1. It is an electrochemical cell.
2. It is used as a source of energy.
3. The stored energy is released during the redox reaction.
4. All of these

Answer: 4. All of these

Question 35. Which of the following reactions is used to make a fuel cell?

1. Cd(s) + 2Ni(OH)₃(s) → CdO(S) + 2Ni(OH)₂(s) + H₂O(1)
2. Pb(s) + PbO2(s) + 2H₂SO₄(aq) → 2PbSO₄ + 2H₂O(1)
3. 2H₂(g) + O₂(g) → 2H₂O(1)
4. 2Fe(s) + O₂(g) + 4H₂(aq) → 2Fe²⁺(aq) + 2H₂O(1)

Answer: 3. 2H₂(g) + O₂(g) → 2H₂O(1)

The Solid State

You know from your previous classes that all three states Of matter are made of postludes, wSScfc osar be the molecules or ions. However, the arrangement of particles differs in each <m fe a gas, this precedes ece apart whereas in a liquid they are closer.

But in both gases and liquids, the oatstoenipszixSes-haver? arrangement and they are in motion. In contrast, in a solid, the particles are tightly packed and; are net free he move due to the strong forces of interaction, between them—they only oscillate about their fared pcasSkaas.

Therefore, solids are rigid and have a definite volume. As you know, liquids too have a despite voters. But they do not have a fixed shape, which solids do.

Crystalline And Amorphous Solids

Based on the order of arrangement of particles, solids can be categorised as crystalline or amorphous. Cryste-fre solids are those in which the particles (atoms, molecules or ions) are arranged in a pattern, that repeats itself over a long range.

Examples are sugar, ice, salt and metals. The particles of amorphous solids, on the other hand, are arranged in a pattern that repeats itself over a very short range. Examples are wax, butter, skin powder, glass and plastics.

Let us now look at a few properties of crystalline and amorphous solids. The differences in their presences arise due to the difference in the arrangement of their constituent particles.

Crystalline solids are anisotropic, meaning that their physical properties, such as electrical resistance, thermal expansion and refractive index, differ when measured along different directions in the same crystal This is because the arrangement of particles differs in different directions.

However, amorphous solids are isotropic, meaning that their physical properties are the same when measured along am. directions. lids are because their constituent particles are more or less randomly arranged.

A crystalline solid has a sharp melting point. In contrast, an amorphous solid liquefies over a range of temperatures. For example, glass and plastics soften over a range of temperatures and can be moulded or Hewn into various shapes, which they retain on cooling. Like liquids, amorphous solids have a tendency to Sow, although very slowly.

Therefore, they are called pseudo solids or supercooled liquids. Glass panes of old buildings are thicker at the bottom than at the top because of this tendency of amorphous solids to Bow.

When you cut a crystalline solid with a sharp object, it breaks into pieces which have smooth and regular surfaces. In other words, you can cut a crystalline solid clearly. If you were to cut an amorphous solid with a sharp 1 object, it would break into pieces with irregular surfaces.

Another important difference between crystalline and amorphous solids is that the former have definite heats of fusion whereas the latter do not.

Types Of Crystalline Solids

Crystalline solids are grouped into different types depending on the nature of the constituent particles and the bond formed between them. Such solids can be classified as ionic, covalent, metallic and molecular.

Ionic solids:

They are composed of anions and cations that are held together by electrostatic forces. For example, common salt, NaCl contains NaT and Cl- ions. Some other examples of ionic solids are CsCl, CaF₂ and ZnS.

Ionic solids have high melting points; they are hard and brittle, and conduct electricity when molten or in solution. Many compounds formed by the reaction of a metallic element with a nonmetallic element are ionic.

Covalent solids:

They are made up of atoms of the same or different elements held together by a network of covalent bonds.

Diamond is the most common example of a covalent solid. Silicon and silicon dioxide are also covalent solids.

These solids are very hard, and strong and have high melting points due to the presence of strong covalent bonds.

Diamond melts above 3500. The oxide of silicon, SiO₂ (commonly called silica), exists in several forms with different crystal structures. Such different forms of the same compound are called polymorphs and the phenomenon, of polymorphism.

SiO₂ does not contain discrete SiO₂ molecules. The large size of Si, together with the decreased tendency of Si to form multiple bonds, leads to the formation of Si—O single bonds in Si04 tetrahedral units. The tetrahedra are linked together in a three-dimensional network by the sharing of each O between two Si.

In a single crystal of quartz, SiO₂, all SiO₄ units are cross-linked into a single unit resulting in a giant molecule or covalent crystal. The strong covalent bonds account for the hardness, very high melting point, and electrical and thermal insulating properties of silica (the localisation of electrons in the covalent bonds prevents their moving freely under an applied electric field).

Metallic solids:

They comprise a network of positive ions whose positions are fixed. The valence electrons are free to move. The force of attraction between the valence electrons and the positive ions is called a metallic bond. Iron, copper and aluminium are examples of metallic solids.

Most metallic solids are very hard and strong, and all of them exhibit high thermal and electrical conductivity; they are malleable and ductile. The melting points of metallic solids vary depending on the strength of metallic bonding. Alkali metals melt below 473 K, most transition metals melt at temperatures above 1273 K. Mercury is a liquid at room temperature.

Molecular solids:

They consist of molecules held together by intermolecular forces. They are classified as polar or nonpolar on the basis of tire type of tire intermolecular force. The physical properties of molecular solids depend on the strength of the intermolecular forces.

In polar molecular solids, the intermolecular forces are dipole-dipole forces. Examples are solid NH₃ and solid HCl. They are soft and do not conduct electricity. Although gases or liquids at room temperature, they melt at higher temperatures than nonpolar molecular solids do. Some polar molecular solids are hydrogen-bonded. Examples are ice and solid ammonia.

Nonpolar molecular solids are bound by van der Waals forces (or dispersion forces), which are very weak. They may comprise small symmetric molecules such as H₂, N₂, O and F₂. Because of the weak intermolecular forces, their melting points are very low (below -73 K). They are soft and do not conduct electricity.

Crystal And Space Lattice

Now let us study the structure of crystalline solids, in which the particles making up the crystal have a basic, periodic arrangement.

A crystal may be defined as a three-dimensional pattern in which a structural motif is repeated in such a way that the environment of every motif is the same throughout the crystal. In other words, the crystal appears exactly the same at one point as it does at a series of other equivalent points. The motif or the repeat unit may be an atom, a molecule or a group of such entities.

To understand the arrangement of atoms/molecules in a crystal, first let us consider a set of parallel motifs separated by a distance say a (the structural motif is represented by a point)

This is a one-dimensional pattern. Now, if we consider the placement of such motifs at a distance b in the second direction, i.e., the y-direction, we can build a two-dimensional pattern by simply arranging the motifs in such a way that in the x-direction, they are separated by a distance a and in the y-direction, by a distance.

If the z-direction is also considered, we get a three-dimensional pattern. If we arrange the points in space so as to represent the actual disposition of the motifs in the crystal, we get what is called a space lattice.

A space lattice is thus a pattern formed by points representing the structural disposition of particles—atoms, of these. Each particle occupies a latte point in the array, Vie npm? the lattice of a crystal into identical parallelepiped by joining the lattice paints with straight Each such pM&fktepip&h with h ’Sat Bask repealing mat of the arrangement of atoms, form or molecule called a unit edh of a nail -cdi in three dimensions generates the entire pattern of a crystal.

Shows the possible unit cells- in a particular two-dimensional array, This arrangement can be extended in a parallel way to a lattice. The choice of a unit cell for a particular crystal depends on various factors such as symmetry and volume since there can be more than one way of placing atoms in a volume. At the present level of learning, we do not go into further details.

Primitive Unit Cell

A space lattice can be divided into unit cells in different ways. The lattice points In a unit cell could be placed only si the comers, or at the comers and inside the body of the cell. The former kind of unit cell Is called a primitive ve trust .cell (Hgnre 1.6).

Wherever the arrangement of particles an a crystal, in each case the lattice contains a volume represented by a unit cell which is regularly repeated throughout the crystal.

In order to describe the geometry of a unit cell for a frre^dimensional lattice, and we must specify the lengths F and c along the three axes, x, y and z, and also the three angles a,p and yin a three-dimensional parallelepiped.

There are 7 types of primitive unit cells and based on these we have seven crystal systems. The simplest unit cell is a cubic unit cell, where all sides are equal and all angles are 901

Cubic Unit Cell

There are three types of cubic unit cells—primitive, body-centred and face-centred. In the primitive unit cell (or a simple cubic unit cell), the atoms are located at the corners of the cell. In the body-centred unit cell, apart from the comers, the atoms are placed at the centre of the cell also. In the face-centred cubic (fee) structure, the atoms are placed at the centre of each face apart from the corners of the cube.

The number of atoms per unit cell varies according to the type of the cell. Assume a three-dimensional arrangement of cubes in space. Imagine four cubes arranged in such a manner that their faces form a square and four cubes arranged similarly above them.

Consider an atom that is at the centre of this arrangement. It is in fact an atom in the comer of a cube. It is being shared by eight cubes. Only 1/8 of the atom belongs to one cube. The same is the case with the other 7 atoms situated at the comers.

Hence, the number of atoms in a primitive cubic unit cell is \(8 \times \frac{1}{8}=1\).

In the case of the body-centred cubic (bcc) structure, the central atom belongs only to one cube. One atom from the comers and one at the centre of the body amount to two atoms per body-centred cubic unit cell. In the face-centred cubic (fee) arrangement, there are four atoms per unit cell. One atom is obtained from the comers as in the case of a primitive cubic unit cell.

Each atom at the face is shared by two cubes; there are six faces and hence \(6 \times \frac{1}{2}=3\) atoms are obtained from the faces.

We have seen that three types of cubic unit cells are possible. Regular stacking of each one of these unit cells results in a particular type of space lattice. Taking into account symmetry considerations (which we will not go into here), it is seen that for the tetragonal crystal system, there are two types of unit cells possible—primitive and body-centred.

In the other crystal systems, one or more than one type of unit cell is possible, amounting to a total of 14 types of unit cells. In 1848, Auguste Bravais showed that the arrangement of these 14 types results in 14 different lattices. They are therefore known as Bravais lattices.

Close Packing In Metallic Crystals

Particles of the same type, like atoms in a metal or anions of an ionic solid, can pack closely resulting in a stable structure of crystals. The particles of a crystal pack in a manner that gives the maximum possible density. Let us consider the various possibilities of this arrangement in a crystal, where the constituent particles are assumed to be spheres and of the same size.

First of all, let us consider the formation of one edge of a crystal. The closest packing in a row of spheres will be one where the spheres are simply touching each other.

In this type of arrangement, each sphere is in contact with two of its neighbours. The coordination number of a particle is the number of its nearest neighbours. In Fluence, in a one-dimensional arrangement of spheres, the coordination number is two.

Now, another row can be aligned below this in any one of the following two ways in order to obtain one face or plane of the crystal. One possibility is that the second row is placed in such a way that each sphere lies exactly above the sphere in the first row; this gives rise to an arrangement of spheres in a square. Hence this arrangement is called square packing.

The coordination number in square close packing in two dimensions is four since each sphere is in contact with the four nearest neighbours. In another type of arrangement, the spheres of the second row can lie in the depressions between the spheres in the first row. The same thing can be repeated for the third row to obtain a hexagonal packing arrangement where one sphere is surrounded by six other spheres in a plane.

The coordination number in the hexagonal close packing in two dimensions is six. Among these arrangements, we find that hexagonal packing results in a close-packed structure leaving less space between the particles of the crystal. If such planes of spheres are stacked one over the other we obtain the complete three-dimensional structure of the crystal.

Let us see how such a structure can be formed. Consider the hexagonal packing of spheres to form the first layer. In this case, a sphere can be surrounded at the maximum by six other spheres of the same size.

Then each of these spheres can again be surrounded by six such other spheres. This arrangement called closest packing can extend indefinitely in a single layer, say layer A.

Now a second layer, B, can be formed above the first layer A still maintaining close packing. If you look at, you will find that there are two types of holes, one type which lies on the same straight horizontal line and the other type which is diagonally opposite to the first ones.

It is not possible to keep spheres of the second layer together on both types of holes. The spheres of the second layer are kept on top of the holes in the first layer so that each particle in layer B is in contact with three particles in layer A.

The third layer can now be stacked above layer B in two ways. In the first arrangement, the spheres are placed exactly as in layer A, so that the third layer resembles layer A and the fourth layer resembles layer B. Thus, the stacking continues as ABABAB …

This arrangement is called the hexagonal closest packing (hep). Examples include Cd, Co, Li, Mg, Na and Zn from the metallic solids. A noncubic unit cell is formed from this arrangement.

In the second type of arrangement, which is called cubic closest packing, the third layer is formed by placing the spheres in the voids or interstices between tire spheres in layer B such that the newly formed layer C does not resemble either layer A or layer B.

Then, the fourth layer is formed by keeping spheres in the voids of layer C and exactly above the spheres in layer A. This is followed by a pattern which is the same as in layer B and then by one which is the same as in layer C.

This type of packing is therefore of the ABCABC … type and is called cubic closest packing or face-centred cubic arrangement (fee) (the smallest unit cell that can describe this arrangement is the face-centred cubic). Examples include Ag, Al, Ca, Cu, Ni and Pb.

Coordination number:

The coordination number of an atom or ion in a crystal is the number of neighbouring atoms or ions it touches. In both the hexagonal closest packed and cubic closest packed structures each particle touches 12 nearest neighbours. This can be understood if we consider three layers at a time and consider a sphere at the centre of a layer.

In the middle layer, there are six nearest neighbours for both hep and cup structures. In the layers above and below, three spheres from each layer touch this central sphere. Thus, in all, there are 12 spheres in close proximity, touching the central sphere. Hence, the coordination number is 12 in both cases.

Packing efficiency in CCP and hep arrangements:

In the help arrangement, there are six atoms belonging to a hexagonal prism, Let us see how. Assume a three-dimensional arrangement of such prisms. There are three spheres in layer B belonging exclusively to this prism. The central sphere at the top of the prism is shared by two prisms (one prism placed exactly above the one shown in the figure). Hence, only half of this sphere belongs to this prism.

There is another sphere of the same kind at the bottom; it also contributes only 1/2 to this prism. Then, the third type of sphere we must consider is placed at the corners. Each sphere in a corner is shared by 6 prisms; 3 in the same plane and 3 above it. There are 12 (6 on top, 6 at bottom) such spheres.

Hence, the spheres from corners contribute to the extent of \(12 \times \frac{1}{6}=2\) for the prism under consideration.

Therefore, the total number of spheres per prism = \(3+\frac{1}{2}+\frac{1}{2}+2=6\)

As you know, the unit cell in a ccp structure is face-centred cubic. You also know that there are four atoms per fee unit cell.

Even when the spheres are closely packed, there is a gap between them due to their shape. Packing efficiency or packing fraction is the fraction of the total space of a unit cell which is occupied by the spheres. To determine packing efficiency, we must first find the total volume of the cell and then the total volume of the spheres. The volume of spheres divided by the volume of the unit cell gives the packing fraction. For a cup structure, the unit cell is the volume of a cube where 7 is the edge length of the cube.

= a³

As there are four atoms per unit cell, the volume occupied by 4 atoms

⇒ \(4 \times \frac{4}{3} \pi r^3\)

where r is the radius of each sphere. Now, we have to relate r and a. As the spheres are in close contact along the face diagonal in a ccp structure, the face diagonal = 4r = √2 a

⇒ \(r=\frac{\sqrt{2}}{4} a\)

Substituting the expression for r from Equation (3) in Equation (2), we get

volume occupied by atoms ⇒ \(4 \times \frac{4}{3} \times \frac{22}{7} \times\left(\frac{\sqrt{2}}{4} a\right)=0.74 a^3\)

Hence, packing fraction ⇒ \(0.74 \frac{a^3}{a^3}=0.74 \text { or } 74 \%\)

Only 74% of the space in a cube is occupied by spheres/atoms in a ccp structure.

Even in a hep structure, the packing efficiency is 74%.

Other structures:

Most metals, other than those cited as forming a hep or a fee structure, crystallise in a bcc structure, where there is one particle in the centre of a cube and one in each corner of the cube. Those particles which are in the same plane do not touch each other.

Here the central particle touches four particles in the upper layer and four in the lower layer, resulting in a coordination number of 8. The crystal structure consists of such repeating units. The particles are not closely packed. Examples of elements with a bcc structure are Cr, Ba, Mo, W and Fe.

There is yet another arrangement possible though again it is not close-packed. It is called the simple cubic structure and is exhibited by polonium.

Here each sphere in a plane touches four of its nearest neighbours and the next layer is stacked in such a manner that each sphere in the second layer is exactly above a sphere in the first layer. This results in an arrangement where the spheres occupy all the comers of a cube and none is at the centre.

The coordination number can be determined by considering three layers at a time. In the middle layer, the central sphere is surrounded by four other spheres and by one sphere each in the upper and lower layers. Thus, the total number of nearest neighbours is 6. In the simple cubic structure, the coordination number of each atom is thus 6.

Packing efficiency In other cubic structures:

In a simple cubic unit cell, the atoms touch each other along the side of the tire cube. If the radius of the atom is jR, the volume of the cell is (2R)³.

The volume of each atom is = \(\frac{4}{3} \pi R^3\)

Therefore, packing efficiency = \(=\frac{\frac{4}{3} \pi R^3}{(2 R)^3}=\frac{\pi}{6}=0.52\)

Only 52% of the total space of the cube is occupied.

In a body-centred cubic unit cell, the atoms on the main body diagonal, and not at the sides, touch each other.

As can be seen, the length of the main body diagonal in terms of the radius of the sphere is 4R. Let us now find the side of the cube.

Let the side of the cube be s. Hence, AB =BC = s.

Consider A ABC. AB and BC are sides of the cube and AC is the face diagonal. ∠B = 90º.

We know that,

⇒ \((A B)^2+(B C)^2=(A-C)^2\)

or, \(s^2+s^2=(A C)^2\)

or, \((A C)^2=2 s^2\)

or, \(A C=\sqrt{2} s,\)

Since BD is also a face diagonal, AC = BD = √2s.

Now, consider A ABD. Again ZB = 90º

⇒ \((A B)^2+(B D)^2=(A D)^2\)

or, \(s^2+(\sqrt{2} s)^2=(A D)^2\)

or, \((A D)^2=s^2+2 s^2=3 s^2 .\)

But AD = 4R.

Hence, \(16 R^2=3 s^2 \text { or } s^2=\frac{16}{3} R^2\)

or, \(s=\frac{4}{\sqrt{3}} R\)

Hence, one side of the cube is \(\left(\frac{4}{\sqrt{3}}\right)^R\)

Volume of the cube = \(\left(\frac{4}{\sqrt{3}} R\right)^3=\frac{64 R^3}{3 \sqrt{3}}\)

Volume of one atom = \(\frac{4}{3} \pi R^3\)

Packing efficiency \(\frac{2 \times \frac{4}{3} \pi R^3}{\frac{64 R^3}{3 \sqrt{3}}}=0.68=68 \%.\)

Voids Or Holes In Close-Packed Structures

In the closest packed arrangement of spheres, the depressions formed by placing the spheres are called voids or holes. We find two types of holes or voids, the octahedral hole and the tetrahedral hole.

An octahedral hole is created when six spheres are in contact while a tetrahedral hole is one formed by four spheres in contact, When the spheres of layer B are placed in such a manner that they occupy the holes in the first layer A, there is a void below every sphere of layer B and above every sphere of layer A.

There are three spheres forming a triangle in one layer and one sphere in another layer, resulting in a total of four spheres around this void. Hence, it is called a tetrahedral void. When the triangles formed in the two layers are placed with their vertices in opposite directions, an octahedral void is obtained. This is obtained when the arrangement in the two layers A and B is such that there is a void in layer B above a void in layer A.

Both the hep and ccp structures have tetrahedral as well as octahedral holes. There are two tetrahedral holes for each atom in both structures. And for each atom, there is only one octahedral hole in the two structures. Shows the octahedral and tetrahedral holes in a ccp unit cell. Such units when repeated give rise to the three-dimensional structure of the crystal.

Location of voids:

Tetrahedral void Consider the fee unit cell. There are spheres at the corners of the cell and also at the centre of each face. Imagine this cube to be divided into eight smaller cubes of equal dimensions.

A close look at each of these smaller cubes shows that all the corners do not contain spheres; in fact, oriy alternate corners do Neither the resulting faces nor the centre contain any sphere.

Out of the eight corners of the smaller cubes, only four are occupied. On joining these four comers containing spheres a regular tetrahedron is obtained It is a Vacant space and the spheres surrounding it form a tetrahedron.

Hence, the void is called a tetrahedral void or tetrahedral hole Since one unit cell can be divided into eight cubes, each of which contains a tetrahedral hole, there are eight such holes per fee unit cell. We know that there are four atoms per fee unit cell and so the number of tetrahedral holes per atom in a fee unit cell is 8 holes/4 atoms = 2 holes per atom.

Octahedral void Again consider a face-centred cubic unit cell. It does not contain a sphere at the centre However, there are six faces, each of which contains a sphere. If the spheres at these faces are connected an octahedron is formed. Hence, we have an octahedral hole at the centre of a fee unit cell. There is an octahedral hole at the edge of the cube also.

In order to locate the hole, consider four cubes of fee unit cells arranged in the form of a rectangular parallelepiped. At the centre of this structure, there is an edge which is shared by all four cubes The spheres at the end of this edge, and those at the faces surrounding this edge, can be connected to form an octahedron. Since there is no sphere at the centre of this octahedron, it is an octahedral hole or octahedral void.

This hole, as can be, is shared by four fee unit cells. So only one-fourth of it belongs to a unit cell. There are twelve edges in a fee unit cell and each of these contains an octahedral hole.

So there are in all \(12 \times \frac{1}{4}=3\) holes per unit cell which are present at the edges.

Apart from this, there is one octahedral hole at the 4 centre which belongs completely to one fee unit cell. In all, there are 3 + 1 = 4 octahedral holes per fee unit cell. Since there are four atoms in a ccp structure, the number of octahedral holes per atom is one.

Ionic Compounds

By now you know that in a crystalline solid, the atoms, ions or molecules are arranged in a definite repeating pattern. The particles behave as though they are spheres and arrange themselves as closely as possible.

The type of packing depends upon of the interaction between the different kinds of spheres. This type of packing of the particles is the basis of the definite structure of crystalline solids.

In the case of ionic compounds, crystals consist of two different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure: therefore depends on the geometry of the ions.

Any ion formed of a single atom behaves as a charged sphere and is surrounded by oppositely charged ions in all directions. Anions tend to be surrounded by cations and vice versa.

The force of attraction between the ions is the same in all directions. Stable ionic crystals result if each ion is surrounded by as many ions as possible of opposite charge and the oppositely charged ions (cations and anions) are in contact with each other. The final structure and the extent of packing is decided by the relative sizes of anions and cations and their relative numbers.

In simple ionic structures, anions, which are generally large, are arranged in a close-packed manner, touching each other with the smallest possible space between them. And cations are found to occupy the voids or holes (the space between the anions).

The holes can be either octahedral or tetrahedral. If the anions that surround the hole are arranged in a tetrahedral manner then the hole is called a tetrahedral hole and the anions are hence four in number. If the anions that surround the hole are arranged in an octahedral manner, an octahedral hole is obtained and we have six anions.

If the cations are small, they tend to occupy tetrahedral holes and those that are relatively bigger occupy octahedral holes. Sometimes the cations may be too large to be able to occupy either of these holes. In such a case, the anions assume a simple cubic structure occupying all the comers of a cube and the cations occupy the cubic holes thus formed, resulting in a body-centred cubic structure.

Let us consider the case where the anions occupy the comers of a cube and the cations occupy all the tetrahedral holes. As the number of one type of atoms/ions per unit cell in a ccp structure is four, the number of anions per unit cell is four. As the tetrahedral holes are eight in number, the number of cations per unit cell is eight.

Hence, the ratio of cations to anions is 8: 4, i.e., 2:1. The formula of the compound is then M₂X, where M is the cation and X is the anion. Similarly, if the cations occupy octahedral holes, the ratio is 1:1 and the formula of the compound will be MX.

Structures Of Ionic Compounds

The relative number of cations and anions varies in different ionic compounds. For the sake of convenience, ionic compounds are divided into four types—MX, MX₂, MX₃ and M₂X.

Ionic compounds of the formula MX generally possess any one of the following structures NaCl or rocksalt structure, caesium chloride or CsCl structure and zinc sulphide, ZnS structure. All these are cubic structures.

Shows the NaCl structure, where both Na+ and Cl^– ions form a fee (face-centred cubic) structure. You can imagine two separate fee lattices formed by Na⁺ and Cl^– ions, which then interpenetrate each other. The coordination number of the cations as well as the anions is 6.

In the structure, the Cs⁺ ion is bigger than the Na⁺ ion. Therefore, the C^– ions occupy the corners of the cube and the Cs⁺ ion is positioned at the centre, resulting in a body-centred cubic structure.

The ionic radius of Cs⁺ is 169 pm and that of Cl^– is 181 pm. When we place Cl^– ions at the comers, each of them touching the adjacent one, a cubic hole is formed and Cs⁺ can occupy this hole, resulting in a bcc structure. The CsCI unit cell can also be drawn with Cs⁺ at the comers and CP at the centre. The bee structure is a result of two interpenetrating simple cubic structures. As against the 6-6 coordination in NaCl, in the CsCI structure, there is an 8-8 coordination.

In the zinc sulphide (ZnS) structure, the large sulphide ions form a face-centred cubic structure that is almost close-packed. The zinc ions occupy alternate tetrahedral holes. Out of the eight tetrahedral sites available, only four sites are occupied.

Only half the tetrahedral sites are occupied as the compound has 1; 1 stoichiometry and for each sulphide ion there are 2 tetrahedral holes.

Both zinc and sulphur have a coordination number of four. ZnS is also known to form a hexagonal structure called the wurtzite structure. Here the sulphide ions form an hep array and zinc ions occupy half of the tetrahedral holes.

Compounds of formula MX₂, which have a 1: 2 stoichiometry (i.e., two anions for every cation), crystallise in the fluorite (CaF₂) structure and the tetragonal rutile (TiO₂) structure. The Ca²⁺ ions in fluorite are in a face-centred cubic arrangement and the fluoride (F^–) ions occupy the tetrahedral holes.

There are, thus, four cations per unit cell and eight anions (equal to the number of tetrahedral holes) resulting in an 8-4 coordination or 1-2 valence type of compound. The coordination of the cation is eight and that of the anion is four.

In the rutile (TiOz) structure, each Ti⁴⁺ ion is attached to 6O^2- ions and each O^2- ion is in turn attached to three Ti⁴⁺ ions. Hence, Ti⁴⁺ ions have a coordination number of 6 and the O^2- ions have a coordination number of 3.

Radius Ratio

The structure of an ionic compound depends largely on the relative sizes of the cation(s) and anion(s) constituting it. Let us see how the sizes of the ions determine the coordination number. Since cations are smaller than anions, let us consider the size requirements for a cation to fit into the interstitial space between anions.

In Δ ABC, cos BCA = cos (30°) = \(\frac{\sqrt{3}}{2}\)

But \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{r^{-}}{r^{+}+r^{-}},\)

where r+ and r- are the radii of the cation and anion respectively.

From Equations 1 and 2, we get

⇒ \(\frac{r^{-}}{r^{+}+r^{-}}=\frac{\sqrt{3}}{2}\)

or, \(\frac{r^{+}+r^{-}}{r^{-}}=\frac{2}{\sqrt{3}}\)

⇒ \(\frac{r^{+}}{r^{-}}+1=\frac{2}{\sqrt{3}}\)

⇒ \(\frac{r^{+}}{r^{-}}=\frac{2}{\sqrt{3}}-1=0.155\)

The radii must satisfy the condition

⇒ \(r^{+}=0.155 r^{-}\)

Based on such calculations, Pauling laid down some rules as criteria for close packing in ionic crystals. These are known as radius ratio rules.

The radius ratio of any ionic compound is given by p = \(=\frac{r_s}{r_1}\), the ratio of the radius h of the smaller ion to the radius of the larger ion. In general, since cations are smaller than anions, the radius ratio is \(\frac{r^{+}}{r^{-}}\).

The cations and anions get as close to each other as possible as determined by their ratio. The larger the radius ratio, the larger is the coordination number of the cation. When the radius ratio is larger, the size of the cation is large and more anions can surround the cation.

Hence, the coordination number is also larger. According to Pauling’s radius ratio rules, if p is between 0.225 and 0.414, then a 4-4 coordination (zinc blende or wurtzite, tetrahedral coordination) is possible, where a cation touches 4 anions], if it is between 1,414 and 0.732 then the compound adepts a 6-6 coordination structure (NaCl, octahedral coordination) and If coordination structure is formed (CsCb cubic holes occupied),

Ionic Radii

X-ray studies of ionic crystals give us information about the dimensions of unit cells from which a of ionic radii can be obtained, assuming that the ions are spheres. Table 1,4 gives some of the Ionic radius values,

Note: pm denotes picometre, which is 1 x 10 ¯¹² metre.

Example 1. On the basis of data given in Tables 1.3 and 1.4, predict the structures of NaCl, KCl and
Solution:

NaCl: From Table 1.4, r_Na+ = 102 pm and r_cl- = 181 pm.

∴ \(\frac{r^{+}}{r^{-}}=\frac{102}{181}=0.564\)

As you can see in Table 1.3, the structure is octahedral. Therefore, there is 6-6 coordination, KCl:  rK = r+ = 138 pm, ra. = r¯ =181 pm,

∴ \(\frac{r^{+}}{r^{-}}=\frac{102}{181}=0.564\)

Li₂S: From Table 1.4, r+ =59 pm and r- =184 pm

∴ \(\frac{r^{+}}{r^{-}}=0.32\) the structure is tetrahedral.

Example 2. The edge length of the unit cell of NaCl is 5,66 A., Assuming close contact of ts a’ and ^– ions, calculate the ionic radius of the Cl Ion given that the ionic radius of the NaT ion is 10² pm.
Solution:

Given

The edge length of the unit cell of NaCl is 5,66 A., Assuming close contact of ts a’ and ^– ions

Along the edge, two Na+ ions and one Cl¯ ion are in close contact. While the two Na+ ions are centred at the corners, each one contributing only a length equal to its radius to the edge length, the Cl ion is in the middle and 2r¯ is its contribution to edge length. If the edge length is a, then

⇒ \(a=2 r^{+}+2 r^{-}\)

= 5.66Å

= 566 pm

r⁺ =102 pm  (given)

∴ \(2 r^{-}=566-2 r^{+}\)

2r¯ = 566 – (2×102)

⇒ \(r^{-}=\frac{(566-204)}{2}=\frac{362}{2}\)

r¯ = 181 pm.

The ionic radius of the Cl” ion is 181 pm.

Calculations Involving Unit Cell Dimensions

Knowledge of the edge length of the unit cell helps us to calculate the density of the crystals and also the mass of atoms in a unit cell. Let us see how density can be calculated.

If a is the edge length of a unit cell of the crystal, the volume is given by the cube of the edge length,

i.e., the volume of the unit cell = a³.

If the mass and volume are known, the density of a unit cell can be determined, which is the same as the density of the substance.

Mass of unit cell = sum of masses of all atoms present per unit cell since in a crystal of a single element all the atoms are of the same type.

Mass of unit cell = number of atoms per unit cell x mass of a single atom = \(n \times \frac{M}{N_{\mathrm{A}}}\)

where M is the atomic mass of the element and NA is the Avogadro constant.

Now,

density of until cell, d= \(n \times \frac{M}{\left(a^3 N_{\mathrm{A}}\right)}\)

If M is in grams and a is in centimetres, then the unit of density is g/cm³.

In the case of molecules, n is the number of molecules per unit cell and M is the molar mass.

Example 1. The density of sodium chloride (NaCl) is 2,163 g cm¯³. Find the edge length of a unit cell of NaCl.
Solution:

Given

The density of sodium chloride (NaCl) is 2,163 g cm¯³.

The molar mass of NaCl = 23 + 35.5 = 58.5 g mol¯¹

Molar volume of NaCl = \(\frac{58.5}{2.163}=27 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\)

Volume of a molecule of NaCl = \(\frac{27}{6.023 \times 10^{23}}=4.48 \times 10^{-23} \mathrm{~cm}^3\) (∵ one mole contains 6.0^-23 x 10^-23 number of molecules).

One unit cell of a fee structure contains 4 molecules.

Hence, volume of unit cell = 4 x 4.48 x 10^-23 = 17.9 x 10^-23 cm³.

The volume of the unit cell in terms of edge length a = a3.

⇒ \(a^3=17.9 \times 10^{-23}\)

∴ \(a=\sqrt[3]{17.9 \times 10^{-23}}=\left(17.9 \times 10^{-23}\right)^{1 / 3} \mathrm{~cm}\)

Example 2. Compound CsCl crystallises in a bcc lattice with a unit cell edge length of 405 pm. Calculate the radius of Ca+ if that of Cr^– is 181 pm.
Solution:

Given

Compound CsCl crystallises in a bcc lattice with a unit cell edge length of 405 pm.

In a body-centred cubic structure, the ions are in close contact along the main body diagonal. In the figure given below the dotted line BC represents the main body diagonal.

Consider the triangle ADB.

If the edge length is a, by Pythagoras’ theorem, a² + a² = square of face diagonal = AB².

∴ AB = √2a.

The face diagonal, one edge and the main body diagonal form a right-angled triangle with the main diagonal as a hypotenuse Face diagonal (AG4B).

Hence, a² + {AB)² -(BC)²

⇒ a² +2a² – (BC)²

⇒ BC = √3a.

But along the main diagonal, we have two anions and one cation in contact; its length is \(2 r^{+}+2 r^{-}\)

⇒ \(\sqrt{3} a=2 r^{+}+2 r^{-}\)

⇒ \(\sqrt{3} \times 405 \mathrm{pm}=2 \times r^{+}+2 \times 181\)

⇒ \(\frac{(\sqrt{3} \times 405)-(2 \times 181)}{2}=r^{+}\)

r⁺=169.7 pm

The ionic radius of the Cs+ ion ≈ 170 pm.

Example 3. NaCl forms a fee structure. If the ionic radii of Na⁺ and Cl^– are 10² pm and 180 pm respectively, find the edge length and hence the volume of a unit cell. Also, calculate the density (the atomic masses of Na and Cl are 23 and 35.5 respectively).
Solution:

Given

NaCl forms a fee structure. If the ionic radii of Na⁺ and Cl^– are 10² pm and 180 pm respectively

Since the ions are closely packed and touching each other, edge length

a = 2r⁺ +2r^– =2×102 + 2×180 = 204+ 360

a =564 pm = 564 x 10^-10 cm.

Volume of unit cell = \(a^3=(564)^3 \times\left(10^{-10}\right)^3=1.79 \times 10^{-28} \mathrm{~cm}^3\)

Molar mass of NaCl = 23 + 35.5 = 58.5 g.

Mass of one molecule of NaCl = \(\frac{58.5}{N_{\mathrm{A}}}\)

There are 4 molecules of NaCl per unit cell.

Hence total mass of unit cell = \(\frac{4 \times 58.5}{6.023 \times 10^{23}} \mathrm{~g}=3.885 \times 10^{-22} \mathrm{~g}\)

Density = \(\frac{\text { mass }}{\text { volume }}=2.16 \mathrm{~g} \mathrm{~cm}^{-3}\)

Example 4. Europium (Eu) is the block element present in the lanthanide series with atomic mass 152. It crystallises in a body-centred cubic structure. Its density is 5.26 g cm¯³. Calculate the radius of an Eu atom.
Solution:

Given

Europium (Eu) is the block element present in the lanthanide series with atomic mass 152. It crystallises in a body-centred cubic structure. Its density is 5.26 g cm¯³.

The body diagonal of the cube = √3 a.

There are three atoms of Eu along this diagonal (one full and 2 half)

Hence, 4K = √3a

⇒ \(R=\frac{\sqrt{3} a}{4}\) (R = radius of atom).

There are 2 atoms in a unit cell of a bcc structure.

∴ \(\rho=\frac{2 \times M}{a^3 \times N_A}\)

⇒ \(5.26=\frac{2 \times 152}{a^3 \times 6.023 \times 10^{23}},\)

⇒ \(a=\sqrt[3]{\frac{2 \times 152}{5.26 \times 6.023 \times 10^{23}}} \mathrm{~cm}=4.58 \times 10^{-8} \mathrm{~cm}\)

= 485 pm.

∴ \(R=\frac{\sqrt{3} \times 458}{4}=198.3 \mathrm{pm}\)

This problem can be tackled in another way.

Alternative method:

Molar volume (volume occupied by 1 mole) = \(\frac{1}{5.26} \times 152=28.9 \mathrm{~cm}^3\)

Volume actually occupied by the Eu atoms without the empty space between them = 28.9 x 0.68 = 19.7 cm³ mol¯¹,

where 0.68 is the packing efficiency.

Volume of one atom = \(\frac{19.7}{6.023 \times 10^{23}}=3.26 \times 10^{-23} \mathrm{~cm}^3=\frac{4}{3} \pi R^3\)

Radius \(\left(\frac{3 V}{4 \pi}\right)^{1 / 3}=1.98 \times 10^{-8} \mathrm{~cm}=198 \mathrm{pm}\)

Example 5. KCI forms a fee structure. Its density is 1.984 g cm³. If the edge length of its unit cell is 629 pm, find the molar mass of KCI.
Solution:

Given

KCI forms a fee structure. Its density is 1.984 g cm³. If the edge length of its unit cell is 629 pm

Edge length = 629 pm = 629 x 10^-12 m = 629 x 10^-10cm.

Density = \(\frac{\text { mass }}{\text { volume }}\)

⇒ \(1.984=\frac{n \times M}{a^3 N_{\mathrm{A}}}\)

⇒ \(1.984=\frac{4 \times M}{\left(629 \times 10^{-10}\right)^3 \times 6.023 \times 10^{23}}\)

∴ M = 74.3.

the molar mass of KCI = 74.3 g mol-1

Example 6. The compound CuCl has a structure similar to that of ZnS. The edge length of the unit cell is 500 pm. Calculate the density (atomic masses: Cu = 63, Cl = 35.5).
Solution:

Given

The compound CuCl has a structure similar to that of ZnS. The edge length of the unit cell is 500 pm.

As you already know p = \(\frac{n M}{a^3 N_{\mathrm{A}}}\)

where p is the density, n is the number of atoms per unit cell, M is the molar mass, a is the edge length and NA is the Avogadro constant.

Therefore, p = \(\frac{4 \times 98.5}{125 \times 10^{-24} \times 6.02 \times 10^{23}}=5.22 \mathrm{~g} \mathrm{~cm}^{-3}\)

(In case of an fcc lattice, n=4).

Example 7. Chromium has a monatomic body-centred cubic structure. The edge length of its unit cell is 300 pm. What is i density (atomic mass of Cr = 52)?
Solution:

Given

Chromium has a monatomic body-centred cubic structure. The edge length of its unit cell is 300 pm.

a = 300 pm = 300 x 10^-10 cm.

⇒ \(\rho=\frac{n M}{a^3 N_A}=\frac{2 \times 52}{\left(300 \times 10^{-10}\right)^3 \times 6.023 \times 10^{23}}=6.39 \mathrm{~g} \mathrm{~cm}^{-3}\)

Example 8. An element has a ccp lattice with a cell edge length of 500 pm. The density of the element is 10 g/cm³. How many atoms are present in 270 g of the element?
Solution:

Given

An element has a ccp lattice with a cell edge length of 500 pm. The density of the element is 10 g/cm³.

Volume of one unit cell = \(=(300 \mathrm{pm})^3=\left(300 \times 10^{-10} \mathrm{~cm}\right)^3\)

⇒ \(=27 \times 10^6 \times 10^{-30} \mathrm{~cm}^3=27 \times 10^{-24} \mathrm{~cm}^3\)

⇒ \(2.7 \times 10^{-23} \mathrm{~cm}^3\)

Volume of 270 g of element = \(\frac{\text { mass }}{\text { density }}=\frac{270 \mathrm{~g}}{10 \mathrm{~g} / \mathrm{cm}^3}=27 \mathrm{~cm}^3 \text {. }\)

Number of unit cells in this volume = \(\frac{27}{2.7 \times 10^{-23}}=10^{24}\)

Since there are 4 atoms in one unit cell of a ccp structure, in 10²⁴ unit cells there will be 4 x 10²⁴ atoms.

Example 9. An element crystallises in a cubic lattice and is found to have a density of 705 g/cm³. The edge length of a unit cell is 250 pm and there are 4 x 10²⁴  atmos in 234 g of the element. Find the number of atoms in a unit cell.
Solution:

Given

An element crystallises in a cubic lattice and is found to have a density of 705 g/cm³. The edge length of a unit cell is 250 pm and there are 4 x 10²⁴  atmos in 234 g of the element

Proceeding as in the previous example, the volume of one unit cell=\(\left(250 \times 10^{-10} \mathrm{~cm}\right)^3=15625 \times 10^{-2} \mathrm{~cm}^3.\)

Number of unit cell in this volume = \(\frac{\text { mass }}{\text { density }}=\frac{234}{7.5}=31.2 \mathrm{~cm}^3\)

Given that the number of atoms in 234 g of the elements is 4 x 10²⁴, the number of atoms in one unit cell = \(\frac{4 \times 10^{24}}{2 \times 10^{24}}=2\).

Hence, there are 2 atmos per unit cell.

Imperfections In Solids

We have studied different types of lattices and you may think that all crystalline solids are perfect in that all unit cells consist of a perfect arrangement of atoms/ions/molecules and the unit cells line up sequentially to form a three-dimensional space lattice with no distortion.

However, only a few crystals have a complete and perfect arrangement of their entities. There may be fewer or more entities placed randomly in a unit cell.

Most crystals suffer from such types of imperfections, which are called defects. Crystal defects occur in points, along lines or along planes. Based on the number of dimensions to which they extend they may be classified as point defects (0 dimension, 0 D defect) involving one or two lattice sites, line defects (1 D defect, involving a row of a lattice), plane defects (2 D defect, when a plane of sites is imperfect) or bulk or volume defects extending in all the three dimensions.

These defects have an important effect on the mechanical, electrical, magnetic and optical properties of solids. Here we will confine ourselves to point defects. Point defects arise due to imperfections at one or more points in a lattice.

Point Defect

A point defect in a crystal may result from any of the following:

1. The creation of the vacancy
2. The presence of an atom/ion of the parent compound in the interstice
3. Substitution of an atom/ion by another atom/ion (the presence of an impurity)
4. The presence of interstitial impurities

The first two are called stoichiometric defects since the}- do not disturb the stoichiometry of the compound, the number of positive and negative ions are in the same ratio as indicated by their chemical formula.

The third one is classified as an impurity defect while the fourth one results in a nonstoichiometric defect due to deviation from the ideal stoichiometric composition of the compound.

Stoichiometric defect:

1. Vacancy defect f creation of vacancy) An atom may be missing from the place where it ought to be present, resulting in the creation of vacancy. Such a defect is called a vacancy defect- The density of the substance decreases due to this defect

2. Interstitial defect When an atom (or ion or molecule) occupies an interstitial site (space between atom/ Ions in the crystal), It results in an interstitial defect in the crystal. There is an increase in the density of the crystal in this case.

