Vasantha

The S-Block Elements

You have already learnt that elements in the periodic table are arranged in order of increasing atomic number.

The long form of the periodic table is divided into four blocks—s, p, d and f—depending on the type of the orbitals (s, p, d or f) that are being filled with valence electrons.

The elements belonging to the s block can have one or a maximum of two s-electrons in their outermost shell.

The s-block of the periodic table comprises two groups—Group 1 and Group 2. The elements of Group 1 have the general ground state electronic configuration of ns¹ and are called alkali metals.

The elements of Group 2 have the general ground state electronic configuration of ns² and are called alkaline earth metals.

Elements in which the p orbitals are being filled with valence electrons fall in the p block.

This block constitutes Groups 13, 14, 15, 16, 17 and 18 with 1, 2, 3, 4, 5 and 6 electrons, respectively, in the p orbitals of the outermost shell, while the s orbitals are already filled.

General Characteristics of s-Block Elements

Some important characteristics of s-block elements are given below. A comparison with the p-block elements is also made, which will be helpful.

1. The s-block elements have the general outermost electronic configuration ns¹ (for alkali metals) or ns² (for alkaline earth metals). However, in the case of p-block elements the electronic configuration of the outermost shell is ns² np^1-6.
2. The s-block elements are metals and form ionic compounds with nonmetals whereas p-block elements are nonmetals and form predominantly covalent compounds with each other.
3. The s-block elements show only one oxidation state, which is equal to their group number.
4. Thus, alkali metals (Group 1) and alkaline earth metals (Group 2) show oxidation states of 1 and 2 respectively. On the other hand, most of the p-block elements show more than one positive as well as a negative oxidation state.
5. Diagonal relationship The first member of each group of the s-block and those of the other groups differ markedly from the rest of the members. Consider the elements of periods 2 and 3.

The first element of a group in the second period shows similarities with the second element of the neighbouring group on the right (diagonally opposite element). This is referred to as a diagonal relationship.

Thus, lithium resembles magnesium, beryllium resembles aluminium and boron resembles silicon.

This diagonal relationship is attributed to similarity in the size of the ions (e.g., Li+ =76 pm and Mg²⁺ =72 pm), and in the electropositive character and polarising power of the elements.

Alkali Metals (Group 1 Elements)

The elements of Group 1 are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs) and francium (Fr).

All these elements are metallic in nature and are known as alkali metals. The last element of this group, francium, is radioactive and has a very short half-life and so, very little is known about it. All the alkali metals are silvery-white, soft and light.

Occurrence

Alkali metals are very reactive and react with oxygen and water. Therefore, they do not occur in the free state.

However, they occur in the combined state in the form of salts like NaCl and KC1, which occur in large amounts in seawater.

Electronic Configuration

Alkali metals have only one electron (ns¹) in the s-orbital of their valence shell. The valence-shell electronic configuration of the alkali metals can be generalised as ns¹, where n = 2 to 7.

Atomic And Ionic Radii

The atoms of alkali metals have the largest size in their respective periods. The atomic radii increase as we go down the group (from Li to Cs) because there is a progressive addition of new energy shells.

This implies that the distance between the nucleus and the last shell increases and the valence electrons are farther from the nucleus. So the atomic radius increases with an increase in atomic number.

However, with the increase in atomic number the nuclear charge also increases. This tends to decrease the atomic radius by attracting the electron cloud inward with greater force.

However, the effective nuclear charge experienced by valence electrons is less than the actual nuclear charge because of the screening of the outermost electrons by the inner core of electrons.

For example, in the case of the sodium atom, the outermost 3s electron is screened by the inner Is and 2s electrons.

The alkali metals tend to lose their valence electrons and change to monovalent cations in order to achieve the noble-gas configuration.

So the radius of the positive ion is smaller than that of the parent atom. However, within the group, the ionic radius increases with an increase in atomic number.

This is accounted for by the addition of electron shells at every step as we go down the group.

Ionisation enthalpies

The first ionisation enthalpies of the elements of Group 1 are the lowest. This is because the valence electrons are loosely held, and can be easily lost by the atom to achieve the nearest noble-gas configuration.

The second ionisation enthalpies of alkali metals are very high. This is because it is difficult to remove an electron from a monovalent cation (formed by the removal of an electron from an atom of an alkali metal) having a noble-gas configuration.

Hydration Enthalpies

The standard enthalpy of hydration of an element is the energy released when 1 mol of gaseous ions of that element become hydrated (surrounded by water molecules) under standard conditions.

The higher the charge on the ions and the smaller their size, the more negative is the hydration enthalpy.

The hydration enthalpies of metal cations increase on moving down the group. Lithium has the most negative hydration enthalpy. ft Therefore, it mostly forms hydrated salts, e.g., LiCl₂H₂O.

Electropositive Character And Oxidation State

Elements that are electropositive tend to lose electrons and form positive ions. The alkali metals are typical electropositive elements.

The electropositive character increases from Li to Cs as the tendency to lose an electron increases on moving down the group.

While combining with other elements in a reaction, alkali metals have a strong tendency to lose the single valence electron to form a unipositive ion.

Thus, they show an oxidation state of +1 in their compounds and are strongly electropositive.

⇒ \(\mathrm{M} \rightarrow \mathrm{M}^{+}+\mathrm{e}^{-}\)

Melting And Boiling Points

The alkali metals have low melting and boiling points, which decrease down the group. This indicates weak inter¬ atomic bonding or metallic bonding.

In a metallic bond negatively charged electrons hold the positive ions together. The bonding is weak in alkali metals due to the presence of only a single valence electron for each atom. On moving down the group, atomic size increases, which progressively weakens intermolecular bonds.

Density

The densities of alkali metals are low and increase on moving down the group. However, potassium is an exception and is lighter than sodium.

The low density is attributed to the large atomic size. On moving down the group, atomic size as well as atomic mass increases.

However, the corresponding increase in atomic mass is not compensated by an increase in atomic volume. Thus, the density, which is the ratio of mass to volume, gradually increases.

Flame Colouration

All alkali metals impart a characteristic colour to a Bunsen flame.

When a sample of any metal salt is heated in a Bunsen burner (a laboratory gas burner) flame it imparts a characteristic colour to the flame.

The orbital electrons of the metal atoms absorb the heat energy supplied and get excited to higher energy states. When the excited electron loses the extra energy, it falls back to its ground state.

The extra energy lost is emitted in the form of radiation, which falls in the visible region, imparting a characteristic colouration to the flame.

Photoelectric effect

All alkali metals except lithium exhibit a photoelectric effect. Due to the low ionisation enthalpies of the metals, the metal surfaces emit electrons when exposed to visible light. Lithium has high ionisation enthalpy values and so does not exhibit a photoelectric effect.

Chemical properties

Alkali metals are very reactive. This is attributed to their low first ionisation enthalpies. We will now discuss some important chemical properties of alkali metals.

Reducing character

Alkali metals are good reducing agents. This is because they have a strong tendency to lose electrons and have large negative values of reduction potential. The reducing character increases from Na to Cs.

However, lithium, in spite of its highest ionisation enthalpy, is the strongest reducing agent among all alkali metals, as indicated by its reduction potential value (3.03 V).

The unusual behaviour of lithium is due to the exceptionally small size of the lithium atom and the high charge-to-radius ratio of the ion.

Being reducing agents, alkali metals react with compounds containing acidic hydrogen atoms (hydrogen atoms generally attached to oxygen or a triple bond are acidic and they can be replaced by metals), such as water, alcohol and acetylene, liberating hydrogen gas.

The standard electrode potential (Ee’)/ which measures the reducing power and is expressed as M+(aq)/M(s), represents the overall change as follows

⇒ \(\begin{aligned}
& \mathrm{M}(\mathrm{s}) \rightarrow \mathrm{M}(\mathrm{g}) \quad \text { (Sublimation enthalpy) } \\
& \mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-} \quad \text { (Ionisation enthalpy) } \\
& \mathrm{M}^{+}(\mathrm{g})+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{M}^{+}(\mathrm{aq}) \quad \text { (Hydration enthalpy) } \\
&
\end{aligned}\)

The reactivity of alkali metals towards compounds containing acidic hydrogen increases from Li to Cs

⇒ \(\begin{aligned}
2 \mathrm{Li}+2 \mathrm{H}_2 \mathrm{O} & \rightarrow 2 \mathrm{LiOH}+\mathrm{H}_2 \\
2 \mathrm{Na}+2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} & \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}+\mathrm{H}_2
\end{aligned}\)

⇒ \(2 \mathrm{Na}+\underset{\text { acetylene }}{\mathrm{HC}} \equiv \mathrm{CH} \rightarrow \underset{\text { sodium acetylide }}{\mathrm{HC}} \equiv \mathrm{C}-\mathrm{Na}+\mathrm{H}_2 acetylene sodium acetylide\)

Reaction With Oxygen

Alkali metals tarnish rapidly in dry air. Lithium forms a mixture of its oxide and nitride on the surface.

When burnt in air or oxygen, alkali metals form different types of oxides. For example, on a reaction with a limited quantity of oxygen, these metals form normal oxides (M₂O).

⇒ \(\begin{gathered}
2 \mathrm{M}+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{M}_2 \mathrm{O} \\
(\mathrm{M}=\mathrm{L}, \mathrm{Na}, \mathrm{K}, \mathrm{Kb}, \mathrm{Cs})
\end{gathered}\)

However, on being heated with an excess of air or oxygen, lithium forms the normal oxide or monoxide, sodium forms its peroxide (Na202) whereas potassium, rubidium and caesium form superoxides (MO₂).

⇒ \(2 \mathrm{Li}+\frac{1}{2} \mathrm{O}_2 \rightarrow \underset{\text { lithium oxide }}{\mathrm{Li}_2 \mathrm{O}}\)

⇒ \( 2 \mathrm{Na}+\mathrm{O}_2 \rightarrow \mathrm{Na}_2 \mathrm{O}_2 sodium peroxide\)

⇒ \(\mathrm{K}+\mathrm{O}_2 \rightarrow \quad \mathrm{KO}_2 potassium superoxide\)

Being the smallest in size, Li+ has a small, strong positive field around it. This implies that Li+ can stabilise only a small anion, O^2–. However, Na+, being larger in size has a weak positive field around it.

Therefore, it can stabilise a bigger peroxide ion, O²²⁺-, which has a weak negative field around it.

On the other hand, K+, Rb+, and Cs+ being still larger, stabilise the still bigger superoxide ion (O²⁺) to form superoxides.

The normal oxides and peroxides are colourless solids. However, superoxides are paramagnetic in nature (due to the presence of one impaired electron in the O^2–ion) and are yellow solids.

The oxides of alkali metals dissolve readily in water to form hydroxides. The reaction is exothermic.

⇒ \( \mathrm{M}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{MOH}+\text { heat } normal oxide \mathrm{Na}_2 \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O}_2+\text { heat } \)

These reactions show that the oxides of alkali metals are basic.

Reaction With Hydrogen

In general, alkali metals react with dihydrogen at about 673 K to give hydrides of the type MH. Lithium, however, reacts at 1073 K.

⇒ \(2 \mathrm{M}+\mathrm{H}_2 \rightarrow \underset{\text { metal hydride }}{2 \mathrm{MH}}\)

The reactivity of alkali metals with hydrogen decreases from Li to Cs. Also, the stability of the hydrides decreases from Li to Cs because, as the size of the alkali metal increases, the M—H bond becomes weak. The hydrides of alkali metals are strong reducing agents and are used in organic synthesis.

Reaction with halogens

Alkali metals react with halogens to form the corresponding halides, M+X

⇒ \(2 \mathrm{M}+\mathrm{X}_2 \rightarrow 2 \mathrm{MX}(\mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br} \text { or } \mathrm{I})\)

The reactivity of alkali metals towards a particular halogen increase from Li to Cs. The reactivity of the halogens towards a particular alkali metal is of the order F2 >C12 > Br2 >I2.

Tire alkali metal halides (MX) are colourless, crystalline solids with high melting points. They can also be prepared conveniently from the corresponding oxide, hydroxide or carbonate by reaction with an aqueous hydrohalic acid (HX)

⇒ \(\begin{gathered}
\mathrm{Li}_2 \mathrm{O}+2 \mathrm{HCl} \rightarrow 2 \mathrm{LiCl}+\mathrm{H}_2 \mathrm{O} \\
\mathrm{Li}_2 \mathrm{CO}_3+2 \mathrm{HCl} \rightarrow 2 \mathrm{LiCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \\
\mathrm{LiOH}+\mathrm{HCl} \rightarrow \mathrm{LiCl}+\mathrm{H}_2 \mathrm{O}
\end{gathered}\)

Alternatively, the nitrates of alkali metals can be decomposed on heating to give the corresponding oxide, which is then converted into a halide.

⇒ \(4 \mathrm{LiNO}_3 \rightarrow 2 \mathrm{Li}_2 \mathrm{O}+4 \mathrm{NO}_2+\mathrm{O}_2\)

Alkali metal halides are ionic compounds. However, Lil is slightly covalent due to polarisation (Li has the maximum polarising power as it is the smallest cation and the iodide ion can be polarised to the maximum extent as it is the largest anion).

Except for LiF, alkali metal halides are soluble in water. The insolubility of LiF is attributed to its high lattice enthalpy, as a result of the combination of the small anion (F-) and the small cation (Li+).

Melting and boiling points of alkali metal halides All the halides of alkali metals are colourless crystalline solids with high melting points because they contain ionic bonds. The melting and boiling points of these halides follow certain trends.

1. For the same alkali metal, the melting and boiling points decrease in the following order:

fluoride > chloride > bromide > iodide

This Can Be Explained On The Basis Of The Lattice Enthalpies Of These Halides. For The Same Alkali Metal Ion, The Lattice Enthalpies Decrease As The Size Of The Halide Ion Increases.

As The Lattice Enthalpies Decrease, The Energy Required To Break The Lattice (Melt The Solid) Decreases. Thus The Melting Points Of Sodium Halides Decrease From Naf To I.

2. For The Same Halide Ion, The Melting Point Of The Lithium Halide Is Lower Than That Of The Sodium Halide. Then There Is A Progressive Decrease In Melting Point From Na To Cs.

The low melting point of LiCl compared to that of NaCl, for example, is due to the fact that LiCl is covalent in nature, while NaCl is held together by ionic bonding.

The lattice enthalpy decreases from NaCl to CsCl as the size of the alkali metal ion increases. So the melting point decreases from NaCl to CsCl.

Reaction With Water

In reaction with water, alkali metals form hydroxides, liberating hydrogen gas.

⇒ \(\begin{aligned}
2 \mathrm{Na} & +2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \\
2 \mathrm{~K}+2 \mathrm{H}_2 \mathrm{O} & \rightarrow 2 \mathrm{KOH}+\mathrm{H}_2
\end{aligned}\)

The reaction is so vigorous that the hydrogen evolved catches fire. The reaction becomes more violent on descending down the group. Therefore, alkali metals should be stored in an inert medium, usually kerosene oil.

Lithium is the only alkali metal which reacts gently with water. This is because it has the highest charge-to-radius ratio.

Owing to this it can polarise the water molecule more and with the highest hydration energy form a stronger metal-to-oxygen bond.

But its gentle behaviour towards water lies not in thermodynamics but in the kinetics of the reaction. Potassium, which liberates less energy, has a low melting point.

So it melts by the heat of the reaction, spreads and exposes a larger surface to the water, thus catching fire.

The hydroxides of the alkali metals are strongly basic and the strength of the base increases down the group.

On descending the group, the decrease in ionisation enthalpy results in weaker M —OH bonds from LiOH to C₂OH, and this accounts for the increase in the basic strength of the hydroxides. Alkali metal hydroxides in solution ionise easily to form M+ and OH ions due to the weak M—OH bond. This accounts for the strong basic nature of the hydroxides.

Reaction with oxoacids

Being electropositive in nature, alkali metals form oxo salts by reacting with oxoacids. For example, HOC₁, H₂CO₃ and HNO₃ react with alkali metals to give salts of oxoacids like MOCI, M₂CO₃ and MNO₃ respectively.

These salts are soluble in water and stable to heat. Li₂CO₃ is considerably less stable. Being strongly basic, Group 1 metals form solid bicarbonates.

Solutions of metals in liquid ammonia

All Group 1 metals and a few Group 2 metals dissolve in liquid ammonia, forming a deep blue solution. One reason why metals dissolve in liquid ammonia is ionisation.

⇒ \(\mathrm{M} \rightarrow \mathrm{M}^{+}+\mathrm{e}^{-}\)

The cations and electrons combine with ammonia to form ammoniated cations and ammoniated electrons respectively. The cations and electrons actually interact with the molecules of the solvent (ammonia). The overall reaction is

⇒ \(\mathrm{M}+(x+y) \mathrm{NH}_3 \rightarrow\left[\mathrm{M}\left(\mathrm{NH}_3\right)_x\right]^{+}+\left[\mathrm{e}\left(\mathrm{NH}_3\right)_y\right]^{-}
ammoniated cation ammoniated electron\)

The ammoniated electron is responsible for the blue colour of the solution. The solution absorbs energy corresponding to the red region of visible light and the ammoniated electrons get excited to higher energy levels.

When the electrons return to lower energy levels they transmit light which imparts a blue colour to the solution.

The solution is also conducting mainly due to the presence of solvated or ammoniated electrons. The solution fades on standing, slowly liberating hydrogen and forming a metal amide

⇒ \(\mathrm{M}^{+}+\mathrm{e}^{-}+\mathrm{NH}_3(\mathrm{l}) \rightarrow \mathrm{MNH}_2+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})\)

The ammoniated electron is responsible for the blue colour of the solution. The solution absorbs energy corresponding to the red region of visible light and the ammoniated electrons get excited to higher energy levels.

When the electrons return to lower energy levels they transmit light which imparts a blue colour to the solution.

The solution is also conducting mainly due to the presence of solvated or ammoniated electrons. The solution fades on standing, slowly liberating hydrogen and forming a metal amide.

⇒ \(\mathrm{M}^{+}+\mathrm{e}^{-}+\mathrm{NH}_3(\mathrm{l}) \rightarrow \mathrm{MNH}_2+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})\)

If the concentration of the solution is increased, the blue colour changes to copper-bronze with a metallic lustre.

This is because, with increased concentration, metal ion clusters are formed. The solutions of alkali metals in liquid ammonia act as powerful reducing agents and are used in organic reactions.

Solubility In Mercury

The alkali metals dissolve readily in mercury to form amalgams. The process is highly exothermic. The amalgams are also used as reducing agents in organic synthesis.

Anomalous Behaviour Of Lithium

The first element, lithium, of Group 1 differs from the other members of this group in many respects. The main reasons for the anomalous behaviour of lithium are as follows.

1. The exceptionally small size of the lithium atom and its ion.
2. The higher polarising power (charge to radius ratio) of Li+ is due to its smaller size
   and the covalent character of its compounds.
3. The high ionisation enthalpy and low electropositivity of lithium as compared to other alkali metals.
4. absence of vacant d-orbitals in its valence shell.

Some of the important properties in respect of which lithium differs from other members of the group are as follows.

1. Lithium is harder than the other alkali metals.
2. The melting point and boiling point of lithium are much higher than those of the other alkali metals.
3. Lithium is the least reactive as observed by its gentle behaviour towards water. In contrast/ the other alkali metals react violently with water.
4. In reaction ‘with oxygen, lithium forms only the monoxide (Li₂O) while other alkali metals form the peroxide (as in the case of sodium) or the superoxide (as in the case of potassium).
5. The ionic character of lithium salts is less pronounced than that of the salts of other alkali metals. This is because of the high polarising power of Li+. (The partial covalent character of ionic bonds. Due to its small size, Li+ has the maximum tendency to draw electrons towards itself from the negative ion in the salt molecule. This results in the distortion of the electronic cloud of the anion.
6. This distortion is known as polarisation. When the degree of polarization is large, the concentration of electrons increases between the two atoms. Therefore, the covalent character of the molecule increases.
7. Lithium is the only alkali metal that combines directly with the nitrogen of air to form the corresponding nitride.
   • \(6 \mathrm{Li}+\mathrm{N}_2 \rightarrow 2 \mathrm{Li}_3 \mathrm{~N}\)
8. The hydroxides, carbonates and nitrates of lithium decompose on heating to form lithium oxide. (Other alkali metal hydroxides and carbonates are stable and the nitrates decompose to give the corresponding nitrite.)
   • \(2 \mathrm{LiOH} \stackrel{\text { heat }}{\longrightarrow} \mathrm{Li}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O}\)
   • \(\mathrm{LiCO}_3 \stackrel{\text { heat }}{\longrightarrow} \mathrm{Li}_2 \mathrm{O}+\mathrm{CO}_2\)
   • \(4 \mathrm{LiNO}_3 \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{Li}_2 \mathrm{O}+4 \mathrm{NO}_2+\mathrm{O}_2\)
   • \(2 \mathrm{NaNO}_3 \longrightarrow 2 \mathrm{NaNO}_2+\mathrm{O}_2\)
9. The hydroxide of lithium (LiOH) is a weak base, whereas the hydroxides of other alkali metals are strong bases.
10. Due to their covalent nature, lithium halides (e.g., LiCl) are soluble in organic solvents, unlike the halides of other alkali metals. Also, LiCl is deliquescent and crystallises as LiCl.₂H₂0.
11. Due to its smaller size, the lithium-ion (Li⁺) is more strongly hydrated in an aqueous solution than other alkali metal ions.