In the case of ionic solids, one or a combination of both of the above two defects may be observed. The imperfections in such solids are mainly classified as Schottky defects and Frenkel defects.

Schottky defect In an ionic solid, when a vacancy at an anionic site is compensated for by a vacancy at a cationic site, thus maintaining electrical neutrality, the defect is termed a Schottky defect.

The greater the number of Schottky pairs, the lower the density of the solid. This is the principal defect in alkali halides. Such defects are responsible for the optical and electrical properties of Nad 1 in KT pairs are vacant in Nad at room temperature.

Frenkel defect When an ion occupies an interstitial site (in between the ions) causing a vacancy at the original place, the defect is called a Frenkel defect A vacancy defect occurs at the original site resulting in an interstitial defect at the new site.

Here the density and the electrical neutrality of the solid are the same as that of the normal solid. Such defects occur when the anion and cation differ greatly in size so that the smaller ion can fit into the interstitial sites. A Frenkel defect in AgCl (which also has a Nad structure).

CaF₂ also has predominantly Frenkel defects but here it is the anion which occupies the interstitial site. SrF₂, PbF₂, ThO₂, UO₂ and ZrO, are other compounds shouting Frenkel defects similar to those in CaF¯.

Impurity defects (presence of substitutional impurity):

In ionic compounds, a cation may be replaced by another cation of similar size. Hus is what is called a substitutional impurity. For example, during the precipitation of BaSO₄ from a solution, if some Sr²⁺ ions are also present, then the Sr^2- ions occupy the lattice of BaS04 crystals. Similarly, some of the sites of Na⁺ ions in NaCl may be occupied by Sr²⁺ ions if molten Nad is crystallised in the presence of SrCl₂.

Nonstoichiometric defects:

When there is a change in the stoichiometry of the solid due to imperfections, the defect is called a nonstoichiometric defect There are two types of such defects—the metal-excess defect and the metal-deficiency defect.

Metal-excess defect This defect may arise due to the creation of anionic vacancies or dun-to (ho presence of extra cations at the interstitial sites, An example of the former type Is shown when Mt in cut rifle of Na vapour, This results in the formation of a colour centre, which imparls yellow Colour to Mat I crystals, 1 his is discussed in the section on colour centres.

The second type of metal-excess defect is observed in nQ, Here (the extra %w’ cations occupy the jMfefsHlidi sites and electrons occupy other interstitial sites to maintain electrical neutrality.

This defers is observed in chrysalis exhibiting Frenkel defect, When nO is healed, it loses oxygen and electrons to form it 31 and turns yellow in colour. These excess cations are trapped in interstitial sites with the electrons occupying the nearly interstitial sites.

Effects Of Imperfections In Solids

ionic conductivity in solids:

The most important aspect of point defects is that they make 11 possible for atoms or ions to move through the structure. If there were no defects it would be difficult to imagine the movement of ions in the la I life, Two mechanisms are possible—the vacancy mechanism (which may lie described as the movement of Him vacancy rather than the ion) and the interstitial mechanism, These form what is called Die hopping model.

Colour centres:

Sodium chloride is colourless when pure at room temperature, It is interesting to note that when it is healed in sodium vapour (for that matter any alkali halide in the alkali metal vapour) it allows a greenish-yellow colour.

In this process, NnCl becomes slightly nonstoichiometric to give \(\mathrm{Na}_{1+\delta} \mathrm{Cl} \text { with } \delta \ll 1\) What happens Is that,

When alkali metal halides are heated in Hie alkali metal vapour, the anions migrate to the surface of the crystal to combine with Hie neutral metal atoms. The metal atoms ionise by losing an electron and then combine with the anions to form the salt, As a result, anion vacancies are produced and the electrons released by the metal occupy the anionic vacancies.

The anionic sites occupied by electrons are called N’enlies, (F stands for Farbenzentre, which means colour centre in Herman.) A series of energy levels are available for Hie alec! runs in the colour centres and the energy required to transfer from one level to another falls hi the visible range of electromagnetism radiation.

When visible light falls on the crystal, the excitation of the electrons imparts a characteristic colour (yellow in ease of NaCl) to the crystal. Similarly, KCl heated in potassium vapour exhibits a violet colour and LiCl heated in Li vapour shows a pink colour.

Electrical Properties Of Solids

Solids can be conductors, semiconductors or insulators depending upon whether and to what extent they conduct electricity. Electrical conductivity is determined by the ease of movement of electrons past the atoms under the influence of an electric field.

It is the reciprocal of electrical resistivity and is measured in Sm^– (siemens per metre). Metals are good conductors of electricity with conductivity in the order of 106 to 108 Sm~ The conductivity in metals depends on the number of electrons available for conduction.

Semiconductors have conductivity in the range of 1Q^– to 10′ Sm^-1 while insulators have a conductivity of 10^-13 Sm^-1. A semiconductor is a substance with a conductivity which can be varied by several orders of magnitude by altering its chemical composition or by increasing electrical potential.

One important characteristic of metals is their ability to conduct electricity which increases as the temperature decreases. In contrast, insulators and semiconductors show a decrease in v conductivity as the temperature is decreased.

Ionic compounds are ordinarily nonconductors in the solid state at room temperature. However, there is a significant increase in conductivity on increasing the temperature or if an impurity is present.

Sometimes, a nonstoichiometric defect can make an insulator a semiconductor. For example, NiO is known to have variable stoichiometry. When prepared at a low temperature (1100 K), it is an insulator, pale green and its stoichiometry is 1: 1.

If nickel is oxidised in the presence of an excess of oxygen at 1500 K, the stoichiometry of the compound changes and is obtained. In this process, oxygen molecules are absorbed onto the surface, dissociate and undergo a redox reaction with some Ni2+ ions to form Ni3+ and O2- ions.

To restore electrical neutrality, some Ni2+ ions diffuse out to the surface and leave cation vacancies inside tire crystals. The colour of the compound so formed is black and it is a moderately good electrical conductor.

NiO can be represented as \(\mathrm{Ni}_{1-3 x}^{2+} \mathrm{Ni}_{2 x}^{3+} \mathrm{v}_x \mathrm{O}\), where v denotes vacancies and x is the number of vacancies. Accordingly, the number of Ni²⁺ and Ni³⁺ ions are indicated in NiO. In this case, there can be movement of electrons from Ni²⁺ to Ni³⁺; the Ni³⁺ ion is thus converted to Ni²⁺.

This leads to the formation of Ni³⁺ at another ‘ point. This ion (Ni³⁺) can again be converted into Ni²⁺ by the movement of an electron. This phenomenon involving the movement of electrons is responsible for the conduction of electricity. NiO is also called a hopping semiconductor.

The tire difference between the conductance of a conductor, that of a semiconductor and that of an insulator is explained based on the band theory of solids. According to the molecular orbital theory (MO theory), when atomic orbitals (AOs) on two atoms combine or mix, they form sets of higher-energy (called antibonding) and lower-energy (called bonding) molecular orbitals (MOs).

The total number of MOs obtained is equal to the number of AOs that combine. As the number of atoms in a molecule increases, so does that of MOs. With the rise in the number of MOs, their energy differences become small and we obtain a continuous band of MOs or energy levels.

Consequently, the MO theory for metals is called band theory. Let us take the example of sodium metal. The valence orbital in sodium is 3s. In diatomic sodium, the 3s orbitals of the two Na atoms combine to form one bonding and one antibonding molecular orbital.

When it is triatomic, three 3s orbitals will combine and when it is Na, n orbitals will combine. The difference in energy between successive MOs in a Na molecule decreases with the increase in the number of sodium atoms forming a continuous band.

The lower band is made of bonding MOs while the upper one is made of antibonding MOs. The rules for filling the MOs are the same as for AOs, with each MO accommodating 2 electrons and the lower-energy MO being filled first followed by the higher-energy MOs. Hence, the lower band is filled with electrons but the upper band is empty. This may be seen from the arrangement of electrons.

In the presence of electrical potential, the electrons can be shifted from one set of energy levels to the other or more precisely from the lower band to the upper band. If the upper band is filled, no such shifting can

In the case of magnesium (electronic configuration [Ar] 3s²) the 3s and 3p orbitals form bands, which overlap in energy, and the resulting composite band is only partially filled. Magnesium is thus a conductor of electricity.

As the valence shells of many metal atoms are only partly filled, all MOs will not be filled; some of the higher MOs will therefore be only partly filled or even empty. The filled band is generally referred to as the tire valence band, and the higher energy band is called the conduction band, In hi case of metals, the two bands overlap and easy electron transfer from the valence to the conduction band can occur, facilitating the conduction of electricity.

Metals are thus good conductors of electricity. However, the electrical conductivity of a metal decreases with increasing temperature. This is because the particles undergo increased vibrational motion about their lattice sites and this vibration disrupts the flow of electrons through the crystal.

In the case of semiconductors, the gap between the valence and tire conduction bands is small. Thus, some thermally excited electrons can move into their conduction band and tire substances can conduct electricity. An insulator, however, has a filled valence band and an appreciable gap between tire valence and the conduction bands

Semiconductors

Silicon and germanium are the most common examples of semiconductors; the width of the forbidden band (band gap which contains no allowed energy states) in them is very small and hence the electrons can easily move from the valence to the conduction band.

These are called intrinsic semiconductors. Their conductivity increases with an increase in temperature because electrons gain thermal energy with the rise in temperature and jump from the valence band to the conduction hand. However, their conductivity is too low for practical purposes.

If small amounts of impurities are incorporated into the lattice of these elements their conductivity increases. Crystal defects arising due to the incorporation of impurities can facilitate the movement of electrons through the solid under the influence of an electric field.

The impurity is called a dopant and the process is called doping- Such materials whose conductivity is controlled by the addition of dopants are called extrinsic semiconductors. Doping alters the number of charge carriers and increases conductivity to a large degree.

Two types of conduction mechanisms may be observed in semiconductors. Any electron that is promoted to the conduction band is a negative charge carrier and moves towards the positive electrode under an applied potential.

The valence electron levels that are left behind in the valence band may be regarded as positive holes. Positive holes move when electrons enter them leaving their own positions. Effectively, therefore, positive holes move in a direction opposite to that of electrons.

If doping increases tire conduction-band electron population, an n-type (negative charge carrier) semiconductor is produced. If doping removes electrons from the valence band, the semiconductor is of the p-type (positive charge carrier). These defects increase with temperature rise and hence the conductivity also increases with temperature.

In order to understand tire concepts better, let us take the example of Si, which is a Group 14 element. Alloys formed from the elements of Group 14 (say Si) and Group 15 are n-type semiconductors. Each atom of the element from Group 15 adds one extra valence electron to the alloy.

Let us see how this happens. Pure, crystalline silicon has four valence electrons at room temperature, forming four covalent bonds with the adjacent silicon atoms in a tetrahedral arrangement. If this crystal is now doped with a Group 15 element, say P, the added P atoms occupy normal Si positions in the structure.

But each P atom has five valence electrons. Of the five valence electrons, four are used to form covalent bonds (as if the dopant were a silicon atom) and one is extra not needed for bonding. According to the band theory, this extra electron occupies a discrete level which is slightly below the bottom of the conduction band.

This level acts as a donor level, meaning that the electrons here have sufficient thermal energy to move into the conduction band, where they are free to move. Since the number of electrons in the conduction band of silicon doped with phosphorus is greater than that in pure silicon, the conductivity of silicon doped in this manner is greater than that of pure silicon.

This type of semiconductor in which atoms with more valence electrons occupy sites in a crystal structure, adding extra electrons (negatively charged particles) to the structure, is called an n-type semiconductor. Germanium doped with arsenic also produces an n-type semiconductor.

If a p-type semiconductor has to be formed from Si, elements of Group 13 (for example Ga) are added to a silicon crystal. (We say that Si is doped with Ga.) Group 13 elements have one valence electron less than the elements of Group 14. Thus, in the case of Si-doped with Ga, the dopant atom has only three valence electrons when it is expected to form four covalent bonds.

Thus, one of the Ga-Si bonds must be deficient by one electron and as per the band theory, the energy level associated with this Ga-Si bond does not form part of the valence band but forms a discrete level just above the valence band. This level is known as the acceptor level since it is capable of accepting an electron.

The gap between the acceptor level and the top of the valence band is small. Consequently, electrons from the valence band have sufficient thermal energy to move to the acceptor level. This leaves behind positive holes in the valence band. These holes can move about in the crystal, contributing to electrical conduction. Shows a comparison of the bands in the three types of semiconductors.

Applications of semiconductors:

Semiconductors find use in transistors, silicon chips, photocells, etc. Doped semiconductors are essential components in the modem solid-state electronic devices, found in radios, televisions, calculators and computers.

The simplest example is the pn junction. In this, silicon is doped in such a way that half is n-type and the other half is p-type. The p-type region has positive charge carriers (holes) while the n-type region has negative charge carriers (electrons).

The electrons in this case can flow spontaneously from the n-type to the p-type region across the junction (band gap) combining with the holes in the p-type region to form negative ions. As electrons diffuse, they leave positively charged ions (donors) in the n region.

Similarly, holes near the pn interface begin to diffuse into the n-type region leaving localised ions (acceptors) with a negative charge. The regions near the pn interface lose their neutrality and become charged forming the space charge region, or depletion layer. The space charge then acts as a barrier to further electron flow.

If an external potential difference is applied to the sample such that the p-type end is positive and the n-type end is negative, current flows through the circuit. Electrons enter the sample from the right-hand electrode, flow through the conduction band of the n-type region, drop into the valence band of the p-type region at the pn junction and then flow through the valence band via the positive holes to leave at the left-hand electrode. A continuous current cannot flow in the opposite direction. The pn junction is a rectifier in the sense that

Gallium arsenide, GaAs, is a semiconductor used in solar cells, LEOs, etc, On applying an external difference, the electrons flow through the pn junction arid and finally fall ink) the holes of the p-type context, During this process they emit energy which is in the form of light (in case of silicon semiconductors tuggy fa released in the form of heat).

Thus, these act as light-emitting diodes (LEDs) in electronic displays, The red light in red laser pointers, bar code readers and CD players is all due to the presence of this semi/x/ruictetor, Partially sub stituting gallium with aluminium or arsenic with phosphorus changes the band gap energy faulting in diodes exhibiting various colours such as yellow, orange and green.

The general formula for such semiconductors is Transistors are typically single crystals of silicon doped to give three zones, They may contain pop or npn junctions and act as voltage amplifiers and oscillators in radios, televisions, hi-fi circuits and in amputees.

There are also controlled-valency semiconductors (in which the valency of the combining atoms is modified m per the need) which find application as thermisters—temperature-sensitive sensors. An example is Li_0.05 Ni_0.05O, It is a hopping semiconductor and its conductivity depends on

Some semiconductors are photoconductive, Le, their conductivity increases greatly on irradiation with light, Amorphous selenium is an example and forms an essential component of the photocopying process.

Superconductivity

In 1911 Kamerlingh Onnes discovered that mercury offers no resistance to the flow of electric current at the very low temperature of 4 K, This phenomenon of offering no resistance to the flow of electricity is called superconductivity. Most metals exhibit superconductivity between 2 K and 5 K,

Since 5 K is a temperature at which one cannot work normally, attempts were made to discover materials offering superconductivity at higher temperatures. Examples of such materials are Tl₂Ca₂ha₂Cu₃O₁₀; (125 K); Bi₂Ca₂Sr₂Cu₃O₁₀ (105 K) and YBa₂Cu₂O₇ (90 K). (The values in brackets indicate the temperatures at which time substances become superconducting,)

Magnetic Properties Of Solids

Before studying the magnetic properties of solids Let us see how a substance behaves under the influence of a magnetic field. A magnetic field produces lines of force that penetrate the medium in which it is applied.

The density of these lines of force is called magnetic flux density, B. The magnetic field H and the magnetic flux density are related as B~p0H in a vacuum, where p() is the permeability of free space.

If a magnetic material is placed in the field, it can change the magnetic flux density. The field of the sample in the applied field is known as its magnetisation, M., Now, the magnetic flux density is given by

⇒ \(B=\mu_0 H\)

There is another parameter, magnetic susceptibility, y (% ~ Magnetisation is usually discussed in terms of χ (chi).

Solids can be broadly classified into two types—(1) those materials which when placed in a magnetic field are drawn away from it and (2) those materials which are drawn towards a magnetic field.

The first type of materials are called diamagnetic materials, In such materials, all the electrons are paired and the magnetic moment of one of a pair is compensated by the equal and opposite moment of the other. When such a material is kept in a magnetic field, the number of lines of force emanating from the magnet is reduced, and the magnetic flux density is reduced.

They are weakly repelled by a magnetic field. Examples of diamagnetic substances are H₂O and C₆H₆, ‘Flic magnetism in the second type of materials is a clue to the presence of Urns, terms or molecules containing unpaired electrons. Magnetic behaviour is mainly exhibited by compounds of transition metals and lanthanides/ many of which possess unpaired d and f electrons respectively.

The unpaired electrons may be oriented M random on the different atoms, when the material is called paramagnetic f, both substances have permanent dipoles and are weakly attracted to a magnetic field. Examples are O₂ and CuO,

Substances which are strongly attracted to a magnetic field are said to be ferromagnetic. They can be permanently magnetised. Examples are iron, cobalt and nickel.

The metal ions of such substances, In the solid state, cluster together in small areas called domains. Each domain behaves like a small magnet. When the sample is demagnetised, the domains are oriented at random and their magnetic moments get cancelled. In the presence of a magnetic field, the domains are permanently aligned in one direction, that of the field, and the sample becomes permanently magnetised.

The orientation of domains in opposite directions leads to antiferromagnetism. The magnetic moments cancel each other out. Examples of antiferromagnetic substances are MnO and NiO, When the magnetic dipoles of the domains are aligned in parallel and antiparallel directions in unequal numbers, a net magnetic moment results and the substance is ferrimagnetic. Examples are ferrites (AFe₂O), where A is a divalent metal) and Fe₃O₄. Ferrimagnetic materials are not so strongly attracted to a magnetic field as ferromagnetic substances.

The magnetic properties are the outcome of the magnetic moments associated with the electrons. The magnetic moment of an unpaired electron arises from two causes, one due to orbital motion around the nucleus and the other due to the spin of the electron. The spin component is of more importance. If the electron is visualised as a bundle of negative charge spinning on its axis, the magnitude of the resulting spin moment pB is 1.73 Bohr magnetons. The Bohr magneton is the fundamental unit of magnetic moment.

It is defined as \(1 \mathrm{BM}=\frac{e h}{(4 \pi m c)}\), where e is the electronic charge, h is the Planck constant, m is the electronic mass and c is the velocity of light. It is equal to 9.27 x 10^-24 A m²,

The spin magnetic moment of an electron is ±ps. The orbital magnetic moment is given by the product of the magnetic quantum number and the Bohr magneton.

The effect of temperature on magnetic behaviour

The temperature dependence of the magnetic susceptibility of a paramagnetic substance is given by Curie’s law.

⇒ \(\chi=\frac{\dot{C}}{T},\)

where C is a constant called Curie constant and T is the temperature in kelvin. Curie’s law takes different forms in the case of ferromagnetic and antiferromagnetic substances as the variation of χ is observed to be different

Two temperatures called the Curie temperature and the N6el temperature, are significant. Ferromagnetic materials change to paramagnetic substances above a temperature called the Curie temperature (Tc). Materials which are antiferromagnetic at low temperatures become paramagnetic above a certain temperature called the Neel temperature (TN).

Metals and alloys:

Iron, cobalt and nickel are ferromagnetic while chromium and manganese are antiferromagnetic at low temperatures. The oxides MnO, FeO, CoO and NiO are all antiferromagnetic at low temperatures and change to paramagnetic above the Neel temperature.

Spinels owe their name to the mineral spinel, MgAl₂O₄.

They have the general formula AB₂O₄, A being a divalent ion and B being a trivalent ion. The commercially important spinels known as ferrites are of the type MFe204; where M is a divalent ion such as Fe²⁺, Ni²⁺, Cu²⁺ and Mg²⁺. They are all either antiferromagnetic or ferrimagnetic. Some garnets [M₃M₂ (SiO₄)₃; where M₁₁ = Ca²⁺, Mg²⁺ or Fe³⁺ and Mm = Al³⁺, Ca³⁺ or Fe³⁺ ] are important ferrimagnetic materials. For example, pyrope (Mg₃Al₂Si₃O₁₂) and andradite (Cr₃Fe₂Si₃O₁₂) are ferrimagnetic.

A major application of ferro- and ferrimagnetic materials is in transformers and motor cores. Magnetic bubble memory devices for storing information use thin films of garnets deposited on nonmagnetic substances.

The Solid State Multiple Choice Questions

Question 1. Bonding in diamonds is

1. Covalent
2. Ionic
3. Dipole
4. Metallic

Answer: 1. Covalent

Question 2. During the formation of a solid,

1. Some Energy Is Lost
2. Some Energy Is Gained
3. Energy Remains Constant
4. Some Energy May Be Gained Or Lost Depending On The System

Answer: 1. Some Energy Is Lost

Question 3. Tetrahedral bonding is characteristic of

1. Ionic Bonds
2. Molecular Bonds
3. Metallic Bonds
4. Covalent Bonds

Answer: 4. Covalent Bonds

Question 4. Ionic solids have

1. A Low Melting Point
2. A High Melting Point
3. A Moderate Melting Point
4. None Of These

Answer: 3. A Moderate Melting Point

Question 5. The bond between ice molecules is

1. Ionic
2. Covalent
3. A Hydrogen
4. Metallic

Answer: 3. A Hydrogen

Question 6. Among the following, the strongest bond is the

1. Ionic Bond
2. Hydrogen Bond
3. Metallic Bond
4. Covalent Bond

Answer: 1. Ionic Bond

Question 7. Ionic bonds are mainly formed in

1. Inorganic Compounds
2. Metals
3. Organic Compounds
4. None Of These

Answer: 1. Inorganic Compounds

Question 8. Molecular solids have

1. Very Low Melting Points
2. Fairly Low Melting Points
3. Very High Melting Points
4. None Of These

Answer: 3. Very High Melting Points

Question 9. Metallic solids are generally

1. Hard And Brittle
2. Soft And Plastically Deformable
3. Malleable And Ductile
4. None Of These

Answer: 2. Soft And Plastically Deformable

Question 10. Among the following, the element which has a covalently bonded crystal structure is

1. Al
2. Pb
3. Ge
4. Bi

Answer: 3. Ge

Question 11. Materials having different properties along different directions are called

1. Isotropic
2. Anisotropic
3. Amorphous
4. None Of These

Answer: 2. Anisotropic

Question 12. The tiny fundamental block which, when repeated in space indefinitely, generates a crystal is called

1. A Primitive Cell
2. A Lattice
3. A Unit Cell
4. None Of These

Answer: 3. A Unit Cell

Question 13. How many basic crystal systems are possible?

1. Four
2. Five
3. Six
4. Seven

Answer: 4. Seven

Question 14. How many types of Bravais lattices are possible in crystals?

1. 7
2. 14
3. 8
4. 5

Answer: 2. 14

Question 15. The number of lattice points in a primitive cell is

1. 4
2. 2
3. 8
4. 1

Answer: 4. 1

Question 16. A unit cell with crystallographic dimensions abc, a \(a \neq b \neq c, \alpha=\gamma=90^{\circ}, \beta \neq 90^{\circ}\), B 90° corresponds to

1. Calcite
2. Graphite
3. Rhombic Sulphur
4. Monoclinic Sulphur

Answer: 4. Monoclinic Sulphur

Question 17. Which of the following structures does a crystal, with a b c and interfacial angles a = B = y = 90°, have?

1. Cubic
2. Tetragonal
3. Trigonal
4. Orthorhombic

Answer: 4. Orthorhombic

Question 18. Cubic close packing of equal-sized spheres is described by

1. AB AB AB AB…
2. AB AC AC AB…
3. ABC ACB ABC ACB…
4. ABC ABC ABC…

Answer: 4. ABC ABC ABC…

Question 19. Which of the following metals has an FCC structure?

1. Al
2. Cu
3. Pb
4. All Of These

Answer: 4. All Of These

Question 20. The percentage of total volume occupied by the particles in a bcc structure is

1. 32
2. 68
3. 50
4. 74

Answer: 2. 68

Question 21. The FCC structure is often called

1. Cubic Close Packed
2. Hexagonal close-packed
3. The Graphite Structure
4. The diamond structure

Answer: 1. Cubic Close Packed

Question 22. The correct order of sizes of voids (holes) is

1. Trigonal > Tetrahedral > Octahedral
2. Hexagonal Close Packed
3. Octahedral > Tetrahedral > Trigonal
4. The Diamond Structure

Answer: 1. Trigonal > Tetrahedral > Octahedral

Question 23. The tetrahedral void has a coordination number of

1. Two
2. Three
3. Four
4. Eight

Answer: 3. Four

Question 24. The octahedral void has a coordination number of

1. Two
2. Six
3. Eight
4. Four

Answer: 2. Six

Question 25. In close-packed (hcp, ccp) arrangements of lattices comprising n atoms of a kind, the number of tetrahedral and octahedral voids present respectively are

1. 2n and n
2. n and 2n
3. n and n
4. 2n and 2n

Answer: 1. 2n and n

Question 26. The coordination number of the simple cubic structure is

1. 6
2. 8
3. 12
4. 4

Answer: 1. 6

Question 27. The effective number of atoms belonging to the unit cell of a simple cubic structure is

1. 8
2. 1
3. 4
4. 6

Answer: 2. 1

Question 28. The atomic packing efficiency of a simple cubic structure is

1. 0.68
2. 0.74
3. 1.00
4. 0.52

Answer: 4. 0.52

Question 29. The only element with a simple cubic structure is

1. Silver
2. Polonium
3. Zinc
4. Iron

Answer: 2. Polonium

Question 30. In the crystal structure of NaCl, the arrangement of Clions is

1. Fcc
2. Bcc
3. Both Fcc And Bcc
4. None Of These

Answer: 1. Fcc

Question 31. Interstitial impurities are a

1. Surface Defect
2. Point Defect
3. Line Defect
4. Volume Defect

Answer: 2. Point Defect

Question 32. In ionic crystals, an ion displaced from a regular site to an interstitial site results in what is called a/an

1. Electronic Defect
2. Schottky Defect
3. Frenkel Defect
4. None Of These

Answer: 3. Frenkel Defect

Question 33. In ionic crystals, the missing of a cation-anion pair results in a/an

1. Electronic Defect
2. Schottky Defect
3. Frenkel Defect
4. None Of These

Answer: 2. Schottky Defect

Question 34. In a solid-crystal lattice, a cation leaves its original site and moves to an interstitial position. The lattice defect is called a/an

1. Interstitial Defect
2. Vacancy Defect
3. Frenkel Defect
4. Schottky Defect

Answer: 3. Frenkel Defect

Question 35. An ionic solid with the Schottky defect contained in its structure

1. An Equal Number Of Cation And Anion Vacancies
2. Anion Vacancies And Interstitial Anions
3. Cation Vacancies
4. Cation Vacancies And Interstitial Cations

Answer: 1. An Equal Number Of Cation And Anion Vacancies

Question 36. For \(\mathrm{NaCl}, r_{\mathrm{Na}^{\prime}} / r_{\mathrm{Cl}}=0.325\) The ratio of the coordination numbers of the ions is

1. 6:6
2. 4:4
3. 8:4
4. 6:12

Answer: 1. 6:6

Question 37. Which of these is found in AgCI?

1. Frenkel defect involving cations
2. Frenkel defect involving anions
3. Schottky defect
4. Interstitial defect

Answer: 1. Frenkel defect involving cations

Question 38. The point defect found in Call Is called

1. Frenkel Defect Involving Cations
2. Schottky Defect
3. Frenkel Defect Involving Anions
4. Edge Dislocation Defect

Answer: 3. Frenkel Defect Involving Anions

Question 39. To prepare an n-type semiconductor, the element to be added to Si is

1. Germanium
2. Arsenic
3. Aluminum
4. Indium

Answer: 2. Arsenic

Question 40. To prepare a p-type semiconductor, the element to be added to Ge is

1. Silicon
2. Arsenic
3. Aluminum
4. Antimony

Answer: 3. Aluminum

Question 41. Which type of atoms are added to Ge to prepare an n-type semiconductor?

1. Trivalent
2. Pentavalent
3. Tetravalent
4. Divalent

Answer: 2. Pentavalent

Question 42. Which type of atoms are added to Si to prepare a p-type semiconductor?

1. Trivalent
2. Pentavalent
3. Tetravalent
4. None of these

Answer: 1. Trivalent

Question 43. An example of a superconductor Is

1. Cu
2. Si
3. Ge
4. Hg

Answer: 4. Hg

Question 44. A superconductor exhibits…… resistance.

1. Small
2. Large
3. Zero
4. Infinite

Answer: 3. Zero

Question 45. One Bohr magneton equals

1. 9.27 × 10^-24 Am²
2. 9.1 x 10^-31 Am²
3. 9.27 x 10^-16 Am²
4. 9.1 x 10^-24 Am²

Answer: 1. 9.27 × 10^-24 Am²

Question 46. The temperature at which an antiferromagnetic substance changes to a paramagnetic substance is called the

1. Curie-Weiss temperature
2. Curie temperature
3. Debye temperature
4. Néel temperature

Answer: 4. Néel temperature

Question 47. The transition from the ferromagnetic to the paramagnetic state takes place at the

1. Curie temperature
2. Curie-Weiss temperature
3. Néel temperature
4. Debye temperature

Answer: 1. Curie temperature

Surface Chemistry

Surface Chemistry Definition: 

Surface chemistry deals with phenomena that occur at surfaces, or interfaces. A surface is an interface where one phase ends and another begins. It is a region in which the properties vary from one phase to another.

The interface is represented by putting a hyphen or a slash between the two bulk phases. For example, the interface between a solid and a gas may be represented as solid gas or solid/gas. An interface may be liquid/vapor, solid/liquid, or solid/gas, among others. No interface exists between gases as they are completely miscible.

The study of surface chemistry is important since many chemical reactions in industry and in biological systems take place at interfaces. Important phenomena that take place at interfaces are corrosion, heterogeneous catalysis, electrode processes, dissolution and crystallization, and so on.

For a proper study of this important branch of chemistry, an extremely clean surface is required. For example, if a metal is involved, a very clean sample stored in a vacuum (at a pressure of 10-8 -10-9 pascal) is required. Otherwise, the surface of the metal will be covered by atmospheric nitrogen and oxygen.

The topics we will cover in this chapter are the adsorption of gases and solutes on solid surfaces, and colloids and catalysis.

Adsorption

The molecules of a gas or a solute are attracted to and retained by the surface of the solid with which they come in contact. For example, if a piece of charcoal is introduced into a closed vessel containing ammonia gas, an appreciable quantity of ammonia is quickly taken up by the charcoal and the pressure of the gas in the enclosed vessel decreases. Not only ammonia but almost all gases (such as H₂, O₂, CO, Cl₂, NH₃, and SO₂) are taken up by charcoal to a greater or lesser degree.

The gas that is thus taken up remains on the surface of the charcoal and does not pass into the interior. The simplest proof of this is that if the same sample of charcoal is more finely divided to produce a greater surface area per unit mass, it can take up more gas. The process of accumulation of any substance on the surface of another is called adsorption.

It is a fairly rapid process, as contrasted with absorption, which is a slow process as it involves diffusion into the interior of the material. The substance on the surface of which the concentration occurs is called the adsorbent and that taken up on the surface is called the adsorbate.

For example, in the example of charcoal adsorbing ammonia, ammonia is the adsorbate, and charcoal is the adsorbent.

How adsorption is different from absorption:

As you know, adsorption is purely a surface phenomenon. However, in absorption, the substance passes through the surface and is distributed throughout the bulk of the solid. For example, when a piece of chalk is dipped in ink, the surface of the chalk retains the color of the ink while the solvent of the ink goes into the bulk due to absorption.

On breaking the chalk, it is found to be white from the inside. Also, anhydrous CaCl₂ absorbs water to form a hydrate while acetic acid is adsorbed from its solution by charcoal. Sometimes the word sorption is used to describe the phenomenon of adsorption and absorption taking place simultaneously.

Examples of adsorption:

1. When a hot crucible is allowed to cool in the open, a film of moisture is formed on the die surface. This is an example of the adsorption of water vapor on the surface of the crucible.
2. Acetic add is adsorbed from its solution by charcoal.
3. When animal charcoal is stirred in a dilute solution of methylene blue, die color of the latter decreases in intensity because the charcoal adsorbs the coloured material and thus decolourises the solution.
4. Water vapor gets adsorbed on the surface of silica gel. Hence, silica gel is used to keep the air in a confined space dry.
5. When a raw sugar solution (yellowish brown) is stirred with animal charcoal, the solution becomes colorless as the coloring matter is adsorbed by the charcoal.

Mechanism Of Adsorption

All particles in the bulk of the adsorbent are surrounded by atoms or molecules of their own kind. Thus, the forces acting between the particles are balanced. However, on the surface, the particles are not surrounded by atoms or molecules of their own kind on all sides.

Therefore, these particles have imbalanced attractive forces. As a result of these forces, the surface of the adsorbent has a tendency to attract and retain the adsorbate molecules. Therefore, the magnitude of adsorption increases with an increase in the surface area of the adsorbent.

When adsorption takes place, there is always a reduction in the attractive forces on the surface. As a result, there is a decrease in surface energy, which appears in the form of heat. This proves that adsorption is an exothermic process.

In other words, adsorption is always accompanied by a decrease in the enthalpy of the system, i.e., is negative whenever a gas is adsorbed; the freedom of movement of its molecules is reduced. As a result, there is a decrease in the entropy of the gas, i.e., AS is negative.

Thus, adsorption results in a decrease in both enthalpy and entropy of the system.

The free energy change (AG) for the process of adsorption is given by

ΔG = ΔH -TΔS

The thermodynamic requirement for a spontaneous (or feasible) process is that, at constant temperature and pressure, there has to be a decrease in Gibbs energy, i.e., AG has to be negative. In the equation AG = AH- TAS, it is possible for AG to be negative only when AH has a sufficiently high negative value since -TAS is positive.

As the process of adsorption continues, AH becomes less and less negative and finally equals TAS. At this point, AG becomes zero and equilibrium is attained. This is called adsorption equilibrium.

Types Of Adsorption

Adsorption is of two types—physisorption (physical adsorption) and chemisorption (chemical adsorption).

Physisorption:

If a gas is held on a solid surface by weak van der Waals forces, the adsorption is termed physisorption or physical adsorption. It is also known as van der Waals adsorption since the forces involved are of the van der Waals type. Absorption of gases by charcoal is an example of physisorption. Let us now discuss some characteristics of physisorption.

Lack of specificity:

Physical adsorption is caused by nonspecific van der Waals forces which act universally between any two molecules. The solid surface (adsorbent) does not show a stronger attraction for permanent gases like H₂, O₂, and N₂.

Nature of adsorbate:

The nature of the gas determines the quantity adsorbed by a solid. It has been found that H₂, O₂, N₂, etc., are adsorbed to a lesser extent and less readily than the more easily liquefiable gases such as NH₃, HCl, and CO₂. The more easily a gas is liquefied (i.e., the higher the critical temperature), the more is physical adsorption. This is because van der Waals forces are stronger near the critical temperature.

Volumes of various gases adsorbed by 1 g of activated charcoal:

A look at Table 5.1 shows that 1 g of activated charcoal adsorbs more SO₂ (critical temperature 630 K) than methane (critical temperature 190 K) and more methane than hydrogen (critical temperature 33 K).

Reversible nature The physical adsorption of a gas by a solid is reversible.

⇒ \(\text { Solid + gas } \rightleftharpoons \text { gas } / \text { solid + heat }\)

According to Le Chatelier’s principle, physical adsorption increases at low temperatures and high pressure and decreases with an increase in temperature and decrease in pressure.

Le Chatelier’s Principle:

Definition:

When a system at equilibrium is subjected to a change, the equilibrium shifts in such a direction as to annul the effect of the change.

Effect of temperature:

Let us consider the reaction

⇒ \(\text { gas }+ \text { solid } \rightleftharpoons \text { gas } / \text { solid }+ \text { heat }\)

Here, the gas is adsorbed on the solid with evolution of heat (exothermic reaction) and desorption takes place1 by absorption of heat (endothermic reaction).

If the temperature of the reaction is increased, the system will absorb heat, and therefore desorption will take! place. Hence in order to effect more adsorption, the temperature of the system will have to be reduced.

Therefore, if a reaction is exothermic then a low temperature will shift the equilibrium in the forward direction. A high temperature favors an endothermic reaction.

An increase in temperature favors endothermic reactions.

A decrease in temperature favors exothermic reactions.

Effect of pressure:

The adsorption of a gas leads to a decrease in pressure. Therefore, according to Le Chatelier’s principle, the magnitude of adsorption will increase with the increase in pressure and vice versa.

Surface area of adsorbent The magnitude of adsorption is directly proportional to the surface area of the adsorbent. So, finely divided metals and porous substances, which have a large surface area for a given mass, are good adsorbents.

Enthalpy of adsorption Since the attraction between gas molecules and a solid surface is only due to weak van der Waals forces, the enthalpy of adsorption is low (20-40 kJ mol^-1 ).

Chemisorption:

When gas molecules or atoms are attracted to and retained by a solid surface through chemical bonds (ionic or covalent), the adsorption is known as chemical adsorption or chemisorption. Many chemisorption processes involve high activation energy.

Chemisorption is, therefore, often referred to as activated adsorption. Sometimes physisorption and chemisorption occur simultaneously. In such cases, it is very difficult to determine the exact kind of adsorption taking place. The physical adsorption of a gas at a low temperature may change to chemisorption at a high temperature.

We shall now discuss some characteristics of chemisorption. High specificity Just like chemical bond formation, chemisorption is highly specific and is caused by the same type of forces that help atoms combine chemically, i.e., by valency forces. Gases that can form compounds with the adsorbent are adsorbed chemically.

Examples are the adsorption of oxygen on a metal surface leading to the formation of an oxide, that of hydrogen on a transition element (such as Ni, Pt, or Pd) with unpaired d-orbitals, leading to the formation of a hydride, and the adsorption of oxygen on activated charcoal, resulting in the production of oxides of carbon.

Irreversibility Chemisorption is usually irreversible as it involves compound formation. Though exothermic, the process is very slow at low temperatures. The energy of activation being high, the rate of adsorption falls off rapidly, making chemisorption too slow to be observed at low temperatures.

So chemical adsorption generally takes place at comparatively high temperatures. As is also true for a chemical combination, an increase in pressure increases chemisorption. This is one of the reasons why high pressure is often used in industrial catalysis.

Surface area Chemical adsorption, too, increases with the increase of the surface area of Jae adsorbent.

Enthalpy of adsorption The enthalpy of chemisorption ranges from 80-240 kJ mol^-1. The high values can be attributed to the fact that chemisorption involves chemical bond formation.

Comparison of physisorption and chemisorption:

Adsorption Isotherm

The amount of gas adsorbed on the surface of the adsorbent depends on pressure and temperature. In other words, the amount of gas adsorbed is a function of temperature and pressure only. Mathematically, this can be represented as

⇒ \(\frac{x}{m}=f(p, T)\)

where x = amount of gas adsorbed on mass m of adsorbent at pressure p and temperature T.

If T is kept constant, then Equation 5.2 becomes

⇒ \(\frac{x}{m}=f(p)\)

Equation (5.3) gives the variation of \(\frac{x}{m}\) with pressure at constant temperature. When \(\frac{x}{m}\) is plotted against p at a constant temperature, the curve obtained is called an adsorption isotherm. Thus, an adsorption isotherm is a mathematical expression or a graphical curve, which represents the variation of adsorption with pressure, at constant temperature.

An increased pressure of a gas causes increased adsorption. The increase of adsorption with increased pressure is experimentally found not to be proportional to pressure but somewhat less. So, Freundlich put the adsorption proportional to a fractional power of pressure. The resulting equation which gives the relationship between the amount adsorbed and the pressure is known as the Freundlich adsorption isotherm.

⇒ \(\frac{x}{m}=k p^{1 / n} \quad \text { or } \quad\left(\frac{x}{m}\right)^n=k p\)

where x is the amount adsorbed by m grams of adsorbent at a pressure p and k and n are constants that depend on the nature of the adsorbent and the gas at a particular temperature.

The relationship is generally represented in the form of a curve the mass of the gas adsorbed per gram of the adsorbent is plotted against pressure. A study of these curves reveals that, at a fixed pressure, there is a decrease in physical adsorption with an increase in temperature.

These curves tend to approach a saturation or limiting value at high pressure. This shows that the Freundlich equation (or Freundlich adsorption isotherm) fails at high pressure.

On taking the logarithms of both sides of Equation 5.4, we have

⇒ \(\log \frac{x}{m}=\log k p^{1 / n}\)

or, \(\log \frac{x}{m}=\log k+\frac{1}{n} \log p\)

This is the equation of a straight line.

The validity of the Freundlich isotherm can be tested by plotting log \(\frac{x}{m}\) along the y-axis (ordinate) and log p along the x-axis (abscissa). If the plot is a straight line, then the Freundlich isotherm is valid. The slope of the straight line = \(\tan \theta=\frac{b}{a}=\frac{1}{n}\) The intercept on the y-axis \(\) gives the value of log k. E

The Freundlich isotherm only gives an approximation of the behavior of adsorption. The values of \(\frac{1}{n}\) lie between 0 and 1 (the probable range being 0.1-0.5).

Thus Equation 5.4 holds good over a limited range of pressure.

When \(\frac{1}{n}\) = 0

⇒ \(\frac{x}{m}=k=\text { constant }\)

Equation 5.6 shows that the magnitude of adsorption is independent of pressure when

⇒ \(\frac{1}{n}\) = 0

When, \(\frac{1}{n}=1\)

⇒ \(\frac{x}{m}=k p\)

or \(\frac{x}{m} \propto p\)

Equation 5.7 shows that the magnitude of adsorption is directly proportional to pressure when \(\frac{1}{n}\) = 1. Both the equations (Equations 5.6 and 5.7) are supported by experimental results.

Adsorption From The Solution Phase

Solid surfaces can also adsorb solutes from solutions. For example, when an aqueous solution of acetic acid is agitated with charcoal, a part of the acetic acid is adsorbed on the charcoal, resulting in a decrease in the concentration of acetic acid.

Also, when a litmus solution is shaken with charcoal, it turns colorless because its coloring matter is adsorbed by the charcoal. To take another example, Mg(OH)₂ becomes blue when it is precipitated in the presence of magnesium reagent (a blue dye). The color so obtained is due to the adsorption of the dye on the precipitate of Mg(OH)₂.

However, adsorption from solutions is much less understood than that of gases by solids.

Adsorption of solutes has the following features:

1. The magnitude of adsorption comes down with the increase in temperature.
2. It rises with an increase in the surface area of the adsorbent.
3. The magnitude of adsorption depends on the concentration of the solute.
4. It also depends on the nature of the adsorbent and the solute.

Hie Freundlich equation approximately describes the behavior of adsorption from solutions. In such cases instead of pressure p we have to use concentration C. Thus the relevant equation becomes

⇒ \(\frac{x}{m}=k C^{1 / n}\)

Taking the logarithm of both sides, we get,

⇒ \(\log \frac{x}{m}=\log k+\frac{1}{n} \log C\)

It has been experimentally shown that the plot of log \(\frac{x}{m}\) against log C is a straight line, which shows the validity of the Freundlich isotherm.

Experimental verification:

Equal volumes of solutions of different concentrations of acetic acid are added to equal amounts of charcoal in four different flasks. The final concentration is determined in each flask after adsorption. The difference in the initial and final concentrations gives the values of x. log \(\frac{x}{m}\) values for the different bottles can be plotted against log C.

Applications of adsorption:

Adsorption finds innumerable applications both in the laboratory and in the industry. We shall now discuss some of the important applications of adsorption.

Creation of high vacuum A partially evacuated vessel is connected to another vessel containing activated charcoal cooled in liquid air. The charcoal adsorbs all the gas molecules from the first vessel, thus creating an almost total vacuum.

Gas masks are used for breathing in areas that contain toxic gases. Such areas include coal mines and places where a major fire has broken out. A gas mask consists of activated charcoal or a mixture of adsorbents which adsorbs all toxic gases and vapours and allows only pure air to pass through the pores.

Removal of colouring matter from solutions Animal charcoal can remove colouring matter from solutions. This property is used for decolorizing sugar solutions.

Heterogeneous catalysis A number of industrial processes involve heterogeneously catalyzed reactions. Most of these reactions take place through the adsorption of reactants on the surface of solid catalysts (adsorbents). Examples include the manufacture of ammonia using an iron catalyst, that of sulphuric acid by the contact process using V₂O₅ as a catalyst, and the hydrogenation of oils employing finely divided nickel.

Separation of inert gases Charcoal adsorbs different gases in varying degrees. This property is used to separate a mixture of inert gases on coconut charcoal at different temperatures.

Curing diseases Many drugs used to kill germs are adsorbed on the surface of such organisms.

Froth floatation In froth floatation, the ore is mixed with pine oil and agitated with water. Air is passed through this solution. The ore particles are adsorbed on the air-oil interface while the impurities remain in water. The adsorbed ore particles come to the surface in the form of foam.

Adsorption indicators Surfaces of silver halide precipitate have the property of adsorbing dyes like eosin and fluorescein. In the case of precipitation titrations (AgNO₃ vs NaCl) the indicator is adsorbed at the endpoint producing a characteristic colour on the precipitate.

Chromatographic analysis based on adsorption is used for the separation, isolation, purification, and identification of the components of a mixture. This is achieved by passing the mixture (dissolved in petroleum ether, benzene, ethyl acetate, or some other solvent) through a column of a suitable adsorbent (alumina or silica gel).

Example 1. What is activated charcoal? How is the extent of absorption by charcoal increased?
Solution:

Activated charcoal:

The extent of absorption by charcoal can be increased by subjecting charcoal to a process of activation. This involves the heating of wood charcoal between 625 K and 1275 K in a vacuum.

During activation, hydrocarbons and other impurities are removed from charcoal and render it more porous, making available a very large surface area for adsorption. The resulting porous charcoal is called activated charcoal.

Example 2. What are the factors on which the amount of gas adsorbed by a solid depends?
Solution:

The factors on which the amount of gas adsorbed by a solid depends

1. The amount of a gas adsorbed by a solid depends upon the
2. Nature of the gas and the adsorbent, the surface area of the adsorbent
3. Temperature and pressure of the adsorbent-adsorbate system.

Example 3. What is desorption?
Solution:

Desorption

Desorption is the process of release of the adsorbed molecules. Desorption may also be called the evaporation of the adsorbed molecules.

Example 4. Why are powdered substances more effective adsorbents than their crystalline forms?
Solution:

Powdered substances have greater surface area than their crystalline forms. The greater the surface area, the greater the adsorption.

Catalysis

For a given temperature and pressure, every chemical reaction occurs at a characteristic rate. If the rate is rapid, the products are formed in a short period of time. If the rate is slow, it takes a long time for the products to be formed. Scientists are interested in discovering ways to increase the rates of slow reactions, particularly when they are important to industrial processes.

In 1835, Berzelius observed that the speed of a number of reactions was enhanced by the mere presence of a foreign substance, which did not apparently take part in the chemical reaction. Let us consider the example of potassium chlorate decomposing to potassium chloride and oxygen when heated to 653-873 K.

2KClO₃ → 2KCl+ 3O₂

The decomposition happens at a lower temperature (473-633 K) if the potassium chlorate is mixed with manganese dioxide at a much-accelerated rate. The amount of manganese dioxide remains unchanged at the end of the reaction and the compound may be used again and again. A substance that can be used to alter the speed of a reaction (usually to speed it up) but itself remains unchanged chemically is called a catalyst and the phenomenon of the speeding up of a reaction using a catalyst is called catalysis.

The speed of a reaction is very important in the chemical industry. (The greater the speed, the more is the productivity.) Catalysts play a major role in this context.

In the manufacture of ammonia, iron is used as a catalyst in the presence of molybdenum to increase the speed of the reaction between nitrogen and hydrogen. Molybdenum acts as a promoter for the iron catalyst. Promoters increase the activity of a catalyst while poisons bring it down.

In the petroleum industry, methods have been found to break up the less useful larger molecules into the smaller, more useful ones. The process is known as cracking. Cracking was first done simply by heating the heavy fractions to a very high temperature (thermal cracking).

More recently it has been found that by using a catalyst, a mixture of aluminum oxide and silicon oxide, cracking can be achieved at a temperature of only 773 K. This is known as catalytic cracking.

Catalysts are also very important in life processes. Photosynthesis is a chemical change that Is carried out in all green plants because of the catalyst chlorophyll which green plants contain.

Countless other chemical changes are carried out in living cells of all kinds. They all depend on biological catalysts called enzymes. Enzymes in our saliva, stomach, and intestines control the digestion of our food.

Though most catalysts help speed up reactions, some slow reactions down. For example, in the decomposition of ammonia on a platinum surface, hydrogen (a product) is strongly adsorbed and inhibits the reaction. Such a condition is called inhibition of the catalyst.

One of the reactants or the products that gets strongly adsorbed on the surface of the catalyst and thereby decreases the reaction rate is called the inhibitor. It is also possible for a reaction to be inhibited by a foreign molecule that does not take part in the reaction.

This type of inhibition is called catalytic poisoning. For example, in the manufacture of sulphuric acid by the contact process, arsenic compounds act as poison for the platinum catalyst in the conversion of SO₂ into SO₃.

Types Of Catalysis

Catalysis may be either homogeneous or heterogeneous.

Homogeneous catalysis:

In homogeneous catalysis, the catalyst and the reactants are present in one single phase (gaseous or liquid). The following are some classical examples of homogeneous catalysis.

1. The oxidation of sulphur dioxide to sulphur trioxide in the presence of nitric oxide as the catalyst in the lead-chamber process for the manufacture of sulphuric acid

⇒ \(2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \stackrel{\mathrm{NO}(\mathrm{g})}{\longrightarrow} 2 \mathrm{SO}_3(\mathrm{~g})\)

All the reactants (SO₂ and O₂) and the catalyst (NO) are in the same phase (gaseous).

2. In the stratosphere, the decomposition of ozone occurs in the presence of Cl atoms as a catalyst

⇒ \(\mathrm{O}_3+\mathrm{O} \stackrel{\mathrm{Cl}}{\longrightarrow} 2 \mathrm{O}_2\)

Cl atoms are formed upon the decomposition of CFCl₃ when the latter is exposed to UV radiation from sunlight.

3. The hydrolysis of ethyl acetate is catalyzed by hydrochloric acid.

⇒ \(\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{l})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \stackrel{\mathrm{H}_3 \mathrm{O}(\mathrm{aq})}{\longrightarrow} \mathrm{CH}_3 \mathrm{COOH}(\mathrm{l})+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})\)

The reactants and the catalyst are in the same phase (liquid).

4. The process of esterification is catalyzed by concentrated

⇒ \(\mathrm{CH}_3 \mathrm{COOH}(\mathrm{l})+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l}) \underset{\mathrm{H}_2 \mathrm{SO}_4(\mathrm{l})}{\stackrel{\text { conc. }}{\longrightarrow}} \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{l})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

The reactants and the catalyst are in the same phase (liquid).

5. The hydrolysis of cane sugar into glucose and fructose is catalyzed by a mineral acid (H₂SO₄)

⇒ \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \stackrel{\mathrm{H}_2 \mathrm{SO}_4(\mathrm{l})}{\longrightarrow} \underbrace{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{aq})+\mathrm{Clucose}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{aq})}_{\text {in solution }}\)

The reactants and the catalyst in this reaction are in the same phase (liquid).

Heterogeneous catalysis:

In heterogeneous catalysis, the catalyst and the reactants are present in different phases.

The following are some classical examples of heterogeneous catalysis:

1. The oxidation of sulfur dioxide into sulfur trioxide in the presence of platinum

⇒ \(2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \stackrel{\mathrm{Pr}(\mathrm{s})}{\longrightarrow} 2 \mathrm{SO}_3(\mathrm{~g})\)

Both the reactants are in a gaseous state while the catalyst is in a solid state.

2. The combination of nitrogen and hydrogen to form ammonia in the presence of iron in the Haber process

⇒ \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \stackrel{\mathrm{Fe}(\mathrm{s})}{\longrightarrow} 2 \mathrm{NH}_3(\mathrm{~g})\)

The reactants are in the gaseous state while the catalyst is in the solid state.

3. The hydrogenation of vegetable oils (unsaturated fatty acids) in the presence of finely divided nickel as a catalyst

⇒ \(\text { vegetable oil }(\mathrm{l})+\mathrm{H}_2(\mathrm{~g}) \stackrel{\mathrm{Ni}(\mathrm{s})}{\longrightarrow} \text { vegetable ghee (dalda) }\)

The two reactants are in different states (liquid and gaseous). The catalyst is in the solid state.

4. The dehydration of formic acid in the presence of Al₂O₃ as a catalyst

⇒ \(\mathrm{HCOOH}(\mathrm{g}) \stackrel{\mathrm{Al}_2 \mathrm{O}_3(\mathrm{~s})}{\longrightarrow} \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\)

The reactant is in the gaseous state while the catalyst is in the solid state.

5. The oxidation of ammonia into nitric oxide in the presence of platinum gauze (catalyst) in Ostwald’s process.

⇒ \(4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \stackrel{\mathrm{Pt}(\mathrm{s})}{\longrightarrow} 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Here, the reactants are in the gaseous state, and the catalyst is in the solid state.

Theories of heterogeneous catalysis:

The two old theories that explain the mechanism of heterogeneous catalysis are the intermediate compound theory and the adsorption theory.

Intermediate compound theory According to this theory, catalysis occurs due to the formation of a reaction intermediate between the reactant and the product. The catalyst is supposed to provide an alternate pathway b) reducing the activation energy between the reactants and products and hence lowering the potential energy barrier as shown in the figure given below.

In view of Arrhenius’s equation

⇒ \(k=A e^{-E_a / R T}\)

it is clear that the lower the value of activation energy (E_a), the greater will be the rate of the reaction.

Adsorption theory The adsorption theory applies only to heterogeneous catalysis. According to this theory, the solid catalyst adsorbs the reactants, which are either in the gaseous state or in solution.

The rise in concentration of the reacting molecules on the surface of the catalyst increases the rate of the reaction. As you know, adsorption is an exothermic process. The heat evolved due to adsorption is used in increasing the rate of the reaction.

The modem adsorption theory is a combination of the intermediate compound theory and the old adsorption theory. According to it, catalytic activity is localized on the surface of the solid catalyst, which has certain active spots at which there are free valencies. The mechanism of catalysis involves the following five steps.

1. The reactants diffuse to the surface of the catalyst.
2. They are adsorbed on the surface of the catalyst.
3. A chemical reaction occurs on the surface of the catalyst through the formation of an intermediate.
4. The products leave the surface of the catalyst to make room for fresh reactant molecules.
5. The reaction products diffuse away from the surface of the catalyst.

When a gas comes in contact with the catalyst’s surface (adsorbent), its molecules form loose chemical bonds with the latter. If different molecules form loose chemical bonds with the catalyst’s surface side by side, they may react with each other, resulting in the formation of new molecules. These new molecules leave the catalyst’s surface to make the surface available for more reactant molecules.

The modem theory explains why the catalyst remains unchanged in mass and chemical composition at the end of the reaction.

Important features of solid catalysts:

Activity The ability of a catalyst to accelerate a chemical reaction is called its activity. The activity of a catalyst largely depends upon the strength of chemical adsorption (chemisorption).

The reactant molecules must get chemically absorbed fairly strongly onto the catalyst. However, they must not be adsorbed so strongly that they stick to surface of the catalyst, leaving no room for other reactant molecules to take their place.

Selectivity Catalysts are highly selective—each of them catalyzes only a particular reaction. Carbon monoxide, on treatment with hydrogen, yields different products with different catalysts.

⇒ \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) \stackrel{\mathrm{Ni}}{\longrightarrow} \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

⇒ \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2(\mathrm{~g}) \stackrel{\mathrm{Cu} / \mathrm{ZnO}-\mathrm{Cr}_2 \mathrm{O}_3}{\longrightarrow} \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})\)

⇒ \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g}) \stackrel{\mathrm{Cu}}{\longrightarrow} \mathrm{HCHO}(\mathrm{g})\)

Shape-selective catalysis by zeolites:

A catalytic reaction that depends upon the size of the pores and the cavities of the catalyst and the size of the molecules of the reactants and products is called shape-selective catalysis.

On account of their honeycomb-like structures, zeolites are good shape-selective catalysts. They are microporous aluminosilicates which have a three-dimensional network of silicates in which aluminum atoms replace some silicon atoms, resulting In an Al-O-Si framework.

Zeolites (natural or synthetic) are healed in a vacuum so that the water trapped in the pores boils off. As a result, the zeolites become porous. The size of the pores ranges from 260 pm to 740 pm. Only those molecules which are small enough can be adsorbed in these pores.

In the petrochemical industry, zeolites are widely used as catalysts for the cracking of hydrocarbons and isomerization. An important zeolite catalyst used in the petroleum industry is ZSM-5 (zeolite sieve of molecular porosity 5). It converts alcohols directly into petrol (a mixture of hydrocarbons) by the process of dehydration.

Enzyme Catalysis

Living plants and animals produce complex nitrogenous organic compounds called enzymes. Enzymes are proteins with molecular weights ranging from 13000 to several million. They form colloidal solutions in water.

Enzymes are effective catalysts that are involved in a wide variety of biological processes. For example, enzymes in the saliva begin the process of breaking starch in food. Some enzymes catalyze the formation of blood clots when necessary; others dissolve the clots after a wound has healed.

Certain enzymes act to enable the body to fight infection or disease. Enzymes can disrupt the cell walls of certain bacteria. Fruits ripen because of the action of enzymes. Enzymes catalyze several reactions that occur in animals and plants. So they are termed biochemical catalysts.

Many enzymes are isolated in pure crystalline forms from living cells. The structures of some of them, e.g., insulin, have been determined and a few enzymes have recently been synthesized. The first enzyme was synthesized in the laboratory in 1969.

Let us now touch upon a few enzyme-catalyzed reactions:

1. Inversion of cane sugar The enzyme invertase converts sucrose (cane sugar) into glucose and fructose.

⇒ \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \stackrel{\text { invertase }}{\longrightarrow} \underset{\text { glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\text { fructose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6^{-}}\)

2. Conversion of glucose into ethyl alcohol The enzyme zymase converts glucose into ethyl alcohol and

⇒ \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{aq}) \stackrel{\text { zymase }}{\longrightarrow} \underset{\text { ethyl alcohol }}{2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{aq})}+2 \mathrm{CO}_2(\mathrm{~g})\)

This process is used in making wine (which contains alcohol) from grapes.

3. Conversion of starch into maltose The enzyme diastase catalyses starch into maltose by partial enzymatic hydrolysis.

⇒ \(\underset{\text { starch }}{2\left(\mathrm{C}_6 \mathrm{H}_{10} \mathrm{O}_5\right)_n(\mathrm{aq})+n \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \stackrel{\text { diastase }}{\longrightarrow}} \underset{\text { maltose }}{n \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})}\)

4. Conversion of maltose into glucose The enzyme maltase catalyses the formation of glucose from maltose and water.

⇒ \(\underset{\text { maltose }}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})}+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \stackrel{\text { maltase }}{\longrightarrow} \underset{\text { glucose }}{2 \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{aq})}\)

5. Decomposition of urea into ammonia and carbon dioxide The enzyme urease catalyzes the formation of ammonia and carbon dioxide from urea and water.

⇒ \(\mathrm{NH}_2 \mathrm{CONH}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \stackrel{\text { urease }}{\longrightarrow} 2 \mathrm{NH}_3(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})\)

6. Digestive enzymes (such as pepsin and trypsin) hydrolyze large protein molecules into smaller groups of proteins. In the stomach, pepsin hydrolyses proteins into peptides while in the intestine, trypsin hydrolyses proteins into amino acids.

7. Conversion of milk into curd Lactobacilli enzymes present in curd convert milk into curd

8. Mycoderma acetic enzymes convert ethyl alcohol into acetic acid.

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CH}_3 \mathrm{COOH}(\mathrm{l})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

9. At 310 K (body temperature), sucrose can be hydrolyzed using an enzyme, suocharase.

⇒ \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \stackrel{\text { saccharase }}{\longrightarrow} \underset{6}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\text { glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6} \text { fractose }\)

10. An enzyme found in the mouth converts sucrose into dextran (a polysaccharide made of glucose units). About 10% of dental plaque is composed of dextran. That is why your dentist tells you not to eat candy.

Summary of some important enzymatic reactions:

Characteristics of enzyme catalysis:

Efficiency Enzymes are extremely efficient. Some operate on as many as 25000 molecules in one second.

Highly specific An enzyme, like any other catalyst, is highly specific—it catalyzes only a certain reaction and not any other. The enzyme urease, for instance, catalyzes the hydrolysis of urea only. The hydrolysis of other amides is not catalyzed by urease.

The enzyme decomposing cane sugar does not decompose malt sugar. Lactic dehydrogenase oxidizes 1-lactic add but not d-lactic add.

Highly active at 298-310 K Enzymes are highly active at 298-310 K. At temperatures above 310 K, enzyme activity is reduced and extreme temperatures can even stop enzyme activity. Below 298 K too, enzyme activity decreases.

Enzymes in the body exhibit the greatest activity at normal body temperature (310 K). At much higher body temperatures, all physiological reactions cease due to loss of enzymatic activity. This is one reason why a high fever is dangerous.

Highly active under optimum pH Each enzyme exhibits maximum activity at its own optimum pH. The optimum pH for most enzymes is in the pH range 5-7.

On either side of that particular pH (or range), the activity of an enzyme is markedly decreased. Amylase in the saliva is the most active at a neutral pH and becomes inactive in the stomach where the pH is 1.6.

Activity increases in the presence of activators and coenzymes Some enzymes are initially in an inactive form. These are called proenzymes. They become active in the presence of another substance called an activatory Activators are generally inorganic ions (such as Ca²⁺, Zn²⁺, Mn²⁺, Mg²⁺, Na⁺ or K⁺, CO²⁺, Cu²⁺).

The catalytic activity of an enzyme is increased when these metal ions are weakly bonded to the enzyme molecule. For example, calcium activates prothrombin. Amylase in the presence of Na^– ions catalyzes normal blood clotting.

A coenzyme is a nonprotein that increases the catalytic activity of the enzyme. Vitamins frequently form part of coenzyme molecules.

Inhibitors, poisons, and promotors Inhibitors reduce enzyme activity. They may be either organic or inorganic molecules. They react either directly with the enzyme or with the activator to prevent the activation of the enzyme. Many poisons are fatal because they inhibit vital enzyme activity

There are two types of inhibitors—competitive and noncompetitive. Since they resemble the substrate in their chemical nature, competitive inhibitors compete directly with the substrate. After linking with such an inhibitor, the enzyme is no longer able to perform its normal function.

Noncompetitive inhibitors react with certain nonspecific functional groups of the enzyme, altering the shape of the active site so that the substrate is not able to fit in.

An example is the cyanide ion. It combines with iron to form a very stable complex, interfering with cytochrome oxidase, which is essential for respiration. Respiration ceases and the person dies.

Mechanism of enzyme catalysis:

The catalytic activity of an enzyme is due to the presence of a specific site on its surface, called the active site. This site is characterized by the presence of functional groups (such as —NH₂, —COOH, —SH, and —OH) which form weak bonds such as ionic bonds, hydrogen bonds, or van der Waals bonds with the molecules of the reactant (substrate).

A substrate that has a complementary shape fits into the active site like a key fits into a lock forming an enzyme-substrate complex. This complex intermediate then breaks apart into the products of the reaction, liberating the original enzyme.

As the enzyme surface has no affinity for the product molecules, the latter leaves the enzyme surface quickly to make room for fresh molecules of substrates to be bound at the active site. Because of this recycling, only small amounts of enzymes are needed.

Example 1. What is the root of desorption in catalysis?
Solution:

The root of desorption in catalysis

Desorption makes room on the surface of the solid catalyst for fresh molecules of reactants to be adsorbed.

Example 2. Why is it necessary to remove carbon monoxide when ammonia is manufactured by the Haber process?
Solution:

Carbon monoxide acts as a poison for the iron catalyst used in the manufacture of ammonia by the Haber process. Carbon monoxide forms pentacarbonyl iron [Fe(CO)₅] with iron at the reaction temperature.

Example 3. Why is a heterogeneous catalyst used in the form of a finely divided powder rather than as something with a smooth surface?
Solution:

For a heterogeneous catalyst to be effective, a large surface area is required. Since a finely divided powder has a large surface area per unit mass, a heterogeneous catalyst is employed in this form.

Example 4. Why is the hydrolysis of an ester slow in the beginning? Why does it become faster after some time?
Solution:

The acid hydrolysis of an ester may be expressed as

The add produced in the reaction acts as a catalyst (autocatalyst). The concentration of H⁺ ions increases due to the formation of the add.

Colloids

Fine particles of sand in water form what is known as a suspension. These particles settle down after a while. In a true solution, such as that of salt in water, the particles do not settle down.

In between the true solution and the suspension, there is a type of system that normally never separates; it is called a colloidal dispersion or sometimes just a colloid. A colloidal solution is also termed a sol in short.

A colloid can be defined as a stable two-phase (heterogeneous) system consisting of finely divided particles (usually a solid) dispersed in a continuous medium (often a liquid).

The finely divided particles form what is called the dispersed phase and the medium is called the dispersion medium. For example, in colloidal gold (also called gold sol), gold is the dispersed phase and water is the dispersion medium.

The basic difference between a true solution and a colloid involves particle size. In a true solution, the particles are ions or small molecules and in a colloid, the dispersed phase may comprise particles of very large individual molecules (e.g., a protein or a synthetic polymer) or an aggregate of many atoms, ions or small molecules.

The size of the dispersed particles in a colloidal system is more than that of the solute particles in a true solution and smaller than that of the particles in a suspension.

The particle size of the dispersed phase in a colloid lies in the range of 1 to 1000 nm (10^-9 to 10^-6 m). In a suspension, the size of the suspended particles is always larger than 1000 nm. In a true solution, the size of the solute molecules is less than 1 nm.

Colloids have a very large surface area per unit mass because they consist of many extremely small particles. Consider a cube of side 1 cm. The total surface area of the cube = 6a² = 6 cm². If the cube is divided into cubes of equal size, then

Volume of each small cube = \(\frac{1}{10^{12}} \mathrm{~cm}^3=10^{-12} \mathrm{~cm}^3\)

The side of each small cube = 10^-4 cm [since volume is (side)³].

surface area of each small cube = \(6 a^2=6 \times\left(10^{-4}\right)^2=6 \times 10^{-8} \mathrm{~cm}^2\).

Total surface area of all the 1012 cubes = \(\left(6 \times 10^{-8}\right) \times 10^{12} \mathrm{~cm}^2=60,000 \mathrm{~cm}^2\)

This large surface area provides them with the property of adsorption, one of their most important properties

Classification Of Colloids

Colloids are classified based on the following criteria:

1. The physical condition of the dispersed phase and the dispersion medium
2. The affinity between the tine dispersed phase and dispersion medium
3. The type of particles of the dispersed phase

One the basic physical condition of the dispersed phase and dispersion medium:

The following eight types of colloidal systems are possible, depending upon whether the dispersed phase and the dispersion medium are solids, liquids, or gases.

Types of colloidal systems:

Many commercial products such as whipped cream and fire-fighting foam are colloids. Whipped cream is a foam (a gas dispersed in a liquid). Most biological fluids are aqueous solids (solids dispersed in water). In a cell, for example, proteins and nucleic acids are particles of colloidal size dispersed in an aqueous solution of ions and small molecules.

Of the eight colloids mentioned in the Table, the most common are sols (solids in liquids), emulsions (liquids in liquids), and gels (liquids in solids). In this chapter, we shall discuss only sols and emulsions.

If water is the dispersion medium, such systems are called hydrosols or simply sols, e.g., ferric hydroxide sol. If alcohol is the dispersion medium, the system is called an alcohol.

On the basis of affinity between the dispersed phase and dispersion medium:

Based on the affinity between colloidal particles and the medium, solid-liquid sols can be classified as lyophilic and lyophobic.

Lyophilic colloids Lyophilic literally means ‘solvent-loving’. Lyophilic colloids include gelatin, gum-arabic, soaps, and starch, all of which have a marked affinity for water. These are simply prepared by mixing the solid (the dispersed phase) with the solvent (the dispersion medium) at a suitable temperature.

If the dispersion medium is separated from the dispersed phase by evaporation, the sol can be made again by simply remixing the dispersed phase with the dispersion medium. That is why these sols are sometimes called reversible sols. These are stable and cannot be easily coagulated.

Lyophobic colloids The word lyophobic means ‘solvent-hating’. When there Is little or no affinity between the dispersed phase and the dispersion medium, the colloid is called a lyophobic colloid.

Examples are gold sols, silver sols, and arsenic sulfide sols. Lyophobic sols cannot be prepared by simply mixing the solid (the dispersed phase) with the solvent (the dispersion medium) at a suitable temperature.

They are made by employing some special methods (as described later). These sols are relatively unstable and can be very easily coagulated by the addition of traces of electrolyte, on evaporation or by shaking.

Sometimes, they are called irreversible sols, because once the dispersed phase is precipitated out, it is not generally easy to get it back in the colloidal state on the addition of the dispersion medium. Lyophobic sols require stabilizing agents for their preservation.

Based on the types of particles of the dispersed phase:

Based on the nature of the particles of the dispersed phase, colloids may be classified as multimolecular, macromolecular, and associated.

Multimolecular colloids When a large number of atoms or molecules are dissolved in a solvent, they tend to form molecular aggregates of colloidal dimensions (diameter < 1 nm). The species formed in this fashion are called multimolecular colloids.

For example, a gold sol may have particles of various sizes possessing several atoms of gold. Similarly, a sulphur sol consists of particles which are aggregates of thousands of S8 molecules held by van der Waals forces. A silver sol (argyle) may contain particles of various sizes, each having several atoms of silver.

Macromolecular colloids True solutions of macromolecules in suitable solvents are called macromolecular colloids because the molecules themselves are large enough to be of colloidal dimensions.

Some naturally occurring macromolecules are starch, cellulose, proteins, and enzymes. Man-made macromolecules such as polyethylene, nylon, polystyrene, and synthetic rubber also belong to this class. These colloids are quite stable.

Associated colloids At low concentrations, some solutions behave as electrolytes. However, at higher concentrations, their solute particles aggregate, and colloids are formed.

Such species are called associated colloids. The molecules of associated colloids have both lyophobic and lyophilic groups. The molecular aggregates are called micelles and usually contain 20-100 molecules. The colloidal properties are due to these micelles.

Micelles are formed above a certain temperature called Kraft temperature (Tfc) and above a particular concentration called critical micelle concentration (CMC). Dyes, soaps, detergents, and many surface-active agents fall into this class. For soaps, the CMC is I ^-4 to 10^-3 mol L^-1.

Mechanism of micelle formation Soaps are sodium or potassium salts of fatty acids and may be represented by a general formula, where R is a straight chain of 10-20 carbon atoms and M⁺ is Na^– or K^–. When a soap, say sodium stearate, is dissolved in water, it dissociates into Na⁺ and RCOCT (carboxylate ion).

The carboxylate ion is composed of a large hydrocarbon group (the hydrophobic (water-repelling) group) called, the tail’ and a polar group (tire hydrophilic (water-loving) group] called the ‘head’.

The \(\mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}^{-}\)ions are, therefore, present on the surface with their —COO^– groups in water and the hydrocarbon group away from water and remaining at the surface.

Long-chain carboxylate ions do not exist as individual ions in aqueous solutions, and at critical micelle concentration, the carboxylate ions are pulled into the bulk of the solution where they arrange themselves in a spherical cluster called a micelle.

Each micelle contains 50-100 long chain carboxylate ions. A micelle resembles a large ball. The polar carboxylate ions are on the outside of the ball because of their affinity for water, and the nonpolar tails are buried in the interior of the ball to minimize their contact with water.

The mechanism of micelle formation in the case of detergents is the same as that in the case of soaps. Detergents, like soaps, have a polar group and a long-chain nonpolar hydrocarbon group that causes the molecules to form micelles in solution.

Cleansing action of soaps and detergents:

Soaps and detergents are composed of molecules containing large hydrocarbon groups [the hydrophobic (water-repelling) group] and one or more polar groups [hydrophilic (water-loving groups). The dirt particles that stick to textile fibers are generally covered by a layer of oil molecules called grease. The dies are and, so, repel water.

Therefore, water cannot wash away such dirt particles from doth by Or* the soap, grease molecules get attached to the nonpolar hydrocarbon tail of the stearate ion, and poor head of stearate ion is directed towards the water.

Since the polar heads of the stearate tons can interact with the wafer., surrounded by the stearate ions is gradually lifted off and pulled in water, The grease finally floats off completely surrounded by detergent (or soap) molecules.

The negatively charged sheath around them prevents them from coming together and forming aggregates. Because of this cleansing action of soap, grease and hence dirt is washed away from doth

Preparation Of Colloids

Preparing lyophilic colloids is fairly simple. Generally, vigorously shaking the dispersed phase in the dispersion medium results in the formation of lyophilic solids. For example, gelatine, starch, and soaps form colloidal solutions by simply dissolving in water.

Due to their low stability, lyophobic colloids present greater difficulty In their preparation. Such colloidal solutions are usually made by the following methods.

Chemical methods:

Many molecules produced in chemical reactions (such as oxidation, reduction, hydrolysis, and double decomposition) aggregate to give particles of colloidal dimensions. For example, sulfur solids can be prepared by the oxidation of hydrogen sulfide.

⇒ \(2 \mathrm{H}_2 \mathrm{~S}+\mathrm{SO}_2 \stackrel{\text { oxidation }}{\longrightarrow} 3 \mathrm{~S}(\mathrm{sol})+2 \mathrm{H}_2 \mathrm{O}\)

The gold sol can be prepared by the reduction of \(\mathrm{AuCl}_4\left(\mathrm{AuCl}_3+\mathrm{HCl}\right)\) using reducing agents like formaldehyde or hydrazine.

⇒ \(2 \mathrm{AuCl}_3+3 \mathrm{HCHO}+3 \mathrm{H}_2 \mathrm{O} \stackrel{\text { reduction }}{\longrightarrow} 2 \mathrm{Au}(\mathrm{sol})+3 \mathrm{HCOOH}+6 \mathrm{HCl}\)

Ferric hydroxide sol is prepared when freshly prepared ferric chloride solution is poured slowly into boiling water.

⇒ \(\mathrm{FeCl}_3+3 \mathrm{H}_2 \mathrm{O} \stackrel{\text { hydrolysis }}{\longrightarrow} \mathrm{Fe}(\mathrm{OH})_3(\mathrm{sol})+3 \mathrm{HCl}\)

Areunions sulfide sol may be prepared by passing H₂S gas through an aqueous solution of arsenic oxide. double

⇒ \(\mathrm{As}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O} \underset{\text { decomposition }}{\stackrel{\text { double }}{\longrightarrow}} \mathrm{As}_2 \mathrm{~S}_3(\mathrm{sol})+3 \mathrm{H}_2 \mathrm{O}\)

Electrical disintegration or Bredig’s Arc method:

This method is particularly suitable for tire preparation of colloidal sols of metals such as gold, silver, and platinum. Two wires of the same metal (say, gold or platinum) used as electrodes are placed in a suitable dispersion medium containing a small amount of alkali.

An electric arc is struck between them. The heat produced by the arc converts the tire’s solid metal into a light vapor state. These vapours condense into liquid form to give lire required colloidal sol.

Peptization:

The conversion of a freshly formed precipitate into a colloidal solution by the addition of a small amount of electrolyte to tire dispersion medium is called peptization.

The electrolyte used is called the peptizing agent. In the tire peptization process, the tire’s freshly formed precipitate adsorbs one of the ions of the electrolyte on its surface.

This causes the development of either positive or negative charged particles on the precipitate, which repel each other and ultimately break up into particles of colloidal size.

For example, a freshly prepared precipitate of Fe(OH)₃ can be changed into a tire colloidal state when the precipitate is treated with water and a small amount of FeCl₃ solution (peptizing agent).

The ferric ions (Fe³⁺) resulting from the ionization of ferric chloride are preferentially adsorbed on the surface of the Fe(OH)₃ particles resulting in [Fe(OH)₃]Fe³⁺.

These positively charged particles of ferric hydroxide repel each other and ultimately break up into particles of colloidal size and are dispersed throughout the medium as a colloidal sol of Fe(OH)₃.

Purification Of Colloidal Solutions

Most lyophobic colloidal solutions prepared by the methods we have discussed contain massive amounts of electrolytes and other soluble impurities.

While the presence of a small amount of electrolytes is essential for the colloidal solution to be stable, larger quantities lead to coagulation. Therefore, the concentration of these soluble impurities must be minimized in order to make the colloidal solution stable. The following methods are generally used for the purification of colloidal solutions.

Ultrafiltration:

Colloidal particles cannot be filtered using ordinary filter papers because the pores are too large. However, the pore size in ordinary filter papers can be reduced by impregnating the filter paper with a solution of gelatin (an animal protein that can form gels) or collodion (a 4% solution of nitrocellulose in a mixture of alcohol and ether) to stop the flow of colloidal particles.

Colloidal particles can be filtered through specially prepared filter papers called ultrafilters. An ultrafilter can be made by dipping ordinary filter paper in a collodion solution, hardening it using formaldehyde, and then drying it.

The process of separating colloidal particles from the solvent and soluble solute present in the colloidal solution by ultrafilters is known as ultrafiltration.

This process is very slow. Sometimes, suction is applied to increase the rate of filtration. The colloidal particles left on the ultrafilter paper are then stirred in a fresh dispersion medium to get a pure colloidal solution.

Dialysis:

When a dissolved substance is separated from a colloidal solution by diffusion through a suitable membrane, the process is called dialysis. Colloidal particles cannot pass through animal membranes (bladder), parchment paper or cellophane sheets while Ions or small molecules can. In dialysis, we use such a membrane to separate ions and small molecules from the colloidal solution.

The apparatus used in dialysis is called a dialyzer. It consists of a parchment or collodion bag containing the colloidal solution which is suspended in a vessel through which water is continuously flowing.

The ions and small molecules (crystalloids) inside the bag migrate out leaving the colloidal solution behind. The process of diffusion can be made faster by using hot water instead of cold water. Remember, excessive dialysis may lead to destabilization of the colloidal solution, and a precipitate may result.

Electrodialysis:

Electrodialysis is especially used when the impurities are ionic. A parchment bag containing the colloidal solution Is suspended in a vessel containing water. Two electrodes are fitted in water. When an electric field is applied across the electrodes, the positive ions move towards the anode through the membrane in water.

By the application of an electric field, the rate of diffusion is increased. This method is utilized in producing pure water from seawater on a large scale.

Artificial kidney (Haemodiaiyser):

The kidney is the filter of the body for waste products; its membrane allows the dissolved waste products to pass through but at the same time, prevents the passage of very large protein molecules. The artificial kidney machine uses tubular coiled cellophane as the dialyzing membrane to purify the blood of patients with renal excretion problems. This process is called hemodialysis.

When the patient’s blood is pumped through the coils, dialysis takes place and the soluble waste products and toxic products in the blood pass through the dialyzing membrane into the dialyzing solutions, the dialyzer thus functioning more or less as a real kidney

Properties Of Colloidal Solutions

Colloidal solutions show characteristic properties which are different from those of true solutions. We shall now discuss some important properties of colloidal solutions.

Colligative properties:

Like true solutions, colloidal solutions also show colligative properties. The four colligative properties, namely lowering of vapor pressure, depression in freezing point, the elevation of boiling point and osmotic pressure depend on the number of particles present in the solution.

In ordinary colloidal solutions, the number of colloidal particles is very small in comparison to the number of molecules or ions of a solute present in a true solution as colloids are aggregates of particles. Hence, the values of colligative properties of colloidal solutions are markedly smaller than those shown by true solutions at the same concentration.

Tyndall effect:

When a beam of light is passed through a clear solution and viewed from a direction at right angles to the direction of the beam of light, it appears completely dark. However, in the case of a colloidal solution, we observe bluish light against this dark background.

This is due to the scattered light from the colloidal particles. The individual particles, which were otherwise invisible, can thus be observed. The above phenomenon was studied in detail by Tyndall and is known as the Tyndall effect.

The apparatus for viewing the colloidal particles is called an ultramicroscope, first constructed and used by Zsigmondy (1903). Using the ultramicroscope, we can distinguish between the different colloidal particles on the basis of light scattered by them.

A common example of the Tyndall effect can be observed in a cinema. You can observe the projection beam all the way from the projector to the screen due to the scattering of light by dust particles.

For the Tyndall effect to be observed, the following conditions must be satisfied.

1. The refractive index of the dispersed phase and that of the dispersion medium must differ appreciably
   in magnitude.
2. The size of the dispersed particles should be comparable to the wavelength of light used

Colour:

The color of a colloidal solution depends upon the wavelength of the light scattered by the dispersed particles, which in turn depends upon the size and nature of the particles. For example, the finest gold sol is red in color but as the size of the particles increases, it becomes purple, then blue, and ultimately golden.

The color also depends upon the manner in which the light is observed. For example, sulfur sols may be colorless, faint yellow, or deep yellow in reflected light and reddish in transmitted light. To cite another example, a mixture of milk and water appears blue when viewed through reflected light and red through transmitted light.

Brownian movement:

When viewed under a powerful ultramicroscope, colloidal particles appear to be in a state of continuous random motion. This ceaseless zigzag motion of the colloidal particles is called Brownian motion after the English botanist Robert Brown who for the first time observed such a movement in the case of the grains of pollen, suspended in water, with the help of a microscope.

Brownian motion is executed by all colloidal particles irrespective of their nature. Colloidal particles suspended in gaseous mediums also exhibit such motion. Brownian motion depends on the size of the colloidal particles and the viscosity of the dispersion medium. Smaller particles, and those in a less viscous medium, move faster.

Brownian motion occurs due to the unbalanced Impacts of the molecules of the dispersion medium colloidal particles. The colloidal units are constantly hit from all sides by the surrounding molecules and are tossed up and down just like a ship in a stormy sea.

Charge on colloidal particles:

Colloidal particles (both lyophobic and lyophilic) are electrically charged, in a given colloidal solution, the nature of the charge is always the same on all the particles of the dispersed phase.

The particles move in an electrical field either towards the cathode or the anode depending on whether they are positively or negatively charged. The following is a list of colloidal solutions along with the nature of charge on their particles.

Some examples of positively charged sols and negatively charged sols:

Sol particles require charge on account of one reason or more, One is electron capture by them during the electrodispersion of metals. Another is the formation of an electrical double layer.

A third, and most important, reason is the preferential adsorption of ions from the solution. Let us discuss the third process. On account of the large surface exposed by the colloidal particles, they preferentially adsorb either positively or negatively charged ions from their dispersion mediums.

When the dispersion medium has two ions or more, the ion is common to the dispersed phase, and the dispersion medium is preferentially adsorbed by the particles of the dispersed phase.

For example, when silver nitrate solution is added to a potassium iodide solution, the precipitated silver iodide sol particles adsorb the iodide ions from the solution and acquire a negative charge. However, when a KI solution is added to an AgNO₃ solution, the precipitated silver iodideol particles adsorb the Ag⁺ ions from the solution and carry a positive charge.

⇒ \(\begin{array}{cc}
\mathrm{AgI} / \mathrm{I}^{-} & \mathrm{AgI} / \mathrm{Ag}^{+} \\
\text {Negatively charged } & \text { Positively charged }
\end{array}\)

Similarly, if FeCl₃ is added to an excess of hot water, a positively charged sol of ferric hydroxide is formed due to the adsorption of Fe²⁺ ions. However, when ferric chloride is added to a NaOH solution, a negatively charged sol is obtained with the adsorption of OH^– ions.

⇒ \(\begin{array}{ll}
\mathrm{Fe}(\mathrm{OH})_3 / \mathrm{Fe}^{3+} & \mathrm{Fe}(\mathrm{OH})_3 / \mathrm{OH}^{-} \\
\text {Positively charged } & \text { Negatively charged }
\end{array}\)

The charged colloidal particles repel one another and save themselves from agglomeration and subsequent precipitation. Electrical charge is thus the cause of their stability.

However, the charged colloidal particles affect the charge distribution In the dispersion medium. The particles attract ions of opposite charge (counter ions) and repel ions having similar charge (coions) present in the medium, forming a second layer ns shown below,

⇒ \(\mathrm{Agl} / \mathrm{I}^{-} / \mathrm{K}^{+} \quad \mathrm{AgI} / \mathrm{Ag}^{+} / \mathrm{I}^{-}\)

This electrical double layer formed at the interface between the dispersed phase and the dispersion medium is called the Helmholtz electrical double layer. According to the modem concept, the first layer of ions is firmly held and is termed the fixed layer while the second layer is mobile and is termed the diffused layer.

The charges of opposite signs on the fixed and diffused parts of the double layer result in a difference in potential between these layers. This potential difference is called the electrokinetic potential or zeta potential.

Electrophoresis or cataphoresis:

The existence of electrical charges on the colloidal particles is shown by the migration of the particles towards either the positive or the negative electrode when they are placed between two charged electrodes.

This migration of colloidal particles towards either the cathode or anode under the influence of an electrical field is known as electrophoresis or cataphoresis. The phenomenon of electrophoresis can be demonstrated by the following experiment.

Two platinum electrodes are fitted, one in each limb of a U-tube. The U-tube is partially filled with water (solvent). A requisite quantity of colloidal solution is placed in the reservoir. The stopcock is opened slightly and the tire reservoir is raised to introduce the colloidal solution into the tire U-tube.

The water is displaced upward. The platinum electrodes must be dipped in the water layer and a direct current voltage in the range of 50 V to 250 V is applied across the electrodes.

If the colloidal particles have a positive charge on their surface, they will move towards the cathode and the level in the cathodic limb will rise whereas the level in the anodic limb will correspondingly move downwards.

On the other hand, if the particles are negatively charged, movement will take place in the opposite direction in the two limbs. Electrophoresis has been used for tire separation of various types of colloidal particles from a mixture. Since different particles move at different velocities, their separation becomes easy.

If the colloidal particles are not allowed to move under the influence of an electric field by some suitable means, then the dispersion medium itself moves. This phenomenon is known as electro-osmosis.