Diagonal Relationship Of Lithium With Magnesium

As you know, the first element (Li) of Group 1 shows similarities in properties with the second element (Mg) of Group 2 (the diagonally opposite element).

This is referred to as a diagonal relationship.

Lithium resembles magnesium with respect to the following important properties.

1. Both lithium and magnesium are equally hard and light.
2. The similarities in lithium and magnesium are mainly due to their similar ionic size (Li+ =76 pm, Mg²⁺ = 72 pm). Both have nearly equal atomic radii, electronegativities and polarising powers.
3. The melting and boiling points of both lithium and magnesium are fairly high.
4. On heating in nitrogen both lithium and magnesium form nitrides, which are ionic compounds.
   • \(\begin{aligned}
     6 \mathrm{Li}+\mathrm{N}_2 & \rightarrow 2 \mathrm{Li}_3 \mathrm{~N} \\
     3 \mathrm{Mg}+\mathrm{N}_2 & \rightarrow \mathrm{Mg}_3 \mathrm{~N}_2
     \end{aligned}\)
5. The hydroxides, carbonates and nitrates of both lithium and magnesium decompose on heating to give the corresponding oxides. The chemical reactions of lithium compounds have been discussed in the previous section.
   • \(\mathrm{Mg}(\mathrm{OH})_2 \stackrel{\text { heat }}{\longrightarrow} \mathrm{MgO}+\mathrm{H}_2 \mathrm{O}\)
   • \(\mathrm{MgCO}_3 \stackrel{\text { heat }}{\longrightarrow} \mathrm{MgO}+\mathrm{CO}_2\)
   • \(2 \mathrm{Mg}\left(\mathrm{NO}_3\right)_2 \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{MgO}+4 \mathrm{NO}_2+\mathrm{O}_2\)
6. The hydroxides of both lithium and magnesium are weak bases.
7. The hydroxides, carbonates and fluorides of both lithium and magnesium are sparingly soluble in water.
8. The halides of lithium and magnesium are soluble in organic solvents (they are covalent compounds).
9. The chlorides of both lithium and magnesium are deliquescent and separate from aqueous solutions as hydrated salts (LiCl^2–2H₂O and MgCl₂-6H₂0).

Uses

Some of the important uses of alkali metals are as follows.

1. Lithium finds use in making useful alloys like the lithium-lead alloy used for making toughened bearings.
2. The lithium-magnesium alloy has high tensile strength and is used to manufacture aircraft components.
3. Lithium is also employed in the production of thermonuclear energy required for propelling rockets and guided missiles.
4. Sodium metal (molten) or its alloys with potassium are used as a coolant in nuclear reactors. The sodium-lead alloy is used for the preparation of lead tetraethyl [Pb(C2H5)4], which is used antiknock agent in petrol. However, its use has nowadays been discouraged as it causes environmental problems. Potassium chloride is used as a fertiliser and potassium hydroxide is used in the manufacture of soft soap.
5. Caesium finds use in the manufacture of photoelectric cells.

Compounds of Sodium

Sodium is the most abundant of the alkali metals. Being extremely reactive, it does not occur in the free state.

Sodium chloride (common salt) is the most common compound of sodium, but many others are also known. Sodium is commercially the most important metal of all alkali metals.

It is used in the manufacture of sodium carbonate, sodium hydroxide, sodium chloride and sodium bicarbonate, which are of great use in the chemical industry.

The process of manufacture, properties and uses of some of these compounds are as follows.

Sodium carbonate

It is also known as washing soda or soda ash. Soda ash is the white powder of sodium carbonate which aggregates on exposure to air due to the formation of monohydrates.

Na₂CO₃-H₂0 is a monohydrate and Na₂CO₃10H₂O is a decahydrate known as washing soda.

Since pre-historic times, it has been found to occur as natural deposits called trona (Na₂CO₃-NaHCO₃-2H₂O) in dried-up lake beds in Egypt. It is also mined in the USA and Kenya. The deposit of trona is converted into sodium carbonate by heating.

⇒ \(2\left(\mathrm{Na}_2 \mathrm{CO}_3 \cdot \mathrm{NaHCO}_3 \cdot 2 \mathrm{H}_2 \mathrm{O}\right) \stackrel{\text { heat }}{\longrightarrow} 3 \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}\)

The Solvay (or ammonia-soda) process

Sodium carbonate is manufactured by the Solvay or the ammonia-soda process. In this process, a purified concentrated solution of sodium chloride (brine) is saturated with ammonia gas.

The ammoniacal brine solution is then carbonated with carbon dioxide, forming sodium bicarbonate, which is insoluble in brine solution due to the common-ion effect and is filtered off.

The sodium bicarbonate on heating decomposes to give anhydrous sodium carbonate. We cannot prepare potassium carbonate by this process as the solubility of KHCO₃ is large in brine solution. The reactions taking place during the process are as follows.

⇒ \(\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \rightarrow \mathrm{NH}_4 \mathrm{HCO}_3\)

⇒ \(\mathrm{NaCl}+\mathrm{NH}_4 \mathrm{HCO}_3 \rightarrow \mathrm{NaHCO}_3 \downarrow+\mathrm{NH}_4 \mathrm{Cl}\)

⇒ \(2 \mathrm{NaHCO}_3 \stackrel{\text { heat }}{\longrightarrow} \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Initially, the carbon dioxide required in the first step is generated by heating limestone.

⇒  \(\mathrm{CaCO}_3 \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_2\)

Subsequently, the CO₂ generated by heating sodium bicarbonate is used. The ammonia gas required for the reaction forming ammonium bicarbonate is obtained from CaO₂ (quicklime), which is shaken with water and then boiled with the ammonium chloride produced when sodium chloride reacts with ammonium bicarbonate.

The only raw materials used in the Solvay process are sodium chloride (common salt), limestone (to produce C02 initially) and ammonia. The by-product obtained in this process is calcium chloride, which is not of much use.

The plant used consists of an ammoniation tower (in which the brine solution is saturated with ammonia and a little carbon dioxide), a filter (in which the ammoniacal brine along with carbon dioxide is filtered), a carbonating tower (in which the ammoniacal liquor is carbonated), a vacuum filter (in which the sodium bicarbonate obtained is filtered) and the ammonia recovery tower (in which calcium hydroxide and ammonium chloride is heated to give ammonia gas).

The diagrammatic representation of the plant used for the Solvay process

Properties

Sodium carbonate is a white, crystalline substance and is obtained as a decahydrate (Na₂CO₃TOH₂O). On heating at 373 K, the decahydrate changes to the monohydrate. At a temperature above 373 K, the monohydrate becomes completely anhydrous.

⇒ [Latex]\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O} \stackrel{373 \mathrm{~K}}{\longrightarrow} \mathrm{Na}_2 \mathrm{CO}_3 \cdot \mathrm{H}_2 \mathrm{O}+9 \mathrm{H}_2 \mathrm{O}[/Latex]

⇒ [Latex]\mathrm{Na}_2 \mathrm{CO}_3 \cdot \mathrm{H}_2 \mathrm{O} \longrightarrow 3 / 3 \mathrm{~K} \longrightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}[/Latex]

The anhydrous sample absorbs moisture from the air and gives Na2C03-H20. It reacts with ads to give carbon dioxide with effervescence.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

Sodium carbonate, when hydrolysed with water, forms an alkaline solution.

⇒ \(\mathrm{CO}_3{ }^{2-}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HCO}_3{ }^{-}+\mathrm{OH}^{-}\)

Sodium carbonate reacts with hot milk of lime to form sodium hydroxide.

⇒ \(\mathrm{Ca}(\mathrm{OH})_2+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow 2 \mathrm{NaOH}+\mathrm{CaCO}_3\)

Uses

1. Sodium carbonate is used in the manufacture of soap, detergents, paper, glass, phosphates and silicates.
2. It is also used in the manufacture of textiles, paints and dyes.
3. Sodium carbonate is employed for water softening, laundering and cleaning.
4. It is used in the laboratory as a reagent and as a primary standard in acid-base titrations. A mixture of sodium carbonate and potassium carbonate is used as a fusion mixture.

Sodium hydroxide

Sodium hydroxide, also known as caustic soda, is the most important alkali used iri the chemical industry.

It is manufactured by the electrolysis of a saturated solution of sodium chloride. In the past, it was also made by the causticising process, front sodium carbonate. This process is not in much use nowadays as other methods are cheaper.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{Ca}(\mathrm{OH})_2 \rightarrow 2 \mathrm{NaOH}+\mathrm{CaCO}_3\)

Electrolysis of sodium chloride solution in a Castner-Kellner Cell

In this cell, a solution of sodium chloride (brine) is electrolysed. The mercury cathode is made to flow along the bottom of the cell. The brine also moves along the cathode and the sodium ions are reduced to form sodium metal.

⇒ \(At cathode:
\begin{aligned}
& \mathrm{Na}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Na} \\
& \mathrm{Na}+\mathrm{Hg} \rightarrow \underset{\text { sodium amalgam }}{\mathrm{Na}-\mathrm{Hg}}
\end{aligned} \)

The sodium metal formed dissolves in mercury to form an amalgam (a loose alloy), which is pumped to a different compartment (called the denuder).

In this compartment H20 trickles over lumps of graphite (here acting as an inert solid); this results in the reaction of water with the amalgam, producing an NaOH solution. The strength of the NaOH collected from this compartment is up to 50%.

⇒ \(\mathrm{Na} \text { (amalgam) }+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{NaOH}+\frac{1}{2} \mathrm{H}_2+\mathrm{Hg}\)

The mercury is recycled back to the electrolysis tank. Hydrogen and chlorine are the two by-products formed. Chlorine is liberated at the anode.

⇒At anode:\(\begin{aligned}
\mathrm{Cl}^{-} & \rightarrow \mathrm{Cl}+\mathrm{e}^{-} \\
\mathrm{Cl} & +\mathrm{Cl} \rightarrow \mathrm{Cl}_2
\end{aligned}\)

In the past, the anodes were made of graphite. However, due to traces of dioxygen produced in a side reaction, the graphite anode became pitted (holes were formed) because of the formation of C02.

Nowadays, in the Castner-Kellner cell, the anodes are made of steel coated with titanium. Titanium is very resistant to corrosion, so the problem of pitting is avoided. An additional advantage is that titanium lowers the electrical resistance.

Nowadays, a nafion membrane is also used in the cell to separate the anolyte and the catholyte. The nation is a 4 polymer of certain organic compounds.

Properties

Sodium hydroxide is a white, deliquescent solid and absorbs moisture from the air. It can even react with traces of carbon dioxide present in the air to form a solid hydrated carbonate.

⇒ \(2 \mathrm{NaOH}+\mathrm{CO}_2 \rightarrow \mathrm{Na}_2 \mathrm{CO}_3 \cdot \mathrm{H}_2 \mathrm{O}\)

It is soluble in water and forms a caustic alkali, which is one of the strongest bases known in an aqueous solution.

(Potassium hydroxide is another strong base.) Its solution is soapy to the touch and being basic in nature gives a pink colour with phenolphthalein. Some of the other properties of sodium hydroxide are as follows.

It reacts with acids to form salt and water.

⇒ \(\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}\)

It precipitates hydroxides of metals from metallic salt solutions.

⇒ \(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3+6 \mathrm{NaOH} \rightarrow 2 \mathrm{Al}(\mathrm{OH})_3+3 \mathrm{Na}_2 \mathrm{SO}_4\)

⇒ \(\underset{\text { ferric sulphate }}{\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3}+6 \mathrm{NaOH} \rightarrow \underset{\text { ferric hydroxide(red ppt) }}{2 \mathrm{Fe}(\mathrm{OH})_3}+3 \mathrm{Na}_2 \mathrm{SO}_4\)

⇒ \(\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+6 \mathrm{NaOH} \rightarrow 2 \mathrm{Cr}(\mathrm{OH})_3+3 \mathrm{Na}_2 \mathrm{SO}_4\)

On being heated with ammonium salts, sodium hydroxide liberates ammonia.

⇒ \(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}\)

On treatment with caustic soda, certain metals like aluminium, zinc, tin and silicon liberate hydrogen.

⇒ \(\mathrm{Zn}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\mathrm{H}_2
sodium zincate\)

⇒ \(2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text { sodium aluminate }}{2 \mathrm{NaAlO}_2}+3 \mathrm{H}_2\)

⇒ \(\mathrm{Si}+2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text { sodium silicate }}{\mathrm{Na}_2 \mathrm{SiO}_3}+2 \mathrm{H}_2\)

Uses

1. Sodium hydroxide is used in the manufacture of soaps, paper, viscose rayon (artificial silk) and dyes.
2. It is used in the petroleum industry for refining crude oils. It is also used as a reagent in laboratories.
3. Sodium hydroxide is also employed in the extraction of metals like aluminium.
4. Its aqueous or alcoholic solution is used for carrying out organic reactions.

Sodium chloride

Sodium chloride (NaCl), also called common salt, is found most abundantly in seawater. It is also found in salt wells and in deposits of rock salt. In countries like India, it is obtained mainly by the evaporation of seawater.

Very pure sodium chloride is prepared by crystallisation of a clear, saturated NaCl solution.

It generally contains sodium sulphate (Na₂SO₄), calcium sulphate (CaSO₄), calcium chloride (CaCl₂) and magnesium chloride (MgCl₂) as impurities. Both CaCl₂ and MgCl₂, being deliquescent, absorb moisture from the atmosphere.

To obtain pure NaCl, the crude salt obtained is dissolved in water and then filtered. The insoluble impurities are removed and the filtrate is then saturated with HC₁ gas when pure crystalline sodium chloride separates out (due to the common-ion effect).

Properties

The melting point of sodium chloride is 801°C. It is soluble in water and glycerol, and slightly soluble in alcohol.

Uses

1. Sodium chloride is an essential constituent of our diet and is also used as a food preservative.
2. On being mixed with water, NaCl gives a mixture that has a freezing point about 10°C lower than that of pure water.
3. Sodium chloride mixed with ice (to achieve low temperatures) is used in the traditional method of making ice cream.
4. It is used in the manufacture of caustic soda, sodium peroxide, sodium carbonate and other sodium compounds.
5. NnCl is also employed in the manufacture of soap for salting out, and ion-exchange resins.

Sodium bicarbonate

It is commonly known as baking soda (NaHC03) and is obtained as the by-product of the ammonia-soda process (Solvay process) for the manufacture of sodium carbonate.

Ordinarily, sodium bicarbonate can be produced from sodium carbonate by passing carbon dioxide gas through its solution in water. Sodium bicarbonate, being sparingly soluble, precipitates out in the reaction.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaHCO}_3\)

Properties

Sodium bicarbonate is a white crystalline substance. Its aqueous solution is alkaline in nature (gives a yellow colour with methyl orange and no colour with phenolphthalein). On heating, it decomposes to form sodium carbonate and carbon dioxide is released.

⇒ \(2 \mathrm{NaHCO}_3 \rightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{CO}_2+\mathrm{H}_2\)

When used in baking powder the carbon dioxide gas released on heating leaves holes in cakes making them light and fluffy. NaHC03 reacts with adds with effervescence caused by to evolution of carbon dioxide.

⇒ \(\mathrm{NaHCO}_3+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

Uses

1. It is used in medicine as an antacid. It is largely used in the treatment of acid spillage.
2. Sodium bicarbonate is also used in fire extinguishers. NaHC03 reacts with add to produce C02, which extinguishes fires.
3. It is also employed in the textiles, tanning, paper and ceramic industries.
4. The hydrogen carbonate ion (HCO^2–) has an important biological role as an intermediate between atmospheric CO₂/H₂CO₃ and the carbonate ion (CO^2–). For aquatic organisms, this is the most important and in some cases the only source of carbon.

Alkaline Earth Metals (Group 2 Elements)

The elements of Group 2 of the periodic table are beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba) and radium (Ra).

All these elements are metallic in nature and are commonly known as alkaline earth metals.

The oxides of the metals magnesium, calcium, barium and strontium were known before the corresponding metals and were named alkaline earths since they were found in the earth’s crust and alkaline in nature.

Subsequently, all the corresponding metals were called alkaline earth metals. These metals are silvery-white and lustrous. They are soft but harder than alkali metals. were

Occurrence

Magnesium is the sixth most abundant element in the earth’s crust whereas calcium is the fifth most. Magnesium salts occur in seawater to the extent of 0.13%.

Calcium occurs as sedimentary deposits of CaCO₃. Strontium and barium are much less abundant whereas beryllium is very rare. Radium, being radioactive, is extremely scarce.

Electronic Configuration

Alkaline earth metals have two electrons in the s orbital of their valence shell. Their general valence-shell electronic configuration is ns², where n = 2 to 7.

The electronic configuration of the individual members.

Alkaline earth metals exhibit similar physical and chemical properties. Beryllium, the first member of the group, differs from the rest of the members.

Let us now discuss the trend of variation in atomic and physical properties of the elements of Group 2. Group 2 elements show the same trend in properties as shown by the elements in Group 1.

Atomic And Ionic Radii

The atomic and ionic radii of elements of Group 2 increase from beryllium to radium because the addition of shells more than compensates for the increase in nuclear charge.

However, the atomic and ionic radii of alkaline earth metals are smaller than those of the corresponding alkali metals. This is because, in the same period, due to the extranuclear charge in the alkaline earth metal atom, the orbital electrons are drawn in.

Ionisation Enthalpy

The ionisation enthalpy of elements decreases within the group as the atomic number increases.

This is due to the increase in atomic size and in the magnitude of the screening effect caused by the electrons of the inner shells. Elements of Group 2 have higher values of first ionisation enthalpies than those of the corresponding elements of Group 1.

This is attributed to the smaller size and higher nuclear charge of alkaline earth metals due to which the electrons in their outermost shells are tightly held.

The values of the second ionisation enthalpies are higher than those of the first ionisation enthalpies in the case of alkaline earth metals since greater energy is required to pull out an electron from a positively charged ion than from a neutral atom.

However, the values of the second ionisation enthalpies of the alkaline earth metals are smaller than those of the corresponding alkali metals.

This is because the second electron is removed from a monovalent cation (in the case of an alkaline earth metal) when the atom has yet to acquire the stable noble-gas configuration.

Electropositive character and oxidation state

The electropositive character increases from beryllium to radium as the tendency of atoms to lose electrons increases.

However, alkaline earth metals are less electropositive than the corresponding alkali metals due to their high ionisation enthalpies.

The elements of Group 2 form bivalent cations and are in the bipositive (M2+) oxidation state, compared to the unipositive (M+) oxidation state of Group 1 elements.

Melting and boiling points

Alkaline earth metals have higher melting and boiling points than those of the corresponding alkali metals. This is because of the small atomic size and stronger metallic bonds due to the presence of two valence electrons in alkaline earth metals.

Density

The alkaline earth metals are denser than the corresponding alkali metals. The density decreases from beryllium to calcium and then increases from calcium to radium.

Flame colouration

Like alkali metals, alkaline earth metals also impart characteristic colours to a flame. Calcium, strontium and barium impart brick red, crimson and apple green colours respectively. The electrons absorb energy from the flame and get excited to higher energy levels.

When they drop back to the ground state, they emit this energy in the form of visible light.

Beryllium and magnesium do not give any flame colouration since their atoms are small in size and their electrons are strongly bound to the nucleus. They need a large amount of energy for excitation to higher energy levels and so much energy is not available in a Bunsen flame.

Photoelectric effect

Unlike alkali metals, alkaline earth metals do not exhibit a photoelectric effect. This is because their ionisation enthalpies are relatively high.

Hydration enthalpies

The hydration enthalpies of the cations of Group 2 elements are greater than those of the corresponding cations of Group 1 elements. This is due to the smaller size and increased charge on the cations formed by Group 2 elements.

The hydration enthalpies, decrease down the group as the size of the ions increases. Due to the high hydration enthalpies, the crystalline compounds of Group 2 elements contain more water of crystallisation than those of the corresponding Group 1 elements. For example, NaCl and KC₁ are anhydrous but MgCl^2–6H₂O, CaG^2–6H₂O and BaCl^2–2H₂O have water of crystallisation.

Chemical properties

Alkaline earth elements are less reactive than alkali metals. Some of the important chemical properties of Group 2 elements are discussed below.

Nature of compounds

Like alkali metals, alkaline earth metals predominantly form ionic compounds. However, their salts are less ionic than those of alkali metals.

The tendency of the elements to form ionic compounds increases down the group due to the decrease in ionisation enthalpies.

The first member of the group, beryllium, forms covalent compounds because of its small size and high ionisation enthalpy. Magnesium sometimes exhibits covalency. All other elements form ionic compounds.

Reducing character

Alkaline earth metals have a lower reducing power than alkali metals. This is due to the higher ionisation enthalpies of alkaline earth metals than those of the corresponding alkali metals.

Reaction with oxygen

All the alkaline earth metals bum in oxygen to form oxides MO, except Sr and Ba, which form dioxides or peroxides (MOz) in the presence of an excess of oxygen.

Beryllium is relatively unreactive and bums brilliantly in the powdered form above 873 K.

Magnesium bums with dazzling brilliance in the air to give magnesium oxide and combine with the nitrogen of the air to give magnesium nitride (Mg₃N₂)

⇒ \(\begin{aligned}
2 \mathrm{Be}+\mathrm{O}_2 & \rightarrow 2 \mathrm{BeO} \\
2 \mathrm{Mg}+\mathrm{O}_2 & \rightarrow 2 \mathrm{MgO} \\
2 \mathrm{Ca}+\mathrm{O}_2 & \rightarrow 2 \mathrm{CaO} \\
\mathrm{Sr}+\mathrm{O}_2 & \rightarrow \mathrm{SrO}_2 \\
\mathrm{Ba}+\mathrm{O}_2 & \rightarrow \mathrm{BaO}_2
\end{aligned}\)

The oxides of alkaline earth metals (except that of beryllium, which is covalent) are basic in nature. The reactivity with oxygen increases from beryllium to barium. The oxide of beryllium (BeO) is amphoteric since it reacts with acids as well as alkalis.

⇒ \(\mathrm{BeO}+2 \mathrm{HCl} \rightarrow \mathrm{BeCl}_2+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{BeO}+2 \mathrm{NaOH} \rightarrow \underset{\text { sodium beryllate }}{\mathrm{Na}_2 \mathrm{BeO}_2}+\mathrm{H}_2 \mathrm{O}\)

Reaction With Water

The reactions of alkaline earth metals with water are less vigorous than those of the corresponding alkali metals.

Beryllium does not react with water, and magnesium reacts with steam or boiling water.

\(\mathrm{Mg}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{MgO}+\mathrm{H}_2\)

The elements Ca, Sr and Ba react with cold water, liberating hydrogen and forming metal hydroxides.

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

Thus, the reactivity of water with alkaline earth metals increases as we move down the group.

The hydroxides of alkaline earth metals can also be obtained by dissolving their oxides (except BeO which is amphoteric) in water.

These hydroxides are strong bases. The hydroxides of alkaline earth metals are relatively less basic than those of the corresponding alkali metals.

This is because they have higher ionisation enthalpies and their ions are smaller in size and bipositive in character.

Consequently, the OH- ions are held more firmly by the M2+ ions and the M —OH bond does not break easily.

The basic character of the hydroxides increases from Be to Ba because of the decreasing ionisation enthalpy of the metal atoms.

⇒ \(\underset{\text { amphoteric }}{\mathrm{Be}(\mathrm{OH})_2}<\underset{\text { weak basc }}{\mathrm{Mg}(\mathrm{OH})_2}<\underset{\text { moderately strong hases }}{\mathrm{Ca}(\mathrm{OH})_2} \mathrm{Sr}(\mathrm{OH})_2<\underset{\text { strong base }}{\mathrm{Ba}(\mathrm{OH})_2}\)

These hydroxides are less soluble in water as compared to alkali metal hydroxides. However, they give strongly alkaline solutions because they readily form OH” ions in aqueous solutions. All these bases neutralise and are added to give salts and water.

⇒ \(\mathrm{M}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{MCl}_2+2 \mathrm{H}_2 \mathrm{O}(\mathrm{M}=\mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba})\)

Ben-Ilium hydroxide, being amphoteric, reacts with alkalis as well to form the beryllate ion.

⇒ \(\mathrm{Be}(\mathrm{OH})_2+2 \mathrm{OH}^{-} \rightarrow\left[\mathrm{Be}(\mathrm{OH})_4\right]^{2-}beryllate ion\)

The solubility of the hydroxides in water increases with the increase in atomic number of the metal atom because of a decrease in lattice enthalpy with the increasing size of the metallic ion.

The hydroxides of beryllium and magnesium are almost insoluble, that of calcium is sparingly soluble while those of strontium and barium are increasingly more soluble.

The solubility of the hydroxides of alkaline earth metals depends on their lattice enthalpy and hydration enthalpy.

The lattice enthalpy of a crystal is the heat that would be released per mole if the ions (atoms/molecules) of the crystal were brought together from an infinite distance apart to form the lattice.

If the lattice enthalpy is higher, the ions are held more tightly in a crystal with the result that the solubility of the crystal in water is less.

The lattice enthalpy of the hydroxides of Group 2 elements decreases down the group due to the increase in the size of the cation.

Also, the greater the charge on the ion or the smaller the size of the ion, the greater the lattice enthalpy. If the hydration enthalpy of a salt is greater, the solubility of the salt in water is more.

The resultant of the lattice and the hydration enthalpies, Le., AÿH = A UlticeH – AhydH, becomes more as we move from Be(OH)2 to Ba(OH)2.

This accounts for the increase in solubility as we move down the group. Generally speaking, a hydroxide dissolves if its hydration enthalpy is more than its lattice enthalpy and vice versa.

The action of carbon dioxide on alkaline earth metal hydroxides On passing carbon dioxide gas through the solutions of alkaline earth metal hydroxides the corresponding carbonates are obtained.

⇒ \(\mathrm{M}(\mathrm{OH})_2+\mathrm{CO}_2 \rightarrow \mathrm{MCO}_3+\mathrm{H}_2 \mathrm{O}\)

The carbonates can also be obtained by the addition of sodium or ammonium carbonate to alkaline earth metal salts.

⇒ \(\mathrm{CaCl}_2+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{CaCO}_3+2 \mathrm{NaCl}\)

On being heated, the carbonates of alkaline earth metals decompose to give the corresponding oxides. The temperature of decomposition increases from beryllium to barium.

⇒ \(\mathrm{MCO}_3 \stackrel{\text { heat }}{\longrightarrow} \mathrm{MO}+\mathrm{CO}_2\)

The solubility of the carbonates of alkaline earth metals decreases in the group. This is due to a decrease in hydration enthalpy as the lattice enthalpy remains almost unchanged.

Reaction with hydrogen

On heating, alkaline earth metals, except beryllium, combine with hydrogen to form metal hydrides.

⇒ \(\mathrm{M}+\mathrm{H}_2 \stackrel{\text { neat }}{\longrightarrow} \mathrm{MH}_2\)

Beryllium hydride can, however, be prepared by the reduction of beryllium chloride with lithium aluminium hydride.

⇒ \(2 \mathrm{BeCl}_2+\mathrm{LiAlH}_4 \rightarrow 2 \mathrm{BeH}_2+\mathrm{LiCl}+\mathrm{AlCl}_3\)

Among all the hydrides of Group 2 elements calcium hydride finds uses in the preparation of dry solvents.

Salts of oxoacids —sulphates, nitrates and carbonates

The sulphates of alkaline earth metals are thermally stable. The solubility of the sulphates in water decreases down the group. Thus, BeS04 and MgS04 are readily soluble but CaS04 is sparingly soluble.

The rest are virtually insoluble. The higher solubilities of BeS04 and MgS04 are due to the high enthalpy of the solution of Be²⁺ and Mg²⁺ ions. The sulphates of Be, Mg and Ca have water of crystallisation, viz BeS04-4H20, Mg$04-7H20 and CaS04-2H20.

The nitrates of alkaline earth metals are prepared by dissolving their carbonates in dilute nitric acid. On heating, the nitrates decompose, forming oxides.

⇒ \(2 \mathrm{M}\left(\mathrm{NO}_3\right)_2 \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{MO}+4 \mathrm{NO}_2+\mathrm{O}_2\)

The carbonates of alkaline earth metals are insoluble in water. The metal carbonates are more stable to heat as we move down the group.

The carbonates decompose on heating to give carbon dioxide and the oxide. Beryllium carbonate is unstable and can be kept only in an atmosphere of C02.

Reaction with halogens

On heating with halogens, alkaline earth metals form the corresponding halides.

⇒ \(\mathrm{M}+\mathrm{X}_2 \rightarrow \mathrm{MX}_2\)

The halides can also be obtained by the action of halogen acids on metals, their oxides, hydroxides and carbonates. Beryllium halides cannot be prepared this way due to the formation of the hydrated ion,[Be(H20)4 ]2

⇒ \(\begin{gathered}
\mathrm{M}+2 \mathrm{HX} \rightarrow \mathrm{MX}_2+\mathrm{H}_2 \\
\mathrm{MO}+2 \mathrm{HX} \rightarrow \mathrm{MX}_2+\mathrm{H}_2 \mathrm{O} \\
\mathrm{M}(\mathrm{OH})_2+2 \mathrm{HX} \rightarrow \mathrm{MX}_2+2 \mathrm{H}_2 \mathrm{O} \\
\mathrm{MCO}_3+2 \mathrm{HX} \rightarrow \mathrm{MX}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
\end{gathered}\)

Beryllium chloride is prepared from the oxide by heating with carbon and chlorine.

⇒ \(\mathrm{BeO}+\mathrm{C}+\mathrm{Cl}_2 \stackrel{870-1070 \mathrm{~K}}{=} \mathrm{BeCl}_2+\mathrm{CO}\)

General characteristics

The halides of alkaline earth metals (except those of beryllium) are ionic in nature; the ionic character increases as the size of the metal ion increases.

Beryllium chloride and fluoride, being covalent (due to the smaller size of Be2*), are soluble in organic solvents whereas the halides of other alkaline earth elements are insoluble. The fluorides of all alkaline earth metals except that of Be are almost insoluble in water. Beryllium fluoride is soluble because it has a high hydration energy. Also, the beryllium halides do not conduct electricity.

Beryllium chloride has low melting as compared to the halides of other alkaline earth metals.

The halides of alkaline earth metals (except those of beryllium) dissolve in water giving acidic solutions from which hydrates such as MgCl 2 -6H20, CaCl ^2–6H20 and BaCl ^2-2H20 crystallise out.

The tendency to form hydrated halides decreases with the increasing size of the metal ions structure of BeCl2 In the solid phase, BeCl 2 has the following polymeric chain structure.

However, in the vapour phase, it forms a chloro-bridged dimer, which dissociates into the linear triatomic monomer at high temperatures (approximately 1200 K).

Anomalous behaviour of beryllium

Beryllium is anomalous in many of its properties. It differs much more from the rest of the members of Group 2 than lithium does from the other elements of Group 1.

The main reasons for the anomalous behaviour of beryllium are the small size of the atom, high ionisation enthalpy and the absence of d orbitals in its valence shell. Beryllium differs from the rest of the members of its group in many of its properties, as listed below.

1. Beryllium is harder than the other elements of its group.
2. It has higher melting and boiling points.
3. It does not react with water even on heating, unlike the other members of the group.
4. It does not react with ads to liberate hydrogen, unlike the other members of the group.
5. Beryllium forms covalent compounds while the other members of Group 2 form ionic compounds.
6. Beryllium oxide is amphoteric whereas the oxides of other metals of Group 2 are basic.
7. Beryllium does not combine directly with hydrogen to form the hydride whereas other metals of this group form hydrides directly.
8. Beryllium carbide reacts with water to give methane. In contrast, the other alkaline earth metals give acetylene.

⇒ \(\mathrm{MgC}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Mg}(\mathrm{OH})_2+\mathrm{C}_2 \mathrm{H}_2\)

Diagonal Relationship Of Beryllium With Aluminium

Beryllium, the first element of Group 2, shows similarities in properties with aluminium, the second member of the next higher group, just as lithium shows a diagonal relationship with magnesium. Some of the properties in which beryllium resembles aluminium are as follows.

1. Both beryllium and aluminium ions have comparable ionic radius and charge-to-radius ratio.
2. Both form covalent compounds, which are soluble in organic solvents, e.g., BeCl2 and A1C13.
3. The oxides of both beryllium and aluminium, viz. BeO and Al203, are high-melting, hard solids. Also, the two oxides are amphoteric and they dissolve both in acids and alkalies. With excess alkalies the hydroxides of Be and A1 form beryllate and aluminate ions respectively.
4. Both beryllium and aluminium do not react readily with acids, due to the presence of an oxide on the surface of the metal.
5. In reaction with water, the carbides of both beryllium and aluminium liberate methane.

⇒ \(\begin{gathered}
\mathrm{Be}_2 \mathrm{C}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{BeO}+\mathrm{CH}_4 \\
\mathrm{Al}_4 \mathrm{C}_3+6 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Al}_2 \mathrm{O}_3+3 \mathrm{CH}_4
\end{gathered}\)

The chlorides of both metals have a bridged polymeric structure

Uses

Some of the important uses of alkaline earth metals are as follows.

1. Beryllium finds use in making alloys like copper-beryllium, which is used in the preparation of high strength springs.
2. Beryllium foil is also used in X-ray tubes to filter out visible light and allow only X-rays to pass through.
3. Magnesium is also used in the preparation of alloys. Duralumin—a magnesium-aluminium-copper alloy—is very light and durable and, therefore, used in the manufacture of aeroplanes and automobile parts. Magnalium, an alloy of aluminium and magnesium, is used for making beam balances.
4. Magnesium in the form of powder is used in incendiary bombs and signals and flash powders.
5. Calcium is used in the extraction of metals from their oxides.
6. Both calcium and barium are used to remove air from vacuum tubes because they can react with oxygen and nitrogen at elevated temperatures.
7. Radium, being radioactive, is used in radiotherapy in the treatment of cancer.

Compounds Of Calcium

Some of the commercially important compounds of calcium are calcium oxide, calcium hydroxide, calcium carbonate, calcium sulphate and cement. The process of manufacture, properties and uses of these compounds are discussed here.

Calcium oxide

Calcium oxide (CaO) is a white, amorphous solid with a high melting point—2845 K. It is commonly known as quicklime and is obtained commercially by heating limestone at 1273 K in specially designed lime kilns.

⇒ \(\mathrm{CaCO}_3 \stackrel{1273 \mathrm{~K}}{\rightleftharpoons} \mathrm{CaO}+\mathrm{CO}_2\)

The reaction is reversible. Maximum yields of calcium oxide are obtained if the carbon dioxide is allowed to escape from the kiln. Carbon dioxide escapes above 1100 K.

Properties

1. Quicklime readily absorbs moisture and carbon dioxide. Treatment of quicklime with water produces calcium hydroxide, Ca(OH)2. The reaction is highly exothermic and occurs with a hissing sound.

⇒ \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(\mathrm{aq}) \quad \Delta H=-645 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Calcium hydroxide is also called slaked lime. Calcium oxide is usually obtained in the form of hard lumps.

The lump gets disintegrated by the addition of a limited amount of water, and slaked lime is formed. The process is known as slaking of lime. Quicklime, slaked with soda (aqueous sodium hydroxide) gives a solid, soda lime.

Soda lime is a mixture of sodium hydroxide (NaOH) and calcium hydroxide, Ca(OH)2. It is much easier to handle soda lime than NaOH, which is corrosive.

Calcium oxide reacts with carbon dioxide to give calcium carbonate (CaO + COz ->CaC03). Though calcium carbonate occurs in nature as limestone, marble, etc., pure CaC03 is obtained by this process

Calcium oxide is a basic oxide and reacts with acids to form salts.

⇒ \(\mathrm{CaO}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}\)

Calcium oxide combines with acidic oxides at a high temperature.

⇒ \(\begin{gathered}
\mathrm{CaO}+\mathrm{SiO}_2 \rightarrow \mathrm{CaSiO}_3 \\
6 \mathrm{CaO}+\mathrm{P}_4 \mathrm{O}_{10} \rightarrow 2 \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2
\end{gathered}\)

⇒ \(
\mathrm{CaO}+\mathrm{SO}_2 \rightarrow \mathrm{CaSO}_3
calcium sulphite\)

On heating with coke in an electric furnace at 2273-3273 K, calcium oxide forms calcium carbide.

⇒ \(\mathrm{CaO}+3 \mathrm{C} \stackrel{2273-3273 \mathrm{~K}}{\longrightarrow} \mathrm{CaC}_2+\mathrm{CO}\)

On heating with ammonium salts, calcium oxide liberates ammonia.

⇒ \(\mathrm{CaO}+2 \mathrm{NH}_4 \mathrm{Cl} \rightarrow \mathrm{CaCl}_2+2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}\)

Uses

1. Calcium oxide is used in the manufacture of bleaching powder, slaked lime, calcium carbide, cement, glass, mortar, etc.
2. It is employed in making glass.
3. It is used in steel-making to remove phosphates and silicates as slag.
4. It is used as the basic lining in furnaces.

Calcium hydroxide

Also known as slaked lime, it is obtained by the action of water on calcium oxide.

⇒ \(\mathrm{CaO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2\)

It can also be obtained by the action of caustic alkalis on a soluble calcium salt.

⇒ \(\mathrm{CaCl}_2+2 \mathrm{NaOH} \rightarrow \mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{NaCl}\)

Properties

1. Calcium hydroxide is a white, amorphous solid and is sparingly soluble in water. The aqueous solution is known as limewater.
2. The suspension of calcium hydroxide in water is known as milk of lime. On heating above 700 K calcium hydroxide loses water to give calcium oxide (quicklime).
3. Carbon dioxide reacts with limewater to give a white precipitate of calcium carbonate. This is the reason why carbon dioxide turns limewater milky. Thus, limewater is used to detect the presence of carbon dioxide gas.

⇒ \(\mathrm{Ca}(\mathrm{OH})_2+\mathrm{CO}_2 \rightarrow \mathrm{CaCO}_3 \downarrow+\mathrm{H}_2 \mathrm{O}\)

However, if carbon dioxide is passed for a longer time, the milkiness disappears because the insoluble CaC03 gets converted into soluble calcium hydrogen carbonate, Ca(HC03 )

⇒ \(
\mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \rightarrow \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2
soluble\)

Like C02, sulphur dioxide also reacts with limewater to give a white precipitate—of calcium sulphite.

⇒ \(\mathrm{Ca}(\mathrm{OH})_2+\mathrm{SO}_2 \rightarrow \mathrm{CaSO}_3 \downarrow+\mathrm{H}_2 \mathrm{O}\)

Chlorine reacts with milk of lime below 308 K to give bleaching powder.

⇒ \(
3 \mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{Cl}_2 \stackrel{<308 \mathrm{~K}}{\longrightarrow} \mathrm{Ca}(\mathrm{OCl})_2 \cdot \mathrm{CaCl}_2 \cdot \mathrm{Ca}(\mathrm{OH})_2 \cdot 2 \mathrm{H}_2 \mathrm{O}
bleaching powder\)

Though the formula for bleaching powder is usually written as Ca (OCl2), it is really a mixture.

On heating with ammonium chloride, calcium hydroxide liberates ammonia and with acids, it forms salts.

⇒ \(\begin{gathered}
\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{NH}_4 \mathrm{Cl} \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaCl}_2+2 \mathrm{NH}_3 \uparrow+2 \mathrm{H}_2 \mathrm{O} \\
\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}
\end{gathered}\)

Uses

1. Calcium hydroxide is used for whitewashing and construction purposes.
2. It is used for softening water. As you have already studied in Chapter 9, the temporary hardness of can be removed by adding slaked lime (the carbonates precipitate). This is called lime softening.
3. It is used in the manufacture of bleaching powder, caustic soda and glass.
4. Soda lime (a mixture of calcium hydroxide and caustic soda) is used in the decarboxylation of sodium salts of fatty acids.

⇒ \(\mathrm{CH}_3 \mathrm{COONa} \underset{\text { hest }}{\stackrel{\text { soda lime }}{\longrightarrow}} \mathrm{CH}_4 \uparrow+\mathrm{Na}_2 \mathrm{CO}_3\)

Plaster of Paris

Plaster of Paris is calcium sulphate hemihydrate, i.e., it has one molecule of water for every two calcium and two sulphate ions.

It is obtained by the controlled heating of gypsum at 393 K. If the heating is not controlled, the anhydrous salt is produced instead of the hemihydrate.

⇒ \(\begin{aligned}
& 2\left[\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}\right] \stackrel{\text { heat }}{\longrightarrow}\left(2 \mathrm{CaSO}_4\right) \cdot \mathrm{H}_2 \mathrm{O}+3 \mathrm{H}_2 \mathrm{O} \\
& \text { gypsum } \\
& \text { plaster of Paris } \\
& \text { (calcium sulphate } \\
& \text { hemilhydrate) } \\
&
\end{aligned}\)

Gypsum is a naturally occurring mineral. Calcium sulphate can be obtained by the reaction of any soluble calcium salt either with dilute sulphuric acid or with sodium sulphate and then used to prepare gypsum.

⇒ \(\begin{gathered}
\mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{HCl} \\
\mathrm{CaCl}_2+\mathrm{Na}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{NaCl}
\end{gathered}\)

Properties

1. Plaster of Paris is a white, powdery substance. A paste of plaster of Paris with one-third of its weight of water sets to a hard mass when allowed to stand for about fifteen minutes.

The setting into a hard mass occurs due to the interlocking of crystals of gypsum. This property is made use of in the setting of broken bones and involves the following two stages.

⇒ \(\underset{\text { plaster of Paris }}{\left(2 \mathrm{CaSO}_4\right) \cdot \mathrm{H}_2 \mathrm{O}} \underset{\text { setting }}{\stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \underset{\substack{\text { gypsum } \\ \text { (orthorhombic crystals) }}}{\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}} \stackrel{\text { hardening }}{\longrightarrow} \underset{\substack{\text { gypsum } \\ \text { (monoclinic crystals) }}}{\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}}\)

The rate of setting of plaster of Paris is made faster by sodium chloride and slower by alum or borax. The air addition of alum to the plaster of Paris makes it set into a hard mass. This mixture is known as Keene’s cement.

On being heated above 475 K, plaster of Paris gives anhydrous calcium sulphate, commonly known as dead burnt plaster. It takes up water very slowly and does not set at all.

Uses

1. Plaster of Paris finds use in surgical bandages, dentistry and orthopaedic plaster for setting broken bones.
2. It is also used in making casts for toys and statues.
3. Plaster of Paris is also used in building materials.

Calcium carbonate

Calcium carbonate is a white solid that occurs naturally in two crystalline forms: calcite and aragonite.

These materials make up the bulk of such rocks as marble, limestone and chalk. Calcium carbonate can be prepared by passing a limited amount of carbon dioxide through slaked lime or by the addition of sodium carbonate to calcium chloride.