Coagulation or precipitation:

Tire electrical charge on colloidal particles is the cause of the stability of lyophobic sols. The charged colloidal particles repel one another and save themselves from agglomeration and subsequent precipitation.

If, by any means, the charge is removed, the particles will come closer to each other to form aggregates (or coagulate) and settle down under the force of gravity. The settling of particles of the dispersed phase is called coagulation or precipitation of the sol.

Lyophobic sols can be coagulated in the following ways. By electrophoresis, The colloidal particles move towards oppositely charged electrodes where they lose their charge and get coagulated.

By mixing oppositely charged sols Two sols having colloidal particles carrying opposite charges on mixing together in equal proportions get neutralized and precipitated.

For example, the mixing of ferric hydroxide sol with arsenious sulfide sol results in the neutralization of the charge on both sols, and both of them are coagulated. Such coagulation is known as mutual coagulation.

By boiling On boiling a sol, the adsorbed layer is disturbed on account of increased collisions with the particles of the dispersion medium. This reduces the charge on the particles of the dispersed phase and ultimately they settle down as a precipitate.

By excessive dialysis On excessive dialysis, traces of the electrolyte present in the sol are almost totally removed, causing destabilization of the sol, which ultimately gets coagulated.

By the addition of electrolyte When an electrolyte is added to a sol, the colloidal particles are precipitated. In the process, the ions of the electrolyte are taken up by the colloidal particles. This results in the neutralisation of the charge on the particles leading to their coagulation.

The ion that causes the charge on the particles of the dispersed phase to be neutralized is called the coagulating ion or flocculating ion. A positive ion causes the precipitation of a negatively charged sol and a negative ion causes the precipitation of a positively charged sol.

The quantity of the electrolyte required to coagulate a definite amount of sol depends upon the valency of the coagulating ion. The higher the valency of the coagulating ion, the smaller the amount of the electrolyte required to bring about coagulation. The smaller the quantity needed, the higher will be the coagulating power of the ion. This is called the Hardy-Schulze rule.

Thus, for the coagulation of the negatively charged arsenious sulfide sol, trivalent cations (Al³⁺ ) are far more effective than the divalent Ba²⁺ ions, which in turn are more effective than the monovalent Na⁺ ions. In the coagulation of the positively charged ferric hydroxide sol, the trivalent PO₃ ions are more effective than the divalent SO₂ ions which in turn are more effective than the monovalent Cl^– anions. The minimum concentration of an electrolyte in millimoles per liter required to cause coagulation in two hours is called the coagulation value.

Coagulation of lyophilic sols:

Lyophilic sols are more stable than lyophobic sols. Two factors that are responsible for the stability of lyophilic sols are the charges and solvation of the particles of the dispersed phase. Lyophilic sols can be coagulated by taking care of these two factors.

This is done by:

1. Using a suitable second solvent or by
2. Adding an electrolyte.

Introduction of a suitable solvent If a second solvent, which interacts strongly with the dispersion medium, is added to a lyophilic sol, coagulation takes place.

Thus, a hydrophilic sol (e.g., protein in water) can be coagulated by the introduction of alcohol. The water molecules interact strongly with the alcohol molecules, after which the protein molecules are desolvated. This brings about coagulation.

Introduction of an electrolyte When electrolytes are added to lyophilic solids, the ions of the electrolyte compete for the solvent molecules. If a sufficient amount of the electrolyte is added, the sol coagulates. This is called salting out of the sol.

Protection Of Colloids

Lyophobic sols, in contrast to lyophilic sols, are easily coagulated with the addition of a small amount of electrolyte. The relatively higher stability of lyophilic sols is due to the fact that lyophilic colloids are solvated, i.e., the particles of the dispersed phase are covered by a sheath of the liquid (dispersion medium).

The presence of a very small amount of a lyophilic colloid inhibits the precipitating action of an electrolyte on a lyophobic colloid. This is due to its formation of a very thin shell surrounding each particle of the lyophobic colloid through which the oppositely charged ions cannot easily penetrate to neutralize the charge on the particles of the lyophobic colloid.

Lyophilic colloids used for this purpose are called protective colloids. The photographic plate is a good example of protective action, where the lyophobic silver bromide is protected by the lyophilic gelatine.

Emulsions

An emulsion is simply a colloidal suspension of one liquid in another. It may be formed by shaking two immiscible liquids vigorously. Generally, one of the two liquids used is water.

Emulsions are of the following two types:

1. Oil-in-water type (o/w type), e.g., milk and vanishing
2. Water-in-oil type (w/o type), e.g., butter and cold cream

The oil-in-water emulsion has tiny droplets of an oily or waxy substance dispersed throughout a water solution (e.g., milk). In this system, water acts as the dispersion medium.

The water-in-oil emulsion has tiny droplets of water dispersed throughout an oil (e.g., natural petroleum, butter). In this system, oil acts as the dispersion medium. An oil-in-water emulsion can be washed off one’s hands with tap water, while a water-in-oil emulsion gives the hand a greasy water-repellent surface.

Emulsions are generally unstable and separate out into two distinct layers on standing. Thus, to get a stable emulsion, it is necessary to add a stabilizing agent.

This is called an emulsifier or emulsifying agent. Soap, gum, gelatine, and protein are some common emulsifying agents for o/w emulsions and heavy metal salts of fatty acids, long-chain alcohols, and lampblack are some common emulsifying agents for w/o emulsions.

Emulsions can be diluted by adding any amount of the liquid forming the dispersion medium. When an additional amount of the dispersed liquid is mixed with the emulsion, a separate layer is formed. Often, the liquid forming the dispersed phase has particles carrying a negative charge. These partides can be coagulated using a suitable electrolyte.

The stability of an emulsion can be reduced by adding a substance that negates the effect of the emulsifier. Sometimes centrifuging or freezing may also help in destabilizing an emulsion. The colloidal particles of an emulsion exhibit Brownian motion. The Tyndall effect is also observed.

Colloids Around Us

We come across many things which are colloidal in nature. Let us now briefly discuss a few of them.

Why the sky is blue:

The sky looks blue because dust particles, along with those of water suspended in air, scatter blue light, which reaches our eyes.

Fog, mist, and rain:

Air contains dust and moisture. When air is cooled below its dew point, the moisture condenses on the surface of dust particles forming tiny droplets that float in air, forming mist and fog.

Clouds are also a colloidal system in which tiny droplets of water are suspended in the air. As they reach the upper atmosphere (a cool region), the tiny colloidal droplets of water grow bigger and come down as rain. Sometimes, it rains when two oppositely charged clouds meet.

Artificial rain may be caused by spraying a sol carrying a charge opposite to the one on a cloud.

Food:

Much of the food we eat is colloidal in nature. Examples are milk, butter, ice cream, and fruit juices.

Ordinary milk is an emulsion of fat in water, the emulsifier being casein. Artificial beverages are either emulsions or colloidal solutions. Cocoa and coffee are emulsions but tea is a colloidal solution.

Blood:

Blood is a colloidal solution of negatively charged particles (albuminoid substance) in a liquid (blood plasma). The styptic action of alum and ferric chloride solution can be explained on the basis of coagulation. The Al³⁺ (from alum) or Fe³⁺ (from FeCl₃) ions cause the coagulation of the negatively charged albuminoid particles forming a clot, and preventing bleeding.

Soil:

Fertile soil is colloidal. Humus acts as a protective colloid and imparts stability to the soil. Due to the presence of humus, the soil adsorbs moisture and nutrients.

Formation of a delta:

River water is a sol of clay. Sea water contains several electrolytes. When water from the river comes in contact with seawater, the electrolytes present in seawater coagulate the sol of clay. This causes the deposition of clay, forming a delta.

Applications Of Colloids

Colloids find many applications in the industry. We shall now briefly discuss some important ones.

Removal of smoke:

Smoke is a sol of solid particles (carbon, arsenic compounds, dust, etc.) in the air. Air pollution is considerably reduced by fitting what is known as a Cottrell apparatus in the chimneys of factories.

Before it emerges from the chimney, the smoke is made to pass through the Cottrel apparatus, which contains plates that have a charge opposite to that carried by the smoke particles.

On coming in contact with these plates, the smoke particles lose their charge, get precipitated, and settle down on the floor of the chamber, from where they are removed

Purification of drinking water:

Water obtained from natural sources (rivers, lakes, etc.) contains suspended impurities. Impure water is usually purified by the addition of alum, which coagulates the suspended impurities and makes water fit for drinking

Tanning:

Changing hide (the skin removed from a dead animal) into leather by chemical treatment is known as tanning. Hide contains protein in the colloidal state. These colloidal particles in hides are positively charged.

On soaking the hides in tannin, which is a negatively charged sol, mutual coagulation takes place. This results in the hardening of leather. In chrome-tanning, chromium compounds are used in place of tannin.

Medicines:

Many medicines in use are colloidal. For example, colloidal silver (argyrol) is used as an eye lotion. Colloidal antimony is effective in curing kala-azar. Colloidal gold is used for intramuscular injections. Milk of magnesia, an emulsion, is used for stomach disorders. Colloidal medicines are more effective because of their large surface area.

Photographic plates and films: An emulsion of the light-sensitive silver bromide in gelatin is coated over glass plates or celluloid films for preparing photographic plates or films.

Rubber industry: Latex is a colloidal suspension of negatively charged rubber particles in water. To obtain rubber, latex is coagulated.

Industrial products: Many industrial products like paints, varnishes, printing inks, gums, adhesives, and resins are colloidal in nature.

Example 1. Why do lyophobic sols containing gelatin not coagulate with the addition of an electrolyte?
Solution:

Gelatin forms a thin protective layer around particles of the dispersed phase and prevents the neutralization of charge by electrolytes.

Example 2. Why is it essential to wash a precipitate with water before estimating it quantitatively?
Solution:

Colloidal particles are precipitated by the addition of an electrolyte. However, some electrolytes remain adsorbed on the surface of the particles of the precipitate. Therefore, it is necessary to wash the precipitate with water repeatedly to remove any electrolytes from the precipitate before estimating it quantitatively.

Surface Chemistry Multiple-Choice Questions

Question 1. The rate of physisorption increases with

1. Decrease in temperature
2. Increase in temperature
3. Decrease in pressure
4. Decrease in surface area

Answer: 1. Decrease in temperature

Question 2. Which one of the following statements is incorrect?

1. Chemical adsorption is reversible in nature.
2. Physical adsorption is due to van der Waals forces.
3. Physical adsorption is reversible in nature.
4. Activation energy plays an important role in chemical adsorption.

Answer: 1. Chemical adsorption is reversible in nature.

Question 3. Which of the following equations represent (s) Freundlich’s adsorption isotherm?

1. \(\frac{x}{m}=\left(\frac{1}{a}\right)+\left(\frac{b p}{a}\right)\)
2. \(\frac{x}{m}=k C^{1 / n}\)
3. \(\log \frac{x}{m}=\log k+n \log p\)
4. \(\log \frac{x}{m}=\log k+\frac{1}{n} \log C\)

Answer:

2. \(\frac{x}{m}=k C^{1 / n}\)

4. \(\log \frac{x}{m}=\log k+\frac{1}{n} \log C\)

Question 4. Which of the following gases will be adsorbed most easily?

1. N₂
2. H₂
3. O₂
4. CO₂

Answer: 4. CO₂

Question 5. According to the adsorption theory of catalysis, the speed of the reaction increases because

1. The Concentration of the reactant molecules at the active centers of the catalyst becomes high due to adsorption
2. In The Process of adsorption, the activation of like molecules becomes high
3. Adsorption produces heat, which increases the speed of the reactions
4. Adsorption lowers the activation energy of the reaction

Answer: 4. Adsorption lowers the activation energy of the reaction

Question 6. The function of an enzyme in an enzyme-catalyzed reaction is to

1. Transport oxygen
2. Conduct catalytic biochemical reactions
3. Provide immunity
4. Provide energy

Answer: 2. Conduct catalytic biochemical reactions

Question 7. Zeolites are

1. Enzyme catalysts
2. Shape-selective catalysts
3. Liquid catalysts
4. Nonspecific catalysts

Answer: 3. Liquid catalysts

Question 8. Which of the following is a true solution?

1. Cement
2. Muddy water
3. CuSO₄ solution
4. Milk

Answer: 3. CuSO₄ solution

Question 9. Which of the following is not a colloid?

1. Chlorophyll
2. Smoke
3. Protein
4. Blood

Answer: 1. Chlorophyll

Question 10. Substances that diffuse rapidly through water and certain membranes arc called

1. Colloids
2. Crystalloids
3. Gels
4. Emulsions

Answer: 2. Crystalloids

Question 11. The size of a particle in a colloidal solution is

1. Greater Than 1000 Nm
2. Within The Range of 1 Nm-1000 Nm
3. Less Than 1 Nm
4. None of these

Answer: 2. Within The Range of 1 Nm-1000 Nm

Question 12. A colloidal system consists of

1. NaCl and water
2. A Dispersion Medium Only
3. A Dispersed Phase Only
4. A Dispersed phase and a dispersion medium

Answer: 4. A Dispersed phase and a dispersion medium

Question 13. Sol is a general term usually applied to

1. A Solid dispersed in a liquid
2. A Solid dispersed in a solid
3. A Solid dispersed in a gas
4. All of these

Answer: 4. All of these

Question 14. Depending upon the types of the particles of the dispersed phase, colloids are classified as

1. Multimolecular colloids
2. Associated colloids
3. Macromolecular colloids
4. All of these

Answer: 4. All of these

Question 15. Fog is an example of a colloidal system of a

1. Solid dispersed in a gas
2. Gas dispersed in a gas
3. Gas dispersed in a liquid
4. Liquid dispersed in a gas

Answer: 4. Liquid dispersed in a gas

Question 16. Which of the following is/are lyophilic colloid(s)?

1. Starch
2. Gum arabic
3. Gelatin
4. Gold

Answer:

1. Starch

2. Gum arabic

3. Gelatin

Question 17. Lyophilic sols are more stable than lyophobic sols because

1. The Colloidal particles have a positive charge
2. The Colloidal particles have a negative charge
3. The Colloidal particles are solvated
4. There is strong electrostatic repulsion between the negatively charged colloidal particles

Answer: 3. The Colloidal particles are solvated

Question 18. The extra stability of lyophilic colloids is due to

1. The charge on the particles
2. A Protective film of the dispersion medium on the particles
3. The smaller size of the particles
4. The larger size of the particles

Answer: 2. A Protective film of the dispersion medium on the particles

Question 19. Lyophilic sols are

1. Irreversible
2. Coagulated by adding an electrolyte
3. Prepared from inorganic compounds
4. Self-stabilizing

Answer: 4. Self-stabilizing

Question 20. Which of the following statements is correct in the context of sodium stearate, \(\mathrm{CH}_3\left(\mathrm{CH}_{12}\right)_{16} \mathrm{COO}^{-} \mathrm{Na}^{+}\)?

1. It is a major component of many bar soaps.
2. The R-group is the nonpolar tail and is hydrophobic.
3. The —COO group is the polar ionic head and is hydrophilic.
4. All of these

Answer: 4. All of these

Question 21. When soap water interacts with grease, the grease

1. Forms micelles, which are removed
2. Forms macromolecular colloids, which are removed
3. Forms colloids of low molecular mass, which are removed
4. Forms multimolecular colloids, which are removed

Answer: 1. Forms micelles, which are removed

Question 22. Which of the following statements is correct?

1. Micelles are formal above critical micelle concentration.
2. Micelles are not associated with colloids.
3. Micelles are formed below critical micelle concentration.
4. Micelles are true solutions.

Answer: 1. Micelles are formal above critical micelle concentration.

Question 23. Micelles are also called

1. Associated colloids
2. Macromolecular colloids
3. Macromolecular colloids
4. Peptizing agents

Answer: 1. Associated colloids

Question 24. The cleansing action of soaps and detergents is due to

1. Their dissociation into ions in water
2. Their emulsifying nature in water
3. Their decrease in viscosity in water
4. The presence of bulky R-groups in them

Answer: 3. Their decrease in viscosity in water

Question 25. Which of the following method(s) is/are used in the preparation of colloidal solutions?

1. Peptization
2. Bredig’s arc method
3. Chemical methods
4. Electrophoresis

Answer:

1. Peptization

2. Bredig’s arc method

3. Chemical methods

Question 26. Peptization is a process

1. Of precipitation of colloidal particles
2. Of purification of colloids
3. Of dispersing precipitates into colloidal solutions
4. In which colloidal particles move in an electric field

Answer: 3. Of dispersing precipitates into colloidal solutions

Question 27. Adding a few drops of dilute HCl to freshly precipitated ferric hydroxide produces a red colloidal solution. This phenomenon is known as

1. Protective action
2. Peptization
3. Dialysis
4. Dissociation

Answer: 2. Peptization

Question 28. Colloidal solutions are purified by

1. Dialysis
2. Peptization
3. Coagulation
4. Flocculation

Answer: 1. Dialysis

Question 29. Tyndall effect can be observed in a

1. True solution
2. Solvent
3. Colloidal solution
4. Precipitate

Answer: 3. Colloidal solution

Question 30. Brownian motion is caused by

1. Heat change in the liquid state
2. Convection currents
3. The collision of the molecules of the dispersion medium with the colloidal particles
4. The attractive force between the colloidal particles and the molecules of the dispersion medium

Answer: 3. The collision of the molecules of the dispersion medium with the colloidal particles

Question 31. AS₂S₃ and Fe(OH)₃ colloidal solutions are

1. Respectively, positively and negatively charged
2. Respectively, negatively and positively charged
3. Both positively charged
4. Both negatively charged

Answer: 2. Respectively, negatively and positively charged

Question 32. The movement of colloidal particles under an applied electric field is known as

1. Dialysis
2. Electrophoresis
3. Electrodialysis
4. None of these

Answer: 2. Electrophoresis

Question 33. The AS₂S₃ sol has a negative charge. Which of the following has the maximum power to precipitate it?

1. H₂SO₄
2. Na₃PO₄
3. CaCl₂
4. AlCl₃

Answer: 4. AlCl₃

Question 34. Which of the following is arranged in order of decreasing coagulating power?

1. NaCl > BaCl₂ > AlCl₃
2. BaC₂2 > AlCl₃ > NaCl
3. AlCl₃ > BaCl₂ > NaCl
4. BaCl₂ > NaCl > AlCl₃

Answer: 3. AlCl₃ > BaCl₂ > NaCl

Question 35. In the context of an AS₂S₃ sol, which of the following has minimum coagulating value?

1. NaCl
2. KCl
3. BaCl₂
4. AlCl₃

Answer: 1. NaCl

Question 36. The method usually employed for the precipitation of a colloidal solution is

1. Dialysis
2. Addition of an electrolyte
3. Diffusion through an animal membrane
4. Condensation

Answer: 2. Addition of an electrolyte

Question 37. In making ice cream, gelatine is used mainly to

1. Prevent the formation of a colloidal sol
2. Enrich fragrance
3. Prevent crystallization and stabilize the mixture
4. Modify the taste

Answer: 3. Prevent crystallization and stabilize the mixture

Question 38. Which of the following is/are correct?

1. An Emulsion is a colloidal solution of a liquid in a liquid.
2. A Gel is a colloidal solution of a solid in a liquid.
3. An Aerosol is a colloidal solution of a liquid in a gas.
4. Foam is a colloidal solution of a gas in a liquid.

Answer:

1. An Emulsion is a colloidal solution of a liquid in a liquid.

3. An Aerosol is a colloidal solution of a liquid in a gas.

4. Foam is a colloidal solution of a gas in a liquid.

Question 39. Which of the following is used to treat diseases of the eye?

1. Colloidal gold
2. Colloidal silver
3. Colloidal antimony
4. None of these

Answer: 3. Colloidal antimony

The d-and f-block Elements

d-Block Elements:

The d block of the periodic table lies between the s and the p block. It comprises rows which are conveniently referred to as series. The first three rows or series are formed by the progressive filling of 3d, 4d and 5d orbitals respectively. The first row of the d block is Period 4, known as the 3d series and so on. The fourth row, which begins with the filling of the 6d series is still incomplete.μ

The terms d-block elements and transition elements are used interchangeably. However, they do not mean the same thing. Transition elements are those which have an incomplete d orbital or those which give rise to cations having partially filled d subshells. Thus elements of Group 12—Zn, Cd and Hg—are d-block elements but not transition elements.

f-Block Elements:

The f block comprises elements in which 4f and 5f orbitals are progressively filled. The members of this block are actually part of Group 3. The two series of the flock are known as lanthanoids and actinoids respectively.

In the case of the d-block elements, the last electron enters the penultimate orbital whereas in the f-block elements (also called inner transition elements), the last electron enters the ante-penultimate orbital, i.e., the f orbital.

The transition and inner transition elements are thus characterised by the presence of partially filled d and f orbitals respectively. This confers some specific properties of those elements and merits a separate study of them along with some of their compounds. However, the normal rules of valence, as applicable to the main group elements, can be used here too.

The d-block Elements

The transition elements exhibit certain characteristic properties. They are all metals exhibiting high tensile strength, malleability and ductility. They are excellent conductors of heat and electricity and have lustre.

They have high melting and boiling points, exhibit variable oxidation states and paramagnetic behaviour in many compounds and form coloured compounds. They form complexes with different anions and neutral species and may also form alloys and interstitial compounds with other metals.

Many transition metals and their compounds display catalytic properties. Some of the transition metals exhibiting variable valency form unstable intermediate compounds. Also, some transition metals provide a suitable reaction surface.

The lattice structures of transition metals vary. The predominant structures displayed are hexagonal close-packed, cubic close-packed, body-centred cubic, etc. Mercury, being the exception, is a liquid at room temperature.

Electronic Configuration

The general electronic configuration of d-block elements is (n- 1)d^1-10ns^1-2. There may be one or two electrons in the s orbital of the outermost shell and one to ten electrons in the d orbitals of the inner shell (n – 1). There are exceptions to these generalizations which arise due to very little difference in the energies of (n -l)d and ns orbitals.

In addition to this, the half-filled or completely filled subshells are relatively more stable. Consequently, the electronic configurations of Cr and Cu are 3d⁵4s¹ (not 3d⁴4s₂) and 3d¹⁰4s¹ (not 3d9⁴s²) respectively. There are also many anomalies in the electronic configurations of the elements in the 4d and 5d series.

The sequence of filling up of orbitals according to the energy levels in multi-electron atoms not only depends on the electron-nuclear attraction, but also on inter-electron repulsion, but a detailed discussion is beyond the scope of this presentation. The outer electronic configurations of the d-block elements are shown in.

The electronic configurations of zinc, cadmium and mercury (which are also the last members of the three respective series) are represented as (n- 1)d¹⁰ns₂. As already stated, they are not regarded as transfer dozens since the d orbitals are completely filled in the ground state and also in their common oxidation state

The d-block elements display certain characteristic properties, which we will study in this chapter- Yo also observes a high degree of similarity in the properties of d-block elements horizontally, in contrast to those s- and p-block elements, where elements of the group display similarities in properties vertically However Group similarities do existing d-block elements also.

Atomic And Ionic Radii

We shall discuss the trends in atomic and ionic sizes across the period and down the group. The size of the atom initially decreases across a period after which there is no appreciable change and finally at the end of the series there is an increase in size.

On moving across a period in the periodic table, the nuclear charge increases and the size of the atom decreases. In the case of the transition elements, each time the nuclear charge increases by unity the extra electron enters the d orbital.

The d-orbital electrons screen the outermost s-orbital electrons from the nucleus incompletely because of their relatively less efficient shielding power, in comparison to that of the electrons in s- and p-orbitals.

As the number of d electrons increases across the period, the screening effect also increases and this, to some extent, counteracts the effect of increased nuclear charge. This is prominent in the middle of the series and thus in the region, a fairly constant size is observed.

As you know, the 3d series constitutes elements with atomic numbers 21 – 30, In the series, electron pairing in d orbitals is not observed till Mn (atomic number = 25). So electron-electron repulsions are not significant.

The pairing of electrons starts from Fe (atomic number = 26) and this brings about inter-electron repulsions which cause the electron cloud to expand resulting in an increase in atomic size, This effect works in opposition to the effect of increased nuclear charge, which tends to reduce the size, and the atomic radii remain practically the same after Cr.

The atomic size of elements increases down the group as extra shells of electrons are added. This trend is observed strictly for Sc, Y and La (Group 3). However, in the subsequent groups, an increase in size is observed between the corresponding members of the 3d and 4d series, but there is hardly any increase between the 4d and 5d series.

There are 14 lanthanoid elements between La and Hf in the 5d series. They all have two outermost s electrons and are classified together because an increase in the proton number corresponds to an increase in the number of 4f electrons (here the antepenultimate 4f shell is filled).

There is a gradual decrease in the atomic size of the lanthanoid elements and this is called lanthanoid contraction. This cancels the normal increase in atomic size on moving down a group. As a result the elements of 4d- and 5d-series have similar sizes and display similar properties.

For a given charge the ionic radius decreases slowly with the increase in atomic number. The ionic radii decrease with an increase in oxidation state, as the nuclear charge increases.

Density

The d-block elements have high density. The atomic volumes of these elements are low as compared to those of the elements of groups 1 and 2. Thus, the densities of the transition elements are high and generally exceed 5 gcm^-3 except for Sc (3.0 g cm^-3), Y and Ti (4.5 gems^-3).

The densities of the elements of the 4d and 5d series are very high, for example, osmium and iridium have densities of 22.57 g cm^-3 and 22.61 g cm^-3 respectively.

Melting And Boiling Points

The melting and boiling points of the transition elements are very high. Most of them melt above 1273 K and the elements Ta, W and Re melt above 3273 K. Notable exceptions are Zn (693 K), Cd (594 K) and Hg which is a liquid at room temperature.

The melting points of transition metals are high because of unpaired d electrons which participate in metallic bond formation. In any row, metals with d5 configuration (exceptions Mn, Tc) have high melting points, indicating that a large number of unpaired d electrons give rise to strong metallic bonds.

The great strengths of the metal-metal bonds are also manifested in high enthalpies of atomisation. In zinc, cadmium and mercury the d shell is completely filled and the d electrons do not participate in bond formation.

Ionisation Enthalpy

A comparison of ionisation enthalpies of elements gives an idea of the ease with which they form ions, or how electropositive they are. Transition elements have high ionisation enthalpies due to the small size of the atoms and high nuclear charge. The ionisation enthalpies of the first series of transition elements are given in.

Generally, ionisation enthalpy increases across a period. However, some deviations are observed from this trend. On moving across a period, due to the increase in nuclear charge, the nucleus tends to pull the outermost electrons inwards.

In transition elements, the electrons are filled in the penultimate shell (n-l)d. The d subshell shields the outermost, ns electrons from the pull of the nucleus. Thus the effect of increased nuclear charge is nullified to an extent by the screening effect of the inner d-orbital electrons and consequently, the ionisation enthalpy increases only gradually across a series. There is a sudden increase in the first ionisation enthalpy of zinc.

This is due to its stable outermost electronic configuration—3d10 4s2. The other exceptions are chromium and copper whose second ionisation enthalpies are notably higher than those of their neighbours.

This is because of the stable outer electronic configuration of Cr⁺(3d⁵) and Cu⁺(3d¹⁰) so the removal of an electron becomes difficult. As you already know, half-filled and completely filled orbitals are stable.

Apart from these exceptions, the values in Table 8.4 show that there is a similar trend of increase in the first and the second ionisation enthalpies across the period. The third ionisation enthalpies are, however, quite high and there is a deviation from the normal trend in the case of Mn²⁺ and Fe²⁺.

The third ionisation enthalpy of manganese is high as Mn²⁺ has a stable d⁵ configuration. Also, the high values of third ionisation enthalpies of nickel, copper and zinc indicate that the highest oxidation state attainable for these is two.

Oxidation States

One of the most striking features of transition elements is that they exist in several oxidation states, which change in units of one. For example, iron exists as Fe²⁺ and Fe³⁺ and copper as Cu⁺ and Cu²⁺.

In contrast, the oxidation states of p-block elements differ by units of two, for example, Sn2²⁺ Sn⁴⁺ (inert pair effect). The common oxidation states of the elements of the 3d series are shown in.

Transition elements exhibit variable oxidation states in their compounds as both (n-l)d and ns electrons of atoms participate in bond formation. As you can see in up to manganese (starting from titanium across the series) the lowest oxidation state of an element is equal to the number of s electrons and the highest oxidation state is equal to the sum of the number of s and d electrons.

Once we go beyond the d5 electronic configuration in the series, the tendency of all d electrons to participate in bond formation decreases. This is because as the number of d electrons increases, they tend to pair up, leaving fewer orbitals available for the sharing of electrons.

The number of oxidation states exhibited by the respective elements is less at both the extreme ends of the series. At the top end (Sc, Ti) there are few electrons to lose or share during bond formation.

The higher oxidation states exist in the form of fluorides, oxides and oxoanions (MnO₄^–, CrO₄^2-). In addition to this, some elements exhibit zero and negative oxidation states with some ligands like carbon monoxide. For example, in Fe(CO)₅, Ni(CO)₄ and Co₂(CO)₈ the metals exhibit zero oxidation state, whereas the oxidation state of cobalt in Na₂[Co(CO)₄]is negative.

Stability of the various oxidation states

Compounds are regarded as stable if they exist at room temperature and remain unaffected by air and water. In general higher oxidation states are more stable for the elements of the 4d and 5d series in comparison to those of the 3d series.

For example, CrO₃ (3d series) is unstable and therefore strongly oxidising whereas MoO₃(4d series) and WO₃ (5d series) are quite stable. The halides and oxides formed by the elements of the 3d series are shown in. A few interesting observations can be made.

Note:

• X denotes all halides.
• X’ denotes fluoride, chloride, bromide.
• X” denotes fluoride, chloride.
• X”‘ denotes chloride, bromide, iodide

It may be noted that the highest oxidation state for titanium is represented by all halides whereas for vanadium and chromium, it is represented in only fluorides. Fluorine is able to stabilise higher oxidation states because it is highly electronegative and the M-F bond has high bond enthalpy.

However, fluorides in low oxidation states are not known. This is presumably because fluorine is highly oxidising. Oxohalides of the type VOX₃ and MnO₃F (representing the metal in the highest oxidation state) are known.

Transition metals in stable oxidation states from all halides. Those in strongly reducing states form heavier halides (not fluorides), whereas those in strongly oxidising states do not form iodides. For example, Cul is known but not CuI₂ as Cu²⁺ oxidises I^– to I₂

⇒\(2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \rightarrow \mathrm{Cu}_2 \mathrm{I}_2+\mathrm{I}_2\)

Note: Mixed oxides are also known—Mn₃O4, Fe₃O₄, Co₃O₄

As you can see, oxides in the highest oxidation states of various metals are known, for example, Sc₂O₃, TiO₂, V₂O₅, CrO₃, Mn₂O₇. Beyond manganese, the stability of the higher oxidation state decreases. In addition to oxides, oxoanions in high oxidation states are also known, for example VO₄^3-, CrO₄^3-, MnO₄^2-and MnO4^–.

Sometimes the magnitude of ionisation enthalpies gives an idea of the relative stabilities of oxidation states. This can be well understood by considering the case of nickel and platinum. Pt(4) species are more stable than Ni(4), but Ni (2) species are more stable than Pt(2). The ionisation enthalpies of the metals are as follows.

Since the sum of the first two ionisation enthalpies of nickel is less than that of platinum, Ni(2) compounds are more stable. However, since the sum of the first four ionisation enthalpies of platinum is less than that of nickel Pt(4) species are more stable.

Standard Electrode Potentials, E°

The standard electrode potentials (E^θ) of the elements of the 3d series for the M²⁺/M couple are shown.

As you can see, except copper all elements have negative E^θ& values, i.e., they can liberate hydrogen from acids and in the process themselves get oxidised, though the rate at which these metals are oxidised is sometimes slow.

However, metals like titanium and vanadium are passive to dilute nonoxidising acids. The standard electrode potential values are related to various thermochemical parameters like enthalpy of sublimation, ionisation enthalpy and enthalpy of hydration. The relationship between these can be shown in the cycle depicted below.

The E^θ values indicate a reduced tendency to form M2+ species across the series. This is attributed to increasing first and second ionisation enthalpy values across the series. The E^θ values of manganese, nickel and zinc are more negative than expected.

This is due to the extra stability of half-filled (d⁵ ) and completely filled (d¹⁰) configurations of Mn²⁺ and Zn²⁺ respectively and a high enthalpy of hydration of Ni²⁺. The positive E^θ value for copper could be attributed to high ionisation enthalpy and low hydration enthalpy.

The E^θ values for M³⁺/M²⁺ redox couple for Mn³⁺/Mn²⁺ (1.57 V), Fe³⁺/Fe²⁺ (0.77 V) and Co³⁺ /Co2²⁺ (1.97 V) are positive whereas those for Ti³⁺/Ti²⁺ (-0.37 V), V³⁺ / V²⁺ (-0.26 V) and Mn³⁺/Mn²⁺ (-0.41 V) are negative. This indicates that Mn3+, Co³⁺ and Fe³⁺ are oxidising andTi²⁺, V²⁺ and Cr²⁺ are reducing.

Formation Of Coloured Species

Many ionic and covalent compounds of transition metals are coloured. lists the colours of the ions in aqueous solutions where water molecules are ligands.

In contrast, compounds of the s- and p-block elements are generally white or colourless except for a few in which colour arises due to the anion, for example, KMnO₄, K₂CrO₄, K₂Cr₂O₇ or where other factors like charge transfer, polarisation, etc., come into play. For example, Agl and SnI₄ are coloured. Sometimes colour arises due to nonstoichiometry.

When light passes through a material, a part of it is absorbed. If the frequency of the light absorbed is in the visible region, the transmitted light is coloured and the colour observed is complementary to the colour absorbed. The colour of transition metals arises due to incomplete 3d subshells.

The d orbitals of an isolated gaseous atom are degenerate. This degeneracy is lifted when the metal ion is surrounded by anions or a solvent. This is because the surrounding groups affect the energy of some d orbitals more than the others so two groups of d orbitals are formed with different energies.

This is called crystal field splitting. Due to this splitting of orbitals an electron may move from the set of lower-energy d orbitals to the set of higher-energy d orbitals by absorbing minute energy.

Such transitions are called d-d transitions and are mainly responsible for the colour of the compounds, d-d transitions are not possible in d° and d¹⁰ configurations and therefore Sc³⁺ (d⁰), Ti⁴⁺ (d⁰), Cu’ (d¹⁰) and Zn2’ (d¹⁰) are colourless.

Magnetic Properties

When different materials are placed in a magnetic field they exhibit different characteristics. The two main types of magnetic behaviour are paramagnetism and diamagnetism. Substances which are attracted by the applied magnetic field are paramagnetic whereas those repelled by it are diamagnetic. The magnetic lines of force travel more readily through paramagnetic substances than through a vacuum.

Paramagnetism arises as a result of unpaired electron spins in the atom. Therefore, substances having unpaired electrons are paramagnetic. Many transition metals and their compounds exhibit paramagnetism.

The extent of paramagnetic character is determined by the magnetic moment of a substance which is determined experimentally and expressed in Bohr magneton (BM) units. The magnetic moment of a substance is the sum of the magnetic moments of each of the unpaired electrons.

The magnetic moment of an electron is associated with its spin angular momentum (produced due to its spinning on the axis) and orbital angular momentum (produced due to its motion in the orbital).

The total magnetic moment is called the effective magnetic moment (μ_eff). Generally, the spin contribution outweighs or quenches the orbital contribution. The magnetic moment due to spin (|ispmoniy ) is related to the number of unpaired electrons (n) by the ‘spin only’ formula.

⇒\(\mu_{\mathrm{eff}} \approx \mu_{\mathrm{spin} \text { only }}=\sqrt{n(n+2)}\)

In a diamagnetic substance, since all electrons are paired, the magnetic moment is zero.

The magnetic moment increases as the number of unpaired electrons increases. Therefore, the observed magnetic moment of an atom, molecule or ion may give an idea of the number of unpaired electrons present. A single 1 s electron has a magnetic moment of 1.73 Bohr magnetons (BM).

Iron, cobalt and nickel are ferromagnetic. This is an extreme case of paramagnetism. Such substances are very strongly attracted by magnets and can be magnetised.

Catalytic Properties

Many transition metals and their compounds exhibit catalytic properties. Some examples are as follows.

1. Titanium(4) chloride is used as the Natta catalyst in the production of polythene. The compound,s generally used along with triethylaluminium and is called the Ziegler-Natta catalyst.
2. Finely divided iron is used in the Haber process for the manufacture of ammonia.
3. Finely divided nickel is used for the hydrogenation of fats and oils.
4. Copper is used in the manufacture of alkylchlorosilanes, for example(CH₃)₂SiCI₂, which is the starting material for the preparation of a class of industrially important polymers—silicones.
5. Platinum is used in the Ostwald process for the catalytic oxidation of ammonia to nitric oxide.
6. V(5) oxide is used in the contact process for the manufacture of sulphuric acid.

Transition metals and their compounds serve as effective catalysts because the metal ions can change their oxidation states, forming unstable intermediate compounds.

For example, in the oxidation of SO₂ to SO₃ V(V) oxide, the catalyst reacts with SO₂ to form SO₃ and the unstable V(4) oxide, which reacts with oxygen to regenerate the V(5) oxide.

The reaction may be represented as

⇒\(\begin{aligned}
& 2 \mathrm{~V}_2 \mathrm{O}_5+2 \mathrm{SO}_2 \longrightarrow 2 \mathrm{~V}_2 \mathrm{O}_4+2 \mathrm{SO}_3 \\
& 2 \mathrm{~V}_2 \mathrm{O}_4+\mathrm{O}_2 \longrightarrow 2 \mathrm{~V}_2 \mathrm{O}_5 \\
& \hline 2 \mathrm{SO}_2+\mathrm{O}_2 \stackrel{\mathrm{V}_2 \mathrm{O}_5}{\longrightarrow} 2 \mathrm{SO}_3 \quad \text { (overall reaction) }
\end{aligned}\)

Another example is the reaction between iodide and persulphate ions, which is catalysed by the change in the oxidation state of iron.

⇒\(2 \mathrm{I}^{-}+\mathrm{S}_2 \mathrm{O}_8^{2-} \stackrel{\mathrm{Fe}^{3+}}{\longrightarrow} \mathrm{I}_2+2 \mathrm{SO}_4^{2-}\)

The catalytic action of iron is explained as

⇒\(2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2\)

⇒\(2 \mathrm{Fe}^{2+}+\mathrm{S}_2 \mathrm{O}_8^{2-} \longrightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{SO}_4^{2-}\)

In addition to this, transition metals also provide a large surface area for the reactants to be absorbed by the formation of bonds between reactant molecules and surface atoms of the catalyst utilising 3d and 4s electrons (in the case of 3d series of transition metals).

This increases the concentration of the reactant molecules on the surface of the catalyst and also lowers the activation energy by weakening the bonds in the reactant molecules.

Tendency To Form Complexes

Transition metals form a unique class of complex compounds. In such compounds, the central metal ion (cation) is linked by coordinate bonds to species which can donate an electron pair. These are called ligands, and they may be anions or neutral molecules containing unshared electron pairs.

The transition elements have a strong tendency to form complexes because of the small size of their metal ions, high ionic charges and the availability of vacant d orbitals of the right energy to accept lone pairs of electrons from ligands. The metal ion utilises(n-l)d orbitals for the bond formation with the ligands.

Tendency To Form Interstitial Compounds

Transition metals form interstitial compounds when small atoms like boron, nitrogen and hydrogen are trapped in empty spaces or interstices in the crystal lattices of metals. Such substances are hard, but less malleable and ductile than pure metals.

Their melting and boiling points are higher than those of the pure metal, the compounds retain metallic conductivity and are chemically inert. They may be stoichiometric as well as nonstoichiometric.

Most metallic carbides of d-block elements are interstitial compounds where certain atoms occupy octahedral voids in the metallic lattice. Some of these are economically useful materials. For example, tungsten carbide (WC) is used for cutting tools and cementite (Fe3C) is a major constituent of steel and cast iron.

Tendency To Form Alloys

The atomic radii of transition metals are comparable. Therefore, it is possible to replace one metal with another in the crystal lattice. This results in the formation of alloys which are actually a blend of metals (atoms of one metal distributed randomly among those of the other in the lattice).

Alloys are harder and more resistant to corrosion than the individual metals. The common ones are ferrous alloys. Chromium, vanadium, manganese, tungsten and molybdenum are used in the production of stainless steel and other varieties of steel.

Alloys of transition metals with nontransition metals, for example, brass (an alloy of copper and zinc) and bronze (an alloy of copper and tin) are important industrial processes.

Comparison of metals of 3d series

The transition metals of the first row display certain similarities. These metals exhibit variable oxidation states. Therefore, it is possible that different metals in various oxidation states show identical electronic configurations.

This results in similarity in certain properties of the metals like colour and magnetic behaviour. A comparison of the metals of the 3d series on the basis of their d-electron. configuration is depicted.

Some Important Compounds of Transition Metals

Oxides

A large number of oxides are known as transition metals due to their variable oxidation states. You already have an idea about the vast number of oxides formed. In general, metals in the higher oxidation states form acidic and covalent oxides while those in the lower oxidation states form basic and ionic oxides. The oxides formed in the intermediate oxidation states of metals are amphoteric.

From Group 3 to Group 7 the highest oxidation state in the respective oxide is the same as the group number. Thus, the highest oxidation state is attained in Sc₂O₃, TiO₂, V₂O₅, CrO₃ and Mn₂O₇. Beyond Group 7, the stability of the highest oxidation state decreases and the highest oxide of iron is Fe₂O₃.

In the transition metals up to those of Group 7, the highest oxidation state is also represented in the form of oxocations (VO₂⁺, TiO²⁺) and oxoanions (CrO₄^2-, MnO₄^–).

Potassium Dichromate (K₂Cr₂O₇

Preparation

It is obtained from the chromite ore (FeO Cr₂O₃ or FeCr₂O₄). The first step involves the fusion of the ore with sodium carbonate in an excess of air, whereby sodium chromate is formed.

⇒ \(4 \mathrm{FeCr}_2 \mathrm{O}_4+8 \mathrm{Na}_2 \mathrm{CO}_3+7 \mathrm{O}_2 \rightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4+2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{CO}_2\)

The fused mass is extracted with water in which sodium chromate dissolves, forming a yellow solution. The solution is then treated with dilute sulphuric acid, and sodium chromate is converted to sodium dichromate.

⇒ \(2 \mathrm{Na}_2 \mathrm{CrO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}\)

On concentrating the solution, sodium sulphate, being less soluble, crystallises out. It is filtered, the solution is further concentrated, and potassium chloride is added to obtain potassium dichromate.

⇒ \(\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{KCl} \rightarrow \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{NaCl}\)

Since sodium dichromate is more soluble than potassium dichromate, on increasing the concentration of the solution, the orange crystals of potassium dichromate are obtained.

Properties

Potassium dichromate is an orange-red solid (melting point 671 K). It is moderately soluble in cold water. When an alkali is added to an aqueous solution of potassium dichromate, the colour changes from orange to yellow due to the formation of potassium chromate.

⇒ \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{KOH} \rightarrow 2 \mathrm{~K}_2 \mathrm{CrO}_4+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O}\)

Acidification of the chromate solution leads to the formation of potassium dichromate and the colour changes again from yellow to orange.