⇒ \(\begin{gathered}
\mathrm{Ca}(\mathrm{OH})_2+\mathrm{CO}_2 \rightarrow \mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{O} \\
\mathrm{CaCl}_2+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{CaCO}_3+2 \mathrm{NaCl}
\end{gathered}\)

The use of excess carbon dioxide to prepare calcium carbonate results in the formation of calcium bicarbonate.

Properties

1. On heating, calcium carbonate decomposes and carbon dioxide is released

⇒ \(\mathrm{CaCO}_3 \stackrel{1200 \mathrm{~K}}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_2\)

2. It reacts with acids to liberate CO2

⇒ \(\begin{gathered}
\mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \\
\mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2
\end{gathered}\)

Uses

1. Calcium carbonate is used as a building material in the form of marble.
2. It is employed in the manufacture of quicklime and also of sodium carbonate by the Solvay process.
3. With magnesium carbonate, calcium carbonate is used as a flux in the extraction of metals like iron.
4. Calcium carbonate is used in toothpaste, face powders, antacids, adhesives, paints and distempers.

Cement

Cement is one of the most important building materials. When mixed with water and allowed to stand, it sets to a very hard mass, which resembles Portland rock—a natural limestone in the Isle of Portland, England.

Hence, cement was given the name Portland cement in 1824 by Joseph Aspidin, a mason who lived in Leeds, England.

Composition of cement The composition of cement is given in terms of oxides. The average composition is:

• CaO 60-65%
• Fe203 2.5%
• Si02 22-25%
• MgO 2-3%
• A1203 6-8%

A lesser proportion of lime than that given above results in a decrease in the strength of the cement, while a higher proportion causes the cement to crack after setting.

Cement which sets slowly contains an excess of silica while quick-setting cement contains an excess of alumina.

At higher temperatures, the calcium oxide reacts with the aluminosilicates and silicates to form a mixture of various silicates and aluminates of calcium, chiefly tricalcium silicate (Ca3Si05), dicalcium silicate (Ca2Si04) and tricalcium aluminate (Ca3 A1206). Of these, tricalcium silicate is the most important constituent.

Manufacture

The most important raw materials needed for the manufacture of cement are limestone (which provides lime) and clay (which provides silica, along with oxides of aluminium, iron and magnesium). When clay and lime are strongly heated together in a rotatory kiln, they fuse and react to form cement clinker.

The following reactions take place.

⇒ \(\begin{aligned}
2 \mathrm{CaO}+\mathrm{SiO}_2 & \rightarrow 2 \mathrm{CaO} \cdot \mathrm{SiO}_2 \\
3 \mathrm{CaO}+\mathrm{SiO}_2 & \rightarrow 3 \mathrm{CaO} \cdot \mathrm{SiO}_2 \\
3 \mathrm{CaO}+\mathrm{Al}_2 \mathrm{O}_3 & \rightarrow 3 \mathrm{CaO} \cdot \mathrm{Al}_2 \mathrm{O}_3 \\
2 \mathrm{CaO}+\mathrm{Al}_2 \mathrm{O}_3 & \rightarrow 2 \mathrm{CaO} \cdot \mathrm{Al}_2 \mathrm{O}_3 \\
4 \mathrm{CaO}+\mathrm{Al}_2 \mathrm{O}_3+\mathrm{Fe}_2 \mathrm{O}_3 & \rightarrow 4 \mathrm{CaO} \cdot \mathrm{Al}_2 \mathrm{O}_3 \cdot \mathrm{Fe}_2 \mathrm{O}_3
\end{aligned}\)

The mixture (cement clinker) obtained is composed of silicates and aluminates. It is powdered and mixed with gypsum (1.5-2%) (CoS04-2H20) to slow down the process of setting. The final mixture is ground to a fine powder which is grey in colour and filled in airtight bags.

Setting Of Cement

When cement is to be used, it is mixed with water. The cement reacts with water to form a gelatinous mass (the reaction is exothermic), which slowly sets into a hard mass which has —Si—O—Si— and —Si—O—Al— chains.

The setting time depends on the final composition of the cement. Tricalcium silicate sets quickly, within 2-3 days.

Dicalcium silicate sets slowly and develops strength in 3-4 weeks. However, tricalcium aluminate sets instantaneously in the presence of water.

The demand for cement has increased considerably due to the increase in construction. Attempts are being made to find a substitute for cement. When mixed with cement, fly ash, a waste product of the steel industry, reduces the cost of cement without affecting its setting quality.

Uses

A mixture of cement and sand (in the ratio of 1:3) is mixed with water (the required amount) and is used as mortar for the construction and plastering of brick walls.

A mixture of cement, sand and gravel (in the ratio 1: 2: 4) is mixed with water (required amount) and is known as concrete, which is used as a building material.

Concrete used in conjunction with iron frameworks is known as reinforced cement concrete (RCC). It is used for the construction of roofs and pillars, and also for the construction of dams, bridges, etc.

High-alumina cement is obtained by fusing limestone and bauxite with small amounts of SiOz and Ti02 at 1750-1900 K in a rotary kiln.

It is more expensive than the usual Portland cement. However, it has the advantage that it is much quicker and acquires high strength in a short time (24 hours). It is used for making beams for bridges and buildings.

It can withstand temperatures up to 1800 K and is, therefore, used with refractory bricks in furnaces. An additional advantage of high-alumina cement is its resistance to seawater and dilute mineral acids.

Biological Role of Sodium, Potassium, Magnesium and Calcium

A number of elements play a very important role in biological systems. Of the 27 essential elements, 15 are metals.

The metals required in major quantities are Na, K, Mg and Ca. Minor quantities of Mn, Fe, Co, Cu, Zn and Mo, and trace amounts of V, Cr, Sn, Ni and Al are also required in biological systems.

Sodium and potassium ions balance the electrical charges associated with the negatively charged organic macromolecules in the cell. They also help to maintain the osmotic pressure inside the cell in order to keep it turgid and prevent its collapse.

Though sodium and potassium are similar in their chemical properties, their biological functions are quite different. Sodium ions are actively expelled from cells into the extracellular fluid whereas potassium ions are not.

In red blood cells, the ratio of potassium to sodium is 7:1 in human beings, rabbits, rats and horses, and 1:15 in cats and dogs. The difference in the amounts of each ion (Na⁺ and K⁺) in biological components is due to the quantitative difference in the ability of Na⁺ and K⁺ to penetrate cell membranes.

However, the different ratios of Na⁺ and K⁺ inside and outside the cell produce an electrical potential across the cell membrane; this is essential for the functioning of nerve and muscle cells. The movement of glucose into the cells is associated with Na ions. They enter the cell together.

The K+ ions inside the cell are essential for the metabolism of glucose, the synthesis of proteins and the activation of enzymes.

Thus, a sodium-potassium pump operates across the cell membranes, which is fuelled by the hydrolysis of ATP (adenosine triphosphate—the energy-currency molecule of living systems) to ADP (adenosine diphosphate).

Magnesium ions are concentrated inside animal cells and calcium ions are concentrated in the body fluids outside the cell, in the same way as K⁺ concentrates inside the cell and Na+ outside it.

Both Mg²⁺ and Ca²⁺ ions play a vital role in biological systems for the storage of energy. They are also essential for the transmission of impulses along nerve fibres. Magnesium is an important constituent of chlorophyll in green plants.

Calcium is an important constituent of bones and teeth. The amount of calcium in bones is nearly 30 grams at birth and builds up to about 1200 grams in an adult. Hence the individual daily amounts should reach 400 milligrams during the adolescent growth spurt.

Calcium ions are also important in blood clotting and are required to trigger the contraction of muscles and to maintain the regular beating of the heart.

The S-Block Elements Multiple Choice Questions

Question 1. The reducing characteristics of alkali metals follow the order:

1. Na < K < Rb < Cs < Li
2. Li < Cs < Rb < K < Na
3. K < Na < Eb < Li < Cs
4. Rb < Li < Cs < K < N a

Answer: 1. Na < K < Rb < Cs < Li

Question 2. What happens when CO₂ is passed into limewater?

1. The limewater becomes turbid due to the formation of calcium bicarbonate.
2. The limewater becomes turbid due to the formation of calcium carbonate.
3. The turbidity formed disappears on passing C02 for a long time.
4. There is no change.

Answer: 2. The limewater becomes turbid due to the formation of calcium carbonate

Question 3. Which of the following reacts slowly with water?

1. Na
2. Ca
3. Li
4. k

Answer: 3. Ca

Question 4. A salt imparts a brick-red colour to a flame. The salt can be one of

1. Na
2. Ca
3. Mg
4. Li

Answer: 2. Ca

Question 5. The chloride of a metal is soluble in an organic solvent. The chloride can be

1. CaCl₂
2. NaCl
3. MgCl₂
4. BeCl₂

Answer: 4. BeCl₂

Question 6. Alkali metals do not occur in nature because they

1. Have A Small Size
2. Are very reactive
3. Are Monovalent
4. Are Radioactive

Answer: 2. Are very reactive

Question 7. From among the following, name the metal, which and its salts will impart a characteristic blue colour to a Bunsen flame.

1. Na
2. K
3. Rb
4. Cs

Answer: 4. Cs

Question 8. Slaked lime reacts with chlorine to give

1. CaCl₂
2. CaO
3. CaOCl₂
4. None Of These

Answer: 3. CaOCl2

Question 9. Gypsum on heating to 393 K gives

1. CaSO₄-2H₂O
2. CaSO₄
3. CaSO₄.1/2 H₂O
4. None Of These

Answer: 3. CaSO₄.1/2 H₂O

Question 10. The by-product of the Solvay process is

1. CO₂
2. NH₃
3. CaCl₂
4. None Of These

Answer: 3. CaCl₂

Question 11. The thermal stability of alkaline earth metal carbonates decreases in the order

1. BaCO₃ > SrCO₃> MgCO₃> CaCO₃
2. BaCO₃> SrCO₃> CaCO₃> MgCO₃
3. MgCO₃> CaCO₃> SrCO₃> BaCO₃
4. CaCO₃> SrCO₃> MgCO₃> BaCO₃

Answer: 2. BaCO₃> SrCO₃> CaCO₃> MgCO₃

Question 12. The raw materials used in the manufacture of Na2C03 by the Solvay process are

1. Sodium Chloride, Limestone And Ammonia
2. Sodium chloride, limestone and carbon dioxide
3. Sodium chloride and carbon dioxide
4. Limestone and calcium chloride

Answer: 1. Sodium Chloride, Limestone And Ammonia

Question 13. Among the following, the correct order of increasing ionic character is

1. MgCl₂ < BeCI₂ < BaCl₂ < CaCl₂
2. BeCl₂ < MgCl₂ < CaCl₂ < BaCl₂
3. MgCl₂ < BaCl₂ < BeCl₂ < CaCl₂
4. BaCl₂ < CaCl₂ < BeCl₂ < MgCl₂

Answer: 2. BeCl₂ < MgCl₂ < CaCl₂ < BaCl₂

Question 14. In the context of alkali metals, which of the following increases with an increase in atomic number?

1. Solubility of sulphates
2. Solubility of hydroxides
3. Ionisation energy
4. Electronegativity

Answer: 2. Solubility of hydroxides

Question 15. A solution of sodium in liquid ammonia is blue. This blue colour is due to

1. Ammonia
2. Sodium metal
3. Solvated electrons
4. Sodium ion

Answer: 3. Solvated electrons

Question 16. Which of the following gives only the monoxide on heating in an excess of air?

1. Rb
2. K
3. Cs
4. Li

Answer: 4. Li

Question 17. The increasing order of the basic characters of MgO, SrO, Kfi) and Cs20 is

1. CS₂O < K₂O < SrO < MgO
2. MgO < SrO < K₂O < Cs₂0
3. SrO < MgO < CS₂O < K₂O
4. MgO < SrO < Cs₂O < K₂O

Answer: 2. MgO < SrO < K₂O < Cs₂0

Question 18. Which of the following is less stable thermally?

1. LiF
2. CsF
3. NaCl
4. RbF

Answer: 2. CsF

Question 19. Which of the following is most stable to heat?

1. MgCO₃
2. SrCO₃
3. CaCO₃
4. BaCO₃

Answer: 4. BaCO₃

Question 20. An electric potential is produced across the membrane of living cells by the different ratios of certain metal ions inside
and outside cells. The metal ions involved are

1. Ca2⁺ and Na⁺
2. K⁺ and Ba2⁺
3. Na⁺ and k⁺
4. Mg2⁺ and Ca2⁺

Answer: 3. Na⁺ and k⁺

Question 21. When beryllium hydroxide and aluminium hydroxide are dissolved in an excess of an alkali, the ions formed are

1. \(\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-} \text {and }\left[\mathrm{Be}(\mathrm{OH})_4\right]^{-}\)
2. \(\left[\mathrm{Al}^{3+}, \mathrm{Be}^{2+}\right] \text { and } \mathrm{OH}^{-}\)
3. \(\left[\mathrm{Al}(\mathrm{OH})_4\right\}^{2-} \text { and }\left[\mathrm{Be}(\mathrm{OH})_4\right]^{-}\)
4. \(\left[\mathrm{Be}(\mathrm{OH})_4\right]^{2-} \text { and }\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-}\)

Answer: 4. \(\left[\mathrm{Be}(\mathrm{OH})_4\right]^{2-} \text { and }\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-}\)

Hydrocarbons

Hydrocarbons, or compounds which contain only carbon and hydrogen, are the simplest organic compounds, the main types being alkanes, alkenes, alkynes and arenes.

Hydrocarbons are mainly used as fuels for domestic and industrial purposes. Petroleum, kerosene, CNG, LPG, etc., are all mixtures of hydrocarbons.

All other families of organic compounds can be considered to be derived from hydrocarbons by replacing one or more hydrogen atoms with the appropriate functional group.

Hydrocarbons Classification

Hydrocarbons can be classified either on the basis of their structures or on the basis of whether they are saturated or unsaturated. Structurally, they can be of two types—acyclic and cyclic.

Acyclic Hydrocarbons

Hydrocarbons which have open chains of carbon atoms are called acyclic hydrocarbons.

These are also known as aliphatic hydrocarbons. The name aliphatic comes from the Greek word aliphos, meaning fat. Alkanes, alkenes and alkynes are acyclic hydrocarbons.

Alkanes are saturated and have only carbon-carbon single bonds. For example, ethane (CH₃CH₃), propane (CH₃CH₂CH₃) and butane (CH₃CH₂CH₂CH₃).

An alkene is an unsaturated hydrocarbon with at least one carbon-carbon double bond.

Some examples are ethene (CH₂=CH₂), propene (CH₃CH=CH₂) and but-2-ene (CH₃CH=CH—CH₃).

Like alkenes, alkynes too, are unsaturated hydrocarbons, but they have at least one carbon-carbon triple bond. Some examples are ethyne (CH=CH), propyne (CH₃C≡CH) and but-2-yne (CH₃C=CCH₃)

Cyclic Hydrocarbons

Hydrocarbons which contain closed chains or rings of carbon atoms are known as cyclic hydrocarbons. These can be of two types—alicyclic and aromatic.

Alicyclic hydrocarbons

They contain a ring of three or more carbon atoms. They resemble aliphatic hydrocarbons in their properties and are of three types, viz., cycloalkanes, cycloalkenes and cycloalkanes.

Cycloalkanes are saturated hydrocarbons in which all the carbon atoms are joined by single covalent bonds to form a ring. Cyclopropane, cyclobutane, cyclopentane and cyclohexane are a few examples.

Cycloalkenes Unsaturated, alicyclic hydrocarbons containing at least one carbon-carbon double bond are known/ as cycloalkenes.

Some examples of Cyclopropane, cyclobutane, cyclopentane and cyclohexane

Cycloalkynes Unsaturated alicyclic hydrocarbons containing one carbon-carbon triple bond are called cycloalkynes.

Some examples are Cyclopropane, cyclobutane, cyclopentane and cyclohexane are unstable since they are highly strained.

Aromatic Hydrocarbons

These are also called arenes. These compounds were obtained from natural resources like balsam and resins. They all had a pleasant odour and so, obtained their name from the Greek word aroma (a pleasing smell).

However, it was later found that not all aromatic compounds had a pleasant smell.

When subjected to various methods of treatment, the aromatic compounds finally produced benzene or its derivatives.

This led to the conclusion that aromatic compounds are hydrocarbons containing one or more benzene rings. These rings may be fused or isolated.

Arenes are also known as benzenoid compounds since their properties are similar to those of benzene. Some examples of aromatic hydrocarbons are as follows.

The classification discussed above may be represented as follows

As stated earlier, hydrocarbons can also be classified as

1. Saturated And
2. Unsaturated.

Saturated hydrocarbons contain only carbon-carbon single bonds and include alkanes and cycloalkanes.

Unsaturated hydrocarbons contain at least one carbon-carbon double or triple bond. Alkenes, alkynes and arenes fall in this class.

Of these, alkenes and alkynes can be either acyclic or cyclic, while arenes are cyclic.

Here it should be made clear that both aromatic and alicyclic compounds are closed-chain structures but their properties are totally different.

This can be attributed to the benzene-based structure of compounds which contains at least six carbon atoms.

Also, the percentage of carbon is higher in aromatic compounds than in the corresponding aliphatic and alicyclic hydrocarbons. For example,

Alkanes

If we examine the molecular formulae of alkanes which, we will see that the successive members of the family differ constantly by —CH₂.

In other words, if we replace one H atom in methane (CH₄) with CH₃, we get C₂H₆—the next higher member (ethane).

The alkanes, thus, form a homologous series with the general formula C_nH_2n+2, where n is the number of carbon atoms in the molecule.

A series of compounds in which each member differs from the next member by a constant number of carbon and hydrogen atoms is called a homologous series, and the members of the series are called homologues.

Hydrocarbons Conformations

For the sake of convenience, the structures of hydrocarbons are represented in two-dimensional forms or plane-structural formulae as shown below. But you are aware of the fact that most organic molecules have three-dimensional structures.

An important aspect of the three-dimensional shape of molecules is their conformations.

The above structural formulae do not show the spatial arrangement of atoms. This drawback can be overcome by representing the structures in their three-dimensional forms.

[These are discussed under conformations of alkanes. In hydrocarbons, the carbon-carbon single bond is a bond (formed by the overlapping of sp3 hybrid orbitals of carbon atoms) and has cylindrical symmetry.

This leads to the possibility of free rotation about the carbon-carbon single bond and hence, a large number of different spatial arrangements of atoms or groups of atoms attached to the carbon atoms.

These different spatial arrangements are called conformations. Thus, the large number of different arrangements of atoms in a molecule which result from the rotation of carbon-carbon single bonds are called conformers or rotational isomers.

Certain physical properties show that the rotation of carbon atoms around the C—C bond is not quite free. This is due to the difference in the energy of various arrangements. The rotation is hindered by an energy barrier which may vary from 1 to 20 kJ mol-1.

The difference in energy is caused due to the interaction between electron clouds of the carbon-hydrogen bonds. The repulsive interaction between electron clouds is referred to as torsional strain.

The energy required to rotate the molecule about the C—C bond is torsional energy. Torsional strain and torsional energy arise due to the steric hindrance that occurs whenever bulky portions of a molecule repel other molecules or other parts of the same molecule.

The angle of torsion is a dihedral angle between two planes. Actually, the dihedral angle is the one formed by the intersection of two planes. Conformational isomers or conformers are rapidly interconvertible and nonseparable.

X The energy for this interconversion or rotational motion is acquired from the collisions between the molecules and with the walls of the container.

Remember that no bonds are C broken or made during rotational motion or change of conformation, so there is only one compound. The phenomenon of conformation can be best understood by considering the case of ethane.

In the ethane molecule, the two carbon atoms are joined by a bond. If one of the carbon atoms (methyl group) is fixed and the other is rotated about the C —C bond, a large number of arrangements of the hydrogen atoms attached to one carbon atom with respect to the hydrogen atoms of the second carbon atom are possible.

Out of the several arrangements (conformations) possible, only the following two are important. Staggered conformation In this the hydrogen atoms bonded to the two carbon atoms arc staggered with respect to one another.

This implies that the hydrogen atoms of the two methyl groups are a maximum distance apart, making the repulsion between them the minimum possible. Therefore, the conformation is of the lowest potential energy for ethane.

Eclipsed conformation In this the hydrogen atoms bonded to one carbon atom are directly behind those connected to the other, leading to the maximum possible repulsion (maximum energy) between the sets of hydrogen atoms.

Of the two conformations, the staggered conformation is more stable because there is less repulsion between the hydrogen atoms.

All other possibilities between staggered and eclipsed forms are known as skew or gauche forms. Hence the order of stability is staggered > skew > eclipsed.

Conformers are mostly represented in one of two ways. One is called sawhorse projection and the other, is Newman projection.

Sawhorse projection In this method of projecting the three-dimensional structure on paper, the molecule is viewed from above.

The C —C bond is drawn diagonally. The carbon atom shown lower is the one in front and the one drawn above it and to the right represents the carbon atom at the back.

Newman projection Named after M. S. Newman, who first proposed this method of representing the three-dimensional structure on paper, this is easier to visualise than the one described before.

The projection is obtained by viewing the molecule along the C —C bond. The carbon atom closer to the viewer is represented by a circle. The other atom lies behind this and so cannot be seen.

Tire bonds between the first carbon atom and the hydrogen atoms attached to it are shown as equally spaced radial lines from the centre of the circle.

The bonds of the carbon atom which is at the back are shown as lines drawn from the circumference of the circle.

The energy difference between the staggered and eclipsed conformations is about 12.5 kJ mol-1.

This energy difference is not large enough to prevent rotation about the CT bond.

In fact, even at room temperature, the two conformations keep changing from one to the other and it is impossible to isolate either. The variation of energy with rotation about the C —C bond.

Isomerism

There can be only one way in which the carbon and hydrogen atoms combine in the first three members of the alkane homologous series.

But when the carbons and hydrogens in higher members combine, more than one structure can result.

For example, in butane (C₄H₁₀) the carbon atoms can be joined either in a continuous chain or in a branched chain.

Since the two butanes have the same molecular formula but different structures, they exhibit structural isomerism.

u-Butane has a continuous-chain structure and isobutane, has a branched-chain structure. Isomers which differ in the way their carbon chains are arranged are known as chain isomers.

Now consider the next homologue, pentane (C₅H₁₂). The carbon chain in this molecule can be arranged in three different ways.

As you can see, all three structures are chain isomers of pentane. We have already discussed the classification of carbon atoms as primary, secondary, tertiary and quaternary in a molecule. Can you identify the 1°, 2°, 3° and 4° carbon atoms in the above molecular structures?

Example An alkane with the molecular formula C8H18 can have the following structures. Identify which of them represents the same molecule and which is the isomer.

Solution:

Let us first write the IUPAC names for all three structures.

1. 2,4-Dimethylhexane
2. 2,4-Dimethylhexane
3. 2,5-Dimethylhexane

Thus, the IUPAC names suggest that structures represent the same molecule and are the isomers of and.

It is important to note that different isomers are different chemical compounds, and so they have different chemical and physical properties. For example, the b.p. of n-butane is 273 K whereas that of isobutane is 261 K.

We have already discussed the alkyl groups that are attached to carbon atoms in alkanes or other classes of compounds. These substituent alkyl groups have the general formula C_nH_2n+1.

Depending upon which carbon atom (of the alkyl group) is attached to the other group, the alkyl groups can be straight-chain or branched-chain. In other words, different types of alkyl groups result, depending upon which hydrogen is removed from the parent alkane.

For Example, there are four different kinds of butyl groups. Two (n-butyl and sec-butyl) are derived from the l straight-drain butane and two (isobutyl and ferf-butyl) are derived from the branched-chain butane.

The alkyl groups themselves are not stable compounds and are simply parts of molecules that help us name compounds.

The prefixes sec- and tert- refer to the kind of carbon (2° and 3° respectively) of the alkyl group which gets attached to the other group. We have already discussed the nomenclature of alkanes and other hydrocarbons.

The examples given in this chapter in the subsequent sections will help you to further understand the IUPAC 2 nomenclature.

Example Write the structure of 3-lsopropyl-2-methylhexane.
Solution:

1. Draw the carbon chain of the parent alkane (hexane) and number the carbon atoms.

⇒ \(\stackrel{1}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{6}{\mathrm{C}}\)

2. Attach isopropyl at C-3 and a methyl group at C-2

3. Add the requisite number of hydrogen atoms to each carbon of the parent chain.

Thus, we get the correct structure of 3-isopropyl-2-methylhexane.

Example Draw the structural formula of the following compounds:

1. 3,3-Diethyl-5-isopropyl-4-methyl octane
2. 2,2,4-Trimethylpentane

Solution:

1. Draw the carbon chain of the parent alkane (octane) and number the carbon atoms in it

⇒ \(\underset{1}{\mathrm{C}}-\mathrm{C}_2-\mathrm{C}_3-\mathrm{C}_4-\mathrm{C}_5-\underset{6}{\mathrm{C}}-\underset{7}{\mathrm{C}}-{ }_8^{\mathrm{C}}\)

Attach two ethyl groups at C-3, one isopropyl at C-5 and one methyl group at C-4.

Add the requisite number of hydrogen atoms to each carbon of the parent chain.

This is the structure of 3,3-Diethyl-5-isopropyl-4-methyloctane.

In this case, the parent alkane is pentane with three substituent methyl groups attached.

Thus, the structure of the compound will be:

Preparation

The main source of alkanes is petroleum, together with the accompanying natural gas.

Natural gas, of course, contains low molecular weight (volatile) alkanes, i.e., 80 per cent of methane and 10 per cent of ethane, the remaining 10 per cent being a mixture of next higher homologues.

Crude petroleum, on the other hand, is a mixture of higher alkanes. Alkanes can be prepared by the following methods.

From Unsaturated Hydrocarbons

Alkanes are obtained by passing a mixture of unsaturated hydrocarbons (e.g., alkenes and alkynes) and hydrogen over a catalyst.

The catalyst is a finely divided metal, usually platinum, palladium or nickel.

A relatively higher temperature and pressure are required with nickel. The catalyst causes the addition of molecular hydrogen to the unsaturated bond.

⇒ \(\underset{\text { Ethene }}{\mathrm{CH}_2=\mathrm{CH}_2}+\mathrm{H}_2 \stackrel{\mathrm{Pt} / \mathrm{Pd} / \mathrm{Ni}}{\longrightarrow} \underset{\text { Ethane }}{\mathrm{CH}_3-\mathrm{CH}_3}\)

⇒ \(\underset{\text { Propene }}{\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2}+\mathrm{H}_2 \stackrel{\mathrm{Pt} / \mathrm{Pd} / \mathrm{Ni}}{\longrightarrow} \underset{\text { Propane }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3}\)

⇒ \(\underset{\text { Propyne }}{\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}}+2 \mathrm{H}_2 \stackrel{\mathrm{Pt} / \mathrm{Pd} / \mathrm{Ni}}{\longrightarrow} \underset{\text { Propane }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3}\)

From alkyl halides

Reduction of alkyl halides (except fluorides) with zinc and dilute acid (viz., hydrochloric acid) leads to the replacement of a halogen atom by a hydrogen atom.

Alkyl halides, on treatment with sodium metal in dry ether, give higher, symmetrical alkanes

⇒ \(\underset{\text { Bromomethane }}{\mathrm{CH}_3 \mathrm{Br}}+2 \mathrm{Na}+\mathrm{CH}_3 \mathrm{Br} \stackrel{\text { dryether }}{\longrightarrow} \underset{\text { Ethane }}{\mathrm{CH}_3-\mathrm{CH}_3+2 \mathrm{NaBr}}\)

This reaction is called the Wurtz reaction. If two different alkyl halides are used in this reaction (e.g., CH₃Br and C₂H₅Br), a mixture of alkanes (CH₃CH₃C₂H₅C₂H₅, CH₃C₂H₅) is obtained.

Decarboxylation of carboxylic acids

Sodium salts of carboxylic acids, on being heated strongly with soda lime (NaOH + CaO), give alkanes.

Since the carbon dioxide is removed, the corresponding alkane obtained has one carbon atom less than the carboxylate salt.

⇒ \(\underset{\text { Sodium acetate }}{\mathrm{CH}_3 \mathrm{COONa}}+\mathrm{NaOH} \underset{\text { heat }}{\stackrel{\mathrm{CaO}}{\longrightarrow}} \mathrm{CH}_4+\mathrm{Na}_2 \mathrm{CO}_3\)

Such a reaction in which carbon dioxide is eliminated from a carboxylic group is called decarboxylation.

Kolbe’s electrolytic method

This method involves the electrolysis of an aqueous solution of the sodium or potassium salt of a carboxylic acid.

A higher alkane containing an even number of carbon atoms is obtained.

⇒ \(\underset{\text { Sodium acrtate }}{2 \mathrm{CH}_3 \mathrm{COONa}}+2 \mathrm{H}_2 \mathrm{O} \stackrel{\text { dectrolysis }}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_3+2 \mathrm{CO}_2+2 \mathrm{NaOH}+\mathrm{H}_2\)

The reaction proceeds as follows.

⇒ \(2 \mathrm{CH}_3 \mathrm{COO} \mathrm{Na}^{+} \stackrel{\text { electrolysis }}{\rightleftharpoons} 2 \mathrm{CH}_3-\stackrel{\mathrm{O}}{\rightleftharpoons}-\overline{\mathrm{O}}+2 \mathrm{Na}^{+}\)

At anode:

At cathode:

⇒ \(2 \mathrm{Na}^{+} \stackrel{+2 \mathrm{e}}{\longrightarrow} 2 \mathrm{Na} \stackrel{2 \mathrm{H}_2 \mathrm{O}}{\longrightarrow} 2 \mathrm{NaOH}+\mathrm{H}_2 \uparrow\)

Since the two free radicals produced as a result of the decarboxylation of the acetate radical combine to form alkanes, methane, which contains only one carbon atom, cannot be prepared by this method.

Physical Properties

Alkanes, being covalent in nature, are held together by weak van der Waals forces. The intermolecular forces are strong in the large molecules.

Thus, the first four members of the alkane homologous series are colourless gases, and the next thirteen members from C-5 to C-17 are colourless liquids. The higher members, from C-18 onwards, are waxy solids.

The alkane molecule is nonpolar or very weakly polar due to the very low electronegativity difference between carbon and the hydrogen atom.

Being nonpolar, alkanes are insoluble in water, which is polar. However, they are soluble in nonpolar solvents like carbon tetrachloride and benzene.

The boiling point of n-alkanes increases with an increase in molecular weight.

In the case of alkanes which exist in straight-chain as well as branched-chain forms, the branched-chain isomer has a lower boiling point than the corresponding straight-chain isomer.

Thus, n-butane has a higher boiling point than that of isobutane. The boiling point is still lower if the branching is more.

Thus, neopentane boils at a temperature lower than the boiling point of isopentane.

Why does the increase in branching of the carbon chain lead to the lowering of the boiling point of the alkane?

Because branching in a carbon chain reduces the surface area of the molecule and therefore brings down intermolecular attractive forces, resulting in a decrease in boiling point.

The melting points of alkanes also increase with the increase in molecular weight but the increase is not so regular.

This is because the intermolecular forces in a crystal depend not only on the size of the molecule but also on how it fits within a crystal lattice.

However, alkanes with an even number of carbon atoms have a higher melting point than those with an odd number of carbon atoms.

This is largely due to the better crystal lattice packing of the alkanes with an even number of carbon atoms.

Chemical Properties

Alkanes are inert in nature and do not undergo any chemical reactions under ordinary conditions with adds, alkalis, oxidising and reducing agents.

However, under certain conditions, alkanes undergo substitution reactions and some others like halogenation, combustion and isomerisation.

The nonreactivity of alkanes with ionic reagents can be attributed to the nonpolar nature of alkanes. We will now discuss some of the reactions of alkanes.

Halogenation

Alkanes react with halogens at 523-673 K or in the presence of sunlight to form halogen-substituted alkanes and hydrogen halide.

Reactions in which the hydrogen atoms of alkanes are substituted by other groups are known as substitution reactions.

The higher alkanes, apart from halogenation, also undergo nitration and sulphonation substitution reactions.

Methane reacts with chlorine to form chloromethane, which, on further reaction with chlorine, forms dichloromethane.

⇒ \(\mathrm{CH}_4+\mathrm{Cl}_2 \longrightarrow \underset{\text { Chloromethane }}{\mathrm{CH}_3 \mathrm{Cl}}+\mathrm{HCl}\)

⇒ \(\mathrm{CH}_3 \mathrm{Cl}+\mathrm{Cl}_2 \longrightarrow \underset{\text { Dichloromethane }}{\mathrm{CH}_2 \mathrm{Cl}_2}+\mathrm{HCl}\)

When the chlorination is continued further, trichloromethane (chloroform) and tetrachloromethane (carbon tetrachloride) are formed.

⇒ \(\mathrm{CH}_2 \mathrm{Cl}_2+\mathrm{Cl}_2 \longrightarrow \underset{{\text { Trichloromethane } \\ \text { (chloroform) }}}{\mathrm{CHCl}_3}+\mathrm{HCl}\)

⇒ \(\mathrm{CHCl}_3+\mathrm{Cl}_2 \longrightarrow \underset{{\text { Tetrachlomomethane } \\ \text { (earbon tetrachloride) }}}{\mathrm{CCl}_4}+\mathrm{HCl}\)

Thus, the chlorination of methane gives a mixture of products. The composition of the product mixture depends on the ratio in which the reactants (methane and chlorine) are used.

If chlorine is used in excess then the percentage of carbon tetrachloride obtained is more. The mixture obtained can be separated into its constituents by fractional distillation.

The chlorination of methane is an example of a chain reaction, a reaction that involves a series of steps. Each step in the chain reaction generates a reactive substance that brings about the next step.

The reactivity of bromine with alkanes is less than that of chlorine. Unlike chlorination and bromination of alkanes, the iodination of alkanes is a reversible process.

However, the reaction can be made to proceed in one direction in the presence of an oxidising agent, which decomposes the hydrogen iodide produced.

⇒ \(\mathrm{CH}_4+\mathrm{I}_2 \rightleftharpoons \mathrm{CH}_3 \mathrm{I}+\mathrm{HI}\)

⇒ \(5 \mathrm{HI}+\underset{\text { lodic acid }}{\mathrm{HIO}_3} \longrightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}\)

The fluorination of alkanes is explosive and even in the dark the reaction needs to be controlled by mixing the reactants in an inert solvent.

The reactivity of alkanes with halogens is of the order: F2 > Cl2 > Br2 > I2. The ease of abstraction of hydrogen atoms follows the sequence 3°> 2°> 1°> CH₃ —H.

Mechanism Let us discuss the mechanism of the chlorination of methane as an example.

The halogenation of other alkanes proceeds according to the same mechanism as that of methane.

The chlorination of methane is believed to proceed via a free-radical mechanism involving three steps—initiation, propagation and termination.

Initiation The reaction is initiated by the homolysis of the chlorine molecule to form free radicals. The Cl—Cl bond is weaker than the C —C and C—H bonds and breaks up easily in the presence of sunlight or heat.

Propagation The chlorine free radical attacks the C—H bond of the methane molecule to generate a methyl radical along with the formation of a molecule of hydrogen chloride.

The methyl free radical thus produced attacks the second molecule of chlorine to yield methyl chloride and another chlorine free radical.

Each of the two chain-propagation steps 1 and 2 consumes a reactive substance and generates another.

Termination Finally, when the chain reaction terminates, the reactive intermediates are not generated but consumed.

The two reactive intermediates like two chlorine radicals or two methyl radicals or one methyl and one chlorine radical combine.

⇒ \(\begin{aligned}
& \mathrm{Cl} \cdot+\mathrm{Cl} \cdot \longrightarrow \mathrm{Cl}_2 \\
& \mathrm{CH}_3+\mathrm{CH}_3+\mathrm{CH}_3 \mathrm{CH}_3 \\
& \mathrm{CH}_3+\mathrm{Cl} \cdot \longrightarrow \mathrm{CH}_3 \mathrm{Cl}
\end{aligned}\)

Oxidation

Combustion On being heated, alkanes completely oxidise in the presence of air or oxygen, producing C02, H2O and a large amount of heat.

⇒ \(\mathrm{CH}_4(\mathrm{~g})+(\text { excess }) 2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \Delta_c \mathrm{H}^{\ominus}=-890 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The more the molecular weight of the alkane, the more is the heat produced on combustion. Alkanes (methane, ethane, propane, butane, etc.) are used as fuel, e.g., LPG.

In an insufficient amount of air, alkanes undergo incomplete combustion, producing carbon in a very finely divided state —carbon black.

Carbon black is used to make paints and printers’ ink. It is also used in the rubber industry to make tyres.

⇒ \(\mathrm{CH}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

Controlled oxidation The controlled oxidation, i.e., oxidation in a regulated supply of air at high temperature and pressure, under various conditions, of alkanes leads to different products. Some examples are as follows.

⇒ \(2 \mathrm{CH}_4+\mathrm{O}_2 \underset{\text { heat }}{\stackrel{\mathrm{Cu} .523 \mathrm{~K}, 1000 \mathrm{~atm}}{\longrightarrow}} \underset{\text { Methanol }}{2 \mathrm{CH}_3 \mathrm{OH}}\)

⇒ \(\mathrm{CH}_4+\mathrm{O}_2 \underset{\text { heat }}{\stackrel{\mathrm{Mo}_2 \mathrm{O}_3}{\longrightarrow}} \underset{\text { Methanal }}{\mathrm{HCHO}}+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_3+3 \mathrm{O}_2 \stackrel{\left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \mathrm{Mn}}{\longrightarrow} \underset{\text { heat }}{2} \underset{\text { Ethanoic adid }}{2 \mathrm{CH}_3 \mathrm{COOH}}+2 \mathrm{H}_2 \mathrm{O}\)

Normally, alkanes are resistant to strong oxidising agents but alkanes with a tertiary hydrogen atom can be oxidised to the corresponding alcohol by potassium permanganate.

⇒ \(\underset{\text { 2-Methylpropane }}{\left(\mathrm{CH}_3\right)_3 \mathrm{CH}}+[\mathrm{O}] \stackrel{\mathrm{KMnO}_4}{\longrightarrow} \underset{\text { 2-Methyl-propan-2-ol }}{\left(\mathrm{CH}_3\right)_3 \mathrm{COH}}\)

Isomerisation On heating in the presence of aluminium chloride and hydrogen chloride, normal alkanes get converted to their corresponding branched-chain isomers. For example,

The process is used to increase the branched-chain content of lower alkanes produced by cracking since branched-chain isomers in petrol are more valuable than n-alkanes.

Aromatisation Alkanes containing six to ten carbon atoms can be converted into benzene and its homologues at a high temperature in the presence of a catalyst.

The process is known as aromatisation and takes place through the simultaneous dehydrogenation and cyclisation of the alkane to give an aromatic compound containing a number of carbon atoms.

Reaction with steam Methane reacts with steam at 1273 K in the presence of nickel, which acts as a catalyst, to give carbon monoxide and dihydrogen. This mixture of carbon monoxide and hydrogen is known as water gas.

⇒ \(\mathrm{CH}_4+\mathrm{H}_2 \mathrm{O} \underset{\text { heat }}{\stackrel{\mathrm{Ni}}{\longrightarrow}} \mathrm{CO}+3 \mathrm{H}_2\)

This procedure is used for the industrial preparation of dihydrogen

Pyrolysis (Cracking)

When heated to high temperatures (773 K-873 K), alkanes decompose into smaller molecules.

This is known as cracking. The thermal decomposition of organic compounds is known as pyrolysis.

The products obtained from the cracking of alkanes depend on the structure of the alkane, pressure conditions and the catalyst present.

For example, dodecane, a constituent of kerosene oil, when heated to 973 K, in the presence of catalysts platinum palladium or nickel, gives a mixture of products (mainly heptane and pentene)

⇒ \(\underset{\text { Dodecane }}{\mathrm{C}_{12} \mathrm{H}_{26}} \underset{\mathrm{P} / \mathrm{Pd} / \mathrm{Ni}}{\stackrel{973 \mathrm{~K}}{\longrightarrow}} \mathrm{C}_7 \mathrm{H}_{16}+\mathrm{C}_5 \mathrm{H}_{10}+\text { other products }\)

Alkenes

Alkenes are unsaturated hydrocarbons with at least one carbon-carbon double bond. Their general formula is CnH2n. They are also called olefins (Greek: oil forming) since the lower members of this series produce oily products in reaction with halogens.

Structure Of Double Bond

The presence of the double bond in alkenes gives rise to certain special structural features which can be better understood if we consider the structure of the simplest alkene, ethene (ethylene).

Each carbon atom in ethene is sp² hybridised, which means that each carbon atom has three sp² orbitals lying in the same plane and inclined to one another at an angle of 120°, and an unhybridised 2p² orbital which is perpendicular to the plane of the sp² orbitals.

Two of the three sp² hybrid orbitals of each carbon atom overlap with the Is orbitals of two hydrogen atoms (forming C—H bonds).

The remaining sp² orbital of each carbon atom overlaps along the internuclear axis to form a (C —C) bond.

The unhybridised orbitals of the two carbon atoms (dash lines), which are perpendicular to the plane of the sp² orbitals, overlap sideways to form a π bond.

Thus, the C—C double bond consists of an o bond and an π bond. The n bond is weaker and makes alkenes reactive.

What makes TC a weaker bond? The n electrons are less involved in holding together the two carbon nuclei than the CT electrons are.

They form a cloud of π electrons above and below the plane of the atoms

During a reaction, these loosely held n electrons are particularly available to a reagent that is seeking electrons.

The electron-deficient reagents or compounds which are seeking a pair of electrons are called electrophilic reagents.

Electrophilic addition is the typical reaction of an alkene. The C —C bond length in ethene is 134 pm, which is less than that in ethane (154 pm). The C—C bond length in ethyne is still shorter (120 pm).

However the bond enthalpy of C=C (ethene) is greater (678 kJ mol-1) than that of C—C (ethane), which is 370 kJ mol-1.

Hence, the alkene molecule is more reactive than its corresponding alkane molecule, due to the presence of the double bond.

Example Calculate the number of it bonds in the following structures and write their IUPAC names

Solution:

1. The structure can be drawn as:

The IUPAC name is 2,5-Dimethylhex-2-ene.

2. The structure can be drawn as:

The IUPAC name of the structure is 7-Methyl-l,3,5-octatriene.

Isomerism

Alkenes normally show the following types of isomerism.

Position isomerism The first two members of the alkene series (viz., ethene, propene) do not show isomerism.

However, butene exhibits position isomerism as but-l-ene and but-2-ene. They differ in the position of the double bond.