⇒ \(2 \mathrm{~K}_2 \mathrm{CrO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{K}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}\)

⇒ \(2 \mathrm{CrO}_4^{2-}+2 \mathrm{H}^{+} \rightarrow \mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{H}_2 \mathrm{O}\)

Hence, there is a pH-dependent equilibrium between dichromate and chromate.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{H}_2 \rightleftharpoons 2 \mathrm{CrO}_4^{2-}+2 \mathrm{H}^{+}\)

The oxidation state of chromium in chromate and dichromate is the same.

Potassium dichromate is a powerful oxidising agent in an acidic medium.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \quad\left(E^{\Theta}=1.33 \mathrm{~V}\right)\)

Acidified potassium dichromate can oxidise iodide to iodine, hydrogen sulphide to sulphur, iron(II) to iron(m) and tin(II) to tin(IV).

The half-reactions for these examples may be given as follows.

2I^–→I₂+2e^–
H₂S→S+2H⁺+2e^–
Fe²⁺→Fe³⁺+2e^–
Sn²⁺→ Sn⁴⁺+2e^–

The complete ionic equation may be obtained by multiplying the half equations for reducing agents with suitable integers and adding to the half equations of the oxidising agent such that the number of electrons cancels out.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{I}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}\)

Sodium dichromate is also a very powerful oxidising agent and is highly soluble in water. It is widely used in organic synthesis.

Structure

The structures of the chromate ion (CrO₄² ) and the dichromate (Cr₂O₇² ) are shown below.

In a chromate ion, the chromium atom is tetrahedrally surrounded by four oxygen atoms. In a dichromate ion, two such tetrahedrons are linked through an oxygen atom.

Uses

Potassium dichromate is widely used as a primary standard in volumetric analysis (the sodium salt is unsuitable as it is deliquescent). It is used in the leather industry for tanning and in textile dyeing. Potassium dichromate is also employed as an oxidising agent in organic and inorganic synthesis.

Potassium Permanganate (KMnO₄)

Preparation

It is prepared commercially by the fusion of the mineral, pyrolusite (MnO₂), with potassium carbonate or potassium hydroxide in the presence of air. Sometimes an oxidising agent like potassium nitrate or potassium chlorate is added to aid the oxidation process. During oxidative fusion, MnO₂ gets oxidised to potassium manganate, which is green in colour.

⇒ \(2 \mathrm{MnO}_2+4 \mathrm{KOH}+\mathrm{O}_2 \rightarrow 2 \mathrm{~K}_2 \mathrm{MnO}_4+2 \mathrm{H}_2 \mathrm{O}\)

The fused mass is then extracted with water and the green solution is treated with chlorine, ozone or carbon dioxide to convert manganate into permanganate.

⇒ \(2 \mathrm{~K}_2 \mathrm{MnO}_4+\mathrm{Cl}_2 \rightarrow 2 \mathrm{KMnO}_4+2 \mathrm{KCl}\)

⇒ \(2 \mathrm{~K}_2 \mathrm{MnO}_4+\mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KMnO}_4+2 \mathrm{KOH}+\mathrm{O}_2\)

⇒ \(3 \mathrm{~K}_2 \mathrm{MnO}_4+2 \mathrm{CO}_2 \rightarrow 2 \mathrm{KMnO}_4+2 \mathrm{~K}_2 \mathrm{CO}_3+\mathrm{MnO}_2\)

Alternatively, the green solution of potassium manganate is allowed to undergo disproportionation in an acidic or neutral medium.

⇒ \(3 \mathrm{~K}_2 \mathrm{MnO}_4+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{KMnO}_4+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{~K}^{+}\)

The oxidation of manganate to permanganate can also be carried out electrolytically using a nickel anode and an iron cathode. The cathodic and anodic compartments are separated by a diaphragm. The K₂MnO₄ solution is put in the cathodic compartment and potassium hydroxide in the other. During electrolysis, the manganate ion undergoes oxidation at the anode to give the permanganate ion.

⇒ \(
\underset{\text { Manganate ion }}{\mathrm{MnO}_4^{2-}} \rightarrow \underset{\text { Permanganate ion }}{\mathrm{MnO}_4^{-}}+\mathrm{e}^{-}\)

⇒ \(\text { Cathode: } 2 \mathrm{H}^{+} \text {(from water) }+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2\)

Potassium permanganate can be prepared in the laboratory by the oxidation of a manganese(II) salt by potassium peroxodisulphate.

⇒ \(2 \mathrm{Mn}^{2+}+5 \mathrm{~S}_2 \mathrm{O}_8^{2-}+8 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{MnO}_4^{-}+10 \mathrm{SO}_4^{2-}+16 \mathrm{H}^{+}\)

Properties

Potassium permanganate is a dark purple, crystalline solid (melting point 523 K) and is soluble in water. It is thermally unstable and decomposes as

⇒ \(2 \mathrm{KMnO}_4 \stackrel{\Delta}{\longrightarrow} \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2\)

It is a powerful oxidising agent in alkaline, neutral and acidic mediums. In an alkaline medium gets reduced to MnO_2, in a neutral medium to manganate and in an acidic medium to manganese(2). The half-reactions of the reduction of permanganate to manganate, manganese dioxide and manganese(2) respectively may be represented as.

⇒ \(\mathrm{MnO}_4^{-}+\mathrm{e}^{-} \rightarrow \mathrm{MnO}_4^{2-} \quad\left(E^\theta=0.56 \mathrm{~V}\right)\)

⇒ \(\mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{-}\left(E^{\ominus}=1.52 \mathrm{~V}\right)\)

⇒ \(\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \quad\left(E^{\ominus}=1.52 \mathrm{~V}\right)\)

Thus the oxidising power of potassium permanganate is pH dependent. Although the course of a reaction is obvious from its redox potential, some reactions may seem to be thermodynamically allowed (for example oxidation of water at pH = 1) but actually do not occur as the rate of the reaction is too slow.

In an alkaline medium, potassium permanganate oxidises iodide to iodate, thiosulphate to sulphate and manganese) salts to manganese(2) oxide. The half-reactions for the oxidation processes are as follows.

⇒ \(\underset{\text { Iodide ion }}{\mathrm{I}^{-}}+6 \mathrm{OH}^{-} \rightarrow \underset{\text { Iodate ion }}{\mathrm{IO}_3^{-}}+3 \mathrm{H}_2 \mathrm{O}+6 \mathrm{e}^{-}\)

⇒ \(\underset{\text { Thiosulphate ion }}{\mathrm{S}_2 \mathrm{O}_3^{2-}}+10 \mathrm{OH}^{-} \rightarrow \underset{\text { Sulphate ion }}{2 \mathrm{SO}_4^{2-}}+5 \mathrm{H}_2 \mathrm{O}+8 \mathrm{e}^{-}\)

⇒ \(\mathrm{Mn}^{2+}+4 \mathrm{OH}^{-} \rightarrow \mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-}\)

The overall redox reaction can be written by adding the half-reaction of the reductant to the half-reaction for KMnO₄, Le.,

⇒ \(\mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{-}\)

After multiplying with suitable integers, the balanced overall reactions are obtained as follows.

⇒ \(2 \mathrm{MnO}_4^{-}+\mathrm{I}^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{MnO}_2+\mathrm{IO}_3^{-}+2 \mathrm{OH}^{-}\)

⇒ \(8 \mathrm{MnO}_4^{-}+3 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{H}_2 \mathrm{O} \rightarrow 8 \mathrm{MnO}_2+6 \mathrm{SO}_4^{2-}+2 \mathrm{OH}^{-}\)

⇒ \(2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}\)

As already stated potassium permanganate gets reduced to manganese(2) in an acidic medium.

An acidified potassium permanganate solution oxidises oxalates to carbon dioxide, iodide to iodine, iron(2) salts to iron(3) salts, nitrites to nitrates, sulphites to sulphates and sulphides to sulphur.

The half-reactions of the same are as follows.

C₂O₄^-2→2CO₂+2_e^–
2I^–→I₂+2e^–
Fe²⁺→Fe³⁺+e^–
NO₂^–+H₂O→SO₄^-2+2H⁺+2e^–
SO₃^2-+H₂O→SO₄^2-+2H⁺+2e^–
S²→S+2e^–

By using techniques already known to you, you can easily write the complete ionic equations of the overall reactions. For example,

⇒ \(2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^* \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

Structure

In a permanganate ion, the manganese atom is tetrahedrally surrounded by four oxygen atoms. The structure of the permanganate ion is similar to that of the CIO₄ or SO₂ ions.

The double bonds arise due to the overlap of the p orbitals of oxygen with the d orbitals of manganese. The permanganate ion is diamagnetic in nature.

Potassium permanganate contains manganese(7), i.e., a d° electron configuration, and its crystals are an intense purple in colour.

The structure of the manganate ion, MnO₄^2-, is somewhat similar to that of the permanganate ion. It has an unpaired electron and is paramagnetic in nature.

Uses

Potassium permanganate is mainly used as an oxidising agent in the laboratory and industry. It is employed as a volumetric agent in redox titrations. Potassium permanganate is also used as a disinfectant and a bleaching agent for cotton, silk and wool.

The bleaching action of potassium permanganate is due to its oxidising power. It is also used for the decolourisation of oil. Alkaline potassium permanganate is referred to as Baeyer’s reagent and is widely used as an oxidation organic synthesis. It is used to check unsaturation in organic compounds.

The f-block Elements

As already stated earlier in the chapter the f-block elements have partially filled f orbitals and they are also known as inner transition elements. The f-block elements are divided into two groups lanthanoids and actinoids.

The lanthanoids comprise lanthanum (La) and the fourteen elements that follow it in the series. The elements closely resemble lanthanum and hence the latter is included in any discussion of the lanthanoids which are generally collectively represented by the symbol Ln. The lanthanoids show very close similarity in properties with each other. Also, they exhibit only one stable oxidation state.

The Lanthanoids

In the series, lanthanum is the preceding d-block element with the outermost electronic configuration 5d 6s.In the remaining fourteen elements the electrons are filled up in the antepenultimate 4f orbital on moving across the period. The names, atomic numbers, symbols and electronic configurations of the element and the tripositive cation (most common oxidation state), as well as atomic and ionic radii, are given.

Note: The electronic configuration outside the xenon core is indicated.

We will now discuss the important general characteristics of the lanthanoids.

Electronic configuration

If you carefully study the outermost electronic configurations of lanthanoids, you will find certain irregularities. For instance, the electronic configuration of Pr is 4f³ 6s² and not 4f² 5d¹ 6s² as expected.

The reason is that the energies of the 5d and 4f orbitals are very similar and therefore it is energetically favourable to move the single 5d electron to the 4f level in most of the elements. However, this does not happen in gadolinium (Gd), as the half-filled 4f orbital gives increased stability.

The electronic configurations of the trivalent ions (most stable oxidation state), however, are of the form 4fⁿ (n =1 to 14) for all lanthanoids.

Atomic and ionic radii

There is a progressive decrease in atomic and ionic radii of elements from lanthanum to lutetium. This is referred to as the lanthanoid contraction. Usually, the atomic and ionic radii decrease on moving from left to right period. This is because of the incomplete shielding of the extranuclear charge.

In lanthanoids, an increase in nuclear charge is accompanied by a simultaneous increase in the number of 4f electrons. The 4f electrons shielders of nuclear charge. The shielding effect of electrons increases in the order f < d < p < s.

Consequences of lanthanoid contraction

1. Due to the contraction in atomic size across the lanthanide series, the elements that follow in the 5d series are considerably smaller than expected. As a result, there is a close similarity in the atomic size of the elements in the 4d and the 5d series.
   For example, the pairs of elements Zr-Hf, Nb-Ta and Mo-W have almost identical sizes. Thus Zr and Hf occur together in nature and are difficult to separate from each other.
2. Because of their similarity in size, the properties of the lanthanoids are very similar.
3. As the size decreases from La³⁺ to Lu³⁺, the ionic character of the hydroxide, Ln(OH)₃, decreases. In other words, the basicity decreases on moving across the series. So La(OH₃ and Ce(OH)₃ are the strongest bases.

Oxidation states

All lanthanoids exhibit a stable common oxidation state of +3. In addition, some elements show +2 and +4 oxidation states. These oxidation states are particularly exhibited when attaining their results in a noble-gas configuration, for example, Ce⁴⁺ (f⁰) or a half-filled f orbital, for example, Eu²⁺ (f⁷) or a completely-filled f orbital, for example, Yb²⁺ (f¹⁴ )

Note: The oxidation states shown in bold typeface are the most stable ones.

Ce(4) is a strong oxidising agent (E^θ for Ce⁴⁺/Ce³⁺ is + 1.74 V) and therefore it can thermodynamically oxidise water, but the reaction rate is very slow and Ce(4) is a good analytical reagent. Tb(4) too is an oxidant.

Praseodymium, dysprosium, neodymium and terbium form oxides of the type LnO₂ where the metal shows a +4 oxidation state. (In these cases Ln4+ does not have an f⁰, f⁷ or f¹⁴ configuration.) Europium(2) has an f⁷ configuration.

However, it is a reducing agent and changes to the more stable +3 state. Ytterbium(II) has an f¹⁴ configuration and is a reductant. These examples highlight the fact that the most stable oxidation state for the lanthanoids is +3.

Ionisation enthalpies

Due to the similarity in size, the first and second ionisation enthalpies do not show a marked variation. The first ionisation enthalpies of the lanthanoids are roughly 600 kJ mol^-1 while the second ionisation enthalpies are about 1200 kJ mol^-1, which is comparable to that of calcium (the sizes of Ca⁺ and Ln⁺ are quite similar).

Variations are observed in the third ionisation enthalpy where it is seen that the values for lanthanum, gadolinium and lutetium are abnormally low. This is due to the extra stability associated with empty, half-filled or completely filled orbitals.

The outer electronic configurations of La, Gd and Lu are 5d¹ 6s², 4f⁷ 5d¹ 6s² and 4f¹⁴ 5dl 6s² respectively. The bivalent cation easily loses the third electron as this results in a very stable configuration (that of La is 5d⁰, that of Ga³⁺is 4f⁷ and the configuration of Lu³⁺is 4f¹⁴)

Chemical reactivity

The lanthanoids are silvery white electropositive metals which tarnish in the air. The hardness increases with atomic number. Samariums are very hard and have a high melting point. The elements are good conductors of heat and electricity.

The metals are reactive, the earlier members of the series resemble calcium whereas the later members behave more like aluminium. They react with almost all elements including nitrogen and liberate hydrogen from acids. The metals form carbides when heated with carbon. They bum in halogens to form halides and form hydroxides and oxides.

Uses

Lanthanoids find little use in the pure state. They are mostly used in alloys. An important alloy is mischmetal which consists of lanthanoid metals (-95% lanthanoids) along with the other rare earths. It is alloyed with boron in lighter flints and is used in small quantities to increase the malleability of iron.

It is also used in magnesium alloys, which are employed in aircraft and bullets. Pyrophoric alloys are used in ignition devices and lighter flints. Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking. Some oxides are used in television screens and fluorescent glasses.

The Actinoids

The actinoids comprise the fourteen elements that follow actinium, i.e., from thorium to lawrencium. Many of the members of this series are radioactive and the later members have short half-lives, which makes their study difficult. shows the atomic numbers, electronic configurations and atomic and ionic sizes of actinoids.

Electronic configuration

The outermost electronic configurations of the preceding elements Fr and Ra are 7s¹ and 7s² respectively. For the following element actinium (Ac), the 7s orbital is completely filled and the filling of the penultimate shell (d) begins so that its outer electronic configuration is 6d¹7s².

Thus the actinoids have a filled 7s subshell and the 5f subshell is filled after thorium, i.e., from protactinium. The number of electrons in the 5f and 6d subshells varies. Certain irregularities in the configuration are observed due to the extra stability associated with the f⁰, f⁷ and f¹⁴ configurations as you have studied in the case of lanthanoids.

The 5f orbitals extend into space beyond the 6s and 6p orbitals and participate in bonding. This is in contrast to the lanthanoids where the 4f orbitals are totally shielded by the outer orbitals and therefore do not participate in bonding.

Oxidation states

Since the 5f, 6d and 7s orbitals are comparable in energy and since electrons in the 5f orbital participate in bond formation, a wide range of oxidation states is displayed, particularly by the earlier elements in the series after thorium.

The +3 oxidation state is the most common and important for the later elements. The highest oxidation state is +7 for neptunium and plutonium after which the higher oxidation states become less stable. The M3+ and M4+ ions tend to hydrolyse.

Atomic and ionic sizes

As in the case of lanthanoids, there is a gradual decrease in the size of atoms and M3+ ions across the series. This is due to the poor shielding of the extra charge on the nucleus by 5f electrons.

This is referred to as actinoid contraction. A comparison of the ionic radii of trivalent ions of lanthanoids and actinoids shows that the ions are of comparable size; hence their properties are also somewhat similar.

Ionisation enthalpy

Actinoids have lower ionisation enthalpies than those of lanthanoids. The 5f electrons are effectively shielded from the nuclear charge and the outer electrons are less firmly held.

Physical and chemical reactivity

The actinoids are silvery metals, highly reactive in nature. They exist in different structural forms because of their irregular metallic radii. They react with hot water and tarnish in the air, forming an oxide coating.

They react readily with hydrochloric acid but turn passive by the action of concentrated nitric acid due to the formation of a protective oxide layer on the metal surface. These metals react with oxygen, hydrogen and halogens but not with sodium hydroxide.

Comparison of lanthanoids and actinoids

Some of the main similarities and differences between the lanthanoids and actinoids are listed as follows.

Similarities

1. Both have partially filled f orbitals.
2. They are electropositive metals.
3. The main oxidation state displayed is +3.
4. They exhibit magnetic properties and form coloured compounds.

Differences

The d- and f-block Elements Multiple-Choice Questions

Question 1. The electronic configuration of Lu²⁺ is

1. [Xe]4f¹⁴ Sd¹ 6²
2. [Xe]4f¹⁴5d¹
3. [Xe]4f⁹6s₂
4. [Xe]4f⁷ d¹

Answer: 2. [Xe]4f¹⁴5d¹

Question 2. Paramagnetism arises due to

1. Paired electrons
2. A lone pair of electrons
3. An unpaired electron
4. None of the above

Answer: 3. An unpaired electron

Question 3. Misch metal is an alloy of

1. Mn, Fe and C
2. Mn, Cr and W
3. Cu, Zn and Ni
4. Lanthanoids and rare earths

Answer: 4. Lanthanoids and rare earth

Question 4. Green vitriol is

1. FeSO₄ .7H₂ O
2. Fe₂ (SO₄) 3
3. CuSO₄ -5H₂ 0
4. CuSO₄

Answer: 1. FeSO₄ .7H₂ O

Question 5. The reducing agent for iron oxide in the blast furnace is

1. Carbon
2. Limestone
3. Carbon dioxide
4. Carbon monoxide

Answer: 4. Carbon monoxide

Question 6. The silver salt widely used in photography is

1. AgCl
2. AgBr
3. Agl
4. AgF

Answer: 2. AgBr

Question 7. Galvanisation of iron is done by

1. Cu plating
2. Zn plating
3. Ag plating
4. Sn plating

Answer: 2. Zn plating

Question 8. Zn reacts with excess NaOH to give

1. ZnH₂
2. ZnO
3. Zn(OH)₂
4. Na₂ [Zn(OH)₄]

Answer: 4. Na₂ [Zn(OH)₄]

Question 9. Which among the following gives a colourless aqueous solution?

1. Ni²⁺
2. Fe²⁺
3. Cu⁺
4. Cu2⁺

Answer: 3. Cu⁺

Question 10. K2[HgI₄] is used to detect

1. NH4⁺
2. NH2
3. Na⁺
4. NO3

Answer: 1. NH4⁺

Question 11. Which of the following is not coloured?

1. Na₂[CuCl₄]
2. Na₂[CdCl₄]
3. K4[Fe(CN)₆]
4. K₃[Fe(CN)₆]

Answer: 2. Na₂[CdCl₄]

Question 12. In the 3D transition series, as the nuclear charge increases, the screening effect

1. Increases
2. Decreases
3. First decreases and then increases
4. Does not change

Answer: 1. Increases

Question 13. Which of these is not a lanthanoid?

1. Lu
2. Eu
3. Mo
4. Gd

Answer: 3. Mo

Question 14. Manganese exhibits the maximum oxidation state in

1. K₂ Mn04
2. KMnO₄
3. Mn3O₄
4. MnO₂

Answer: 2. KMnO₄

Question 15. Chromium exhibits the maximum oxidation state in

1. KCr₂O₇
2. Cr₂O₃
3. CrO
4. Cr₂(SO₄)₃

Answer: 1. KCr₂O₇

Question 16. Which of the following is coloured?

1. ZnSO₄
2. Ag₂SO₄
3. CuCl
4. CuCl₂

Answer: 4. CuCl₂

Question 17. What is the highest oxidation state of Ti?

1. +5
2. +6
3. +4
4. +3

Answer: 3. +4

Question 18 Due to the lanthanoid contraction, which of the following pairs of elements have similar sizes?

1. Zr and Y
2. Zr and Hf
3. Zr and Z_n
4. Zr and Nb

Answer: 2. Zr and Hf

Question 19. Maximum number of oxidation states is displayed by

1. Mn(Z=25)
2. Fe(Z=26)
3. Cr(Z=24)
4. Co(Z=27)

Answer: 1. Mn(Z=25)

Question 20. Which of the following dissolves in hot NaOH?

1. Fe
2. Zn
3. Cu
4. Ag

Answer: 2. Zn

Question 21. Which of these forms a colourless solution in aqueous medium?

1. Cr³⁺
2. Ti³⁺
3. Sc³⁺
4. V^3-

Answer: 3. Sc³⁺

Question 22. Which statement is not correct?

1. La(OH), is less basic than Lu(OH)₃.
2. The atomic radii of Zr and Hf are the same.
3. The ionic radius decreases from La³⁺ to Lu³⁺.
4. The +3 oxidation state is commonly displayed by lanthanoids.

Answer: 1. La(OH), is less basic than Lu(OH)₃.

Alcohols Phenols And Ethers

Alcohols:

Alcohols can be regarded as derivatives of hydrocarbons, saturated or unsaturated, where one or more of the hydrogens have been replaced by hydroxyl group(s).

Ethyl alcohol (C₂H₅OH) has been known since ancient times. It is an important constituent of alcoholic beverages, like beer, wine and brandy. It is also used in making tincture iodine (I₂+ C₂H₅OH), cough syrups and tonics.

Isopropyl alcohol [(CH₃)₂ CHOH] is the common ‘rubbing alcohol’ which is used as a 70% solution in water for its antibacterial properties. Methyl alcohol is widely used as an industrial solvent. Glycol is widely used as a solvent and antifreeze for fuels. Glycerol is used in medicines and cosmetics.

When the hydroxyl group is connected to a carbon atom of a benzene ring, the compound is known as a phenol. Phenols possess the general formula Ar-OH where Ar is a phenyl or substituted phenyl group. Phenol containing a small amount of water is known as carbolic acid, which is used as a disinfectant. Phenol was the firs compound to be used as an antiseptic (1867).

Phenols are considered to be different from alcohols because their chemical properties are rather different.

Ethers: Ethers are compounds in which two carbon atoms are connected to a single oxygen (C-O-C). Diethyl eth has since long been used as a general anaesthetic. It is also used as a solvent.

⇒ CH₃– O- CH₃
(Dimethyl ether)

⇒ CH₃-CH₂OCH₃
(Ethyl Methyl Ether)

If the C-O-C unit is part of a ring, the molecule is known as a cyclic ether, examples being (tetrahydrofuran) and dioxane. If the two ether-linkage carbons are directly bonded to each other to form a t-membered ring, the molecule is known as an oxirane (or epoxide).

Example 1:  Can you what petroleum ether is

Solution: Petroleum ether is in fact not an ether, but a mixture of alkanes.

Classification Alcohols Phenols And Ethers

Depending upon the number of hydroxyl groups, alcohols and phenols are classified into monohydric (one-OH group) dihydric (two-OH groups) and trihydric (three-OH groups) and polyhydric (more than three -OH groups) compounds.

Structures of some monohydric, dihydric, trihydric and polyhydric compounds:

Monohydric alcohols may be classified on the basis of the hybridisation of the carbon atom to which the hydroxyl group is attached.

Compounds Containing An Sp³ C-OH Bond

Primary, secondary and tertiary alcohols:

Monohydric alcohols are classified into primary, secondary and tertiary alcohols, depending upon the number of alkyl groups attached to the carbon atom carrying the hydroxyl group.

Thus, a primary alcohol contains a monovalent —CH₂ OH group, a secondary alcohol contains a bivalent ->CHOH group and a tertiary alcohol has a trivalent ->C-OH group.

Allyl alcohol:

In an allyl alcohol, the hydroxyl group is attached to that sp³-hybridised carbon which is bonded to the carbon-carbon double bond.

Benzyl alcohol:

In benzyl alcohol, the hydroxyl group is attached to that sp³-hybridised carbon which is bonded to the benzene ring.

Alcohols Containing An Sp² C-OH Bond

Vinyl alcohol:

In vinyl alcohol, the hydroxyl group is bonded to the sp² -hybridised carbon atom of a carbon-carbon double bond.

Phenol:

In phenol, the hydroxyl group is attached to the sp²-hybridised carbon atom of a benzene ring.

Example 2: Classify the following as primary (1°), secondary (2°) and tertiary (3°) alcohols.

Solution:  

1. 1 and 3 are primary alcohols.
2. 2 and 6 are secondary alcohols.
3. 4 and 5 are tertiary alcohols.

Ethers:

Ethers are said to be symmetrical or simple when the two alkyl groups are the same and unsymmetrical or mixed when they are different.

Symmetrical Ethers:

⇒ CH₃ – O- CH₃
(Dimethyl ether)

⇒  C₂H₅ – O-  C₂H₅
(Diethyl ether)

Unsymmetrical Ethers:

⇒  CH₃ – O- C₂H₅
(Ethyl methyl ether)

⇒  CH₃ – O- CH₂– CH₂ -CH₃
(Methyl n – propyl ether)

Nomenclature

There are three different systems of naming alcohols, and two systems of naming phenols and ethers.

Trivial name

Simple alcohols are commonly known by their trivial names or common names. According to this system, the name of an alcohol is derived by writing the name of the alkyl group and then the word alcohol. The name is always written as two separate words.

In the common system of nomenclature, the position of an additional substituent is indicated by letters of the Greek alphabet rather than by numbering, the carbon attached to the OH group being labelled α.

Any simple group that has a common name may be used in the alkyl alcohol system, with one exception. The group C₆H₅-is phenyl, but the compound C₆H₅-OH is phenol and not phenyl alcohol.

Substituted phenols are considered to be derivatives of the parent compound phenol.

However, phenyl-substituted alkyl alcohols are normal alcohols. For example,
CH₂OH

The Carbinol System

In the carbinol system, the simplest alcohol, CH₃OH, is called carbinol. More complex alcohols are considered to be alkyl-substituted carbinols. The whole name is written as one word.

The IUPAC System

Alcohols:

In the IUPAC system, alcohols are named by replacing the ‘e’ of the corresponding alkane by the suffix ‘ol’.

CH₃OH – Methane-e + ol = Methanol

CH₃CH₂OH –  Ethane-e+ol Ethanol

The positions of the hydroxyl group and other substituents are indicated by numbers, the lowest possible number being given to the carbon atom attached to the hydroxyl group.

In the IUPAC system, if two-OH groups are present in a compound (as in CH₂OH CH₂OH), the suffix becomes ‘diol’ instead of ‘ol’. If three – OH groups are present (as in CH₂OH – CHOH- CH₂OH), the suffix becomes’triol’ instead of ‘ool’. Thus:

(Note that terminal ‘e’ of the parent alkane has not been removed because the suffix ‘diol’ or triol begins with a consonant.)

Gives the common and IUPAC names of certain typical alcohols:

Monocyclic alcohols are named using the prefix cyclo and considering the -O-H group to be attached to C-1.

Phenols:

In the IUPAC system, a compound with the hydroxyl group directly attached to a benzene ring is called phenol.

Substituted phenols are named as derivatives of the parent compound phenol as illustrated by the following IUPAC names.

The common and IUPAC names of some phenols:

Ethers:

In the common system, ethers are named by first writing the names of the two alkyl groups in alphabetical order and adding the word ether.

⇒ CH₃OCH₂CH
( Ethyl Methyl Ether)

⇒  CH₃OC(CH₃)
(t-Butyl methyl ether)

In symmetrical ethers, the prefix ‘di’ is used.

Diethyl Ether

⇒ (CH₃)₂ CHOCH(CH₃)₂
(Disopropyl ether)

In the IUPAC system ethers are considered to be alkoxyalkanes. For complex molecules, the simplest organic group along with the oxygen atom is named as an alkoxy group and used as a substituent on a more complex chain.

The Common And IUPAC names of some ethers:

A few examples of IUPAC nomenclature:

Cyclic ethers are known as epoxides The term ‘epoxy’ is derived from epoxide

Example 3:  Draw the structures of the folloeing compounds.

1. 2,2,4 – Trimethyl – 3 pentanol
2. 2- Ethoxypropane
3. 5 – Methyl – 2, 4 – Hepatanediol
4. 2 – Ethyl – 6- Methylphenol
5. 2- Fluoro – 3- Methyl – 2 buten-1 – ol

Solution:

Structures of Alcohols, Ethers And Phenols

In alcohol, the hybridisation of carbon is approximately sp³. So is the hybridisation of oxygen. Oxygen share one bond with carbon and one with hydrogen. The two lone pairs of electrons on oxygen occupy orbitals that ar each approximately sp²-hybridised.

The  bond angle in an alcohol is 108.9° (a little less than the tetrahedral angle 109° 28′). This is due to the repulsion between the two lone pairs of electrons on oxygen.

In dimethyl ether the bond angle is 111.7° (a little more than the value of the tetrahedral angle). This is due to repulsion between the hydrogens marked H, in the above structural formula.

The C-O bond length of (136 pm) in phenol is slightly less than the C-O bond length (141 pm) in methanol. This is because the hydroxyl group in phenol is directly attached to the sp²-hybridised carbon of the benzene ring, which acts as an electron-withdrawing group.

General Methods Of Preparing Alcohols

Alcohols can be prepared by several methods. In this chapter, we will discuss methods which are generally used to prepare alcohols in the laboratory.

From Alkenes

By hydration:

Alcohols are prepared commercially by the hydration of alkenes (obtained cheaply from the cracking of crude oil) in the presence of an acid catalyst. The addition of a molecule of water to an alkene takes place according to the Markovnikov rule. Except for the hydration of ethylene, the reaction produces secondary and tertiary alcohols.

By hydroboration oxidation:

Alkenes react with diborane to give trialkyl boron compounds, which yield alcohols on oxidation with alkaline hydrogen peroxide.

Hydroboration Mechanism:

Because of the vacant orbital on boron, borane may be regarded as an electrophile and attacks the electrons of an alkene in accordance with the Markovnikov rule. But the chief product of the overall reaction, n-propyl alcohol, arises from the anti-Markovnikov addition of a molecule of water at the site of the double bond.

By The Hydrolysis Of Alkyl Halides

Alkyl halides can easily be hydrolysed to the corresponding alcohols by boiling with an aqueous alkali or moist silver oxide (AgOH).

The hydrolysis of alkyl halides in an aqueous alkali may occur by either the S_N1 or the S_N2 mechanism. The hydrolysis of most primary halides occurs by the S_N2 mechanism and yields alcohols in appreciable quantities.

Tertiary halides also undergo hydrolysis, but these reactions occur by the S_N1 rather than the S_N2 mechanism.

By The Reduction Of Aldehydes And Ketones

Carbonyl compounds can be reduced to the corresponding alcohols by a number of methods, either catalytically or by the use of chemical reagents.

Catalytic reduction:

Aldehydes and ketones are reduced by hydrogen in the presence of Ni, Pt or Pd catalyst at room temperature and moderate pressure to the corresponding alcohols.

⇒ \(\mathrm{RCHO} \stackrel{\mathrm{H}_2 / \mathrm{Ni}}{\longrightarrow} \mathrm{RCH}_2 \mathrm{OH}\)

During the process, the double bond is converted into a single bond.

⇒ \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CHO} \stackrel{\mathrm{H}_2 / \mathrm{Ni}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\)

Reduction by metal hydrides:

Aldehydes and ketones are easily converted to alcohols in dry ether by LIAIH₄. LIAIH₄ is a specific reagent an does not ordinarily reduce the ethylenic double bond in the molecule to a single bond.

Sodium borohydride is a milder reducing agent and reduces aldehydes and ketones only.

⇒ \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CHO} \stackrel{\mathrm{Li} \mathrm{AlH}_4}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2 \mathrm{OH}\)

Aldehydes are reduced to primary alcohols. Ketones are reduced to secondary alcohols.

By The Reduction Of Carboxylic Acids And Esters

Carboxylic acids and esters are usually reduced by LiAlH₄ in dry ether.

In order to reduce a carboxylic acid to an alcohol, the carboxylic acid is first converted to an ester and then reduced catalytically by hydrogen to yield an alcohol.

⇒ \(\mathrm{RCOOH}+\mathrm{R}^{\prime} \mathrm{OH} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{RCOOR} \stackrel{\mathrm{H}_2}{\longrightarrow} \mathrm{RCH}_2 \mathrm{OH}+\mathrm{R}^{\prime} \mathrm{OH}\)

Grignard Reagents

Primary, secondary and tertiary alcohols can be prepared by the use of Grignard reagents.

The addition of a Grignard reagent to formaldehyde yields a primary alcohol

The addition of a Grignard reagent to formaldehyde yields a Secondary alcohol

The addition of a Grignard reagent to formaldehyde yields a Tertiary alcohol

Physical Properties Of Alcohols

The introduction of a hydroxyl group in a hydrocarbon brings about a marked change in its physical properties like melting point, boiling point and solubility in water.

Boiling point:

The boiling and melting points of alcohols show a regular increase with the increase in the number of carbon atoms. The boiling point rises by about 20 K for each additional carbon atom.

This is due to the increase in van der Waals attraction. In the case of isomeric alcohols, the boiling points show a regular decrease with an increase in branching as there is a decrease in van der Waals attraction due to branching (less surface area).

The boiling points of some alcohols:

When compared to hydrocarbons, alkyl halides and ethers of comparable molecular weights the boiling points of alcohols are unusually high due to the association of their molecules through intermolecular hydrogen bonding. The high boiling point of alcohols may be attributed to the high energy required to break the hydrogen bond.

Solubility:

The lower members of the class of alcohols like methyl, ethyl, n-propyl, t-butyl and many polyhydric alcohols are completely soluble in water. This may be attributed to the intermolecular hydrogen bonding between the molecules of alcohol and water.

The solubility of alcohols decreases in water as the alkyl chain increases in length. Alcohols have both hydrophilic (waterlike, due to the OH group) and hydrophobic (alkane like, due to the carbon chain) moieties. With the increase in molecular weight, the hydrophobic character of alcohols increases. This reduces their solubility in water.

Among isomeric alcohols, tertiary alcohols are more soluble than primary and secondary alcohols. This can be attributed to a decrease in the relative volume of the hydrophobic portion.

Chemical Properties Of Alcohols

Alcohols react both as nucleophiles and electrophiles.

The O-H bond is broken when alcohols react as nucleophiles.

The reactions of alcohol can be divided into the following four types.

1. Those in which the O-H bond is cleaved,
2. Those in which the C-O bond is cleaved,
3.  Those in which the oxygen acts as a base, and
4. Oxidation.

Reactions Due To Fission Of O-H Bond

Reactions with metals-Acidic nature:

Alcohols behave as weak acids. This characteristic can be explained on the basis of the fact that the hydrogen atom is attached to the electronegative oxygen atom, which attracts the pair of electrons of the O-H bond.

Due to this attraction, there is a tendency for the loss of hydrogen as a proton. Thus, alcohols react with strong electropositive metals, evolving hydrogen

The order of acid strength of various types of alcohols is primary > secondary > tertiary.

This is because oxygen is considerably more electronegative than either carbon or hydrogen. Therefore the C-O and O-H bonds are polarised towards the oxygen atom.

The tertiary alcohol is the weakest because of the presence of three electron-releasing alkyl groups on the carbon atom attached to oxygen. This increases the electron density on oxygen tending to decrease the polarity of the O-H bond. This causes a decrease in acid strength.

Reaction with Grignard reagent

Alcohols react with Grignard reagent to form alkanes.

In these reactions, the order of reactivity of alcohols is primary>Secondary> Tertiary

Esterification

Alcohols react with

1. Carboxylic acids
2. Acid halides and
3. Acid anhydrides to form esters

The reaction of an alcohol with carboxylic acid is reversible. It is catalysed by concentrated H₂SO₄. Concentrated H₂SO₄ removes water as soon as it is formed so that the forward reaction is favoured.

Esterification Mechanism:

The reaction of an alcohol with an acid chloride occurs rapidly and does not require an acid catalyst. Pyridine (a base) is usually added to the reaction mixture to neutralise HCl as soon as it is formed during the reaction. Alcohols react with acid anhydrides to form esters in the absence of a catalyst.

Reactions Of Alcohols Involving C-O Bond Cleavage

Formation of an alkyl halide

Alcohols react with a variety of reagents to yield alkyl halides. The most commonly used reagents are hydrogen halides (HCl, HBr or HI), phosphorus tribromide (PBr₃) and thionyl chloride (SOCI₂).

Reactions with hydrogen halides

Alcohols react with dry HCl(g) in the presence of anhydrous ZnCl₂ as a catalyst. CH₃CH₂OH + HCl(g)- anhydrous →CH₃CH₂Cl+ H₂O ZnCl₂ Ethyl chloride (chloroethane)

Hydrogen halides Mechanism

Lucas reagent

Low-molecular-weight primary, secondary and tertiary alcohols react at different rates with a solution of anhydrous zinc chloride in concentrated hydrochloric acid (Lucas reagent) (alcohols are soluble in the Lucas reagent) to give alkyl halides which are not soluble in the reaction mixture and can be seen as a separate oily layer (and on shaking give a cloudy appearance).

The ease with which the alkyl halide forms depends upon the ease with which the alcohol is converted into a carbocation. Tertiary alcohols react instantly, secondary alcohol within two or three minutes, and primary alcohols do not produce turbidity at room temperature.

The difference in reaction rates described above is the basis of the Lucas test by means of which primary, secondary and tertiary alcohols are distinguished.

Alcohols react with HBr (formed by the reaction between KBr and concentrated H₂SO₄) to yield alkyl bromides.

Alcohols give alkyl iodides on reaction with hydrogen iodide.

Reaction with phosphorus halides

Alcohols react with phosphorus halides to form alkyl halides.

⇒ \(\mathrm{R}-\mathrm{OH}+\mathrm{PX}_5 \rightarrow \mathrm{RX}+\mathrm{POX}_3+\mathrm{HX}\)

⇒ \(3 \mathrm{R}-\mathrm{OH}+\mathrm{PX}_3 \rightarrow 3 \mathrm{RX}+\mathrm{H}_3 \mathrm{PO}_3\)

For example:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{PCl}_5 \rightarrow \underset{\text { Ethyl chloride }}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}}+\mathrm{POCl}_3+\mathrm{HCl}\)

⇒ \(3 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\underset{\left(2 \mathrm{P}+3 \mathrm{Br}_2\right)}{\mathrm{PBr}_3} \rightarrow \underset{\text { Ethyl bromide }}{3 \mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}}+\mathrm{H}_3 \mathrm{PO}_3\)

⇒ \(3 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{PI}_3 \rightarrow \underset{\text { Ethyl iodide }}{3 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}}+\mathrm{H}_3 \mathrm{PO}_3\)

Victor Meyer Test (Distinction between primary, secondary and tertiary alcohols):

Reaction with thionyl chloride

Alcohols react with thionyl chloride in the presence of pyridine to form alkyl chlorides.

⇒ \(\mathrm{ROH}+\underset{\begin{array}{c}
\text { Thionyl } \\
\text { chloride }
\end{array}}{\mathrm{SOCl}_2} \stackrel{\text { Pyridine }}{\longrightarrow} \mathrm{RCl}+\underbrace{\mathrm{SO}_2+\mathrm{HCl}}_{\text {gas }}\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{SOCl}_2 \stackrel{\text { Pyridine }}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}+3 \mathrm{SO}_2 \uparrow+\mathrm{HCl} \uparrow\)

Reaction with concentrated H₂SO₄

Like other acids, concentrated H₂SO₄ also reacts with an alcohol to form an ester (alkyl hydrogen sulphate).

Reaction with Ammonia:

A mixture of primary, secondary and tertiary amines is formed when vapours of alcohol and ammonia are passed through heated alumina at 633 K.

Dehydration

Alcohols can easily be dehydrated to olefins at high temperatures in the presence of an acid

This is a ẞ-elimination reaction and is favoured by high temperature.

An elimination reaction in which a proton is lost from one carbon (B-carbon) and a nucleophile is lost from the adjacent carbon (a-carbon) is called a -elimination reaction or 1, 2-elimination reaction. The most common examples of B-elimination reactions include the dehydrohalogenation of alkyl halides and the dehydration of alcohols

In order of decreasing acidity, the acids commonly used in the reaction are H₂SO₄>H₃PO₄ >(COOH)₂ >HCOOH> KHSO₄ > Al₂O₃

The dehydration of primary alcohols requires a higher temperature than is needed for the dehydration of secondary or tertiary alcohols.

Secondary alcohols undergo dehydration in milder conditions.

Tertiary alcohols undergo dehydration at a much lower temperature.

Thus, the relative ease with which alcohols undergo dehydration is in the following order.

Dehydration Mechanism:

In case the olefin concerned has an isomer, the more stable isomer is formed (Saytzev rule).

Some Primary and secondary alcohols undergo rearrangement during dehydration. For example,

Oxidation And Dehydrogenation Of Alcohols

The oxidation of alcohol takes place with the cleavage of O-H and C-H bonds to form the C = O bond.

This type of cleavage and formation of bonds occurs in oxidation and dehydrogenation reactions.

Primary, secondary and tertiary alcohols differ in their behaviour towards oxidising agents. In general, primary alcohols yield aldehydes, which are readily oxidised further to carboxylic acids. Secondary alcohols give ketones, which are relatively resistant to further oxidation.

Tertiary alcohols do not have a hydrogen atom on the carbon attached to the hydroxyl group and are resistant to oxidation.

The orange-red colour of a solution of chromic anhydride (CrO₃ in aqueous sulphuric acid) is immediately discharged when this solution is added dropwise to a solution of a primary or secondary alcohol in acetone. Tertiary alcohols fail to react immediately.

Reactions of primary, secondary and tertiary alcohols with CrO₃ in aqueous H₂SO₄:

Acidic or basic aqueous potassium permanganate is often used to oxidise primary alcohols to carboxylic acids directly.

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH} \frac{\mathrm{KMnO}_4 / \mathrm{H}_2 \mathrm{SO}_4}{[\mathrm{O}]} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}\)

Tertiary alcohols are not easily oxidised by acidified KMnO4. However, at elevated temperatures, the use of strong oxidising agents (acidified KMnO₄ ) leads to the breaking of C-C bonds and the formation of a carboxylic acid containing a fewer number of carbons.

Primary alcohols are mainly oxidised to aldehydes by a weak oxidising agent, pyridinium chloromate (PCC).

⇒ \(\mathrm{R}-\mathrm{CH}_2-\mathrm{OH} \underset{[\mathrm{O}]}{\stackrel{\mathrm{PCC}}{\longrightarrow}} \mathrm{R}-\mathrm{CHO}\)

Pyridinium chlorochromate is a complex compound of chromium trioxide, pyridine and concentrated HCl.

This compound oxidises a primary alcohol to an aldehyde and the reaction stops at this stage. Pyridinium chlorochromate does not attack double bonds.

When alcohol vapour is passed over copper at 573 K, dehydrogenation occurs; a primary alcohol yields an aldehyde and a secondary alcohol yields a ketone. A tertiary alcohol undergoes dehydration to yield an alkene.

⇒ \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH} \underset{573 \mathrm{~K}}{\stackrel{\mathrm{Cu}}{\longrightarrow}} \underset{\text { Acetaldehyde }}{\mathrm{CH}_3-\mathrm{CHO}}+\mathrm{H}_2\)

⇒ \(\underset{\text { Isopropyl alcohol }}{\mathrm{CH}_3 \mathrm{CHOHCH}_3} \underset{573 \mathrm{~K}}{\stackrel{\mathrm{Cu}}{\longrightarrow}} \underset{\text { Acetone }}{\mathrm{CH}_3 \mathrm{COCH}_3}+\mathrm{H}_2\)

This reaction may be used for differentiating primary, secondary and tertiary alcohols.

Example 4: Give the structures of the products of the following reactions. Give reasons for your answer.

Solution:

This is an electrophilic addition reaction. In an acid medium water adds according to the Markovnikov rule.

NaBH₄ is a weak reducing agent>It reduces the ketonic group selectively.

This s the nucleophilic addition reaction of a Grignard regent with a carbonyl compound

Some Commercially Important Alcohols

Methanol

Methanol is known as wood alcohol or wood spirit because it was first obtained by the destructive distillation of wood. Today methanol is produced commercially by passing a mixture of carbon monoxide and hydrogen over a catalyst containing oxides of chromium, copper and zinc at a temperature of 623-723 K and a pressure of 200 atm.

Methanol is highly toxic. Ingestion of even small quantities of methanol can cause blindness; large quantities cause death. It is used as a laboratory reagent and as a solvent for paints and varnishes. It is also used in the preparation of formaldehyde.

Ethanol

Ethanol is manufactured on a large scale from molasses, a brown syrup prepared from raw sugar during the sugar-manufacturing process. Molasses is diluted with water whereby some of the cane sugar dissolves. The diluted solution is fermented with yeast.

The enzymes invertase and zymase are present in yeast. Invertase converts sugar into a mixture of the isomers glucose and fructose. Zymase converts this mixture into ethanol and carbon dioxide. After fermentation is over the alcohol is distilled. The alcohol so obtained is 95% pure, the rest being water. This is referred to as rectified spirit.

⇒  \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_2 \mathrm{O} \stackrel{\text { invertase }}{\longrightarrow} \underset{\text { Glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\text { Fructose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}\)

⇒ \(\underset{\text { Glucose and fructose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6} \stackrel{\text { zymase }}{\longrightarrow} \underset{\text { Ethyl alcohol (ethanol) }}{2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}+2 \mathrm{CO}_2\)

Pure ethanol (absolute alcohol) cannot be obtained by fractionation. It is difficult and expensive to remove the last traces of water. Water can be removed by distilling the rectified spirit with anhydrous benzene. A ternary azeotropic mixture of 7.5% water, 18.5% alcohol and 74% benzene is formed, which distils out first. The residual liquid in the distilling flask is absolute alcohol (99.9%).

The last traces of water can be removed by distilling it over metallic magnesium. Absolute alcohol is hygroscopic and should be carefully preserved away from moisture. Rectified spirit and absolute alcohol, like alcoholic drinks, are taxed at a very high rate. For many industrial purposes pure alcohol is not needed and it is made undrinkable by adding various other chemicals.

This industrial methylated spirit or denatured alcohol is a mixture of 95% rectified spirit and 5% methanol, which is extremely poisonous. Sometimes CuSO₄ is added to give a blue colour to methylated spirit, so that people do not drink it by mistake. Alcohol is mainly used as a solvent for paints, varnishes, perfurmes, and so on. It may be a useful fuel some day, since it burns to form CO₂ and water, producing a considerable amount of heat.

Wine production

• Wine is made from grapes, which contain sugar. The grapes are crushed to squeeze out the juice. A fungus (yeast) grows naturally on the skins of the grapes. Its enzymes start the fermentation of the sugar in the juice as soon as the grapes have been crushed.
• Fermentation is carried out in the absence of air (anaerobic condition). This is because the oxygen of air oxidises ethanol to acetic acid, which destroys the taste of alcoholic drinks.
• Fermentation is complete when all the sugar has been converted into alcohol after several days. The wine is stored for several months, or even years, to mature before being bottled for sale.
• Red wine is produced from black grapes by mixing some of the skins of black grapes into the juice during fermentation so that the colouring matter from the skins passes into the wine. Sparkling wines, like champagne, are made by bottling the wine before fermentation is complete. Fermentation continues inside the bottle and the carbon dioxide produced dissolves in the wine.
• When the bottle is opened, the pressure is released and the carbon dioxide bubbles out slowly. Wine contains between 10% and 12% of alcohol by weight. To make stronger wines like sherry and port, brandy (34% of alcohol by weight) is added to increase the ethanol content to about 17% by weight. The average strengths of some alcoholic drinks are given in the following table.

The alcohol content of some common alcoholic drinks:

The effects of alcohol on the body

In small quantities, ethyl alcohol (generally called alcohol) is a stimulant and can be beneficial. In small quantities, alcohol increases the flow of the digestive juices and stimulates the appetite. In larger quantities, alcohol makes the whole nervous system less sensitive. An excessive intake of alcohol over a long period can cause addiction (alcoholism), affect the liver and kidneys, and eventually cause damage to the brain.

Phenols

Phenol was discovered in the middle oil fraction during the distillation of coal tar sometime in the nineteenth century and was named carbolic acid.

Phenol Methods of Preparation

Phenol is prepared by the following methods.

1. Laboratory methods:

From benzene diazonium salts: Aniline is diazotised with NaNO₂ and HCl at 0-5°C (278 K) and the resulting benzene diazonium salt forms phenol with water.

From benzene sulphonic acid:  The fusion of the sodium salt of benzene sulphonic acid with solid NaOH at 623 K followed by acidification yields phenol.

From chlorobenzene: Chlorobenzene reacts with aqueous sodium hydroxide solution at 623 K and 300 atm to produce sodium phenoxide, which on acidification yields phenol.

2. Commercial method (from cumene):

The oxidation of isopropylbenzene (cumene) with air in the presence of an acid catalyst (H₂SO₄) gives cumene hydroperoxide which on hydrolysis with an acid yields phenol and acetone.

Cumene itself is prepared by the Friedel-Crafts alkylation of benzene with propyl chloride

Physical Properties of Phenols

Phenol is a colourless, crystalline, low-melting (315 K) solid. The O-H bond of phenol is polar and therefore involved in intermolecular hydrogen bonding as shown below.

Due to intermolecular hydrogen bonding, the boiling point of phenol (455 K) is higher than that of toluer (383 K), which has a comparable molecular weight.

Phenol is moderately soluble in water because it forms hydrogen bonds with water.

Chemical Properties of Phenols

Reactions of the phenolic (-OH) group

Acidic nature:  The acidic nature of phenol is indicated by the following reactions.

1. Phenol turns blue litmus red.

2. Phenol reacts with sodium metal to evolve hydrogen gas.

⇒ 2C₆H₅OH + 2Na →2C₆H₅ONa+ H₂

3. Phenol reacts with sodium hydroxide to give sodium phenoxide.

⇒ C₆H₅ OH + NaOH→ C₆H₅ONa+ H₂O

4. Phenols fail to react with Na₂CO₃ or NaHCO₃. In fact, phenols are precipitated from an aq solution of sodium phenoxide by bubbling CO₂ gas.

⇒ C₆H₅ONa + CO₂+ H₂O→ C₆H₅OH+NaHCO₃

The hydroxyl group of phenol is directly attached to the sp²-hybridised carbon of the benzene ring. Due to the higher electronegativity of the sp²-hybridised carbon of phenol, the electron density decreases on the oxygen atom. This increases the polarisation of the O-H bond and results in an increase in the ionisation of phenol.

The negative charge on the oxygen atom of the phenoxide anion gets delocalised in the benzene ring. This delocalisation makes the phenoxide anion more stable and favours the ionisation of phenol.

The delocalisation, no doubt, is also possible in phenol as shown below.

As these resonating structures involve charge separation, phenol is much less stable than the phenoxide anion. The presence of an electron-withdrawing group (NO₂) at the o and p positions enhances the acidic strength of phenol.

Thus, p-nitrophenol is more acidic than phenol. On the other hand, an electron-releasing group attached to the benzene ring decreases the acid strength. Cresols, for example, are less acidic than phenol.

Phenol is more acidic than ethyl alcohol. Let us see how.

Phenol is soluble in aqueous sodium hydroxide solution and forms sodium phenoxide. Ethyl alcohol, on the other hand, does not react at all with sodium hydroxide.

In phenol, the hydroxyl group is bonded directly to an sp²-hybridised carbon atom of the aromatic ring. Being more electronegative than the sp³-hybridised carbon, the sp²-hybridised carbon polarises the O-H bond. This means that the ionisation of phenol is greater than that of ethyl alcohol (in which the OH group is attached to a -hybridised carbon).

The greater acidity of phenol as compared to ethyl alcohol is attributed to the greater stability of the phenoxide anion than that of the ethoxide anion. In the ethoxide (C₂H₅, O^–) anion, the full negative charge is localised on the oxygen atom and therefore the anion is not stable.

As a consequence, this anion is strongly basic and the corresponding acid (C₂ H₅ OH) is very weak. On the other hand, in the phenoxide anion, the negative charge is delocalised over the entire molecule as shown below.

The delocalisation of charge stabilises the phenoxide anion and favours ionisation of the phenol to the phenoxide anion and H.

pk, values of some phenols and ethanol:

The greater the pK, value, the weaker the acid.

Reaction with Zn dust:

On distillation with Zn dust, phenol is converted to benzene.

⇒ C₆H₅OH + Zn→ C₆H₆ + ZnO

Alkylation The sodium salt of phenol (sodium phenoxide) reacts with an alkyl halide to form the corresponding ether.

Acylation Phenol reacts with acetyl chloride in the presence of pyridine (a base) to yield phenyl acetate

Reaction with Ammonia

Phenol reacts with ammonia at 573 K in the presence of anhydrous ZnCI₂ to yield aniline.

Reaction with phosphorus pentachloride:

On treatment with phosphorus pentachloride, phenol gives chlorobenzene (poor yield)

⇒ C₆H₅OH+ PCI₅→ C₆H₅CI+ HCI (Chlorobenzene)

Mainly, triphenyl phosphate is formed in the side reaction.

⇒ 3C₆H₆OH + POCI₆ → (C₆H₅)₃ PO₄ +3HCI Triphenyl phosphate

Substitution reactions in the benzene nucleus

The presence of the OH group in phenols activates the benzene ring and electrophilic substitution becomes possible. Phenols undergo electrophilic substitution reactions more readily than benzene. The hydroxyl group directs the incoming group to ortho- and para-positions as these positions become electron-rich.

Phenol undergoes nitration, halogenation, sulphonation and Friedel-Crafts reaction readily.

Nitration Phenol can be nitrated in dilute aqueous nitric acid even at room temperature. Ortho- as well as para- para-nitrophenols are formed.

The ortho- and para-isomers can be separated by steam distillation. o-nitrophenol is steam volatile. It has higher volatility because of the intramolecular hydrogen bonding between the hydroxyl group and the nitro group.

p-nitrophenol is less volatile because intermolecular hydrogen bonding causes association between its molecules. Thus, o-nitrophenol passes over with the steam, and p-nitrophenol remains in the distillation flask.

Phenol is first sulfonated and then nitrated to form picric acid. It is not treated with nitric acid first because phenol is easily oxidised by nitric acid.

Halogenation: When bromine water is added to phenol, a white precipitate of 2, 4, 6-tribromophenol is formed

Bromination in a nonpolar solvent (carbon disulphide) affords p-bromophenol as the main product.

Halogenation takes place even in the absence of a Lewis acid because the hydroxyl group in phenol activates the benzene ring towards electrophilic substitution.

Sulphonation:

Phenol reacts with concentrated H₂SO₄, to form a mixture of ortho- and para-hydroxybenzene sulphonic acid. At higher temperatures, predominantly the p-isomer is formed because it is less reactive than the ortho-isomer due to less steric interference of the bulky sulphonic acid group.

Friedel-Crafts reaction: 

Phenol reacts with methyl chloride in the presence of anhydrous AICI₃ to give o-cresol and p-cresol.

Kolbe reaction:

On being heated with CO₂ at 393-413 K and 1.5 atm the sodium salt of phenol yields the sodium salt of salicylic acid. On acidification, the latter salt gives salicylic acid. This entire sequence of reactions comprises what is known as the Kolbe reaction.

Reimer-Tiemann reaction:

On being heated with chloroform and caustic alkali, phenol gives salicylaldehyde.

In this reaction, dichlorocarbene (: CCl₂) is a reactive intermediate, which is formed by the alkaline hydrolysis of chloroform.

⇒ \(\stackrel{\ominus}{\mathrm{O}}+\mathrm{H}-\mathrm{CCl}_3 \rightleftharpoons \mathrm{H}_2 \mathrm{O}+: \stackrel{\ominus}{\mathrm{C}} \mathrm{Cl}_3 \rightarrow \stackrel{\ominus}{\mathrm{C}}+: \mathrm{CCl}_2\)

: CCl₂ is an electrophilic reagent. It reacts with the phenoxide anion to form salicylaldehyde

Coupling reaction: In the presence of an alkali, phenol couples with benzene diazonium chloride to form p-hydroxy azobenzene, which is a red dye.

Reaction with phthalic anhydride:

On being heated with phthalic anhydride in the presence of concentrated H₂SO₄, phenol gives phenolphthalein. Phenolphthalein gives a red colour with an alkali. It is used as an indicator in acid-base titration.

Reaction with FeCl₃: On treatment with a neutral FeCl3 solution, phenol gives a violet colouration. The violet colour is due to the formation of a water-soluble iron complex.

6C₆H₅OH + FeCl₃ →3H⁺ [Fe(OC₆H₅)₆]^-3+3HCl [complex ion (violet)]

This reaction is used as a test for the presence of phenol.

Liebermann’s nitroso reaction:

Phenol forms a deep blue solution on treatment with sodium nitrite and concentrated H₂SO₄ in cold conditions.

The colour turns red when the solution is diluted with water. The red solution becomes blue when it is made alkaline. This reaction is known as Liebermann’s nitroso reaction and is used as a test for the presence of phenol.

Oxidation:  On oxidation with chromic acid, phenol gives benzoquinone.

Phenol Uses

Phenol is used in the preparation of

1. Bakelite (synthetic resin),
2. Picric acid (as an explosive),
3. Phenolphthalein (as an indicator), and
4. Salol, aspirin, salicylic acid (as drugs).

Ethers

Ethers Methods of preparation

By the dehydration of alcohol:

Ethers are usually prepared by heating alcohols with concentrated H₂SO₄.

⇒ \(2 \mathrm{ROH} \stackrel{\Delta, \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow} \mathrm{R}-\mathrm{O}-\mathrm{R}+\mathrm{H}_2 \mathrm{O}\)

Diethyl ether is prepared by heating ethyl alcohol with concentrated H₂SO₄ at 413 K. The reaction occurs in two steps. At a low temperature, ethyl alcohol reacts with concentrated H₂SO₄ to form ethyl hydrogen sulphate which again reacts with ethyl alcohol at 413 K to yield diethyl ether

Ethers Mechanism

 

This is a nucleophilic bimolecular substitution (SN₂) reaction.

This method is only used in the preparation of ethers having unhindered primary alkyl group dehydration secondary or tertiary alcohols, yield alkenes as elimination competes with substitution.

From alcohol vapours and heated alumina:

Diethyl ether is formed when vapours of alcohol are passed through heated alumina at 513-533 K. anhydrous

⇒ \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH}+\mathrm{H} \mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3 \frac{\text { anhydrous } \mathrm{Al}_2 \mathrm{O}_3}{513-533 \mathrm{~K}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3+\mathrm{H}_2 \mathrm{O}\)

By Williamson Synthesis

Diethyl ether is formed when ethyl iodide is heated with an alcoholic solution of sodium ethoxide. (produced by the action of sodium on ethyl alcohol).

⇒ \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{I}+\mathrm{Na}-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3 \rightarrow \mathrm{CH}_3-\underset{\text { Diedium ethoxide }}{\mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3+\mathrm{NaI}}\)

This is an example of Williamson’s synthesis.

 Williamson’s synthesis Mechanism

Williamson synthesis involves the displacement of a halide ion from an alkyl halide by an alkoxide. It follows the S_N2 mechanism.

Symmetrical as well as unsymmetrical ethers can be synthesised. However, the alkyl halide must be a primary alkyl halide. For example, sodium f-butoxide reacts with methyl bromide to yield t-butyl methyl ether.

If a tertiary alkyl halide is treated with sodium ethoxide, then an alkene is formed due to the dehydrohalogenation of the alkyl halide. For example, the reaction of CH₃CH₂ONa with (CH₃)₃ C-Br gives 2-methylpropene exclusively.

The ethoxide anion (a strong base) reacts with t-butyl bromide, leading to an elimination reaction. Phenol can also be converted into an ether by Williamson synthesis.

From an alkyl halide and dry silver oxide:

Diethyl ether is formed on heating ethyl iodide with dry silver oxide.

⇒ \(2 \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{I}+\mathrm{Ag}_2 \mathrm{O} \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3+2 \mathrm{AgI}\)

Physical Properties of Ethers

Diethyl ether is a colourless, volatile liquid. The C–O bonds in ethers are polar and ethers do have a dipole noment (for diethyl ether, HD = 1.18), revealing the angular nature of the molecule. The bond angle in diethyl ther is around 110°.

Boiling point:

Ethers have lower boiling points than the corresponding isomeric alcohols since there is no association between the molecules due to hydrogen bonding. The boiling points of ethers are close to those of alkanes of comparable molecular weight.

Solubility:

The rule of thumb which states that compounds having no more than four carbons per oxygen are water-soluble holds for ethers as well as for alcohols. The solubility of ethers in water is comparable to that of alcohols of nearly the same molecular weights.

The solubility of diethyl ether and 1-butanol is about 10 g per 100 g of H₂O at 298 K. This is because of their capability to form hydrogen bonds with water.

Chemical Properties of Ethers

Ethers are relatively inert to most reagents. They are stable to bases, to catalytic hydrogenation, and to most other reducing agents.

Reactions with acids

Ethers are stable to dilute acids but do react with hot concentrated acids. Strong HBr or HI causes cleavage of the C-O bond in ethers.

Reaction with HI:

Two molecules of an alkyl iodide are formed when an ether is boiled with HI. Initially a molecule of an alcohol is also formed, which reacts further to form a second molecule of an alkyl iodide.

⇒ \(\mathrm{C}_2 \mathrm{H}_5-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5+2 \mathrm{HI} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{I}+\mathrm{H}_2 \mathrm{O}\)

 Ether Mechanism:

The ether first dissolves in the acid due to its basic nature with the formation of an oxonium ion. This is followed by an S_N2 reaction with the iodide ion.

In the next step, the ethyl alcohol formed reacts with HI to produce a second mole of ethyl iodide.

In the case of an unsymmetrical ether, the iodide ion (a nucleophile) attacks the least substituted carbon (carbon the smaller alkyl group) of the oxonium ion and displaces an alcohol molecule by the S_N2 mechanism.

At High Temperature, the excess HI reacts with Ethyl Alcohol to yield ethyl iodide

However, when one of the alkyl groups is the tertiary butyl group, the iodide formed is t-butyl iodide through the S 1 mechanism. In this reaction, the nucleophilic iodide ion attacks the more substituted carbon (carbon of the bigger alkyl group i.e.,

Anisole (an aromatic ether) reacts with HI to form methyl iodide and phenol. Alkyl aryl ethers are cleaved at the alkyl-oxygen bond because neither S_N1 nor S_N2 processes can normally occur at the aromatic carbon. The phenyl group is electron-rich and tends to repel any nucleophile.

Phenol does not react further to give a halide as the nucleophilic substitution reaction of phenol is rather difficult.

The C-O bond in ethers can also be cleaved upon their reactions with HBr and with HCl. Since these acids are less reactive than HI (because CI and Br are poorer nucleophiles than I), higher concentrations and temperatures are required for them to be effective.

The cyclic ether tetrahydrofuran is cleaved by HCI in the presence of ZnCl₂ to yield 1, 4 dichlorobutane, a valuable intermediate in the manufacture of nylon.

Reaction due to ethereal oxygen:

Ethers are bases and can react with acids such as sulphuric acid, boron trifluoride, and Grignard reagents.

Auto – Oxidation:

Ethers react with atmospheric oxygen to form peroxides

Peroxides explode violently on heating. During the distillation of an ether, the residue in the distilling flask becomes rich in peroxides. Therefore, ethers should not be distilled to dryness.

The presence of peroxide in an ether can be detected by shaking a small volume of an ether mixed with an aqueous KI solution.  A purple colour confirms the presence of peroxide.

Electrophilic substitution reactions:

The alkoxy group (OR) activates the benzene ring towards electrophilic substitution. It directs the incoming group to ortho- and para-positions as these positions become electron-rich as shown below.

Halogenation:  Bromination of anisole (methyl phenyl ether) in an acetic acid medium mainly gives p-bromoanisole

Nitration:  In the presence of concentrated H₂SO₄, anisole reacts with concentrated HNO₃ at 323-333 K to give ortho- and para- para-nitro anisole.

Friedel-Crafts reaction:  Anisole reacts with an alkyl halide or acyl halide in the presence of anhydrous AICI₃ as a catalyst. The alkyl or acyl groups are introduced at the ortho- and para-positions

Example 5: Give the structures of the products of the following reactions. Explain your answers.

Solution:

This is an S_N2 reaction. The nucleophile Br attacks the less substituted alkyl group.

In this case, the nucleophile (I) cannot attack the electron-rich carbon of benzene. It preferentially attacks the more stable tertiary carbonium ion.

This is an electrophilic aromatic substitution reaction. Since the OCH, group is o- and p- directing, NO₂ (HNO₃ + 2H₂SO₄→2HSO₄ + NO₂ + H₂O) is substituted at the o- and p-positions.

This is an S_N2 reaction. The nucleophile Br attacks the less-substituted alkyl group.

Alcohols Phenols And Ethers Multiple-Choice Questions

Question 1. Glycerol is a

1. Monohydric alcohol
2. Trihydric alcohol
3. Dihydric alcohol
4. None of these

Answer: 3. Dihydric alcohol

Question 2. Methyl alcohol and ethyl alcohol are distinguished by the reaction with

1. I₂/NaOH
2. Na
3. CH₃COOH
4. None of these

Answer: 1. I₂/NaOH

Question 3. C₂H₂OH and CH₃-O-CH, are

1. Position isomers
2. Chain isomers
3. Functional isomers
4. Metamers

Answer: 2. Functional isomers

Question 4. Which of the following compounds will give an ester with an acid?

1. Paraffin
2. Alcohol
3. Alkene
4. Alkyl halide

Answer: 2. Alcohol

Question 5. The first component formed on the oxidation of a primary alcohol is a/an

1. Ketone
2. Ester
3. Carboxylic acid
4. Aldehyde

Answer: 4. Aldehyde

Question 6. Alcohols contains the functional group

1. -OH
2. -CHO
3. C=O
4. -NH₂

Answer: 1. -OH

Question 7. Lucas reagent is

1. Concentrated HCl + anhydrous ZnCl₂
2. Dilute HCl + Hydrated ZnCl₂
3. Concentrated HNO₃+ anhydrous ZnCl₂
4. Concentrated HNO₃ + anhydrous MgCl₂

Answer: 1. Concentrated HCl + anhydrous ZnCl₂

Question 8. Ethanol and methanol are distinguished by

1. The chloroform test
2. The Victor Meyer test
3. Their rates of esterification
4. The iodoform test

Answer: 4. The iodoform test

Question 9. A primary alcohol contains a

1. CHOH group
2. -CH₂OH
3. ->C-OH group
4. None of these

Answer: 2. -CH₂OH

Question 10. On distillation with zinc, phenol gives

1. Nitrobenzene
2. Aniline
3. Benzene
4. Zinc Phenoxide

Answer: 3. Benzene

Question 11. On reaction with bromine-water, phenol gives

1. Bromobenzene
2. Dibromorphenol
3. 2,4,6 – tribromophenol
4. Picric acid

Answer: 3. 2,4,6 – tribromophenol

Question 12. Methyl propyl ether is

1. Asymmetrical ether
2. An unsymmetrical ether
3. A simple ether
4. None of these

Answer: 2. An unsymmetrical ether

Question 13. In reaction with a mineral acid, an ether gives an

1. Oxonium salt
2. Ether peroxide
3. Alkane
4. Aldehyde

Answer: 1. Oxonium salt

Question 14. Diethyl ether and methyl propyl ether are

1. Functional isomers
2. Metamers
3. Position isomers
4. Chain isomers

Answer: 2. Metamers

Question 15. The IUPAC name of CH₃OC₂H₅ is

1. Methoxy ethane
2. Propoxymetane
3. Ethyl methyl ether
4. Diethyl ether

Answer:  1. Methoxy ethane

Question 16. To prepare diethyl ether, bromoethane is heated with which of the following compounds?

1. Ethanol
2. Sodium ethoxide in dry ether
3. C₂H₂OH/Na
4. HI

Answer: 2. Sodium ethoxide in dry ether

Question 17. Diethyl ether and 1-butanol are

1. Position isomers
2. Functional Isomers
3. Optical isomers
4. Metamers

Answer: 2. Functional Isomers

Question 18. The correct order of the boiling points of primary (1), secondary (2°) and tertiary (3°) alcohols is

1. 1°> 2° >3°
2. 3° 2° 1°
3. 2>1° >30
4. 2° >30 > 1°

Answer: 1. 1°> 2° >3°

Question 19. Among the following, which is the most acidic?

1. Phenol
2. Benzyl alcohol
3. m-Chlorophenol
4. Cyclohexanol

Answer: 1. Phenol

Question 20. Which of the following forms hydrogen bonds to a greater extent?

1. Ethanol
2. Diethyl ether
3. Triethyl amine
4. Ethyl chloride

Answer: 1. Ethanol

Question 21. Which of the following will respond positively to the iodoform test?

1. CH₂OH
2. (CH₃)₂CHOH
3. (CH₃), COH
4. CH₃ OH

Answer:  2. (CH₃)₂CHOH

Question 22. Formaldehyde reacts with CH₃MgBr to give

1. C₂H₂OH
2. CH₃COOH
3. CH₃CHO
4. HCHO

Answer: 1. C₂H₂OH

Question 23. On treatment with B₂H₆/H₂O₂, R-CH=CH₂ gives

1. RCOCH₃
2. RCHOH CH₂OH
3. RCH₂ CHO
4. RCH₂CH₂OH

Answer: 4. RCH₂CH₂OH

Question 24. Which of the following will give a yellow precipitate with I₂/NaOH?

1. CH₃COCH₂CH₃
2. CH₃COOCOCH₃
3. CH₃CONH₂
4. CH₃CH(OH)CH₂CH₃

Answer:  1 and 4 – CH₃COCH₂CH₃  and CH₃CH(OH)CH₂CH₃

Question 25. Phenol reacts with CHCl₃/aqueous NaOH. The electrophilic reagent which attacks the benzene nucleus is

1. CHCl₃
2. CHC1₂
3. COCI₂
4. CCl₂

Answer: 4. CCl₂

Question 26. Upon reacting with neutral FeCl₃, phenol gives a

1. Green colour
2. Violet colour
3. Red colour
4. Blue colour

Answer: 2. Violet colour

Question 27. The reaction of

Answer: 1 and 4

Question 28. The central oxygen atom of ether is

1. Sp-hybridised
2. Sp²-hybridised
3. Sp³d²-hybridised
4. sp-hybridised

Answer: 3.  Sp³d²-hybridised

Question 29. What is the order of dehydration of the following?

1. 1 < 2 < 3 <4
2. 2<3<4<1
3. 1< 3< 2<4
4. 1 < 4 < 2 = 3

Answer: 1. 1 < 2 < 3 <4

Question 30. How many isomeric acyclic alcohols and ethers are possible for C₄H₈O?

1. 7
2. 9
3. 5
4. 8

Answer: 4. 8

Question 31. Among the following reagents, phenol can be distinguished from ethyl alcohol by all except

1. NaOH
2. FeCl₃
3. Br₂/H₂O
4. Na

Answer: 4. Na

Question 32. The order of esterification of alcohol is

1. 3° > 2° > 1°
2. 1° > 2° >3°
3. 2° >3° 1°
4. None of these

Answer: 3. 2° >3° 1°

Aldehydes, Ketones and Carboxylic Acids

Aldehydes And Ketones

Aldehydes and ketones are compounds containing the carbonyl group (CO). When two alkyl groups are attached to the carbonyl group, the compound is a ketone. When two hydrogens, or one hydrogen and one alkyl group, are attached to the carbonyl group, the compound is an aldehyde.

The chemistry of the carbonyl group is extremely important. Aldehydes and ketones are reactive in nature. Many useful products such as dyes, resins, perfumes, plastics and cloths are made from them. Many useful compounds containing these functional groups can be obtained from plants.

For example, vanillin (from vanilla beans), salicylaldehyde (from meadow sweet) and cinnamaldehyde (from cinnamon). These are flavouring compounds. Vanillin is used for the vanilla flavour in ice creams. Salicylaldehyde has an odour like bitter almonds and is used in perfumery. Cinnamaldehyde is responsible for the flavour of cinnamon.

Nomenclature Of Aldehydes

Aldehydes are named using either of two different systems.

Trivial name

Simple aldehydes are commonly known by their trivial names or common names. According to this system, the name of an aldehyde is derived from the name of the corresponding carboxylic acid by dropping the suffix ic (oic) acid and adding in its place the suffix aldehyde.

In the common system of nomenclature, the position of an additional substituent is indicated by the Greek letters α, β, γ, etc., a being the carbon attached to the carbonyl group, β being the next carbon, and so on.

IUPAC name

In the IUPAC system, aldehydes are named by adding the suffix al to the name of the corresponding hydrocarbon, the ‘e’ of the hydrocarbon being omitted.

The substituents on the chain are prefixed in alphabetical order along with the numbers indicating their positions. These numbers are allocated by considering the aldehydic carbon to be the first carbon.

The IUPAC name of a simple aromatic aldehyde in which the aldehydic group is directly attached to the benzene ring is benzenecarbaldehyde. However, the common name benzaldehyde is also retained in IUPAC nomenclature. Other aromatic aldehydes are named as substituted benzaldehydes.

The common and IUPAC names of some aldehydes:

The following are examples of some more aldehydes and their IUPAC names:

A molecule with a double bond and a triple bond is called an enyne (ene + yne). A molecule with a double bond, triple bond and an aldehyde group is known as a enynal (ene + yne + al). The term is used as a suffix.

The numbering of carbons begins with the carbon of the aldehyde group. The substituents and the functional are located accordingly.

A molecule with two double bonds and an aldehyde group is known as a dienal (diene + al). The term is used as a suffix. Numbering of carbons begins with the carbon of the aldehyde group. The substituents and the functional groups are located accordingly.

If an unbranched chain is directly linked to more than two aldehydic groups, these aldehydes are named from the parent hydrocarbon by the substitutive use of a suffix, example,tricarbaldehyde. The suffix ‘al’ is not used. Numbering of carbons begins at the end nearest the functional group. The functional groups are located accordingly.

Nomenclature Of Ketones

Ketones too are named using either of two different systems.

Trivial names

Simple ketones may be named by using the names of the alkyl groups attached to the carbonyl group followed by the word ketone. The names of the alkyl groups are written alphabetically.

IUPAC names

In the IUPAC system ketones are named by adding the suffix ‘one’ to the name of the corresponding hydrocarbon and omitting the final ‘e’ of the hydrocarbon.

The position of the carbonyl group is given by a number which can be placed before the parent name as given below or immediately before the suffix (for example, pentane-2-one).

The common and IUPAC names of some ketones:

Ketones Structure

In the IUPAC system, the longest chain containing the carbonyl group is taken as the parent hydrocarbon and the positions of the carbonyl as well as any other substituent present in the molecule are indicated by numbers The longest chain containing the carbonyl group is numbered from the end that gives the carbonyl carbon the lowest number.

A molecule with a double bond and a ketone is called an enone (ene + one) and the term is used as a suffix. The numbering of carbons begins at the end nearest the functional group.

In cyclic ketone the carbonyl carbon is a

Example 1: Draw the structures of the following compounds.

1. β -Ethoxybutyraldehyde
2. 3-Methylcyclopentanecarbaldehyde
3. 3-Oxob utanal
4. Diisopropyl ketone
5. 4-Bromoacetophenone

Solution:

The allocation of Greek letters to the carbon atoms in the common system starts from the carbon next to the aldehyde group, whereas the numbers in the IUPAC system always begin with the carbon of the aldehyde group.

⇒ \(\stackrel{\beta}{\mathrm{C}}-\stackrel{\gamma}{\mathrm{C}}-\stackrel{\beta}{\mathrm{C}}-\stackrel{\alpha}{\mathrm{C}}-\mathrm{CHO}\) (Used In Common names)

⇒  \(\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}} \mathrm{HO}\)(Used In IUPAC names

Structures Of The Carbonyl Group Used In IUPAC Names

The carbon atom of the carbonyl group is sp2-hybridised and three of its electrons form three o bonds. The molecule is planar and the bond angles are close to 120°. The remaining p orbital of the carbon with one electron overlaps a p orbital of oxygen with one electron to form a bond between these atoms. The oxygen atom also has two lone pairs of electrons that occupy the remaining orbital.

The electrons in the bond of the carbonyl group are not equally shared. They are pulled more towards the more electronegative oxygen atom. As a result, the C-O bond is polarised in the direction C-O. Therefore, the electron-deficient carbonyl carbon is electrophilic (a Lewis acid) in nature and the electron-rich carbonyl oxygen is nucleophilic (Lewis base) in nature.

Due to bond polarity, carbonyl compounds have substantial dipole moments.

General Methods Of Preparation Of Aldehydes And Ketones

By the oxidation of alcohol

Aldehydes Primary alcohols yield aldehydes upon oxidation with pyridinium chlorochromate (PCC) in a CH₂ Cl₂ medium. Aqueous methods like Jones oxidation (Na₂Cr₂O₇ and dilute H₂SO₄ in acetone) are not useful for this purpose since the aldehyde that is formed is further oxidised to a carboxylic acid.

Ketones Secondary alcohols yield ketones upon oxidation with K₂Cr₂O₇/H₂SO₄ in an acetone medium.

Tertiary alcohols are stable to oxidation under these conditions.

Oppenauer oxidation

Oppenauer oxidation involves the oxidation of primary alcohols and secondary alcohols to aldehydes and ketones respectively in the presence of aluminium isopropoxide, Al[OCH(CH₃)₂]₃

By the dehydrogenation of alcohol

When the vapours of a primary alcohol are passed over hot copper at about 573 K, the alcohol is readily dehydrogenated to yield an aldehyde.

⇒ \(\underset{\text { (Primary alcohol) }}{\mathrm{R}-\mathrm{CH}_2 \mathrm{OH}} \underset{573 \mathrm{~K}}{\stackrel{\mathrm{Cu}}{\longrightarrow}} \underset{\text { (Aldehyde) }}{\mathrm{R}}-\mathrm{CHO}+\mathrm{H}_2\)

On dehydrogenation, secondary alcohols give ketones.

⇒ \(\underset{\text { (Secondary alcohol) }}{\mathrm{R}-\mathrm{CHOH}-\mathrm{R}^{\prime}} \underset{573 \mathrm{~K}}{\stackrel{\mathrm{Cu}}{\longrightarrow}} \underset{\text { (Ketone) }}{\mathrm{R}}-\mathrm{CO}-\mathrm{R}^{\prime}+\mathrm{H}_2\)

By the pyrolysis of calcium salts of carboxylic acids

When the calcium salts of carboxylic acids are heated to high temperatures, symmetrical ketones are obtained.

Formation of an aldehyde: On dry distillation, calcium formate gives formaldehyde.

On dry distillation, a mixture of calcium acetate and calcium formate yields acetaldehyde.

Formation of a ketone:  On dry distillation, the calcium salt of a carboxylic acid, other than formic acid, gives a ketone. For example, in dry distillation, calcium acetate gives acetone.

By the hydrolysis of gem-dihalides (1, 1-dihalides)

Formation of an aldehyde Boiling a gem-dihalide with aqueous NaOH gives a dihydroxy compound. Being unstable, this compound loses a water molecule readily to form an aldehyde.

Formation of a ketone If the halo groups are not present on the terminal carbon atom, i.e., the gem-dihalide is an internal dihalide, then a ketone is obtained.

By the hydration of alkynes

The hydration of alkynes with 20% H₂SO₄ in the presence of a mercuric salt (HgSO₄) gives aldehydes or ketones. Formation of an aldehyde On hydration with 20% H₂SO₄ in the presence of HgSO₄ at 353 K, acetylene gives an unstable enol, which soon isomerises to acetaldehyde.

Hydration of alkynes Mechanism:

Formation of ketones: On hydration with 20% H2SO4 in the presence of HgSO4 (catalyst) at 353 K, substituted alkynes give ketones.

By ozonolysis

Depending on the structure of an alkene, aldehydes and ketones are obtained by making the alkenes react with ozone and subsequent treatment with zinc dust and water.

The ozonide may also react with oxidising agents such as H₂O₂ to give carboxylic acids or with more powerful reducing agents such as NaBH, to give alcohols.

The ozonolysis of cyclohexene is particularly useful as it gives 1, 6-dicarbonyl compounds that are otherwise difficult to make. In the simplest case we get hexane-1, 6-dioic acid (adipic acid), a monomer used in the manufacture of nylon.

Aldehydes And Ketones

By the reduction of acid chlorides

Aldehydes Acid chlorides are reduced with hydrogen gas in the presence of palladised BaSO₄ or BaCO₃ (Rosenmund reduction) to aldehydes. Ketones are not prepared by this method.

⇒ \(\underset{\text { (Acid chloride) }}{\mathrm{R}-\mathrm{COCl}}+2 \mathrm{H} \stackrel{\mathrm{Pd} / \mathrm{BaSO}_4}{\longrightarrow} \underset{\text { (Aldehyde) }}{\mathrm{R}-\mathrm{CHO}+\mathrm{HCl}}\)

⇒ \(\underset{\text { Acetyl chloride }}{\mathrm{CH}_3-\mathrm{COCl}}+2 \mathrm{H} \stackrel{\mathrm{Pd} / \mathrm{BaSO}_4}{\longrightarrow} \underset{\text { Acetaldehyde }}{\mathrm{CH}_3 \cdot \mathrm{CHO}}+\mathrm{HCl}\)

The function of BaSO₄ is to poison the catalyst. This prevents the reduction of the aldehyde to an alcohol.

Ketones Acid chlorides react with dimethyl cadmium to yield ketones.

Grignard reagents, being more reactive than dimethyl cadmium, react with ketones to yield 3° alcohols. For this reason, Grignard reagents are not used in such reactions. Dimethyl cadmium does not react with ketones. Dimethyl cadmium can be prepared by making a Grignard reagent react with cadmium chloride.

⇒\(\underset{\text { Methylmagnesium bromide }}{2 \mathrm{CH}_3 \mathrm{MgBr}}+\mathrm{CdCl}_2 \rightarrow \underset{\text { Dimethylcadmium }}{\left(\mathrm{CH}_3\right)_2 \mathrm{Cd}}+2 \mathrm{MgBrCl}]\)

From alkyl cyanides

Aldehydes On reduction by stannous chloride and concentrated HCl, an alkyl cyanide (also called an alkyl nitrile) yields an imino chloride, which on hydrolysis with water yields an aldehyde. This reaction is known as the Stephen reaction.

Alkyl cyanides Mechanism:

Nitriles may also be selectively reduced by di-isobutylaluminium hydride (DIBAL-H) to imines, which upon hydrolysis give aldehydes.

DIBAL-H is sterically congested and therefore not very reactive. For this reason, it does not reduce the Di isobutyl aluminium hydride (DIBAL-H).

Ethylenic bond of unsaturated nitriles.

The reaction of an ester with exactly 1 equivalent of DIBAL-H at low temperature gives an aldehyde.

Ketones: On treatment with Grignard reagents, alkyl cyanides give ketones.

Physical Properties Of Aldehydes And Ketones

Formaldehyde is a gas while aldehydes or ketones containing up to 11 carbon atoms are colourless liquid Aldehydes and ketones containing more carbon atoms are solids.

Aldehydes and ketones are polar in nature. Their boiling points are higher than those of alkanes of similar molecular masses due to dipole-dipole interaction.

Because aldehydes and ketones are not associated with hydrogen bonds, their boiling points are lower than those of alcohols and carboxylic acids of comparable molecular weights.

The first few members of aliphatic aldehydes and ketones are soluble in water. For example, formaldehyde, acetaldehyde and acetone are soluble in water in all proportions. As the length of the alkyl group chain increases, the solubility of the compound decreases as in alcohols.

The solubility of lower aldehydes and ketones in water arises from the ability of the oxygen of the carbonyl group to form hydrogen bonds with water molecules.

All aldehydes and ketones are soluble to quite an extent in organic solvents like benzene and methyl alcohol. Formaldehyde and acetaldehyde have an unpleasant odour but higher aldehydes have a fruity smell.

Some of the ketones have a sweet smell. Many naturally occurring aldehydes and ketones are used as additives in perfumes and as flavouring agents.

Chemical Properties Of Aldehydes And Ketones

Aldehydes and ketones undergo similar chemical reactions because they contain the carbonyl functional group.

Nucleophilic addition reactions

The carbonyl group is represented by the two contributing structures

This imparts a substantial dipole moment to the carbonyl group, with the carbon bearing a partial positive harge and the oxygen bearing a partial negative charge.

The polar nature of the carbonyl group and the ability of the oxygen atom to accommodate the extra electron-air facilitates the attack of the nucleophile at the carbonyl carbon. The nucleophile normally adds to the carbony

Carbon from a direction approximately perpendicular to the plane of the sp2-hybridised orbitals of the carbonyl carbon. In this process, the hybridisation of the carbonyl carbon changes from sp’ to sp and a tetrahedral alkoxide anion intermediate is formed. The reaction is completed by the abstraction of a proton from the reaction medium to give the addition product.

Reactivity of aldehydes and ketones

Aldehydes are more reactive than ketones because of two factors.

Inductive effect The carbonyl carbon atom of a ketone carries two electron-donating groups, whereas that of an aldehyde has only one. Thus the ketone carbonyl carbon atom has less tendency to attract a nucleophile.

Steric effect Bulky groups adjacent to >C=O cause more steric strain in the addition product than in the parent carbonyl and reduce reactivity towards addition.