⇒ \(\mathrm{H}_3 \mathrm{CCH}_2 \mathrm{CH}=\mathrm{CH}_2
But-1ene\)

⇒ \(\mathrm{H}_3 \mathrm{CCH}=\mathrm{CHCH}_3
But-2-ne\)

Chain isomerism Butene also shows chain isomerism as isobutene has a branched-chain structure and n-butene has a straight-chain structure

Geometrical isomerism Compounds which have the same structural formula, but in which atoms or groups have a different spatial arrangement about a double bond are called geometrical isomers and the phenomenon is known as geometrical isomerism.

Geometrical isomerism is exhibited by alkenes and their derivatives in which the double-bonded carbon atoms are linked to two different groups. In other words, all compounds containing a carbon-carbon double bond do not exhibit geometrical isomerism.

An alkene molecule exhibits geometrical isomerism only when the atoms or groups attached to each double-bonded carbon atom are different.

That is to say, geometrical isomerism is exhibited by molecules of the kind ABC=CAB or of the kind ABC=CDE but not of the kind AAC=CDE or of the kind ABC=CDD.

You have already studied the infinite number of conformational isomers which arise due to the rotation of carbon atoms connected through bonds in alkanes.

However, this is not the case in alkenes which have a double bond containing a cr bond and π bond. Rotation is possible in the case of a bond but not at all in that of a n bond.

The reason for this simply lies in the formation of π bonds. An π bond is formed by the overlap of two p orbitals.

The p orbitals lie perpendicular (above and below) to the plane of A orbitals(sp2 hybrid orbitals in this case) forming a CT bond.

Conversion of 1 into 2 is possible only when the molecule is twisted, which would definitely break the 71 bond. The n bond is formed by the sideways overlapping of p orbitals, the groups or atoms attached to the sp2 hybridised carbon atom cannot rotate about it.

Rotation is therefore possible by breaking the n bond, which would require energy of the order of 251 kJ mol-1.

This energy is not available at room temperature. Because of the energy barrier, there is hindered rotation about the double bond, which gives rise to geometric isomers.

Using a ball and stick model, geometrical isomers can be represented

Isomers in which similar atoms or groups lie on the same side of the double bond are called cis-isomers and those in which similar atoms or groups lie on opposite sides of the double bond are called trans-isomers.

The trans-isomers are usually more stable than the corresponding cis isomers. This will become clear if we consider cis- and trans-isomers with the formula ABC=CAB, in which A is the bulkier of the two groups A

And B (e.g., CH₃ —CH=CH—CH₃). In the cis-isomer, the two bulky A groups are very close to each other. The repulsion between the overlapping electron clouds of the A groups makes this isomer less stable than the trans isomer, in which the A groups are far apart.

There is no absolute method for determining whether a particular isomer is a cis- or trans-isomer.

However, the following characteristics are useful.

1. The cis-isomer is more polar than the trans-isomer. In other words, the cis-isomer has a greater dipole moment than the trans-isomer.

This will become clear if we consider the cis- and trans-isomers of 1,2-dichloroethene.

The trans-isomer has a small dipole moment. This is because the dipole moments of the two C—Cl bonds are opposed due to the symmetry of the molecule. The cis-isomer, being nonsymmetrical, has a larger dipole moment.

2. The cis-isomer has a lower melting point than the corresponding trans-isomer. For example, the m.p. of maleic acid, cis-HOOCCH=CHCOOH, is 130°C, while that of fumaric add trans-HOOCCH=CHCOOH, is 286º0.

3. The cis-isomer has a higher boiling point than the corresponding trans-isomer. For example, the b.p. of cis-2-butene is 4°C and that of trans-2-butene is 1ºC.

4. If the cis- and trans-isomers of an alkene have a COOH group then the cis compound forms an anhydride on heating while the trans compound does not. To give an example, this is how we distinguish between maleic acid and fumaric acid.

Example Draw cis and trans isomers of those which are capable of exhibiting cis-trans isomerism, of the following compounds.

1. Hex-l-ene
2. Hex-2-ene
3. Hex-3-ene

Solution: Hex-2-ene and Hex-3-ene will exhibit cis-trans isomerism.

Preparation

Alkenes are generally obtained from saturated compounds by the elimination of atoms or groups from the two adjacent carbon atoms.

Some common methods of preparation are discussed in the following.

Dehydration Of Alcohols

The elimination of a water molecule from a compound is called dehydration. On being heated with concentrated sulphuric acid or phosphoric acid, alcohols form the corresponding alkenes by the elimination of a water molecule.

This reaction is since the hydrogen attached to the p-carbon atom is removed. The carbon atom attached to the functional group is and the one next to it is p.

⇒ \(\underset{\text { Ethanol }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}} \underset{-43-453 \mathrm{~K}}{\stackrel{\operatorname{conc} \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}} \underset{\text { Ethere }}{\mathrm{CH}_2}=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\underset{\text { Propanol }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}} \underset{43-453 \mathrm{~K}}{\stackrel{\text { Ponc } \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}} \underset{\text { Prop-1 -ene }}{\mathrm{CH}_3 \mathrm{CH}}=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}\)

The mechanism of the dehydration of alcohols by sulphuric arid is as follows.

The order of reactivity with respect to dehydration is 3° > 2° > 1°. Tertiary alcohols undergo rapid dehydration as they form the most stable carbocations (3°), which in turn form the most stable alkenes.

In the case of secondary or tertiary alcohols, the elimination of the water molecule gives rise to more than one alkene.

For example, but-2-ol on dehydration gives but-2-ene as the major product and but-l-ene as the minor product. As you can see, there are two p carbons in the structure of butane-2-ol.

The course of the reaction in such cases is determined by the Sayiezeff rule, according to which hydrogen is abstracted preferentially from the carbon atom attached to fewer hydrogen atoms.

Dehydrohalogenation of alkyl halides

The elimination of a molecule of hydrogen halide (H—X) from an alkyl halide in the presence of alcoholic potash forms an alkene.

This is dehydrohalogenation; it involves the elimination of one halogen atom and one hydrogen atom from the adjacent carbon atoms of a molecule of an alkyl halide.

This reaction is called p elimination. Ethyl bromide, on being heated with alcoholic potassium hydroxide, undergoes dehydrohalogenation.

⇒ \(\underset{\text { Ethyl bromide }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}} \stackrel{\text { KOH, } \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}{\longrightarrow} \underset{\text { Ethene }}{\mathrm{CH}_2}=\mathrm{CH}_2+\mathrm{KBr}+\mathrm{H}_2 \mathrm{O}\)

The mechanism of dehydrohalogenation of alkyl halides is shown below:

B’ is a base like C₂H₂O^–.

Depending on the structure of the starting alkyl halide, more than one product can be obtained in this case.

For example, 1-chlorobutane yields one product whereas 2-chlorobutane or sec-butyl chloride yields two products.

The ease of elimination of halogen atoms is of the order RF < RC1 < RBr < RI.

Dehalogenation of vicinal dihalides

Vicinal dihalides are those alkyl halides in which halogen atoms are present on adjacent carbon atoms. When treated with zinc, a vicinal dihalide gives an alkene.

Partial reduction of alkynes

On partial reduction with hydrogen in the presence of a catalyst (Lindlar’s catalyst), alkynes give alkenes.

A specially prepared palladium (palladised charcoal partially deactivated with poisons like quinoline) is Lindlar’s catalyst.

Unless the triple bond in the alkyne undergoing reduction is at the end of a chain, the probability of obtaining either cis- or a trans-alkene cannot be ruled out.

The choice of reducing agent decides the isomer which predominates. With Lindlar’s catalyst cis-alkenes are obtained. However, on reduction with sodium in liquid ammonia, alkynes form trans-alkenes.

Electrolysis of salts of carboxylic acids

Alkenes are obtained by the electrolysis of an aqueous solution of potassium salts of dicarboxylic acid.

Thus potassium succinate on electrolysis gives a carboxylate anion and a potassium cation. Carbon dioxide is released at the anode and hydrogen at the cathode.

At Anode: 

At Cathode: 

⇒ \(\begin{aligned}
& 2 \mathrm{~K}^{+}+2 \mathrm{e} \longrightarrow 2 \mathrm{~K} \\
& 2 \mathrm{~K}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{KOH}+\mathrm{H}_2 \uparrow
\end{aligned}\)

Cracking or pyrolysis of alkanes.

On being heated to 773-873 K or in the presence of a catalyst to 498 K, higher alkanes yield alkenes.

Physical Properties

The physical properties of alkenes are similar to those of alkanes. The lower members of this class (up to C₄) are colourless gases, the higher members (C₆-C₁₈) are liquids and the rest are solids at room temperature.

They are less dense than water and insoluble in it but soluble in common organic solvents like benzene and chloroform.

The boiling points of alkenes, as is the case with alkanes, rise with the increase in the number of carbon atoms.

The boiling point increases by 20°C to 30°C for each carbon atom added to the molecule. Also, branching in an alkene lowers the boiling point.

Chemical properties

We know that the C=C double bond in alkenes consists of a bond and a n bond. Alkenes owe their reactivity to the weaker T₂ bonds. They usually undergo addition reactions, the most important being electrophilic and free radical additions.

As discussed earlier in the chapter, the addition of hydrogen to alkenes, in the presence of a catalyst, yields the corresponding alkanes.

Addition of halogens

Alkenes on treatment with halogens yield 1,2-halogenated alkanes (adduct). Alkenes react readily with chlorine and bromine at room temperature without any exposure to ultraviolet light.

The reaction is carried out in an inert solvent like carbon tetrachloride. The reaction of an alkene with bromine is used as a test for unsaturation.

A solution of bromine in carbon tetrachloride is red, which slowly decolourises as a colourless vicinal dihalide is formed.

The reaction is an example of electrophilic addition and involves two steps. In the first step, the electrophile (a positively charged species) adds to the carbon-carbon double bond, resulting in the formation of a carbocation intermediate.

The second step involves the addition of the nucleophile (a negatively charged species) to the positively charged carbon forming the adduct.

⇒ \(\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_2-\mathrm{CH}_2-\mathrm{Br}+\mathrm{Br}^{\ominus} \longrightarrow \underset{\text { 1,2-Dibromoethane }}{\mathrm{CH}}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br}\)

Addition of hydrogen halides

On reaction with hydrogen halides, alkenes give the corresponding alkyl halides.

In the case of symmetrical alkenes (like ethene), only one product is obtained

⇒ \(\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{HBr} \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}
Ethene   Ethyl bromide\)

However, in the case of unsymmetrical alkenes like propene, two products may be formed.

The above reaction shows that out of two possible products, only one product is formed.

Such reactions are called regiospecific reactions. The exclusive formation of one product can be explained by the Markovtiikov rule.

According to this rule (formulated by the Russian chemist Vladimir Markovnikov in 1870), a molecule of a hydrogen halide adds to an unsymmetrical alkene in such a way that the negative part of the reagent (X₆-) goes to that carbon atom of the alkene which is bonded to fewer hydrogen atoms. The Markovnikov rule was explained on the basis of the relative stability of carbocations.

Thus, in the above example, when hydrogen bromide is added to propene there is a possibility of formation of two carbocations — primary and secondary.

However, the secondary carbocation is more stable than the primary one. Therefore, the actual product formed is 2-bromopropane.

Sometimes, it has been seen that the addition of hydrogen bromide to unsymmetrical alkenes is in contradiction to the Markovnikov rule.

Kharasch and Mayo (1933) first observed and found the cause for such addition.

They observed the anti-Markovnikov addition of hydrogen bromide to propene in the presence of benzyl peroxide. This is also known as the Kharasch effect or peroxide effect.

You have already studied that peroxides are compounds with —O—O— linkage. Certain organic peroxides like benzoyl peroxide (C₆H₅COO)² are synthesised and used as reagents.

It is believed that the peroxide dissociates to give two alkoxy free radicals, which attack hydrogen bromide to form a bromine free radical.

The bromine free radical generated can add to any of the two carbon atoms of the double bond in an unsymmetrical alkene giving rise to either a primary or a secondary free radical.

The order of stability of free radicals is tertiary > secondary > primary. Therefore, in the presence of a peroxide the secondary free radical is predominantly formed and reacts with HBr to form an anti-Markovnikov product.

HC1 and HI give the normal Markovnikov addition products even in the presence of peroxides.

The chances of the homolytic cleavage of the H—Cl bond are lower than that of the H—Br bond because it is stronger than the H —Br bond.

The iodine free radical is easily formed since the H—I bond is weak (297 kJ mol-1) and combines with another iodine free radical to form an iodine molecule instead of the relatively unstable carbon-iodine bond.

Addition of sulphuric acid

An alkene reacts with cold concentrated sulphuric acid to form alkyl hydrogen sulphate, which on heating with water gives the corresponding alcohol. The method is used to manufacture ethyl alcohol.

In the case of unsymmetrical alkenes, the addition of sulphuric acid is consistent with the Markovnikov rule. Like the addition of alkyl halides, this reaction is also an example of electrophilic addition.

Addition of water

Water adds to the more reactive alkene only in the presence of an acid catalyst (H3P04) to form alcohols. The addition follows the Markovnikov rule.

Oxidation

Alkenes are more readily oxidised than alkanes are. This is due to the presence of a double bond, which is the site of attack. The nature of the oxidation products depends on the oxidising agent employed.

1. Hydroxylation On treatment with cold aqueous dilute potassium permanganate solution, a 1,2-dihydroxy compound (diol) is obtained. The reaction is known as hydroxylation of alkenes.

2. Oxidative cleavage of alkenes Oxidation by sodium periodate (NalO₄) in the presence of a permanganate solution yields a carboxylic add or a ketone. In the reactions that follow the double bond undergoes oxidative damage.

3. Ozonolysis When ozone gas is passed into a solution of an alkene in an inert solvent like carbon tetrachloride or chloroform at low temperatures, a cyclic unstable compound called ozonide is formed.

Ozonide is then treated with water or hydrogen in the presence of a reducing agent like zinc to give compounds containing the >C=0 group (carbonyl compounds), i.e., aldehydes and ketones.

The process of addition of ozone to an alkene-forming ozonide and then hydrolysis of the ozonide to form carbonyl compounds is called ozonolysis.

In this addition reaction, first, an unstable molozonide is formed, which rearranges spontaneously to form the ozonide. Some other examples of ozonolysis of alkenes are given below.

Ozonolysift is used for locating the position of the double bond in an alkene.

The carbon atoms in the carbonyl group of the two carbonyl compounds obtained after hydrolysis of ozonide are the ones bonded to each other by a double bond in the original alkene.

Polymerisation

A polymer is a substance with large molecules consisting of repeating units called monomers.

Polymers are formed by polymerisation, which is a chemical reaction in which the molecules join each other by addition or condensation (elimination of a small molecule, each time a molecule joins the other).

Some simple alkenes can also be converted into polymers. For example, ethylene (a monomer) polymerises on heating under pressure in the absence of air and the presence of a catalyst to give polyethene, a highly useful industrial polymer.

This polymer is formed by an addition reaction and is, therefore, called an addition polymer or chain-growth polymer. In the same way, propene polymerises into polypropene.

Nowadays, the excessive use of polythene and polypropylene has created environmental hazards, which is a matter of great concern.

Alkynes

Like alkenes, alkynes are also unsaturated hydrocarbons, but they contain a carbon-carbon triple bond (C=C).

The general formula of alkynes is C_nH_2n-2, which shows that they contain two hydrogen atoms less than the corresponding alkenes and four hydrogen atoms less than the corresponding alkanes.

Isomerism

Unlike alkenes, alkynes do not exhibit cis-trans isomerism since the molecules are linear. We normally encounter three types of isomerism in alkynes.

1. Chain isomerism: Alkynes exhibit chain isomerism. For example, Pent-l-yne is a straight-chain and 3-Methylbut-l-yne is a branched-chain isomer.

2. Position isomerism: This arises due to different positions of the triple bond.

⇒ \(\begin{gathered}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH} \\
\text { But-1-yne } \\
\text { (terminal) }
\end{gathered}\)

⇒ \(\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CCH}_3
But-2-yne (nonterminal)\)

3. Functional isomerism Alkynes exhibit functional isomerism with alkadienes. The tire functional group in But- 2-one is a triple bond whereas, in Buta-1,3-diene, it is a double bond.

⇒ \(\underset{\text { But-2-yne }}{\mathrm{CH}_3 \mathrm{C} \text { ant }} \mathrm{CCH}_3\)

⇒ \(\mathrm{CH}_2=\underset{\text { Buta-1,2-diene }}{\mathrm{CH}}-\mathrm{CH}=\mathrm{CH}_2\)

Structure Of Triple Bond

To understand the structure of a triple bond, let us consider acetylene, the first member with the simplest structure in the alkyne homologous series. Tire two carbon atoms in ethyne are sp hybridised.

That is to say, each carbon atom has two sp hybrid orbitals, which are collinear and two unhybridised orbitals (2p_yand 2p_z), which are perpendicular to each other and to the sp orbitals.

One sp hybrid orbital of one carbon atom overlaps with one sp orbital of the second carbon atom, forming a C—C CT bond. The other sp hybrid orbital of each carbon atom forms a bond with a hydrogen atom.

Since the two sp hybrid orbitals of the carbon atoms are collinear and overlapping of the orbitals takes place along the internuclear axis, the two carbon and the two hydrogen atoms lie along a straight line. In other words, the acetylene molecule is linear.

The two unhybridised orbitals of one carbon atom form two n bonds with the two unhybridised orbitals of the other carbon atom.

The orbital structure of acetylene is Thus, the carbon-carbon triple bond in acetylene consists of one strong bond and two weak n bonds.

The H —C—H bond angle is 180°, since die Hacetylene molecule is linear. Two n bonds formed together make a single cylindrical sheath about the line joining the two nuclei.

Preparation

From calcium carbide

Acetylene is manufactured by treating calcium carbide (CaC₂) with water.

⇒ \(\mathrm{CaC}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}+\mathrm{Ca}(\mathrm{OH})_2\)

For this purpose, calcium carbide is obtained by heating a mixture of coke and limestone in an electric furnace.

⇒ \(\begin{gathered}
\mathrm{CaCO}_3 \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_2 \\
\mathrm{CaO}+3 \mathrm{C} \stackrel{2273-3273 \mathrm{~K}}{\longrightarrow} \mathrm{CaC}_2+\mathrm{CO}
\end{gathered}\)

Generally, there can be two processes for the synthesis of alkynes. One of them involves the formation of a triple bond by elimination reactions and the other involves the addition of alkyl groups to molecules containing a triple bond.

Dehydrohalogenation Of Dihalides

Alkynes can be prepared by the elimination of two molecules of hydrogen halides from a dihalide.

The dihalide may be geminal (in which both the halogen atoms are attached to the same carbon atom) or vicinal (in which the two halogen atoms are attached to adjacent carbon atoms).

1. Vicinal dihalides On treatment with suitable bases, vicinal dihalides undergo an elimination reaction whereby two molecules of hydrogen halide are eliminated from the adjacent carbon atoms to give an alkyne. A stronger base like sodamide is used for unreactive vinylic halides.

The dehydrohalogenation of dihalides is a useful method to obtain alkynes as dihalides can be readily obtained by the addition of halogens to the corresponding alkenes.

2. Geminal dihalides (1,1-dihalides) On treatment with a base, geminal dihalides also undergo dehydrohalogenation to give alkynes.

Dehalogenation of tetrahalides

Alkynes may also be obtained by the dehalogenation of tetrahalides.

Physical Properties

Alkynes resemble alkenes and alkanes in their physical properties. They are all colourless and odourless, with the exception of ethyne, which has a characteristic odour.

The first three members are gases, the next eight members are liquids and the higher members are solids at room temperature.

The melting points, boiling points and densities increase gradually with an increase in molecular weight and show the usual effects of chain branching.

Alkynes have slightly higher boiling points than the corresponding alkenes and alkanes.

They are weakly polar and dissolve readily in low-polarity solvents like ether, benzene and carbon tetrachloride. Alkynes are insoluble in water.

Chemical properties

The reactivity of alkynes is mainly due to the presence of two 7t bonds and their acidic nature.

The acidic nature of alkynes

Alkynes are acidic organic compounds but, unlike other acidic inorganic compounds, they do not turn blue litmus red or neutralise aqueous bases.

Yet, they are acidic because they have the tendency to lose hydrogen ions. A hydrogen atom can be released as a positive hydrogen ion by the breaking of a bond if it is bonded with an electronegative atom which can accommodate the bonded electron pair left behind.

If we consider the second period of the p block of the periodic table, the electronegativity of elements decreases as F > O > N > C.

From this order of electronegativity, we may conclude that HF is a strong acid, H₂O is a weak acid, NH₃ is still weaker and CH₄ is not considered an acid at all.

But in the case of a compound containing a triply bonded carbon atom, like acetylene, the relative acidity is of the order H₂O > HC=CH > NH₃.

It seems that the carbon atom in acetylene is a different element than the one in methane, CH₄ (single bond) and ethene (double bond).

Also, the alkynes, which contain hydrogen attached to triply bonded carbon atoms (known as terminal alkynes) show comparable acidity.

Why is the triple bond responsible for the acidity of alkynes? This can be explained on the basis of the hybridisation state of the carbon atom in the compound.

Electrons in the s orbital have lower energy than those in the p orbital. Since the s orbital is spherical the electrons in it are much closer to the nucleus than the electrons in the p orbital, which is lobed on either side of the nucleus.

Therefore, in the case of hybrid orbitals, more s character means more electronegativity and a more stable anion.

The sp hybrid orbital of the triply bonded carbon atom in ethyne has 50 per cent s character whereas it is 33.3 per cent in the sp² hybrid orbital of ethene and 25 per cent in the sp³ hybrid orbital of ethane.

Therefore, the electrons forming the C —H bond in ethyne are closer to carbon, making the carbon electronegative.

So, hydrogen has the tendency to leave as a positive ion, thereby showing acidity.

Acetylene and other terminal alkynes containing acidic hydrogen atoms can be converted into metal acetylides and metal alkynes on treatment with sodium and sodamide respectively in liquid ammonia.

The sodium derivatives obtained above on treatment with alkyl halides give alkynes. By this method, the lower alkynes can be converted to higher alkynes. Alkanes and alkenes do not undergo these reactions.

Acidic alkynes react with an ammoniacal solution of copper sulphate and an ammoniacal solution of silver nitrate (Tollen’s reagent) to give the corresponding copper and silver alkylides, which are red and white precipitates respectively.

The reaction can be used to distinguish between terminal and nonterminal alkynes and is used as a diagnostic test for the —C=CH group.

Addition Reactions

Like alkenes, alkynes undergo electrophilic addition reactions. However, with alkynes, the addition may occur once or twice depending on the number of molar equivalents of the reagents employed.

1. Addition of hydrogen (reduction) Alkynes can be reduced, first to alkenes and then to the corresponding alkanes in the presence of catalysts—finely divided platinum or palladium or nickel.

The above reaction can be stopped at the alkene stage by using Lindlar’s catalyst (Pd-BaSO₄ poisoned with quinoline)

\(\mathrm{HC} \equiv \mathrm{CH} \underset{\mathrm{Pd}-\mathrm{BuSO}_4}{\stackrel{\mathrm{H}_2}{\longrightarrow}} \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2\)

2. Addition of halogens Alkynes add one mole of bromine to give a dibromo and two moles of bromine to give a tetrabromo product.

A solution of bromine in carbon tetrachloride, which is red, gets decolourised in addition to an alkyne. This is used as a test to detect unsaturation.

Chlorine reacts in a similar fashion.

3. Addition of hydrogen halides Alkynes react with two moles of hydrogen bromide or hydrogen chloride to first form the corresponding haloalkene and finally geminal dihalide. Both the electrophilic addition reactions follow the Markovnikov rule.

The anti-Markovnikov addition of hydrogen bromide to alkynes takes place in the presence of peroxides. The reaction occurs through a free-radical mechanism.

\(\underset{\text { Propyne }}{\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}} \frac{\mathrm{HBr}}{\text { peroxide }}-\underset{\text { 1-Bromopropene }}{\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHBr}}\)

4. Addition of water Alkynes, although immiscible with water, can be hydrated in the presence of mercuric sulphate dissolved in dilute sulphuric acid. The triple bond adds a molecule of water to form vinyl alcohol, an unstable product, which isomerises to an aldehyde or a ketone.

In the case of the substituted alkynes, the addition follows the Markovnikov rule and ketones are formed.

Oxidation Of Alkynes

When oxidised with an alkaline potassium permanganate solution, the acetylene or terminal alkyne molecule undergoes cleavage at the site of the triple bond to give carboxylic acids and carbon dioxide. But nonterminal alkynes give only carboxylic acids on oxidation.

⇒ \(\mathrm{R}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}+4[\mathrm{O}] \stackrel{\mathrm{KMnO}_4}{\longrightarrow} \mathrm{R}-\mathrm{COOH}+\mathrm{CO}_2\)

⇒ \(\mathrm{R}-\mathrm{C} \equiv \mathrm{C}-\mathrm{R}+4[\mathrm{O}] \stackrel{\mathrm{KMnO}_4}{\longrightarrow} \mathrm{R}-\mathrm{COOH}+\mathrm{RCOOH}\)

Thus, by identifying the products formed during oxidation by a potassium permanganate solution, it is possible to understand the structure of alkynes.

However, on treatment with ozone, alkynes form ozonides (their structure is different from the ozonides obtained in the case of alkenes), which on treatment with water yield carboxylic acids.

Polymerisation

Alkynes undergo polymerisation to form linear or cyclic polymers. Linear polymers are obtained in the presence of cuprous chloride and ammonium chloride, which, together, catalyse the reaction.

The linear polymer polyacetylene is a conductor of electricity and is used as an electrode in batteries.

When passed through a red hot tube, acetylene polymerises to form a cyclic product, benzene.

Aromatic Hydrocarbons

We already know that aromatic hydrocarbons, also known as arenes or benzenoid compounds, contain one or more benzene rings.

In fact, the scope of the term ‘aromatic’ is now not limited to benzenoid compounds, but also includes non-benzenoid compounds which do not have a carbon sextet and yet possess aromatic character. We will discuss aromaticity.

Isomerism

Position isomerism

Isomerism is not observed in the case of monosubstituted benzene compounds since all the six hydrogen atoms which may be substituted are equivalent.

Disubstituted and trisubstituted benzene compounds exhibit isomerism. Because substituents attached to the benzene ring have different relative positions. For example, dichlorobenzene can have the following three possible isomers.

Similarly, a trisubstituted benzene exhibits three isomers having different relative positions of the substituents.

Ring-chain isomerism

If two compounds have the same molecular formula but one has an open chain and the other a cyclic structure, they are called ring-chain isomers.

Structure Of Benzene

The number of hydrogen atoms in benzene (C6H6) is eight less than that in the corresponding saturated alkane, hexane (C6H14).

Benzene, therefore, should be expected to be an unsaturated compound. This expectation is supported by the fact that benzene adds three molecules of hydrogen in the presence of Raney nickel or platinum at 473-523 K when cyclohexane, C₆H₁₂, is formed.

⇒ \(\underset{\text { Bentene }}{\mathrm{C}_6 \mathrm{H}_6}+3 \mathrm{H}_2 \underset{473-523 \mathrm{~K}}{\stackrel{\text { RaneyNi }}{\longrightarrow}} \underset{\text { Cyclohecane }}{\mathrm{C}_6 \mathrm{H}_{12}}\)

The formation of benzene hexachloride (C₆H₆C₁₆) from benzene in the presence of sunlight is another reaction which proves the unsaturation of benzene.

On the other hand, certain reactions show that benzene unexpectedly behaves like a saturated compound. It undergoes substitution reactions and unlike alkenes and alkynes, it does not decolourise bromine in a solution of carbon tetrachloride. Instead, when treated with Br² in the presence of FeBr³, it forms monobromobenzene, which is a substitution product.

Also, like saturated compounds, benzene is resistant to oxidation. Unlike alkenes and alkynes, benzene does not undergo oxidation readily. Even with strong oxidising agents like chromic acid and potassium permanganate, benzene oxidises only very slowly to CO₂ and H₂O.

Kekule Structure

Until 1865 no scientist could come up with a satisfactory structure for benzene, though they knew its molecular formula.

Then Kekule, a German chemist, proposed that the six carbon atoms of benzene are joined to each other by alternate single and double bonds to form a six-membered ring and that each carbon atom is also bonded to a hydrogen atom.

Strangely enough, Kekuld got this insight into the structure of benzene in a dream. The structure he proposed was a major breakthrough in science, but it could not explain a few characteristics of benzene.

1. It could not explain, for example, why despite containing three double bonds, benzene is not easily oxidised by oxidising agents like KMnO₂ and why it undergoes substitution reactions.
2. If the Kekultf structure is valid, there should be two o-disubstituted products.
3. One with a single bond between the two chlorine atoms and the other with a double bond between the two chlorine atoms. However, in reality, there is only one o-dichlorobenzene.
4. Kekutd tried to account for this by proposing a dynamic equilibrium between the two structures. That is to say, the positions of the single and double bonds are not fixed but oscillate back and forth between the alternate positions.
5. According to Kekule, there should be two kinds of bonds in benzene and the C—C bond length should be greater than the C=C bond length.
6. However, X-ray diffraction studies show that benzene is a regular hexagon with an angle of 120° (all the bonds are alike) and that all the C —C bond lengths are equal (139 pm).
7. The C—C bond length in benzene lies between the single-bond length (154 pm) and the double-bond length (134 pm).
8. Thus, the Kekul6 structure could not explain the fact that the carbon-carbon bond lengths in benzene are equal, or account for the unusual stability of benzene. These were later explained on the basis of molecular orbitals and resonance.

Molecular Orbital Structure

According to the molecular orbital concept, each carbon atom in benzene is sp² hybridised, which means that each carbon atom has three sp² orbitals.

Of these orbitals, two ores are involved in forming a bond with two adjacent carbon atoms and the remaining orbital forms a CT bond with a hydrogen atom, meaning that there are six C—C o bonds and six C—H bonds which lie in one plane. This explains why the angle between any two adjacent a bond (C-C—C) is 120°

Each carbon atom in the molecule, being sp² hybridised, is left with an unhybridised p² orbital.

The six p. orbitals (with one electron each) are parallel to one another and perpendicular to the plane of the a-bonded ring of carbon atoms.

Each of these p orbitals overlaps with the p orbital of the adjacent carbon atom in equally good ways, forming three n bonds in all.

The electron cloud of one p² orbital spreads to the neighbouring p² orbitals equally well.

This type of participation of n electrons in more than one bond is called delocalisation of electrons.

All the six carbon atoms in benzene attract these delocalised electrons equally, thus leading to the formation of two n electron clouds—one above and the other below the plane of the carbon and hydrogen atom.

Due to this all the C—C bond lengths are equal, all the C—H bonds are equivalent and the dipole moment of the molecule is zero. All these factors contribute to the stability of benzene.

Resonance

The concept of resonance can also be used to explain the stability of benzene.

The real structure of benzene can be represented as a resonance hybrid of the two Kekule structures. Structures 1 and 2 are known as resonating or canonical structures.

The resonance structure of benzene explains the following observations.

1. The C—C bond length in benzene is 139 pm which lies between the C—C bond length (154 pm) and the C=C bond length (134 pm).
2. The existence of only one o-dichlorobenzene.
3. The stability of benzene or the fact that it behaves like a saturated hydrocarbon. Because of resonance, the n-electron charge in benzene gets distributed over a large area; in other words, it gets delocalised. This leads to a decrease in the energy of the resonance hybrid by about 50 kJ mol”1 as compared to that of the contributing structures. The difference in energy is called resonance energy.

Aromaticity

Despite being unsaturated, benzene resists addition and oxidation reactions. Thus, benzene shows characteristic stability. Earlier the term ‘aromatic character’ or ‘aromaticity’ was used to signify the characteristic chemical behaviour of benzene and its related compounds.

Robinson was the first to point out that the presence of alternate double and single bonds conferred aromaticity to the benzene ring. This was attributed to the delocalisation of the six n electrons over the planar carbon hexagon.

The modem theory of aromaticity was advanced by Eric Hiickel (1931). The basic concepts of this theory are as follows.

• The complete delocalisation of the n electrons in the ring system is necessary to make the system aromatic I in character.
• Ample or complete delocalisation of the n electrons is possible only if the ring is flat or coplanar so as to allow cyclic overlap of TC electrons. For example, benzene, which has a coplanar ring, is aromatic whereas 1,3,5,7-cyclooctatetraene, being nonplanar, lacks aromaticity.
• Hiickel’s rule or the (4n + 2) rule states that a planar, cyclic system should have n electrons to exhibit aromatic character. Here n is a natural number. Thus, in other words, to be aromatic, a molecule should have 6 7t(n =1), 10 n(n = 2) electrons and so on. Some examples of both aromatic and nonaromatic cyclic hydrocarbons are given below.

Preparation Of Benzene

Enormous quantities of aromatic compounds are obtained from coal and petroleum, petroleum being the chief source of benzene and toluene.

Still, larger quantities of aromatic hydrocarbons are synthesised from alkanes through the process of catalytic reforming. The process involves dehydrogenation, cyclisation and isomerisation.

As stated earlier in the chapter, acetylene (or ethyne) when passed through a heated tube polymerises to a small extent to benzene.

Recently, large-scale production of benzene from acetylene (from petroleum sources) has been achieved by using cobalt carbonyls and other metal complexes.

Decarboxylation of aromatic acids

On being heated with soda lime, sodium salts of benzoic acids give benzene.

By Reduction Of Phenol

When passed over heated zinc dust, phenol in vapour form is reduced to benzene.

Reduction of diazonium salts

Aryldiazonium salts, on reduction with hypophosphorous acid, yield the corresponding arene. The diazo group is replaced by H.

Physical Properties

Being nonpolar, aromatic hydrocarbons are insoluble in water but soluble in organic solvents. Benzene is a colourless liquid, b.p. 353 K. It freezes to a solid at 278.5 K.

It has a characteristic odour. It is highly inflammable and bums with a sooty flame. It is slightly soluble in water. Being toxic and carcinogenic in nature, its use as a solvent has been now prohibited in most advanced countries.

Chemical Properties

Arenes, the derivatives of benzene, do not undergo addition and oxidation reactions under normal conditions, though the benzene ring contains three double bonds. Benzene and its derivatives are characterised by electrophilic substitution reactions.

The stability of benzene has been attributed to resonance. It can also be explained by considering its enthalpies of hydrogenation and combustion, which are lower than expected.

The enthalpy of hydrogenation is the amount of energy that evolves when one mole of an unsaturated compound is hydrogenated. In most alkenes, this value is about 120 kJ mol _1 for each double bond.

On the basis of this value, one expects the enthalpy of hydrogenation for cyclohexatriene or benzene to be 120 x 3 = 360 kJ mol-1 whereas the actual value for benzene is only 208 kJ mol-1, which is 152 kJ mol-1 less than the expected value. This means that benzene is more stable than expected.

Similarly, the enthalpy of combustion of benzene is also lower than expected.

Electrophilic Substitution Reactions

The attacking reagent in the reaction is electron deficient, as the name electrophilic suggests, whereas the benzene ring serves as a source of electrons.

In the reaction, a hydrogen atom of the benzene ring is replaced by an active species, the electrophile (E⁺).

Depending upon the type of electrophile, electrophilic substitution reactions include nitration, halogenation, sulphonation, Friedel-Craft alkylation and acylation reactions.

1. Nitration of Benzene and its homologues, on reaction with a mixture of concentrated nitric acid and sulphuric acid, form nitrobenzene. In the mixture of the two acids, sulphuric acid serves as an acid and nitric acid as a base.

The mechanism of the reaction involves the generation of the electrophilic nitronium ion by protonation and its reaction with benzene to form a resonance-stabilised carbocation. The carbocation rapidly loses a proton to give nitrobenzene. The following steps exhibit the sequence of reactions.

2. Halogenation Arenes react with halogens (bromine or chlorine) in the presence of a Lewis add catalyst—ferric bromide, ferric chloride or aluminium chloride—to give aryl halides.

The mechanism of halogenation follows the same steps as that of nitration. The electrophile X+ is generated with the help of the Lewis acid.

⇒ \(\mathrm{Br}-\mathrm{Br}+\mathrm{FeBr}_3 \longrightarrow \underset{\text { Bromonium ion }}{\mathrm{Br}^{+}}+[\mathrm{FeBr}]_4^{-}\)

If an excess of the electrophilic reagent is used, all the hydrogen atoms can be replaced with the electrophile. For example, on treatment with an excess of anhydrous aluminium chloride in darkness, benzene gives hexachlorobenzene.

3. Sulphonation Benzene reacts with fuming sulphuric acid (H₂S₂O₇) to give benzene sulphonic acid. (Fuming sulphuric add is obtained by passing S03 through 98% H₂SO₄.) The electrophilic reagent in this reaction is the electron-deficient sulphur trioxide.

The mechanism of the reaction involves the generation of the electrophilic reagent SO₃, which attaches itself to the benzene ring to form the carbocation.

The intermediate carbocation formed rapidly loses a proton to give the anion of benzene sulphonic acid. The acid is strong and remains highly dissociated.

4. Friedel-Crafts reaction This electrophilic substitution reaction involves the introduction of an alkyl group in the benzene ring.

In the presence of a Lewis acid, which acts as a catalyst, an alkyl halide reacts with benzene. The products obtained are alkyl benzenes, which are not otherwise easily synthesised. The reaction is known as Friedel-Crafts alkylation.

The mechanism of the Friedel-Crafts reaction involves the following steps:

1. Generation of the electrophile (a carbocation in this case)

⇒ \(\mathrm{RCl}+\mathrm{AlCl}_3 \rightleftharpoons \mathrm{AlCl}_4^{-}+\mathrm{R}^{\oplus}\)

Elimination of a proton (H+)

In Friedel-Crafts alkylation, if in place of alkyl halides (RX), an acyl halide (e.g., acetyl chloride or benzoyl chloride) is used, the product obtained is the corresponding ketone. This reaction is known as Friedel-Crafts acylation.

Acetophenone (C₆H₅COCH₃) and benzophenone (C₆H₅COC₆H₅) can be prepared from acetyl chloride and benzoyl chloride respectively.

Addition Reactions

1. Addition reactions of hydrogen In the presence of finely divided platinum metal,
   benzene adds hydrogen to give cyclohexane.
2. Addition of halogens Benzene adds three molecules of chlorine or bromine under the influence of UV light to form the corresponding hexahalide. The chlorination of benzene gives benzene hexachloride, also called lindane, which has insecticidal properties.

Oxidation

Benzene bums with a sooty flame in oxygen, giving CO₂ and H₂O. A large amount of heat is liberated.

⇒ \(2 \mathrm{C}_6 \mathrm{H}_6+15 \mathrm{O}_2 \longrightarrow 12 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O} ; \quad \Delta H^{\Theta}=-6530 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Benzene can also be oxidised in presence of vanadium pentoxide catalyst at 773 K to give maleic acid and maleic anhydride.

Directive Influence Of Substituents In Monosubstituted Benzene

We have seen that benzene forms only one monosubstituted product. However, when the substituted benzene undergoes electrophilic attack, the group already present on the ring influences the rate of the reaction as well as the site of attack.

We, therefore, say that the substituent group affects the reactivity as well as orientation in electrophilic aromatic substitution reactions.

Depending on the reactivity or how readily the reaction occurs, the substituent groups can be divided into the following two classes.

1. Activating groups These groups cause the ring to be more reactive than benzene. For example, —CH₃ and —NH₂.
2. Deactivating groups These groups cause the ring to be less reactive than benzene. For example, —COOH and —NO₂.

According to orientation, the substituent groups can be divided into the following two classes.

1. Ortho-para directing The substituent groups that release electrons activate the benzene ring and direct the incoming groups to ortho and para positions. For example, —CH₃, —NH2 and —OH. In general, the activating groups are a-, and p-directing.

2. Meta directing The substituent groups that withdraw electrons deactivate the ring and direct the incoming groups to the meta position. For example, —NO₂, —CN and —COOH. In general, the deactivating groups are directed.

Halogens are exceptions—they are deactivating and still o-, p-directing.

Nitration Of Toluene

Toluene on nitration gives a mixture of all three isomers, o,p and m. The yield of o,p-nitrotoluenes is almost 96%, and a very small amount of the m-isomer (4%) is formed.

The example illustrates that the CH₃ group is o-, p-directing. As we have just said, the yield of o-, and p-isomers is much more than that of the m-isomer.

On the other hand, when benzene undergoes nitration, the yield is only 45- 50% which means it is less reactive than toluene.

The greater reactivity of toluene is also indicated by the fact that toluene can be nitrated under milder conditions i.e., lower temperature and lower concentration of HN03.

The alkyl group releases electrons into the ring, thus increasing the nucleophilicity of the ring and stabilising the intermediate carbocation.

To understand the reason behind the o-, p-directing nature of the alkyl group, let us observe the relative stabilities of the different carbocations formed by the electrophile attack at the o- p- and m-positions of toluene.

When the attack occurs at the ortho position, out of three canonical structures, the structure (1) is relatively stable because the positive charge is placed on the carbon atom immediately next to the alkyl group.

The canonical form (2) in case of the para attack, has a positive charge on the carbon atom adjacent to the alkyl group. There is no such equivalent structure formed by the attack at the meta position.

So the carbocations resulting from the attack at the ortho and para positions are more stable than the carbocations formed from the attack at the meta position. Thus, the alkyl groups are ortho- and para-directing. In the structures shown below, E denotes an electrophilic group.

The structures shown below depict the o- and p- p-directing nature of the —OH group in the case of a phenol. The stability of structures is due to the complete octet of electrons in every atom.

Nitration Of Nitrobenzene

Nitration of nitrobenzene gives the m-isomer in a large amount (93%).

Nitrobenzene undergoes nitration at the rate of only 10″4 times that of benzene. This shows that the NO₂ group is a strong deactivating group.

The nitro group is electron-withdrawing, and when present on an aromatic ring, tends to intensify the positive charge and destabilise the intermediate carbocation.

The destabilisation is more prominent in the intermediates arising from ortho and para attacks.

Thus, the meta attack is favoured. If we compare the resonance structures of carbocations formed by attacks at the ortho and para positions of nitrobenzene, the structures (1) and (2) are destabilised, because the positive charge is placed on the carbon atom next to the electron-withdrawing nitro group.

There is no such structure arising from the meta attack and so the meta attack is favoured.

Haloarenes

Halogens deactivate the ring through the inductive effect by direct substitution to the ortho and para positions.

Halogens, being more electronegative than carbon, withdraw (-1 effect) the electrons from the aromatic ring, leaving it less nucleophilic.

However, halogens have unshared electron pairs which can delocalise the electrons in the ring by extended n bonding. This is possible only when the attack occurs at the ortho and para positions.

The resonance structures (1) and (2) formed from ortho and para attacks have the halogens bonded directly to the carbon on which the positive charge is placed.

The halogen atom is able to effect a further delocalisation of positive charge by sharing one of its nonbonding electron pairs to give two resonance structures, (3) for ortho attack and (4) for para attack.