The order of reactivity of different carbonyl compounds towards nucleophilic addition is given below.

Example 2: Arrange the following carbonyl compounds in increasing order of their reactivity in nucleophilic addition reactions. Give reasons.

1. CH₃CHO, CH₃CH₂CHO, CH₃COCH₃, CH₃COCH₂CH₃

2.

Solution: 

1. The reactivity of aldehydes and ketones towards a nucleophile is influenced by the nature of the alkyl group attached to the carbonyl carbon in the following two ways.

• First, an electron-donating alkyl group reduces the positive charge of the carbonyl carbon and makes it less susceptible to nucleophilic attack.
• Secondly, the bulkier alkyl group attached to the carbonyl carbon presents greater steric hindrance than the smaller hydrogen atom to the approaching nucleophile.

From the above considerations, the increasing reactivity of the given compounds towards the nucleophile is as follows.

⇒ CH₃COCH₂CH₃<CH₃COCH₃<CH₃CH₂CHO<CH₃CHO

2. Similar influences are exerted by an aryl substituent as delocalised orbitals of the ring act as an electron source.

In the case of p-nitrobenzaldehyde, the electron-withdrawing nitro group at the p-position increases the positive character of the carbonyl carbon and thus facilitates the attack of nucleophiles.

In p-tolualdehyde, the methyl group at the p-position increases the electron density on the carbon of the carbonyl group by hyperconjugation as shown and makes it less reactive than benzaldehyde towards nucleophilic addition.

Acetophenone (a ketone) is less reactive than all the given aldehydes. The electron-donating methyl group and л-orbital of the benzene ring, acting as an electron source, make the carbonyl carbon electron-rich and less susceptible to nucleophilic attack. Further, the phenyl group and methyl group attached to the carbonyl carbon present greater steric hindrance to the approaching nucleophile.

Ketone Functional Group

On the basis of the above considerations, the order of increasing reactivity of the given compounds towards nucleophiles is as follows.

Ketones and aldehydes undergo very similar reactions. In this chapter, unless mentioned otherwise, each reaction discussed is equally applicable to both aldehydes and ketones and a generic structure is used for both.

Examples of nucleophilic addition reactions

Aldehydes and ketones undergo the following nucleophilic addition reactions.

Reaction with hydrocyanic acid (HCN) HCN adds to aldehydes and ketones to produce cyanohydrins.

The reaction is slow with pure HCN. However, in the presence of a base, CN (a stronger nucleophile) is generated, which readily adds to carbonyl compounds to yield the corresponding cyanohydrin.

⇒ \(\mathrm{HCN}+\mathrm{OH}^{-} \rightarrow: \overline{\mathrm{CN}}+\mathrm{H}_2 \mathrm{O}\)

Nucleophilic Mechanism:

The cyanide anion, the nucleophile, adds to the carbonyl carbon to yield an unstable intermediate oxygen anion, which, being a strong base, abstracts a proton from the solvent or HCN to give cyanohydrin.

Some plants produce cyanohydrins. The seeds of cherries, plums, peaches and apricots contain cyanohydrins. The odour and flavour of almonds is due to benzaldehyde and the cyanohydrin of benzaldehyde.

Addition of sodium bisulphite (NaHSO₄)

Aldehydes and ketones react with an aqueous, saturated solution of sodium bisulphite to yield a bisulphite product. This product is sparingly soluble in water and can be separated by filtration and decomposed by mineral acids to give back the carbonyl compound. The method is often used for separating a carbonyl compound from a noncarbonyl impurity.

Sodium bisulphite Mechanism:

Addition of Grignard reagents

Aldehydes and ketones react with Grignard reagents to produce alcohols. Formaldehyde yields primary (1°) alcohols, other aldehydes give secondary (2°) alcohols and all ketones give tertiary (3°) alcohols.

Grignard reagents Mechanism:

Addition of alcohol

In the presence of dry HCl gas, one mole of an aldehyde reacts with one mole of alcohol to give a hemiacetal (alkoxy alcohol), which again reacts with one mole of the alcohol to yield an acetal (gem-alkoxy compound).

Alcohol Mechanism:

Ketones do not react with monohydric alcohols in the presence of dry HCl but react with ethylene glycol (a dihydric alcohol) to form cyclic ketals.

In the presence of dry HCl gas, the oxygen atom of a carbonyl compound is bonded to H*, leaving a stabilised \(\left(>\mathrm{C}=\mathrm{O}+\mathrm{H}^{+} \rightleftharpoons>\mathrm{C}=\stackrel{+}{\mathrm{O}}-\mathrm{H} \leftrightarrow \stackrel{+}{\mathrm{C}}-\mathrm{OH}\right)\) arbonium ion, which can react with ethylene glycol (a nucleophile) to yield a cyclic ketal.

Reaction with ammonia derivative (NH₂-Y):

Several derivatives of ammonia such as

Can take part in addition reactions with aldehydes and ketones. The reaction is catalysed both by acids and alkalis. Under their influence, a molecule of water is eliminated, introducing a double bond between C and N. The products are crystalline, high-melting solids, very useful as derivatives for identification.

Ammonia derivative Mechanism:

Reaction with PCI₅

When aldehydes and ketones are treated with PCl₅, the oxygen atom of the carbonyl group is substituted by two chlorine atoms and gem-dihalides are formed.

Reduction

Reducing agents such as LiAlH₄, NaBH₄ and H₂/Ni reduce aldehydes to primary alcohols and ketones to secondary alcohols.

LiAlH₄ is a versatile reducing agent. It reduces not only aldehydes and ketones but carboxylic acids, esters and nitriles as well. Sodium borohydride is a weak and selective reducing agent and reduces aldehydes and ketones only, but not carboxylic acids. NaBH₄ is also unreactive towards C = C and C=C bonds.

If a molecule contains both an aldehyde and an ester, only the aldehyde will be reduced by NaBH₄.

Aldehydes and ketones may be reduced to hydrocarbons on treatment with Zn (Hg) and concentrated HCl. In this reaction, the >C=O group is converted into the >CH2 group. This reaction is known as Clemmensen reduction.

⇒ \(>\mathrm{C}=\mathrm{O}+4 \mathrm{H} \frac{\mathrm{Zn}-\mathrm{Hg}}{\mathrm{HCl}}>\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}\)

Aldehydes and ketones are also easily reduced to hydrocarbons in the presence of excess hydrazine and a strong base on heating. This reaction is known as Wolff-Kishner reduction.

Oxidation of aldehydes

Aldehydes are among the most readily oxidised classes of organic compounds. They are converted to carboxylic acid by numerous oxidising agents such as KMnO₄, K₂Cr₂O7 and HNO₃

Aldehydes are also oxidised by relatively weak oxidising agents such as Tollens reagent and Fehling’s solution.

Oxidation by Tollens reagen:

When an aldehyde is warmed with ammoniacal silver nitrate (Tollens reagent), the aldehyde is oxidised to carboxylic acid and the silver ion is reduced to free silver, which deposits in the form of a mirror on the inner wall of the test tube.

Ordinary mirrors are prepared in this way, using formaldehyde.

Tollens reagent is prepared by adding one drop of an aqueous solution of NaOH to a silver nitrate solution (10 mL) and dissolving the resultant precipitate in a minimum quantity of ammonium hydroxide solution.

⇒ \(\mathrm{AgNO}_3+\mathrm{NaOH}+2 \mathrm{NH}_4 \mathrm{OH} \rightarrow \underset{\text { Tollens reagent }}{\mathrm{Ag}\left(\stackrel{+}{\mathrm{NH}_3}\right)_2 \overline{\mathrm{O}} \mathrm{H}}+\mathrm{NaNO}_3+2 \mathrm{H}_2 \mathrm{O}\)

Oxidation by Fehling’s solution:

When heated with Fehling’s solution, an aldehyde is oxidised to a carboxylic acid and the Fehling’s solution is reduced to cuprous oxide (Cu₂O) as a brick-red precipitate.

Fehling’s solution is a deep blue solution prepared by mixing equal volumes of Fehling’s solution (1) and Fehling’s solution (2). Fehling’s solution (1) is an aqueous copper sulphate solution. Fehling’s solution (2) is an alkaline solution of sodium potassium tartrate (Rochelle salt).

When Fehling’s solution (1) is mixed with Fehling’s solution (2), a deep blue soluble complex is formed, which reacts with an aldehyde to form the sodium salt of a carboxylic acid and a red-brown precipitate of Cu₂O.

⇒ \(\mathrm{CuSO}_4+2 \mathrm{NaOH} \longrightarrow \mathrm{Cu}(\mathrm{OH})_2+\mathrm{Na}_2 \mathrm{SO}_4\)

Fehling’s solution can oxidise aliphatic aldehydes only, while Tollens reagent can oxidise aliphatic as well as aromatic aldehydes. These reagents have no effect on ketones.

The aldehyde group undergoes aerial oxidation. For example, crystals of benzoic acid grow inside a bottle filled with benzaldehyde in the presence of sunlight.

Oxidation by Benedict’s reagent:

Benedict’s reagent is an alkaline solution of copper sulphate, sodium carbonate and sodium citrate. On heating with Benedict’s reagent, aliphatic aldehydes yield a red-brown precipitate of Cu₂O

Using Benedict’s reagent we can detect the presence of sugar in urine.

Oxidation of ketones

The oxidation of ketones requires stronger oxidising agents such as permanganate and high temperatures. It involves the cleavage of a carbon-carbon bond adjacent to the carbonyl group, and different carboxylic acids are formed.

Oxidation of methyl ketones (haloform reaction) Aldehydes and ketones containing a keto methyl group are oxidised by sodium hypohalite (NaOX). Sodium salts of carboxylic acids containing one carbon less than the methyl ketone are formed. The methyl group forms a haloform.

For example, acetone is oxidised by I₂/NaOH to give iodoform. The three hydrogens in the methyl group are acidic and are readily displaced by the halogen to yield a triiodo compound which is cleaved under the influence of the alkali to yield iodoform and the corresponding carboxylic acid.

Iodoform is a pale yellow solid and its appearance indicates the presence of a methyl ketone.

Methyl ketones give the corresponding chloroform or bromoform with hypochlorite or hypobromite and so this reaction is known as a haloform reaction.

The hypohalite does not attack a double bond, if present in the molecule.

Compounds containing the methylcarbinol group also respond positively to the iodoform test.

Thus, even ethyl alcohol responds positively to the iodoform test.

Reactions due to α-hydrogen

A carbon atom located next to a carbonyl carbon is known as an a-carbon and hydrogens attached to an a-carbon are called a-hydrogens. These hydrogens are acidic due to the electron-withdrawing nature of the carbonyl group and resonance stabilisation of the conjugate base.

Aldol condensation:

Two molecules of an aldehyde or a ketone containing a-hydrogen undergo condensation in the presence of a dilute alkali to give a β-hydroxy aldehyde (aldol) or a β-hydroxy ketone (ketol). These reactions are called aldol condensation reactions.

Aldols (aldehyde-alcohols) can be dehydrated easily by heating to yield α, and β unsaturated aldehydes. For example, on heating, β-hydroxybutyraldehyde yields crotonaldehyde.

In the presence of Ba(OH)₂, two molecules of acetone condense to yield diacetonyl alcohol.

Aldol Condensation Mechanism:

Cross-aldol condensation:

An aldol condensation between two different carbonyl compounds, each containing a-hydrogen, gives a mixture of four products. This reaction is called a cross-aldol condensation.

It is generally difficult to separate a mixture of four products. A cross-aldol condensation yielding a pure product can, however, be obtained when one of the reactants has no α -hydrogen (for example, benzaldehyde, or maldehyde).

Other reactions

Cannizzaro reaction:

When an aldehyde that has no a-hydrogens is treated with a concentrated aqueous alkali, a disproportionation reaction occurs. One molecule of the aldehyde is reduced to a primary alcohol, and another molecule is oxidised to the corresponding carboxylic acid salt.

This reaction is known as the Cannizzaro reaction.

Cannizzaro reaction Mechanism:

Crossed Cannizzaro reaction: 

When an aldehyde that has no a-hydrogens is treated with formaldehyde and a strong base, it is the formaldehyde rather than the other aldehyde that is oxidised. Such a reaction is known as a crossed Cannizzaro reaction. For example,

Aromatic Aldehydes

Methods of preparation:

Aromatic aldehydes and ketones are prepared by the following methods.

Aromatic aldehydes

By the oxidation of toluene (Etard reaction), The partial oxidation of toluene by chromyl chloride (CrO₂ Cl₂) gives a chromium complex, which on hydrolysis yields benzaldehyde.

This reaction is known as the Etard reaction. If a strong oxidising agent is used, toluene is oxidised to benzoic acid.

Upon treatment with chromic oxide in acetic anhydride, toluene gives benzylidene diacetate. On hydrolysis with a dilute acid, Benzylidene diacetate yields benzaldehyde.

By the hydrolysis of benzal chloride:

On hydrolysis, benzal chloride yields benzaldehyde. Hydrolysis is done by using an aqueous NaOH solution. Benzal chloride is prepared by the chlorination of toluene in the presence of sunlight.

By the Gattermann-Koch synthesis:

Benzaldehyde is prepared by treating benzene or its derivative with CO+ HCl in the presence of anhydrous AlCl₃ and Cu₂Cl₂

Gattermann-Koch synthesis Mechanism:

Aromatic ketones

By the acylation of benzene:

Aromatic hydrocarbons react with acyl chloride in the presence of anhydrous AlCl₃ to yield aromatic ketones. For example, on treatment with acetyl chloride in the presence of anhydrous AlCl₃ benzene or substituted benzene gives acetophenone.

Acylation of benzene Mechanism:

Chemical Reactions

Most of the chemical reactions of benzaldehyde are similar to those of aliphatic aldehydes. We will now discuss some reactions of benzaldehyde which are different from those of aliphatic aldehydes.

Perkin reaction:

On being heated with acetic anhydride in the presence of sodium acetate, benzaldehyde gives cinnamic acid (a, B- unsaturated acid).

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}+\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O} \stackrel{\mathrm{CH}_3 \mathrm{COONa}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CHCOOH}+\mathrm{CH}_3 \mathrm{COOH}\)

This reaction is known as the Perkin reaction.

Perkin reaction Mechanism:

Benzoin condensation

Benzaldehyde and most other aromatic aldehydes undergo a self-condensation, known as benzoin condensation, when treated with potassium cyanide in an alcoholic solution. The product formed from benzaldehyde is called benzoin.

Benzoin condensation Mechanism

Claisen-Schmidt reaction

Benzaldehyde undergoes condensation with aldehydes and ketones in the presence of a dilute alkali at room temperature to yield unsaturated carbonyl compounds.

Reaction with aniline

On being heated with aniline, benzaldehyde yields benzylidene aniline (Schiff base).

Electrophilic substitution reactions

The benzene ring of aromatic aldehydes and ketones undergoes nitration, sulphonation and halogenation to yield the corresponding meta-substituted product.

Example 3: Complete each reaction below by supplying the missing starting material, reagent(s) or product. 

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}\)

2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CN} \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\)

3. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3 \rightarrow \mathrm{CH}_3 \mathrm{CHO}\)

Solution: 

1.  \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH} \stackrel{\mathrm{PCC}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}\)

A primary alcohol is oxidised to an aldehyde by PCC. (i) CH,CH,CN DIBAL-HCH,CH,CHO

2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CN} \stackrel{\text { DIBAL-H }}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\)

A nitrile is converted to an aldehyde by DIBAL-H.

This is an Etard reaction. The partial oxidation of p-fluorotoluene by chromyl chloride (CrO₂Cl₂) gives a chromium complex, which on hydrolysis yields p-fluorobenzaldehyde.

5.  \(\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{OH} \stackrel{\mathrm{PCC}}{\longrightarrow} \mathrm{CH}_2=\mathrm{CH}-\mathrm{CHO}\)

A primary alcohol is oxidised to aldehyde by PCC. The double bond is not affected by oxidation with PCC.

NaBH₄ is a weak reducing agent. It only reduces the ketonic group to a secondary alcoholic group. Double bonds are not affected by NaBH₄. H₂ Pt (cat.)

In catalytic hydrogenation, only ethylenic double bonds are reduced. The ketonic group remains unaffected.

LiAlH is a versatile reducing agent. It reduces the aldehydic group to a primary alcoholic group.

11. \(2 \mathrm{CH}_3 \mathrm{COCl} \stackrel{\left(\mathrm{CH}_3\right)_2 \mathrm{Cd}}{\longrightarrow} 2 \mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{CdCl}_2\)

Acid chlorides react with dimethyl cadmium to give ketones.

This is a Friedel-Crafts acylation reaction.

Mechanism:

Uses Of Aldehydes And Ketones

1. Formalin (40% aqueous solution of formaldehyde) is used to preserve biological specimens.
2. Formaldehyde is used to prepare the most important plastic bakelite (a phenol-formaldehyde plastic).
3. Acetaldehyde is used for manufacturing organic compounds such as acetic acid, ethyl acetate, acetic anhydride and 1-butanol.
4. Benzaldehyde is used in perfumery and the dye industry.
5. Butyraldehyde, vanillin and camphor are used as flavouring agents in the perfume industry.

Carboxylic Acids

Organic compounds which contain a carboxyl group  are termed carboxylic acids. The group is so named because it can be considered as a combination of the carbonyl and hydroxyl groups.

Carboxylic acids may be aliphatic or aromatic. Aliphatic carboxylic acids are compounds in which the carbon of the carboxyl group is attached to an alkyl group and aromatic carboxylic acids are those in which the carbon of the carboxyl group is attached to an aryl group.

The unbranched long-chain monocarboxylic acids (C₁₃ -C₁₈) (monocarboxylic acids are compounds in which only one -COOH group is present) are commonly called fatty acids because many of them are obtained by the hydrolysis of animal fats or vegetable oils. Oils and fats are in fact esters of glycerol and fatty acids. Important fatty acids are stearic acid (C₁₇H₃₃COOH), palmitic acid (C₁₅ H₃₁ COOH) and oleic acid (C₁₇H₃₃COOH).

Formic acid is produced by ants and nettles. The sting of ants and nettles irritates the skin. Acetic acid is responsible for the sour taste of vinegar.

Nomenclature

Two systems of nomenclature are currently in use for carboxylic acids.

Trivial name

Simple carboxylic acids are known by their trivial names or common names derived from their natural sources, e.g., formic acid (Latin: formica, ant), acetic acid (Latin: acetum, vinegar), butyric acid (Latin: butyrum, butter). In substituted acids, the positions of the substituents are indicated by the Greek letters a, ẞ, 7, 8 and so on. The carbon atom next to the carboxyl group is a. For example,

IUPAC name

The IUPAC name of a monocarboxylic acid is derived from the name of the corresponding alkane by dropping the last ‘e’ of the alkane and adding the suffix ‘oic acid’. Positions of substituents on the chain are worked out by considering the carboxylic carbon to be C(1).

Carboxylic acids containing more than one carboxylic acid group are named by retaining the ending’-e’ of the alkane and the number of carboxylic acids are indicated by adding the prefix di, tri, etc., to the term ‘oic acid’. The positions of carboxylic acid groups are located by numbers.

The common and IUPAC names of some carboxylic acids

The following are some formulae and IUPAC names of carboxylic acids

[Note: If an unbranched chain is directly linked to more than two carboxyl groups, these carboxylic acids are named from the parent hydrocarbon by the substitutive used of a suffix such as ‘tricarboxylic acid’, etc., in place of the suffix ‘oic acid’.]

Alicyclic carboxylic acids are also named by adding the suffix ‘carboxylic acid’ to the name of a parent hydrocarbon. For example,

Methods Of Preparation

By the oxidation of primary alcohols or aldehydes

The oxidation of primary alcohols yields a carboxylic acid. The oxidising agent most often used is potassium permanganate in an acidic or alkaline medium, or potassium dichromate or chromium trioxide in an acid medium.

⇒ \(\mathrm{RCH}_2 \mathrm{OH} \underset{\text { 2. } \mathrm{H}_3 \mathrm{O}^{+}}{\stackrel{\text { alkaline } \mathrm{KMnO}_4}{\longrightarrow}} \mathrm{RCOOH}\)

⇒ \(\mathrm{RCH}_2 \mathrm{OH} \stackrel{\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 / \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow} \mathrm{RCOOH}\)

The initial product in the oxidation of a primary alcohol is the corresponding aldehyde. However, the aldehyde undergoes oxidation more rapidly than the primary alcohol. So it is normally not shown in the equation. Both aliphatic and aromatic aldehydes are readily oxidised to carboxylic acids even by mild oxidising agents like Tollens reagent. But usually, acid dichromate or permanganate solutions are used for oxidation.

By the oxidation of alkylbenzenes (Side-chain oxidation)

Aromatic compounds with alkyl side chains are oxidised with alkaline KMnO₄, acidified K₂Cr₂O₇ or chromic acid to carboxylic acids. For example, toluene, propylbenzene and isopropylbenzene are oxidised to benzoic acid on oxidation with alkaline KMnO₄. As t-butylbenzene does not possess hydrogen on the benzyl carbon, it is not oxidised to benzoic acid.

By the hydrolysis of nitriles and amides

Aliphatic and aromatic nitriles give carboxylic acids on hydrolysis upon boiling with acids or alkalis.

Nitriles and Amides Mechanism:

In order to obtain amides rather than carboxylic acids, mild reaction conditions are used. (Then the reaction stops at the amide stage.)

Grignard reagents: Grignard reagents (RMgX) react with carbon dioxide (dry ice is a convenient source of CO₂) to yield salts of carboxylic acids. Treatment of the salt with a mineral acid liberates the carboxylic acid.]

Grignard reagents Mechanism :

By the hydrolysis of esters, amides, acid halides and acid anhydrides

On hydrolysis, esters, amides, acyl halides and acid anhydrides give carboxylic acids.

(Y = OR,X,NH₂,OCOR)

Hydrolysis of esters Esters undergo hydrolysis by refluxing with dilute HCl or dilute alkali to yield carboxylic acids. Acidic hydrolysis gives carboxylic acid directly but alkaline hydrolysis with dilute NaOH gives the sodium salt of a carboxylic acid, which on acidification yields carboxylic acid.

With acid:

⇒ \(\mathrm{CH}_3-\mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O} \stackrel{\mathrm{H}^{+}}{\rightleftharpoons} \mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)

With alkali:

⇒ \(\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O} \stackrel{\mathrm{OH}^{-}}{\rightleftharpoons} \mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_2 \mathrm{O}\)

Mechanism of alkaline hydrolysis of esters:

Hydrolysis of amides: On Hydrolosis with acid or alkaline amides yield carboxylic acids

 Alkaline hydrolysis of amides Mechanism:

Hydrolysis of acid halides and acid anhydrides:

Acid halides and Acid anhydrides Mechanism:  

Physical Properties Of Monocarboxylic Acids

Lower carboxylic acids (with up to nine carbon atoms) are liquids with disagreeable odours. Higher members are waxlike solids. They are less volatile and almost odourless. Formic acid and acetic acid are present in traces in the secretions of our skin.

The ability of a dog to differentiate one person from another depends upon its highly developed sense of smell. Since the metabolic processes of different persons are different, the composition of lower carboxylic acids secreted by the skin is different in case of different people and a dog is able to distinguish one person from another.

The boiling points of carboxylic acids are higher than those of other classes of compounds of comparable molecular weight. For example, acetic acid (molecular weight 60) boils at 393 K, but propanol (molecular weight 60) boils at 370 K, and ethyl chloride (molecular weight 64) boils at only 286 K.

The increase in boiling point is attributed to the hydrogen bonding between the carboxylic acid molecules. More heat is required to break hydrogen bonds in the dimeric structure of carboxylic acids.

Example 4: Arrange te following compounds in increasing order of their boiling points give reasons

C₂H₅–O–C₂H₅, CH₃CH₂COOH, CH₃CH₂CH₂CH₂OH, CH₃CH₂CH₂CHO, CH₃CH₂CH₂ CH₂CH₃.

Solution:  

The boiling points of carboxylic acids are higher than those of other classes of compounds of comparable molecular mass. For example, propanoic acid (molecular mass = 74) boils at 414 K, 1-butanol (molecular mass = 74) boils at 391 K, butanal (molecular mass = 72) boils at 347.7 K, ethoxyethane (molecular mass = 74) boils at 318 K and n-pentane (molecular mass = 72) boils at 319 K. The high boiling point of propanoic (propionic) acid is attributed to the dimeric structure due to hydrogen bonding.

Propionic acid forms even stronger hydrogen bonds than does 1-butanol. The O-H bond of propionic acid is more strongly polarised as O-H due to the adjacent electron-attracting carbonyl group and the hydrogen bridge may be bonded to the more negatively charged carbonyl group rather than to the oxygen of another O-H bond of the carboxylic acid group.

The carbonyl group of butanal  is more polar in nature than the C-O-C group in ethoxyethane (C₂H₅-O-C₂H₅). The dipole-dipole attraction of the molecules of butanal is stronger than in ethoxyethane. Therefore, the boiling point of butanal is higher than that of ethoxyethane.

On the other hand, butanal does not form hydrogen bonds, and its boiling point is lower than those of 1-butanol and propanoic acid. n-pentane is nonpolar in nature, the only attractive forces among the molecules of n-pentane being the weak van der Waals attractive forces.

Therefore, the increasing order of boiling points of the given compounds is as follows.
CH₃CH₂CH₂CH₂CH₃ <CH₃CH₂OCH₂CH₃<CH₃CH₂CH₂CHO <CH₃CH₂CH₂CH₂OH
<CH₃CH₂COOH

Lower members of the carboxylic acids are completely soluble in water. This is because of the ability of the carboxyl group to form hydrogen bonds with water molecules. 

Higher fatty acids are insoluble in water due to the increased hydrophobic nature of long-chain alkyl groups.

Structure Of Carboxylic Acids

A carboxylic acid may be thought of as a resonance hybrid of the following two structures.

This resonance interaction has two important consequences.

1. The O-H bond is weakened by the electron-withdrawing effect of the carbonyl group so that typical carboxylic acids (pK_a =4±1) are much more acidic than alcohol (pK_a 17). The conjugate base of a carboxylic acid is also stabilised by resonance. The negative charge is shared equally by both oxygens, making them completely equivalent.

2. The electrophilicity of the carbonyl group is reduced by the electron-donating effect of the O-H group. Therefore, carboxyl carbon is less reactive towards nucleophiles than the carbonyl group of aldehydes and ketones.

Due to the diminished electron-withdrawing capacity of the carbonyl group in carboxylic acids (and their derivatives), α -hydrogens in such compounds are less acidic than those in ketones and aldehydes.

Strength Of Carboxylic Acids

The strength of a Bronsted acid is usually expressed in terms of its pK_a value. The lower the pKa value, the stronger is the acid. The pK_a values for a large number of carboxylic acids are known.

Table 12.5 pKa values for selected carboxylic acids (R-COOH)

From the table, it is evident that substituted acids (CICH₂-COOH, CI₂ CH-COOH and CI₂ C-COOH) as well as formic acid are stronger than acetic acid.

It is reasonable to expect that electron-donating substituents [Example, CH₃-C₂H₃-,(CH₃)₂, CH-,(CH₃)₃C-1 should be acid weakening while electron-withdrawing substituents (-Cl₂ -NO₂, CN₂ etc.) should be acid strengthening. The closer the substituent group is to the carboxyl group, the greater the effect it will have.

Further, if the number of electronegative halogens on the a-carbon is increased from one to three, the pK value reaches that of a strong mineral acid. The three halogens of trichloroacetic acid exert strong as shown. Due to this, the hydrogen of the -COOH group can readily leave as a proton.

When groups such as C₆H₅, and vinyl (CH₂=CH-) are attached to the carboxyl group then the acid strength should decrease due to resonance, as shown below.

But, in fact, the acid strength increases because an sp² carbon has greater effective electronegativity than an sp carbon.

In the presence of electron-attracting groups (NO₂, Cl, SO₃ H), the acid strength of aromatic carboxylic acids increases and in the presence of electron-donating groups (CH₃, OH, OCH₃, etc.), the acid strength decreases.

Example: Among the following pairs of acids, which is stronger? Give reasons

1. CH₃COOH or CICH₂COOH

2. FCH₂COOH or CICH₂COOH

Solution:

Generally, the acidity of a compound is increased by electron-attracting substituents whi electron-donating substituents decrease acid strength. Chloroacetic acid is more acidic tha acetic acid.

The higher acidity of chloroacetic acid is attributed to inductive effect. The chlori atom on chloroacetic acid assists in loosening the O-H bond and making the proton eas removable by pulling the electron towards itself due to its high electronegativity.

The electron-donating methyl group in acetic acid decreases the acid strength.

The larger the electron-withdrawing inductive effect, the greater is the acidity. As a result, fluoroacetic acid is stronger than chloroacetic acid-fluorine is considerably more electronegative than chlorine.

Inductive effect decreases rapidly with distance. The carboxylic acid in which the electron-attracting nitro group is closer to the carboxyl group is stronger because it can pull electrons more effectively. Therefore, 3-nitrobutanoic acid is stronger than 4-nitrobutanoic acid.

The electronegative CCl₃– group attracts electrons from the benzene ring, resulting in a positive charge on the carbon atom adjacent to the carboxylic group.

This in turn induces a positive charge on the carboxyl carbon. This results in the release of a proton to yield the carboxylate anion. The carboxylate anion is readily stabilised through resonance and increases the acid strength.

The effect in p-methylbenzoic acid is opposite to that in 4-trichloromethylbenzoic acid. The methyl group is able to donate electrons to the benzene ring through hyperconjugation and built up a negative charge on the carboxyl carbon. This additional charge tends to hold the proton and thus decreases the acid character.

Chemical Properties Carboxylic Acids

The chemical reactions of the carboxylic group involve the following.

1. Cleavage of the O-H bond,
2. Cleavage of the C-OH bond,
3. The COOH group, and
4. Substitution reactions in the hydrocarbon part.

Reactions involving cleavage of the O-H bond

Reactions with metals and alkali Like alcohols, carboxylic acids also evolve hydrogen with sodium metal and like phenols, carboxylic acids react even with weak bases, such as sodium carbonate and sodium bicarbonate, to evolve carbon dioxide.

2CH₃COOH + 2Na→2CH₃COONa+ H₂

CH₃COOH + NaOH → CH₃COONa +H₂O

CH₃COOH + NaHCO₃-CH₃COONa +H₂O+CO₂↑

Note that the evolution of CO₂ is from NaHCO₃ and not from the carbxylic acid. This reaction is used to detect the -COOH group in an unknown organic compound. Carboxylic acids evolve CO₂ with effervescence.

Carboxylic acids are only partially ionised in an aqueous solution and thus display weak acidic properties. A convenient way to measure the acidity of an acid is in terms of its ionisation constant (K_a) and PK_avalues. For acetic acid, the expression for K_a is obtained as follows.

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_3 \stackrel{+}{\mathrm{O}}\)

For the above reaction

⇒ \(K_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}_3 \stackrel{+}{\mathrm{O}}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\)

The concentration term for water is neglected since it is not affected to any appreciable extent by the ionisation of the acid. A higher value for K_a implies a strong acid.

The strength of an acid is also indicated by its pK, value.

pK_a = -log K_a

Thus, for acetic acid, whose K_a is 1.8 × 10-5, the pK_a can be calculated.

PK-log (1.8 × 10^-5)=

-0.3+5=4.7.

A smaller value of pK_a implies a stronger acid (excellent proton donor). For hydrochloric acid, the pK, value is -7.0. The corresponding values for trifluoroacetic acid, benzoic acid and acetic acid are 0.23, 4.19 and 4.7 respectively.

Stronger acids have pK, values < 1, moderately strong acids have pK, values between 1 and 5, weak acids have pK, values between 5 and 15, and extremely weak acids have pK, values > 15.

Reactions involving cleavage of the C-OH bond

Esterification Carboxylic acids react readily with alcohols or phenol in the presence of catalytic amounts of mineral acids (concentrated H₂SO₄) to yield esters. The process is called esterification. The conversion of a carboxylic acid into an ester involves nucleophilic substitution at the carbonyl (acyl) carbon.

First of all, the carbonyl oxygen of the -COOH group is protonated, which fascilitates the nucleophilic addition of the alcohol (the alcohol serving as the nucleophile) to the carbonyl group, yielding a tetrahedral intermediate.

Proton transfer of the intermediate converts the acid’s OH group to the protonated form which, being a good leaving group, departs as a water molecule. The protonated ester so formed finally loses a proton to give an ester.

C-OH bond Mechanism:

The reaction has several notable characteristics.

• It is reversible. Completion of the reaction requires either water to be removed by azeotropic distillation o an excess of one of the reagents (usually alcohol) to be used.
• The alcohol’s oxygen is retained in the ester, indicating that the R-O bond does not break during the reaction.
• The reaction requires catalysis by strong acids.

Reactions with PCI₅, PCI₃ and SOCI₂ The reaction of a carboxylic acid with PCl₅, PCI₃ or

SOCI₂ yields an acid chloride.

⇒ RCOOH + PCI₅→ RCOC1 + POCl₃ + HCl

⇒ 3RCOOH+ PCl₃, 3RCOCI+ H₃PO₃

⇒ RCOOH + SOCI₂→ RCOCI+ HCI ↑+ SO₂ ↑

With thionyl chloride, the by-products are both gases. With the phosphorus halide, the by-products are either nonvolatile phosphorus acid or the volatile liquid phosphorus oxychloride (boiling point 378 K). The latter may be difficult to separate from the acid chloride if the boiling points of the two are similar. Therefore, thionyl chloride is preferred for the preparation of acid chlorides.

Reaction with ammonia:

Carboxylic acids react with ammonia to yield ammonium salts, which give amides on being heated. For example,

Formation of anhydrides:

On being heated with concentrated H₂SO₄ or P₂O₅, a carboxylic acid gives an acid anhydride.

Carboxylic acids may be converted to their anhydrides by making their sodium salts react with the corresponding acid chloride.

⇒ RCOONa + RCOC1→ R–COO–COR+NaCl

Mixed anhydrides can also be prepared by this process.

⇒ RCOONa + R’COC1→ RCO—0—COR’ + NaCl

Reactions involving the carboxyl group

Reduction:

Carboxylic acids are conveniently reduced to primary alcohols upon reaction with lithium aluminium hydride in ether or with diborane, which reduces esters, nitro and halo groups with difficulty. Sodium borohydride is a weak reducing agent and does not reduce the carboxyl group.

Decarboxylation: on dry distillation with soda lime, the sodium salt of a carboxylic acid gives an alkane.

The decarboxylation (removal of CO₂) of a carboxylic acid takes place through the carboxylate anion from which the group ^–CH₃ departs along with its bonding electron pair. The carbanion (carbon carrying a negative charge) reacts with a proton to give the alkane.

Decarboxylation Mechanism:

Kolbe’s electrolytic method: In this method, an aqueous solution of a sodium salt of carboxylic acid is subjected to electrolysis to yield an alkane.

⇒ 2CH₃COONa+ 2H₂O→C₂H6 +2CO₂ + 2NaOH + H₂ at anode at cathode

In another reaction, dry distillation with calcium formate, the calcium salt of a carboxylic acid gives an aldehyde. For example, a mixture of calcium acetate and calcium formate yields acetaldehyde on dry distillation.

On dry distillation, the calcium salt of a carboxylic acid gives a ketone. For example, calcium acetate, on dry distillation, yields acetone.

Hunsdiecker reaction:

Carboxylic acids form silver salts when their ammoniacal solutions are treated with silver nitrate. On being refluxed with Br₂, the salt forms an alkyl bromide. This is known as the Hunsdiecker reaction.

⇒  \(\underset{\text { Silver carboxylate }}{\mathrm{RCOOAg}}+\mathrm{Br}_2 \underset{\Delta}{\stackrel{\mathrm{CCl}_4}{\longrightarrow}} \underset{\text { Alkyl bromide }}{\mathrm{R}-\mathrm{Br}}+\mathrm{CO}_2+\mathrm{AgBr}\)

Substitution reactions in the hydrocarbon part

Halogenation:  When treated with chlorine or bromine in the presence of red phosphorus, aliphatic carboxylic acids containing a-hydrogen form a-halogen acids.

Similarly,

This reaction is called the Hell-Volhard-Zelinsky reaction or HVZ reaction.

Halogenation Mechanism:

Reducing properties of formic acid

Formic acid shows reducing properties in the following reactions.

Reaction with Tollens reagent:  On gently warming with Tollens reagent, formic acid gives a grey precipitate of metallic silver.

⇒ HCOOH + 2Ag(NH₃)₂(Tollens reagent)OH → Ag + CO₃ +H₃O+4NH₃

Reaction with Fehling’s reagent: Upon being heated with Fehling’s reagent, formic acid gives a brick-red precipitate of Cu₂O.)

Reaction with mercuric chloride: Formic acid reduces mercuric chloride to mercuric chloride (Hg₂Cl₂) as a white precipitate. Mercurous chloride is further reduced to mercury as a grey precipitate.

⇒ HCOOH+2HgCl₂Hg₂Cl₂ + CO₂ + 2HCl

⇒ HCOOH+Hg₂Cl₂→ 2Hg(grey ppt.) +CO₂+2HCl

Electrophilic substitution reactions of benzoic acid

Nitration:

Aromatic carboxylic acids undergo electrophilic substitution reactions. In benzoic acid, the -COOH group deactivates the benzene ring and is meta-directing. On nitration with concentrated HNO₃ and concentrated H₂SO₄, benzoic acid gives m-nitrobenzoic acid.

Benzoic acid does not undergo Friedel-Crafts reaction (alkylation or acylation) because this reaction is not possible with a deactivated aromatic nucleus and the catalyst anhydrous AICI₃ (Lewis acid) gets bonded with the COOH group as shown below.

Halogenation:

In the presence of ferric bromide, which acts as a catalyst, benzoic acid reacts with bromine to form m-bromobenzoic acid.

Carboxylic Acids Uses 

• Formic acid is used in the rubber, cloth, dye and leather industries.
• Acetic acid is used as a solvent and in the preparation of vinegar.
• .Higher fatty acids are used in the manufacture of soaps and detergents.

Aldehydes, Ketones and Carboxylic Acids Multiple-Choice Questions

Question 1. The functional group of an aldehyde is

Answer: 2.

Question 2. The general formula of aldehydes and ketones is

1. C_n H_2n-2O
2. C_n H_2nO
3. C_nH_2n+1O
4. C_nH_2nO₂

Answer: 2. C_n H_2nO

Question 3. The IUPAC name of acetone is

1. Methanal
2. Ethanal
3. Propanone
4. Ethanone

Answer:  3. Propanone

Question 4. Which of the following is used to distinguish between an aldehyde and a ketone?

1. Concentrated H₂SO₄
2. Hydrazine
3. Tollens reagent
4. Nitrous acid

Answer: 3. Tollens reagent

Question 5. On being heated with NaOH solution, formaldehyde gives

1. Formic acid
2. Acetonw
3. Methyl alcohol
4. Ethyl formate

Answer: 3. Methyl alcohol

Question 6. Which of the following is formed when a mixture of calcium acetate and calcium formate is dry distilled?

1. Methanol
2. Ethanol
3. Ethanal
4. Acetic acid

Answer:  4. Acetic acid

Question 7. Which of the following is involved in the Cannizzaro reaction?

1. CH₃CHO
2. HCHO
3. HCOOH
4. CH₃COCH₃

Answer: 2. HCHO

Question 8. On dry distillation, calcium acetate gives

1. Ethyl acetate
2. Calcium formate
3. Acetone
4. Acetaldehyde

Answer: 3. Acetone

Question 9. Which of the following reagents reacts with aldehydes as well as ketones?

1. Tollens reagent
2. Fehling’s solution
3. Schiff base
4. Grignard reagent

Answer: 4. Grignard reagent

Question 10. Which of the following can be used to distinguish between aldehydes and ketones?

1. Fehling’s solution
2. H₂SO₄
3. NaHSO₃
4. NH₃

Answer: 1. Fehling’s solution

Question 11. Which of the following responds positively to the iodoform test?

1. C₂H₂OH
2. CH₃OH
3. CH₃CHO
4. C₂H₄c

Answer: 1 and 3 Or C₂H₂OH and CH₃CHO

Question 12. Which of the following is oxidised to acetone?

1. CH₃ CHO
2. C₂H₅OH
3. CH₃OH
4. CH₃-CHOH-CH₃

Answer: 3. CH₃OH

Question 13. On reacting with chlorine, acetaldehyde gives

1. Acetyl chloride
2. Chloral
3. Dichloroacetic acid
4. None of these

Answer:  2. Chloral

Question 14. On being heated with ammoniacal silver nitrate, acetaldehyde gives

1. Acetone
2. Silver acetate
3. A silver mirror
4. Formaldehyde

Answer: 3. A silver mirror

Question 15. In the following reaction, what is the appropriate reagent?

1. NH-NH₂, KOH
2. Zn/Hg,
3. H₃/Ni
4. NaBH₄

Answer: 1. NH-NH₂, KOH

Question 16. What is the correct order of reactivity of C₆H₅ MgBr with the following?

1. 1 > 2> 3
2. 3 >2 >1
3. 2>3>1
4. 1 >2 >3

Answer: 3. 2>3>1

Question 17. Which of the following reagents is used to separate acetaldehyde from acetophenone?

1. NaHSO₃
2. C₆H₅NHNH₃
3. NH₂OH
4. NaOH/I₂

Answer: 4. NaOH/I₂

Question 18. On treatment with 1% HgSO₄ and 20% H₂SO₄, but-1-yne gives

1. CH₃CH₂COCH₃
2. CH₃CH₂CH₂CHO
3. CH₃CH₂CHO and HCHO
4. CH₃CH₂CHOO and HCOOH

Answer: 1.

Question 19. The boiling points of the compounds

1. CH₃CH₂CH₂CHO
2. CH₃CH₂COOH and HCOOH
3. CH₃CH₂CH₂OH
4. CH₃COOH I

Answer: 3. CH₃CH₂CH₂OH

Question 20. The formation of cyanohydrin from CH₃COCH₃ is an example of

1. Electrophilic substitution
2. Nucleophilic substitution
3. Nucleophilic addition
4. Electrophilic addition

Answer: 3. Nucleophilic addition

Question 21. Which of the following products is obtained when CH₃MgBr reacts with formaldehyde?

1. C₂H₂OH
2. CH₃COOH
3. HCHO
4. CH₃CHO

Answer: 1. C₂H₂OH

Question 22. In which of the following reactions does an aromatic aldehyde react with acetic anhydride in the presence of sodium acetate to give an unsaturated aromatic acid?

1. Friedel-Crafts reaction
2. Wurtz reaction
3. Perkin reaction
4. None of these

Answer: 2. Wurtz reaction

Question 23. Which of the following does not respond positively to the iodoform test?

1. 2-pentanone
2. 3-pentanone
3. Ethanal
4. Ethonol

Answer: 2. 3-pentanone

Question 24. Which of the following is Tollens reagent?

1. [Ag(NH₃)₂]+ ion
2. Cu(OH)₂
3. CuO
4. Ag₂O

Answer: 1. [Ag(NH₃)₂]+ ion

Question 25. Which of the following is used to distinguish between aliphatic and aromatic aldehydes?

1. Tollens reagent
2. Benedict’s reagent
3. Schiff base
4. Iodoform reaction

Answer: 4. Iodoform reaction

Question 26. Which of the following reactions does benzaldehyde not undergo?

1. Aldol condensation
2. Benzoin condensation
3. Cannizzaro reaction
4. Perkin reaction

Answer: 1. Aldol condensation

Question 27. Which of the following will yield acetaldehyde?

1. The dry distillation of calcium acetate
2. The reduction of acetic acid by LIAIH
3. The oxidation of isopropyl alcohol by K₂Cr₂O₇/H₂SO₄
4. The ozonolysis of 2-butene

Answer: 4. The ozonolysis of 2-butene

Question 28. Which of the following reagents is used to distinguish between formic acid and acetic acid?

1. Phosphorus pentachloride
2. Sodium
3. Grignard reagent
4. Tollens reagent

Answer: 3. Grignard reagent

Question 29. Carbon dioxide is evolved when propanoic acid is treated with a NaHCO₃ solution. The carbon atom of CO₂ results from the

1. Methyl group
2. Carboxylic acid group
3. Methylene group
4. Bicarbonate

Answer: 4. Bicarbonate

Question 30. Which of the following does not reduce Fehling’s solution?

1. Formic acid
2. Acetic acid
3. Formaldehyde
4. Acetaldehyde

Answer: 2. Acetic acid

Question 31. An organic compound (X), on being heated with K₂Cr₂O, and H₂SO₄ gives another compound (Y). The latter on heating with I, and Na₂CO₃ solution forms an iodoform. Among the following, which one could be the compound (X)?

1. CH₃COCH₃
2. CH₃OH
3. CH₃CHO
4. CH₃CHOHCH₃

Answer:  4. CH₃CHO

Question 32. Which of the following is used to distinguish between formic acid and acetic acid?

1. Sodium
2. Mercuric chloride
3. Sodium ethoxide
4. 2,4-Dinitrophenylhydrazine

Answer: 2. Mercuric chloride

Question 33. An ester is hydrolysed by KOH and acidified to get a white precipitate. The ester is

1. Methyl acetate
2. Ethyl acetate
3. Ethyl formate
4. Ethyl benzoate

Answer: 4. Ethyl benzoate

Question 34. What are the products formed in the following reaction? \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOCH}_3 \stackrel{\mathrm{LiAlH}_4}{\longrightarrow}\)

1. C₆H₅COOCH and CH₃OH
2. C₆H₅C₃OOH and CH₃OH
3. C₆H₅CH₂OH and CH₃CHO and CH₃COOH
4. All of these

Answer: 4. C₆H₅CH₂OH and CH₃CHO and CH₃COOH

Biomolecules

Biomolecules:

All living beings are made up of complex molecules called biomolecules. Biomolecules not only help make up the structure of the body but also provide the energy required to carry out life processes.

Examples of biomolecules:

Examples of biomolecules are carbohydrates, proteins, nucleic acids and lipids. Apart from biomolecules, some simple molecules like vitamins and mineral salts play an important part in the functioning of living beings.

The study of these molecules and their interaction is called biochemistry.

Carbohydrates

Carbohydrates are compounds with the general formula C_n (H₂O)_m · Originally, they were thought to be hydrates of carbon, hence the name. However, they are not considered to be so any more for the following reasons. 1. Carbon does not form hydrates.

Some carbohydrates, like rhamnose (C₆H₁₂O₅) and deoxyribose (C₅H₁₀O₄

Several compounds, such as formaldehyde (CH₂O) and acetic acid (C₂H₄O₂), have the same general formula as carbohydrates but differ from them in their properties.

In terms of their functional groups, carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones, or compounds which yield polyhydroxy aldehydes and polyhydroxy ketones on hydrolysis. The simplest carbohydrate is glyceraldehyde

⇒\(\left[\mathrm{C}_3\left(\mathrm{H}_2 \mathrm{O}\right)_3\right]\)

Carbohydrates are mainly produced by plants. In nature, C₆H₁₂O₆ (glucose) is produced by photosynthesis. Cellulose, which makes up the cell wall of plant cells, is a carbohydrate. Carbohydrates are an important constituent of the food we eat (they are found in rice, potatoes and bread, among other things). They provide us with the energy we require to carry out our life processes.

Classification And Nomenclature

Carbohydrates may be classified as monosaccharides (containing 1 sugar molecule), disaccharides (containing 2 sugar molecules) and polysaccharides (containing many sugar molecules). The names of carbohydrates have an ‘ose’ at the end, for example, glucose, fructose and sucrose.

Monosaccharides are the simplest sugars, and cannot be broken down or hydrolysed into simpler ones. They contain three to nine carbon atoms and are further categorised as trioses, tetroses, pentoses, hexoses, etc. Functionally, those containing an aldehydic group are known as aldoses and those with a ketonic group, are ketoses.

Often, the nature of the carbonyl functional group and the number of carbon atoms present in a monosaccharide are also indicated, example , aldopentoses, aldohexoses, ketopentoses and ketohexoses.

Carbohydrates which on hydrolysis give two to nine monosaccharide units are called oligosaccharides. They are further categorised as disaccharides, trisaccharides, etc., depending upon the number of monosaccharides obtained on hydrolysis.

Disaccharides are hydrolysed to give two molecules of monosaccharides, which may be the same or different. For instance, maltose gives two molecules of glucose, while sucrose yields one molecule each of glucose and fructose.

⇒ \(\underset{\text { Maltose }}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}+\mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text { Glucose }}{2 \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}\)

⇒ \(\underset{\text { Sucrose }}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}+\mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text { Glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\text { Fructose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}\)

Trisaccharides (For example, raffinose) are hydrolysed to yield three molecules of monosaccharides.

⇒ \(\underset{\text { Raffinose }}{\mathrm{C}_{18} \mathrm{H}_{32} \mathrm{O}_{16}}+\underset{\text { Galactose }}{2 \mathrm{H}_2 \mathrm{O}} \rightarrow \underset{\text { Glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\text { Fructose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}{\text { Gluction }}\)

Polysaccharides are high-molecular-weight carbohydrates that contain several monosaccharide units. In contrast to monosaccharides and disaccharides, which are water-soluble and sweet in taste, polysaccharides are tasteless, water-insoluble substances. The hydrolysis of a polysaccharide yields many molecules of monosaccharides. For example, starch gives many molecules of glucose.

⇒ \(\underset{\text { Starch }}{\left(\mathrm{C}_6 \mathrm{H}_{10} \mathrm{O}_5\right)_n+n \mathrm{H}_2 \mathrm{O} \rightarrow n \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}\)

Carbohydrates which are sweet in taste are called sugars and those that are not are called nonsugars. Monosaccharides and disaccharides are sugars and polysaccharides are nonsugars.

Carbohydrates may also be classified as reducing and nonreducing sugars. All carbohydrates which reduce Fehling’s solution and Tollens reagent are called reducing sugars. All monosaccharides (aldose or ketose) are reducing sugars.

For example, glucose reduces Fehling’s solution to Cu₂O (red precipitate) and Tollens reagent to metallic silver. Glucose also reduces Benedict’s solution (an aqueous solution of CuSO₄ and sodium citrate) to Cu₂O (red precipitate).

This reaction is the traditional one used for the diabetes test in which urine is tested for glucose. Fructose (an a-hydroxy ketone) also responds positively to these tests because a-hydroxyketones, in general, are oxidised very easily to diketones by these reagents.

Maltose and lactose are reducing sugars as they reduce Fehling’s solution and Tollen’s reagent. These can produce a free aldehydic group in solution.

Sugars in which the carbonyl group is tied up in an acetal linkage are nonreducing sugars. Sucrose is an example of a non-reducing sugar.

Monosaccharides

Monosaccharides are crystalline substances and exhibit many reactions characteristic of the carbonyl and hydroxyl groups. They usually contain asymmetric carbon atoms, and exist as several optical isomers.

Glyceraldehyde is an aldotriose with one asymmetric carbon atom and exists in the (+) and (−) forms. It is chosen as a standard in describing the configurations of higher monosaccharides.

The D and L configurations (1 and 2) of glyceraldehyde are given below.

The sugars related to D-glyceraldehyde (1) form the D-series and those that are derived from L-glyceraldehyde (2) form the L-series.

Glucose

Glucose is the most widely occurring monosaccharide in nature and is found in the free state in sweet fruits and honey. Glucose was originally isolated from grapes. Glucose, because of its origin, is sometimes called grape sugar.

It is also obtained by the hydrolysis of starch and cellulose. The alternative name dextrose originated from the fact that the common form of glucose rotates a plane of polarised light to the right, i.e., it is dextrorotatory.

Glucose is involved in the metabolic activities of living organisms. In the blood stream, a definite concentration of glucose must be maintained because both an excess and a deficiency are harmful. Excess glucose is excreted in urine. The concentration of glucose in the body is maintained by the action of insulin.

Preparation Of Glucose

On hydrolysis with dilute HCl or dilute H₂SO₄ in the presence of alcohol, sucrose (cane sugar) gives glucose and fructose in equal amounts.

⇒ \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_2 \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \underset{\text { Glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\text { Fructose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}\)

Glucose is obtained on a commercial scale from the hydrolysis of starch in the presence of a mineral acid.

⇒ \(\underset{\text { Starch }}{\left(\mathrm{C}_6 \mathrm{H}_{10} \mathrm{O}_5\right)_n}+n \mathrm{H}_2 \mathrm{O} \underset{393 \mathrm{~K}, 2-3 \mathrm{~atm}}{\stackrel{\mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}} n \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\)

Starch is hydrolysed with dilute H₂SO₄ The mixture is heated under pressure. When the hydrolysis is complete, the excess acid is neutralised with Ca(OH)2 and filtered. The filtrate is decolourised with animal charcoal, filtered and finally concentrated and cooled to get crystals of glucose.

Structure Of Glucose

1. Analytical data and molecular weight suggest that the molecular formula of glucose should be C₆H₁₂O₆.

2. On prolonged heating with HI, glucose gives n-hexane, indicating that all six carbon atoms are linked in a straight chain.

3. Glucose reacts with hydroxylamine to form an oxime. This suggests the presence of a carbonyl group.

4. On mild oxidation with bromine water, glucose is converted to carboxylic acid (gluconic acid), which has six carbon atoms. This indicates that the carbonyl group present in glucose is an aldehydic group.

5. With acetic anhydride, glucose yields a pentaacetate derivative.

This shows the presence of five hydroxyl groups in the glucose molecule. Since glucose is stable, all five hydroxyl groups are attached to different carbon atoms.

6. On strong oxidation with HNO₃, glucose yields saccharic acid (a dicarboxylic acid).

This reaction indicates the presence of a primary alcoholic group (-CH₂OH).

On the basis of the above reactions, the following open-chain structure may be assigned to glucose.

In Structure 1, there are four asymmetric carbon atoms indicated by an asterisk.

The spatial arrangement of five hydroxyl groups in glucose (1) was established by Fisher on the basis of his study of different reactions.

The prefix D- before glucose (2) indicates that the hydroxyl group attached to the bottom asymmetric carbon atom is on the right-hand side and it would be L-glucose if the hydroxyl group were on the left. Usually, D- and L- configurations are assigned to sugars by comparing their structures with the configuration of D- of D- and L- glyceraldehyde.

It must be remembered that, here, the symbols D- and L- refer to the configuration of the compound and have no relationship with the sign of rotation.

Cyclic structure of glucose:

The structure of D-glucose (2) accounts for most of its reactions satisfactorily but fa to explain the following facts.

1. The structure shows the presence of an aldehydic group but the compound does not respond to the Schiff test nor does it form a bisulphite addition compound with NaHSO₃.

2. D-glucose forms two isomeric pentacetates which fail to undergo condensation with NH₂OH.

3. D-glucose itself exists in two isomeric forms known as α- and β-D-glucose, which have different specific rotations. This phenomenon of change in specific rotation is termed mutarotation.

Specific rotation

Specific rotation [a] is defined as the rotation in degrees brought about by a solution containing 1 g of a substance in 1 mL of solution, examined in a polarimeter tube 1 decimetre long.

⇒ \([\alpha]=\frac{\text { observed rotation }}{\text { tube length }(\mathrm{dm}) \times \text { concentration }(\mathrm{g} / \mathrm{mL})}\)

α -D-glucose → α -D-glucos e ←β -D-glucose

[α]_D = +112° → [α]_D = +52° → [α]_D = + 19°

α -D-glucose (m.p. 419 K) is obtained by crystallisation from a concentrated solution of glucose at 303 K while β-D-glucose (m.p. 423 K) is obtained by crystallisation from a hot and saturated aqueous solution of glucose at 371 K.

It appears that in D-glucose the aldehydic group is not free. Rather it forms a cyclic hemiacetal with the —OH group originally situated at C₅. As the aldehydic group enters into hemiacetal formation, the aldehydic carbon becomes asymmetric.

Hence, two oxide ring structures differing in their C₁ configuration are possible. Such diastereomers, which differ in their C₁ configuration only, are known as anomers. An equilibrium mixture is obtained containing both the anomers as well as the open-chain form.

This interconversion is a manifestation of the phenomenon of mutarotation.

Haworth proposed a pyranose structure for these two anomeric forms (α- and β-). These structures resemble the six-membered heterocyclic pyran ring.

Fructose

Fructose is a ketohexose. In nature, it is found, along with glucose, principally in fruits and honey. It is also formed by the hydrolysis of table sugar. Fructose forms a furanose ring and is the sweetest of all the sugars. It is also called laevulose, indicating its laevorotatory property.

Structure Of Fructose

1. The molecular formula of fructose is C₆H₁₂O₆
2. It forms an oxime with hydroxylamine. This shows that it contains a carbonyl group.
3. The oxidation of fructose gives a mixture of glycolic acid, tartaric acid and trihydroxyglutaric acid (all of which have fewer carbon atoms than fructose). Therefore, it must be a ketone.
4.  On acetylation, fructose gives a pentaacetate, proving the presence of five hydroxyl groups.
5.  The reduction of fructose (Pd/H₂) yields a hexahydric alcohol, which on further reduction with hot HI gives n-hexane. Therefore, six carbon atoms of fructose form a straight chain.
6. On treatment with HCN, fructose gives cyanohydrin, which on hydrolysis yields the corresponding acid. The resulting acid on reduction with hot HI gives 2 – methylhexanoic acid (3). This shows that the carbonyl group in fructose is adjacent to the terminal carbon.

Considering all the facts, the open-chain structure (4) may be assigned to fructose.

The structure of D-(-)-fructose (4) (the minus sign indicates its laevorotatory nature) accounts for most of the reactions satisfactorily but is unable to explain the following facts.

1. It does not add on to NaHSO₃ as ketones do.
2. It shows mutarotation.
3. It forms two isomeric fructosides.

All these can be accounted for by a ring structure for fructose. The C=O group of fructose reacts with the -OH group at C₅ to form a five-membered ring and is named as furanose. The two furanose structures resemble the five-membered heterocyclic furan ring.

Haworth represented the above two cyclic structures of fructofuranose as follows.

Disaccharides

A disaccharide is a compound that can be hydrolysed to two different monosaccharides or two molecules of the same monosaccharide. Three important disaccharides are sucrose, lactose and maltose. All are isomers of each other and have the empirical formula C₁₂H₂₂O₁₁.

In a disaccharide, the two monosaccharides are joined together by a glycoside linkage. A glycoside bond is formed when two monosaccharides are joined together by an oxide linkage (-O) formed by the loss of a
water molecule.

Sucrose (nonreducing)

Sucrose is the technical name for table sugar (also called cane or beet sugar). Upon hydrolysis, sucrose yields an equimolar mixture of the monosaccharides D-(+)-glucose and D-(-)-fructose.

Sucrose is composed of an α-glucose plus β-fructose. When these two molecules are joined together a water molecule is released.

The OH group on C_n of a-glucose is bonded to the OH group attached to C₂ of β-fructose. This is called a 1- linkage. The oxygen atom bridging the two monosaccharides constitutes a glycoside bond.

Sucrose is dextrorotatory (optical rotation = +66) and the equilibrium mixture of glucose has an optical rotation of +52°, while fructose has a large negative rotation of -92°. At the end of the hydrolysis of sucrose, the equimolar mixture of glucose and fructose has a negative rotation (it is laevorotatory). Thus, the reaction proceeds with an inversion of rotation from a positive to a negative value. This is called the inversion of sucrose and the product mixture is called inverted sugar.

Maltose (reducing)

Maltose, called malt sugar, is found in the germinating seeds of barley or malt. Upon hydrolysis, it yields two molecules of glucose.

Maltose has a 1 → 4 linkage between two a-glucose molecules. The glycosidic 1-hydroxy group of the second molecule of glucose is free and can produce an aldehydic group at C₁ in the solution.

Therefore it shows a reducing property and is called a reducing sugar. Like sucrose, maltose is easily fermented into ethyl alcohol and CO₂

Lactose (reducing)

Lactose, also called milk sugar, is found in the milk of all mammals. Upon hydrolysis, it yields glucose and galactose.

Lactose + H₂O→glucose + galactose

In lactose, the OH group on C₁ of β-D-galactose is bonded to the OH group on C₄ of β-D-glucose. This is a 1 → 4 linkage.

The glycosidic hydroxy group of C₁ of the second molecule of glucose is free to produce an aldehydic group located at C₁. Therefore, lactose shows reducing properties and is said to be a reducing sugar.

Cow’s milk usually contains about 5% lactose by volume whereas human milk contains about 7%. Milk turns sour if kept for some time at 35°C because of the bacteria present in the air. These bacteria convert the lactose of milk into lactic acid, which is sour.

Polysaccharides

Polysaccharides are complex polymers. They have a high molecular weight and are made up of several monosaccharide (D-glucose) units linked together through oxygen atoms. They occur widely in plants and animals. Two important polysaccharides are starch and cellulose, both of which can be represented by the general formula (C₆H₁₀5)_n. Polysaccharides do not exhibit the characteristic reactions of the aldehyde group.

Starch

Plants store energy mainly in the form of starch. It forms the most important source of carbohydrates in the food we eat. It is found in wheat, rice, potatoes and some other vegetables.

Starch is insoluble in water, and forms a colloidal dispersion. Starch hydrolyses to form a number of α-D(+)- glucose molecules. It is not homogeneous. It consists of 15-20% amylose, which has a straight chain, and amylopectin (80-85%), which has a great deal of branching.

The amylose polymer consists of 200-1000 α-D(+)-glucose molecules joined by a (1→4) glycosidic linkages (Figure 14.8). Amylose is soluble in water, and gives a blue colour with iodine. Amylopectin, which is insoluble in water, consists of a number of amylose chains joined by a (1→6) glycosidic linkages.

Starch hydrolyses through the following stages.

Starch → dextrin → maltose → glucose

In the laboratory, the degree of hydrolysis can be observed by testing the solution with an iodine reagent. Starch reacts with this reagent to produce a deep blue-black colour while maltose and glucose produce no colour change.

As a fruit ripens, starch is hydrolysed to glucose and the fruit becomes sweet. Unlike cellulose, starch can be hydrolysed by enzymes. If you chew bread thoroughly before swallowing it, it will taste sweet. The sweetness is due to the sugar formed from the hydrolysis of starch.

Cellulose

Cellulose is present only in plants and is the most widely occurring organic substance found among them. The cell wall of a plant is mainly cellulose. Cellulose also forms a considerable part of cotton, wood and jute.

Cellulose is a polymer of ẞ-glucose. It is insoluble in water. A molecule of cellulose has a linear chain. On complete hydrolysis, cellulose yields D(+) glucose only. The D(+) glucose units in cellulose are 1,4-B-linked.

The human digestive tract breaks down starch to glucose but does not contain the enzymes required to hydrolyse B-glucose linkages and thus cellulose cannot be digested by human beings. Various derivatives of cellulose, such as cellulose nitrate and cellulose acetate, find commercial use. Cellulose nitrate is used to prepare smokeless gunpowder. Cellulose acetate can be spun into yarn or extruded into film (cellophane).

Glycogen

Starch is converted to a-glucose by animal metabolism. Glucose is repolymerised to form glycogen in the liver. Glycogen is also found in muscles and the brain. When exercise depletes blood sugar, the hydrolysis of liver glycogen maintains the normal glucose content of the blood. Glycogen consists of branched chains of glucose molecules. It is also called animal starch because its structure is similar to that of amylopectin. However, there is more branching in glycogen than in amylopectin.

Example 1:

1. Why is glucose soluble in water? Explain.
2.  How do you explain the absence of an aldehydic group in glucose pentaacetate?

Solution:

1. Glucose contains five hydroxyl groups, which form intermolecular hydrogen bonds with wat
2. In an aqueous solution, D-(+)-glucose may be regarded as the equilibrium mixture of the following three forms.

During the acetylation of glucose, the anomeric hydroxyl group at C1 is acylated and hence glucose pentaacetate can no longer attain the open-chain form. Thus, there is no aldehydic group.

Proteins

Approximately 16% of the total weight of a human being is protein. With the exception of water, which comprises 65% of our body weight, no other material constitutes more of the body than does protein. Among the main sources of protein are meat, fish, pulses, cheese and soya.

Many different proteins are responsible for the body’s innumerable functions. For example, the protein myoglobin in muscle tissue stores oxygen until the muscle cells need it. The red blood cells contain the protein haemoglobin, which takes oxygen from the lungs to the tissues.

There are proteins in the saliva, gastric juices, and intestinal juices, all of which help digest food. The pituitary gland secretes a protein, called human growth hormone, which regulates growth. Other hormones (also proteins), control processes in reproduction.

Certain cells in the pancreas secrete insulin, a protein that regulates the amount of sugar in blood. Human skin is made of protein. Thus proteins are involved in all life processes. Since they are fundamental to the living system, these compounds were given the name ‘protein’, from the Greek proteins, which means ‘primary’ or first.

The protein molecules present in animal and plant tissues have large numbers of C, H, O and N atoms and take up a considerable amount of space. Many proteins contain more than ten thousand atoms. Although they vary greatly in size, a protein with a length of 44 Å, a height of 44 Å and a width of 25 Å is considered relatively small.

Amino Acids (The Building Blocks Of Proteins)

How the atoms in a protein molecule are put together is of great concern to scientists. To gain an insight into the nature of proteins, a protein was heated with a solution of hydrochloric acid for a long period of time to break some bonds in the protein molecule.

The reaction mixture was found to contain a variety of smaller molecules. These molecules were isolated and examined, and were given the name amino acids. Amino acids contain an amino group and a carboxyl group. The formula and structure of glycine, a typical amino acid, is NH₂CH₂COOH.

Examination of the reaction mixture revealed other amino acids as well as glycine. There are 20 different amino acids known to exist in the various proteins.

Proteins are condensation polymers of the amino acid monomers. Amino acids which occur in nature have an amino group (-NH₂) attached to an a-carbon atom (the carbon of the first —CH₂– group attached to -COOH).

These amino acids differ from each other-they have distinctive side chains attached to the a-carbon atom replacing the H atom of glycine. An amino acid is often represented by the following general structure.

Where R may be aliphatic, aromatic or heterocyclic.

In all amino acids except glycine, the a-carbon atom is bonded to four different atoms or groups. Any molecule which contains a carbon atom bonded to four different groups is one of a pair of optical isomers.

One optical isomer is designated as the L isomer and the other as the D isomer. The amino acids of proteins are all L isomers.

Naturally occurring amino acids from plant and animal sources have the L configuration and are designated as L(+) or L(-), depending upon their behaviour towards a plane of polarised light.

Amino acids are often represented by a three-letter symbol, for example, Gly for glycine, ala for alanine, and so on.

Classification Of Amino Acids

Amino Acids Are Classified In Three Different Ways.

1. As you know, amino acids are compounds containing an amino as well as an acidic (mostly carboxylic) group. A simple classification is based on the position of the amino group with respect to the carboxylic group.

Thus a-amino acids have an amino group attached to the a-carbon with respect to the carboxylic group, β-amino acids contain the amino group attached to the β-carbon, and so on.

In this chapter, we shall discuss only amino acids as these are the building units of proteins.

2. Most amino acids have only one basic and one acidic group and exhibit amphoteric properties. They are called neutral amino acids. In some cases, however, an amino acid may contain more carboxylic groups than amino groups.

They are termed acidic amino acids, examples being aspartic and glutamic acids. Amino acids with a greater number of basic (amino) groups than acidic groups, for example, lysine, arginine, and histidine, are basic in nature.

3. Certain amino acids can be synthesised in the body while others have to be obtained through the food we eat. The former are called nonessential amino acids while the latter are called essential amino acids.  gives the structures, names and abbreviations of some common amino acids that have been isolated by protein hydrolysis. Essential amino acids are marked with an asterisk.

The most common natural amino acids: 

Electrical Properties-Zwitterion Structure

Amino acids are high-melting solids, which, because of their polar groups, are insoluble in organic solvents but soluble in water. They behave like salts rather than simple amines or carboxylic acids.

Since the carboxylic acid group is acidic and the amino group is basic, amino acids actually exist as dipolar ions (zwitterions), rather than in the un-ionised form.

From Amino Acids To Proteins (The Peptide Bond)

Proteins are formed by the linking together of amino acids. The amino group of one molecule reacts with the carboxyl group of another, with the elimination of a molecule of water. When two amino acids combine, they form a dipeptide.

If three amino acids combine, they form a tripeptide, when four combine, they form a tetrapeptide, and so on. When more than ten combine, they are said to form a polypeptide.

The -CONH group (the amide group), which joins amino acids together, is called a peptide link. A polypeptide molecule might have the structure given in below

A polypeptide has 1 free -NH₂ at one end of the chain and 1 free -COOH group at the other end. The amino end is referred to as the N-terminal and the carboxylic acid end as the C-terminal.

A polypeptide with more than a hundred amino acid residues, with a molecular mass higher than 10,000 u, is called a protein. In a polypeptide or protein, the peptide bond is the strongest chemical bond.

Classification Of Proteins

• Proteins may be categorised as fibrous and globular.
• Fibrous proteins are water-insoluble. They have long threadlike molecules that lie side by side to form fibres. Typical examples are keratin (present in hair, wool, silk) and myosin (present in muscles).
• The polypeptide chains in fibrous proteins are held together by hydrogen and disulphide bonds.
• Globular proteins are soluble in water and dilute acids, bases and salt solutions. Molecules of globular proteins are folded into spherical units. Albumin and plasma proteins are common examples of globular proteins.

Structure Of Proteins

The structure of proteins is studied at four different levels-primary, secondary, tertiary and quaternary. Each successive level is more complex than the previous one.

The Primary Structure

• In the context of the primary structure of a protein, we are concerned with the way amino acid residues join together to form chains. There are two aspects to this.
• The first of these is the geometry of the peptide link. Secondly, the sequence in which the amino acid residues appear in a chain is important.

The Secondary Structure

• In the context of the secondary structure of a protein, we are concerned with the arrangements of polypeptide chains, which results in a particular shape. This shape arises as a consequence of hydrogen bonding.
• The way in which the hydrogen bonds are arranged results in the formation of two possible structures—the a-helix structure and the β-sheet structure (or pleated-sheet structure).
• In the a-helix structure, hydrogen bonding within the chain twists it into a coil.
•   Hydrogen bonding occurs between the C=O group of one turn and the >N-H group of the turn below. An example of a protein with such a structure is keratin, which is found in hair and nails.
• In the pleated-sheet structure, hydrogen bonding occurs between different chains. The chains are arranged in the form of sheets of proteins type of structure is found in silk.

Tertiary structure

• The tertiary structure represents how the protein molecule is folded upon itself. It comes about on account of the folding and superposition of various secondary structures.
• The tertiary structure is found in two most important shapes-the globular and the fibrous. The secondary structure represents fibrous proteins.
• The folded molecule is held together by hydrogen bonding between side chains, salt bridges, disulphide bonds and other weak bonds. Myoglobin has an a-helical coil that is folded in upon itself.

Quaternary Structure

• Several polypeptide units, known as subunits (not always identical), can aggregate to form large complexes. The structure of the protein that results on account of the spatial arrangement of several subunits is known as the quaternary structure.
• Haemoglobin is composed of 4 protein chains that are held together by hydrogen bonding, salt bridges and other weak bonds.

Denaturation Of Proteins

• Under appropriate conditions the delicate three-dimensional structure of globular proteins may be disturbed. This process is called denaturation.
• Denaturation commonly occurs when the protein is subjected to extremes in temperature or when there is a change in pH. It is usually accompanied by a considerable decrease in the water solubility of the protein.
• An example is the coagulation that results in the hardening of the white and the yolk of an egg upon heating.
• Denaturation is essentially a disorganisation of the helical structure of the protein molecule caused by the breaking up of the cross-linked chains in the protein structure.
• Not only the hydrogen bonds but also the disulphide (-S-S-) bonds, salt bridges and other weak bonds are broken. This bond breaking results in loss of biological activity because the unique three-dimensional structure involving secondary and tertiary structures is destroyed.
• This is often accompanied by precipitation and coagulation. Remember that only the weak bonds of the protein molecules are broken and none of the peptide linkages are affected.

Enzymes

All life processes, such as the digestion of food, involve a series of reactions. These reactions occur very rapidly under mild conditions. These reactions, if performed in the laboratory, might take hours or days even under very vigorous conditions.

Somehow the body manages to increase greatly the rate of these reactions without heating. This is achieved by the catalysis of biochemical reactions by a type of molecule known as an enzyme.

• Most enzymes are globular proteins, and a few are nonproteins. Apart from being rapid, the reactions involving enzymes are quantitative and highly specific—a single enzyme will catalyse only a specific metabolic reaction.
• One series of enzymes catalyses the breakage of the peptide bonds in proteins so that amino acids are formed.
• Enzymes in saliva begin the process of breaking down starch into a-glucose.
• Some enzymes catalyse the formation of blood clots when necessary; others dissolve the clots after a wound has healed. Certain enzymes help the body to fight infections. Fruits ripen because of the action of enzymes. As the list suggests, enzymes are involved in a variety of processes.
• An enzyme is commonly named by adding the suffix ‘ase’ to the name of the substrate with which it reacts. For example, the enzyme urease catalyses the hydrolysis of urea into carbon dioxide and ammonia while maltase catalyses the hydrolysis of maltose to D (+) glucose. Some enzymes are popularly known by their trivial names, for example, pepsin.
• Enzymes are classified according to the type of reaction they catalyse.

The major classes of enzymes are as follows:

1. Oxidoreductases catalyse oxidation-reduction reactions.
2. Transferases catalyse the transfer of a characteristic chemical group from one molecule to another.
3. Hydrolases catalyse the reaction of the substrate with water.
4. Isomerases catalyse various types of isomerisation.

An enzyme is a biological catalyst. It is required only in small quantities for the progress of a reaction. Like a chemical catalyst, an enzyme reduces the magnitude of activation energy of a reaction.

For example, the activation energy for the acid hydrolysis of sucrose is 6.22 kJ mol1, while it is only 2.15 kJ mol1 when hydrolysed by the enzyme sucrose.

Mechanism Of Enzyme Action

A number of active sites (cavities) are present on the surface of an enzyme. These active sites are characterised by the presence of functional groups such as -NH2, -COOH, -SH and -OH.

These functional groups form weak bonds, such as hydrogen bonds or van der Waals bonds, with the corresponding substrate. The shape of a substrate is complementary to that of the corresponding enzyme.

So an enzyme fits into a substrate just like a key fits into a lock. Thus, an enzyme-substrate complex is formed, which then decomposes to yield the products.

Coenzymes

• A coenzyme is a nonprotein substance, needed for enzyme activity, that forms part of certain enzymes.
• Vitamins frequently form part of the coenzyme molecule.

Factors Affecting Enzyme Activity

• Since most enzymes are proteins, any of the factors that denature proteins also prevent them from acting.
• Enzymes in the body are the most active at normal body temperature, 98.5°F or 37°C.
• Temperatures above normal reduce enzyme activity and enzymes stop acting at extremes of temperature. Temperatures below normal also reduce enzyme activity.
• Each enzyme is the most active at its own optimum pH. On either side of this pH, its activity is markedly decreased.

Vitamins

• Vitamins are organic compounds which are essential for normal growth in human beings but most of them are not synthesised by the human body.
• However, plants can synthesise nearly all vitamins. Although required in small quantities, a lack of vitamins in the diet causes various diseases, known as deficiency diseases.
• Therefore these substances must be part of the diet along with carbohydrates, fats, proteins and minerals. Remember, however, that an excess of vitamins is harmful and you should not take vitamin pills unless the doctor advises you to.
• Vitamins are denoted by letters of the alphabet. They are classed as either water-soluble or fat-soluble. The B-complex vitamins and vitamin C are water-soluble.
• Vitamins A, D, E and K are fat-soluble. The table lists certain vitamins, their sources and the corresponding deficiency diseases.

Example 2:

1. Why are amino acids high-melting solids and water-soluble?
2. Why can vitamin C not be stored in our body?
3. Where does the water present in an egg go after the egg is boiled?

Solution:

1. Amino acids exist as internal salts called dipolar ions or zwitterions.

Due to the polar nature of amino acids, their molecules are attracted to each other by a strong dipole-dipole force. Therefore, their melting points are high. Because of their dipolar ionic structure, amino acids are soluble in water.

2. Vitamin C is soluble in water and is readily excreted in urine.

3. Upon heating, the proteins in eggs undergo denaturation and then coagulation, causing hardening of the white and the yolk. The water present in the egg gets absorbed/adsorbed in the coagulated proteins through hydrogen bonding.

Nucleic Acids

• Nucleic acids are those substances that are responsible for the passing on of hereditary traits and for the synthesis of proteins. They are high-molecular-weight polymers, consisting of repeating units called nucleotides.
• Nucleotides are made up of three parts—a nitrogen base, a five-carbon sugar and a phosphoric acid residue. The bases of nucleotides are derived from either pyrimidine or purine.
• The bases derived from pyrimidine are thymine (T), cytosine (C) and uracil (U). Those derived from purine are adenine (A) and guanine (G).
• A typical nucleotide how nucleotides combine to form a nucleic acid chain.
• Ribonucleic acid (RNA) is produced in the nucleus and migrates to the cytoplasm. It is involved in protein synthesis. Also, the genetic material of some microorganisms, such as many viruses, is RNA.
• The nucleus also contains deoxyribonucleic acid (DNA), the substance which is responsible for cell replication and is sometimes called the genetic code.

Structure Of Nucleic Acids

Like proteins, RNA and DNA have a high molecular weight; molecular weights of up to 10 million have been observed. On hydrolysis, both types of nucleic acids yield phosphoric acid, a sugar and a mixture of bases derived from purine and those from pyrimidine.

The sugar obtained from RNA is B-D-ribose, while that obtained from DNA is B-D-2-deoxyribose. The major bases obtained from DNA are the purine bases adenine and guanine and the pyrimidine bases cytosine and thymine.

RNA yields mainly adenine, guanine, cytosine and another pyrimidine base, uracil.

The mild degradation of nucleic acid yields nucleotides. Each nucleotide contains one purine or pyrimidine base, one phosphate unit, and one pentose unit.

The phosphate unit may be selectively removed by further careful hydrolysis. Then the nucleotide is converted into a nucleoside, a molecule built up of a pentose joined to a purine or pyrimidine base.

In a nucleotide, C₁ of the sugar is joined to N₁ of a pyrimidine or N₉ of a purine; the phosphoric acid unit is present as an ester at C₅ of the sugar.

Example 3: Name the three products which are formed when a nucleotide from RNA containing adenine is hydrolysed.

Solution:

The three products are

1. Adenine,
2. β-D-ribose and
3. Phosphoric acid.

In a nucleic acid chain, the phosphoric acid is esterified to form a bridge between C₅ of the sugar of one nucleoside and C₃ of the sugar of another nucleoside.

In this way, the sugar-phosphate units can form a long backbone or framework, which bears purine and pyrimidine bases at regular intervals.

A typical segment of a DNA chain is given in below

• The manner in which the sugar, phosphate and bases are linked with one another in nucleic acids determines the primary structure of the nucleic acids.
• Nucleic acids have a secondary structure also. Watson and Crick in 1953 proposed the now-accepted double helical structure of DNA.
• According to their of two DNA analyses, the molecule actually consists of complementary strands which are twisted about a common axis as helices with the same chirality (handedness).
• Each adenine unit of one chain is specially hydrogen-bonded to a thymine of the opposite chain and each guanine of one chain is similarly bonded to a complementary cytosine unit.
• It should be noted that the base pairing is restricted by hydrogen bonding requirements.
• The hydrogen atoms in purine and pyrimidine bases have well-defined positions. Adenine cannot pair with cytosine because there would be two hydrogen atoms near one of the bonding positions and none at the other.
• Similarly, guanine cannot pair with thymine. The G-C bond is stronger by 50% than the A-T bond.
• The double helical structure of DNA is shown schematically in Figure 14.29. The helical strands represent the sugar-phosphate backbones, which are held nicely in place by hydrogen bonding between the complementary base units.
• The order of the bases on the chain of the DNA molecule is extremely significant biologically. It is the fundamental unit of the hereditary information carried by genes.
• There are three different kinds of RNA molecules— messenger RNA (mRNA), transfer RNA (tRNA) and ribosomal RNA (rRNA).
• The RNA molecule is helical and single-stranded but occasionally small portions of it have a double-helical structure.
• This feature is observed when part of the molecule folds back upon itself to form complementary base pairs.

DNA Fingerprinting

Just as every person in the world can be positively identified by his or her fingerprints, so can every individual be identified nowadays by DNA fingerprinting.

This is because the base sequence of the DNA of every individual is unique and cannot be altered by any means.

DNA fingerprinting is now used to:

1.  Identify Criminals In Forensic Laboratories,
2.  Determine The Paternity Of An Individual, And
3.  Identify Dead Bodies By Comparing The Dnas Of Parents Or Children.

Biological Functions Of Nucleic Acids

• The DNA regulates two life processes. First, it can duplicate itself and secondly, it acts as a template for providing RNAs, which carry out protein synthesis.
• A molecule of DNA reproduces (or duplicates) itself by a remarkably simple mechanism. The two strands of the DNA molecule dissociate (an ‘unzipping’ process).
• Each strand then serves as a template for the synthesis of a complementary, new strand (Figure 14.30). The new (daughter) DNA molecules are identical to the original (parent) molecule—they contain all the original genetic information.
• The synthesis of identical copies of DNA is called replication. This process is the reason why children look much more like their relatives than like animals or trees.
• DNA has not only the critical function of reproducing itself, but also the important function of transferring information by causing the synthesis of a second type of nucleic acid called RNA.
• This transfer of information is called transcription. The major function of RNA is to transfer this information from nucleic acids to proteins, a process called translation.
• The three different kinds of RNA molecules-messenger RNA (mRNA), transfer RNA (tRNA) and ribosomal RNA (rRNA)-perform different functions.

Biomolecules Multiple-Choice Questions

Question: 1. A carbohydrate contains at least

1. 6 Carbons
2. 3 Carbons
3. 4 Carbons
4. 2 Carbons

Answer: 2. 3 Carbons

Question: 2. Which of the following is laevorotatory?

1. Glucose
2. Fructose
3. Sucrose
4. None Of These

Answer: 2. Fructose

Question: 3. Which of the following carbohydrates forms a silver mirror on being treated with Tollens reagent?

1. Sucrose
2. Fructose
3. Glucose
4. Starch

Answer: 3. Glucose

Question: 4. Which of the following carbohydrates is an essential component of plant cells?

1. Starch
2. Cellulose
3. Sucrose
4. Vitamins

Answer: 2. Cellulose

Question: 5. The hydrolysis of sucrose leads to

1. Saponification
2. Hydration
3. Esterification
4. Inversion

Answer: 4. Inversion

Question: 6. Which of the following is the sweetest of all the sugars?

1. Sucrose
2. Maltose
3. Fructose
4. Lactose

Answer: 2. Maltose

Question: 7. On hydrolysis, maltose yields

1. Glucose And Mannose
2. Galactose And Glucose
3. Glucose
4. Mannose

Answer: 3. Glucose

Question: 8. The function of enzymes is to

1. Provide Energy
2. Provide Immunity
3. Transport Oxygen
4. Catalyse Biochemical Reactions

Answer: 4. Catalyse Biochemical Reactions

Question: 9. A deficiency of which of the following may cause night blindness?

1. Vitamin B₁₂
2. Vitamin A
3. Vitamin C
4. Vitamin E

Answer: 2. Vitamin A

Question: 10. Which of the following functional groups is/are present in amino acids?

1. -COOH group
2. -NH₂ group
3. -CH₃ group
4. ‘a’ and ‘b’

Answer: 4. ‘a’ and ‘b’

Question: 11. What is the order in which a base, a phosphate and a sugar are arranged in nucleic acids?

1. Base-Phosphate-Sugar
2. Phosphate-Base-Sugar
3. Sugar-Base-Phosphate
4. Base-Sugar-Phosphate

Answer: 4. Base-Sugar-Phosphate

Question: 12. Which of the following is related to steroids?

1. Vitamin E
2. Vitamin K
3. Vitamin B
4. Vitamin D

Answer: 4. Vitamin D

Question: 13. A deficiency of vitamin C causes

1. Beriberi
2. Night Blindness
3. Rickets
4. Scurvy

Answer: 4. Scurvy

Question: 14. Which of the following is capable of forming a zwitterion?

1. H₂NCH₂COOH
2. CH₃COOH
3. CH₃CH₂NH₂
4. CCl₃NO₂

Answer: 1. H₂NCH₂COOH

Question: 15. Which of the following are generally not produced in our body?

1. Enzymes
2. Vitamins
3. Proteins
4. Hormones

Answer: 2. Vitamins

Question: 16. Nucleic acids are

1. Polymers Of Nucleotides
2. Polymers Of Nucleosides
3. Polymers Of Purine Bases
4. Polymers Of Phosphate Esters

Answer: 1. Polymers Of Nucleotides

Question: 17. The bases common to DNA and RNA are

1. Adenine, Cytosine And Uracil
2. Guanine, Adenine And Cytosine
3. Guanine, Uracil And Thymine
4. Adenine, Thymine And Guanine

Answer: 2. Guanine, Adenine And Cytosine

Question: 18. The functions of DNA are

1. To Synthesise Rna
2. To Synthesise Proteins
3. To Carry Genetic Information From Parent To Offspring
4. A, B And C

Answer: 4. A, B And C

Question: 19. The purine base present in RNA is

1. Guanine
2. Thymine
3. Cytosine
4. Uracil

Answer: 4. Uracil

Question: 20. In respect of which base does RNA differ from DNA?

1. Thymine
2. Adenine
3. Cytosine
4. Guanine

Answer: 1. Thymine

Question: 21. Insulin is

1. An Amino Acid
2. A Protein
3. A Carbohydrate
4. A Lipid

Answer: 2. A Protein

Question: 22. The secondary structure of a protein refers to the

1. Fixed Configuration Of The Polypeptide Backbone
2. A-Helical Backbone
3. Hydrophobic Interactions
4. Sequence Of A-Amino Acids

Answer: 2. A-Helical Backbone

Question: 23. Which of the following is/are disaccharides with the molecular formula C₁₂H₂₂O₁₁?

1. Cane Sugar
2. Fruit Sugar
3. Lactose
4. A Ketohexose

Answer: 1. Cane Sugar, 3. Lactose, 4. A Ketohexose

Question: 24. Fructose is

1. Grape Sugar
2. Laevulose
3. Raffinose
4. Maltose

Answer: 2. Laevulose, 3. Raffinose, 4. Maltose