Halogens cannot exert their stabilising influence on the intermediate when the meta position is attacked. The result is that they deactivate the ring by one effect and direct substitution to ortho and para positions by another.

In general, except for the alkyl and phenyl substituents, all of the ortho-para directing groups have at least one pair of nonbonding electrons on the atom adjacent to the benzene ring.

On the basis of studies carried out with a large number of substituted benzene derivatives, the effect of substituents on electrophilic aromatic substitution.

Polynuclear Hydrocarbons

Polynuclear hydrocarbons are divided into two groups—one containing isolated rings and the other containing fused rings.

Biphenyl and diphenylmethane are examples of isolated rings. When two or more rings are fused together they are known as condensed polynuclear hydrocarbons, for example, naphthalene, anthracene, phenanthrene and pyrene.

Toxicity Of Polynuclear Hydrocarbons

Although many aromatic compounds are essential for life (e.g., vitamins, hormones, steroids, purines, pyrimidines), many others are quite toxic, and several benzenoid compounds, including benzene itself, are carcinogenic. Two other examples are benzo [a] pyrene and 7-methyl [a] anthracene.

Certain molecules from the environment become carcinogenic by activation through metabolic processes.

Two such examples are dibenzo (a, 1) pyrene, a polycyclic aromatic hydrocarbon, and aflatoxin BJ, a compound formed during the metabolism of certain fungi.

Some alkyl or alkyl derivatives of 3,4-benzphenanthrene (also called benzoic phenanthrene) have also been found to be carcinogenic. For instance, 2-methyl- 3,4-benzophenanthrene is mildly carcinogenic. People who work in industries dealing with such polynuclear aromatics are at a high risk of cancer. Also, prolonged exposure to coal tar leads to cancer. Though some hydrocarbons are toxic, many of them are useful too.

As already stated, hydrocarbons are used as fuels. In fact, as of now, they are one of the earth’s most important energy resources. For example, petroleum, LPG and CNG. Petroleum The word petroleum is derived from the Greek word ‘petra’ meaning ‘rock’ and ‘oleum’ meaning ‘oil’.

Petroleum is oily, viscous and usually dark. It is found deep below the earth’s crust, entrapped within rocks and below the surface of the oceans.

It is believed to have been formed from the dead remains of animals and trees which got buried deep inside the earth millions of years ago.

The absence of air, presence of moisture, high pressure and high temperature under the earth’s crust facilitates the conversion of dead remains to petroleum.

Petroleum is also formed from the decomposition of marine animals at the bottom of the sea.

In India, petroleum is obtained from the sea bed at Bombay High. It is found trapped in sedimentary rocks, in the oil centres situated at Assam and Gujarat. The crude oil is pumped out from the wells and is called black gold.

The composition of crude petroleum varies with its occurrence. However, they all contain alkanes (straight and branched-chain hydrocarbons from C1 to C40), cycloalkanes, naphthalenes and aromatic hydrocarbons. The low-boiling fractions of petroleum are composed of alkanes.

It is the composition of higher-boiling fractions which differs according to the source of petroleum Besides hydrocarbons, petroleum also contains compounds containing oxygen, nitrogen and sulphur.

LPG Liquefied petroleum gas (LPG) contains hydrocarbons like propane and butane and is obtained as a product from the refining of petroleum. The gas mixture is liquefied by compression and is used mostly as household cooking fuel. LPG is odourless, and small amounts of foul-smelling substances are added to detect any leakage.

It is also used in organic synthesis, especially of synthetic rubber. CNG (Compressed natural gas) Natural gas is a mixture of gases occurring in pockets below the earth’s surface along with petroleum. This mixture usually contains methane and ethane along with small amounts of higher hydrocarbon gases like propane and butane.

Like petroleum, it is believed to have been formed from the decomposition of organic matter trapped in layers of sedimentary rocks. CNG is widely used as a fuel. In India, it occurs in large amounts in Bombay High, the Gulf of Cambay and Ankleshwar in Gujarat.

Earlier, natural gas was considered a hazard in the petroleum industry, as it forms an explosive mixture with air. So it was ‘flared’, or burnt away. Natural gas is now recovered from the source through pipelines.

It is compressed and filled in cylinders for use. In India, CNG is used as a fuel for motor vehicles, buses, etc. The use of CNG, particularly in Delhi, has considerably reduced air pollution. Natural gas also serves as a starting material for making useful petrochemicals.

Hydrocarbons Multiple Choice Questions

Question 1. Which of the following statements is/are correct about methane?

1. The carbon atom is sp³ hybridised.
2. The H-C—H bond angle is109° 28′.
3. It has four bonds.
4. The carbon atom is sp² hybridised.

Answer: 1. The carbon atom is sp³ hybridised.

Question 2. How many 1°, 2°, 36 and 4° carbon atoms are present in 2,2,4-trimethylpentane?

1. Five 1°, one 2°, one 3°, one 4°
2. Four 1°, two 2°, two 3°, zero 4°
3. Four 1°, two 2°, one 3°, one 4°
4. Three 1°, two 2°, two 3°, one 4°

Answer: 1. Five 1°, one 2°, one 3°, one 4°

Question 3. How many a and it bonds are present in ethene?

1. 5ct, 2n
2. 5a, lit
3. 4a, 3it
4. 6a, zero it

Answer: 2. 5a, lit

Question 4. Which of the following statements is/are correct about acetylene?

1. It contains two and three bonds.
2. The H—C—C bond angle is 180°.
3. It contains two it and two bonds.
4. The molecule is nonlinear.

Answer: 1. It contains two and three o bonds.

Question 5. Which of the following statements is/are correct about benzene?

1. Each C atom in benzene is sp² hybridised.
2. All the C—C bond lengths are equal.
3. All the C—Co bonds and C—Ha bonds lie in the same plane.
4. Each C atom in benzene is sp³ hybridised.

Answer: 1. Each C atom in benzene is sp² hybridised.

Question 6. Which of the following is/are correct?

1. The eclipsed conformation of ethane is less stable than the staggered conformation.
2. The eclipsed conformation is more stable than the staggered conformation.
3. Both are equally stable.
4. Both the conformations can be separated.

Answer: 1. The eclipsed conformation of ethane is less stable than the staggered conformation.

Question 7. A compound on ozonolysis gives propanal and benzaldehyde. The compound can be

1. 1-Phenylbut-l-en
2. Pent-2-en
3. 2-Ethylbut-ene
4. None of the above

Answer: 1. 1-Phenylbut-l-en

Question 8. Propanal and Pentan-3-one are the products obtained by the ozonolysis of which of the following alkenes?

1. 3,4-Dimethylhept-3-ene
2. 1-Phenylbut-l-ene
3. 4-Ethylhex-3-ene
4. None Of The Above

Answer: 3. 4-Ethylhex-3-ene

Question 9. But-l-ene and but-2-ene are

1. Geometrical isomer
2. Chain isomers
3. Position isomers
4. Optical isomers

Answer: 2. Chain isomers

Question 10. How many structures can a compound with the molecular formula C₄H₈ have?

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

Answer: 3. 3

Question 11. Which of the following statements is/are true of geometrical isomers?

1. The frans-isomers of alkenes are more stable than the cis-isomers.
2. The cis-isomers of alkenes are more stable than the frans-isomers.
3. The c/s-isomer is more polar than the trans-isomer.
4. The frans-isomer is more polar than the cis-isomer.

Answer: 1. The frans-isomers of alkenes are more stable than the cis-isomers.

Question 12. Which of the following compounds will exhibit geometrical isomerism?

1. Hex-3-ene
2. 3-Ethylpent-2-ene
3. But-2-ene
4. 1,1,2-Tribromoethene

Answer: 1. Hex-3-ene

Question 13. An aqueous solution of a compound gives ethane on electrolysis. The compound can be

1. Sodium acetate
2. Sodium ethoxide
3. Sodium propanoate
4. None of these

Answer: 1. Sodium acetate

Question 14. Among the following, the most reactive compound in the context of electrophilic nitration is

1. Benzene
2. Toluene
3. Benzoic Add
4. Nitrobenzene

Answer: 2. Toluene

Question 15. In a reaction of QH5A, the major product is an m-isomer. So A is

1. -Cl
2. —COOH
3. —OH
4. All

Answer:  4. All

Question 16. Which of the following is aromatic in nature?

Answer: 3.

Question 17. The C—C bond distance is maximum in

1. Ethane
2. Ethyne
3. Benzene
4. All Are Equal

Answer: 1. Ethane

Question 18. The number of n electrons in naphthalene is

1. 3
2. 4
3. 5
4. 10

Answer: 4. 10

Organic Chemistry—Some Basic Principles And Techniques

The famous Swedish chemist, Berzelius, proposed in 1807 that chemicals should be divided into two separate groups—organic and inorganic.

It was believed that organic compounds were those that existed in or were created by living organisms by some ‘vital force’. And that inorganic compounds were simple and could be obtained from minerals.

However, this concept became outmoded when Wohler prepared urea, an organic compound, from an inorganic compound— ammonium cyanate.

The compound (urea) prepared was identical to the one found in urine.

This was followed by the synthesis and analysis of many more organic compounds. The modem definition of organic compounds states that such compounds are those that contain carbon.

However, carbon dioxide, carbon monoxide and carbon disulphide are considered to be inorganic compounds.

It is now well known that most organic compounds contain, apart from carbon and hydrogen, other elements like oxygen, nitrogen, sulphur, phosphorus and halogens.

In view of this, organic chemistry is now regarded as the chemistry of hydrocarbons (compounds containing carbon and hydrogen only) and their derivatives.

It is interesting to know that the total number of organic compounds is much more than that of the compounds of all other elements taken together.

What is so specific about carbon that it forms so many compounds? A unique property to form bonds with other carbon atoms.

This property is referred to as catenation. Owing to this characteristic property, carbon can form straight-chain compounds, branched-chain compounds and rings of different sizes.

Structure Of Organic Compounds

As you already know, carbon being tetravalent, methane has four covalent bonds, one each between a single carbon atom and each of the four hydrogen atoms.

You have already about hybridisation, i.e., the intermixing of atomic orbitals to form new orbitals.

In the case of methane, the three 2p orbitals of the carbon atom intermix with the 2s orbital to produce four sp³ hybrid orbitals which form four covalent bonds with the Is orbital of four hydrogen atoms.

The structure of ethene and ethyne molecules which have been discussed while we studied sp² and sp hybridisation respectively. To understand and predict the properties of organic compounds, it is important to know their molecular structures. sp²

The nature of orbitals depends on the type of hybridisation they have resulted from.

Nature of orbitals

In sp³ hybridisation, the four sp³ hybrid orbitals of carbon are formed by mixing one 2s and three 2p orbitals.

So each sp³ orbital has 25% s and 75% p character. The percentage changes in sp² hybridisation, to 33% s and 66% p character.

The sp hybridised carbon has orbitals with 50% s and 50% p character. With more s character the carbon atom becomes more electronegative in comparison to other carbon atoms (sp² and sp³ hybridised carbon atoms). This is because of the closeness of the s orbital to the nucleus.

Relative bond length and strength of hybrid orbitals

The bond length and the strength of the bonds formed by the combining of hybrid orbitals are influenced by the type of hybridisation in compounds. You know that hybridisation is a theoretical concept.

Therefore, the energy of hybrid orbitals is theoretically estimated. Let us see how bond length and strength are influenced by hybridisation considering the examples of ethane, ethene and ethyne.

The sp hybrid orbital of carbon in ethyne is shorter and stronger than the sp² hybrid orbital in ethene, owing to the 50% s character in ethyne, more than that in ethene.

The sp³ hybrid orbital of carbon in ethane forms longer and weaker bonds than those in ethene and ethyne. This can be attributed to less s character (25%) in the sp hybrid orbital. The distribution of hybrid orbitals in space leads to different bond angles.

The π Bond

A double bond exists between two carbon atoms in a molecule of ethene. Out of two, one is a bond (formed by the overlap of sp² hybrid orbitals, one each from each of the carbon atoms) and the other is it bond (formed by the overlap of the two 2p_z unhybridised orbitals of each carbon atom).

Similarly, the triple bond in a molecule of ethyne comprises one a bond and two n bonds. You have already studied the two types of bonds. Let us discuss some characteristic features of JI bonds.

The pi (it) bond is formed by the overlap of two 2p_z unhybridised orbitals. The overlapping (sidewise or lateral overlap) takes place in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis.

The n bond is weaker than the bond. Due to lateral overlap, the electron charge cloud of the n bond is placed above and below the intermolecular axis.

This makes it easier for the reagent to attack n electrons than electrons. Acetylene, which has a triple bond between the carbon atoms, contains one a and two its bonds, and is, therefore, readily attacked by oxidising reagents.

Structural Representation Of An Organic Molecule

The structures of organic molecules can be represented in a number of ways. The simplest way of representing the structure of an organic compound is by representing a bond by a dash, viz., a single bond by a single dash (—), a double bond by two dashes (=) and a triple bond by three dashes (=).

The lone pair of electrons on atoms like oxygen, nitrogen and sulphur may or may not be shown. Some examples of such complete structural formulas are given below.

Organic molecules can also be represented by condensed structural formulae. In this type of representation, some or all of the covalent bonds in the molecule are omitted and the number of each of the identical atoms is indicated by a subscript.

⇒ \(\begin{array}{ccc}
\mathrm{CH}_3 \mathrm{CH}_3 & \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2 & \mathrm{HC} \equiv \mathrm{CH} \\
\text { or } & \text { or } & \text { or } \\
\mathrm{C}_2 \mathrm{H}_6 & \mathrm{C}_2 \mathrm{H}_4 & \mathrm{C}_2 \mathrm{H}_2 \\
\text { Ethane } & \text { Ethene } & \text { Ethyne }
\end{array}\)

Similarly, CH₃CH₂CH₂CH₂2CH₂CH₂CH₂CH₂CH₂COOH can be condensed to CH₃ (CH₂)₈COOH. In bond line representation only bonds or lines are used to represent a molecule, and carbon and hydrogen atoms are not shown.

The end of each line represents a carbon atom which is understood to be attached to the required number of hydrogen atoms necessary to satisfy the tetravalency of carbon.

The lines are drawn in a zig-zag fashion and represent carbon-carbon bonds. The atoms nitrogen, oxygen and chlorine are specifically written.

The terminals denote the methyl (—CH₃) group unless otherwise indicated. Thus CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂COOH can be represented as

In the same way, the bond line formulas of 2-methyl butadiene and 1,3,5-hexatriene are written as follows

A compound containing one or more rings is called a cyclic compound and is represented by drawing the appropriate ring(s) without showing C and H atoms.

The comers of the ring represent C atoms and the sides I denote carbon-carbon bonds. Any functional group present is also shown in the structure.

The wedge-and-dash representation is used to represent the three-dimensional structure of a molecule.

By convention, a solid wedge indicates a bond projecting from the plane of paper towards the viewer and a dashed wedge shows a bond projecting inwards from the plane of the paper.

A normal line indicates the bond on the plane of the paper. The three-dimensional representation of methane is shown here.

Functional Groups

A functional group is an atom or n group of atoms responsible for the chemical behaviour of an organic compound. Structural features like double and triple bonds are also considered to be functional groups.

In an organic molecule, the functional group is the most reactive part whereas the remaining portion determines the physical properties of the compound. For example, amine).

If we attach the — NH₂ (amino) as a functional group In CH₃CH₂NH₂ (ethyl —NH2 group to some other alkyl radical like CH-,—(methyl) or I7 —(propyl), we would get the compounds CH₃ NH₂ and C₃H₇NH₂.

Compounds containing the same functional group undergo more or less similar reactions and are, therefore, said to belong to n family. There may be exceptions in the case of very large molecules or if more than one functional group is present.

Hie compounds we have just discussed belong to a family called amines. Some other functional groups which you may come across at the present level of learning.

Homologous Series

A family of organic compounds like alkanes and alcohols contains a characteristic functional group and forms what is known as a homologous series.

Each member of the series is known as a homologue. The members differ from each other in molecular formulae by a —CH₂ unit.

The first four members of the homologous series of the alcohol family are as follows.

⇒ \(\begin{array}{ll}
\mathrm{CH}_3 \mathrm{OH} & \text { Methyl alcohol } \\
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} & \text { Ethyl alcohol } \\
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH} & \text { Propyl alcohol } \\
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH} & \text { Butyl alcohol }
\end{array}\)

Characteristics of a homologous series

1. A particular homologous series can be represented by a general formula. For example, in the case of alcohols, the general molecular formula is C_nH_2n+1OH. By putting the value of n as 1, 2, 3, 4,… where n is the number of carbon atoms in the molecule, we get the molecular formula of the corresponding homologues.
2. Any member of the homologous series differs from the next lower or higher by the unit CH₂(methylene group) or by 14 atomic mass units (12 + 2 x 1 = 14).
3. The different homologues of the series can be prepared by employing general methods of preparation.
4. The members of a homologous series show similar chemical properties.
5. The physical properties (such as density, melting point, and boiling point) of the successive members in a homologous series show a more or less regular gradation.

Nomenclature Of Organic Compounds

Hydrocarbons are compounds which contain only hydrogen and carbon. They are the simplest organic compounds.

Organic compounds bearing functional group(s) are derived from hydrocarbons by replacing one or more hydrogen atoms with the functional group.

As more and more carbon compounds were known it became necessary to introduce a system of nomenclature.

A formal system of nomenclature of organic compounds was introduced in 1931, known as the IUC system.

The system was further developed by the International Union of Pure and Applied Chemistry (IUPAC).

Before the IUPAC system of nomenclature was introduced, organic compounds were given common or trivial names. The names were based on the source of the compounds or some other specific features.

For example, acetic add can be obtained from vinegar and was, therefore, named after the Latin word for vinegar, aceluin. Some ants contain formic acid and the Latin word for ant isformica.

Urea was so named since it was first isolated from the urine of mammals. Methyl alcohol was called wood spirit since it could be obtained by the destructive distillation of wood.

Many of the common names are still used by chemists. Common names are useful when the other systematic names are very complicated.

The newly discovered allotrope of carbon (C₆₀) has been given the common name Buckminsterfullerene, after the architect Buckminster Fuller, whose designs are similar in structure to the C₆₀ cluster of carbon.

The IUPAC System Of Nomenclature

The IUPAC system is simple, and any chemist can deduce the structure of an organic compound if he or she knows its IUPAC name. In the IUPAC system of nomenclature, a chemical name has three parts: prefix, parent and suffix The parent name tells us how many carbon atoms are present in the principal chain.

The suffixes are of two types—primary and secondary, and both indicate the functional groups present in the compound.

The primary suffix is added to the root derived from the parent name indicating the nature of the bond between the carbon atoms. For example, the suffix one in ethyne shows the presence of a triple bond.

The secondary suffix is added after the primary suffix to indicate the functional group(s) present in the carbon chain. For example, the IUPAC name of alcohol—alkanol—has two suffixes an and ol.

Here an indicates a single C —C bond and ol indicates the class of the compound—alcohol.

Note While adding the secondary suffix to the primary suffix, the terminal e of the primary suffix is removed if the secondary suffix begins with a vowel but it is retained if the secondary suffix begins with a consonant.

The prefixes indicating substituents on the parent chain are also of two types—primary and secondary.

A primary prefix such as ‘cyclo’ is added before the root word to indicate the cyclic nature of the carbon skeleton. For example, cyclohexane.

In a polyfunctional compound the groups which are not considered the principal functional group in the IUPAC system of nomenclature, but regarded as substituents, are called secondary prefixes.

For example, halogens, alkoxy, alkyl, aryl and nitro groups. The secondary prefix comes before the primary prefix. For example,

To begin with, let us study the systemic IUPAC nomenclature of hydrocarbons. On the basis of their structure, hydrocarbons may be of three types—straight-chain, branched-chain and cyclic.

Alkanes

Primary prefix Secondary prefix Root word Primary suffix IUPAC name cydo Chloro hex ane Chlorocyclohexane The root word for alkanes from CH₄ to C₄H₁₀ are derived from trivial names.

However, the names of alkanes containing five or more carbon atoms are derived by adding prefixes such as pent (five), hex (six), hept (seven), oct (eight) and so on, indicating the number of carbon atoms in the molecule, with the suffix ane. The molecular formula and the IUPAC names of some straight-chain alkanes.

On removing one hydrogen atom from an alkane, a univalent alkyl group is formed. The individual alkyl group is named by replacing the suffix ane of the parent hydrocarbon with yl.

Branched-chain alkanes are named according to the following rules.

1. The longest continuous chain of carbon atoms is the parent chain. For example, the following compound is a hexane since the longest chain contains six carbon atoms.

The following compound is a heptane since the longest chain has seven carbon atoms.

In case a molecule contains two equally long carbon chains, the one attached to a greater number of side chains or substituents is selected. Correct longest chain: Parent chain of seven carbon atoms with one side chain and two substituents.

Incorrect longest chain: Parent chain of seven carbon atoms with only one side drain.

2. For the longest chain from one end to the other, the direction of numbering is chosen to give the lower number(s) to the substituent(s).

3. The location of the substituent groups is designated by using rule 2. The number indicating the position of the substituent should be prefixed to the name of the parent alkane.

The numbers are separated from the groups by hyphens. Note that the following common names are retained by IUPAC for unsubstituted hydrocarbons only.

Precisely speaking, isobutane is 2-methylpropane, isopentane is 2-methylbutane and neopentane is 2, 2-dimethylpropane.

Lowest sum rule: In case the parent chain has two or more side chains or substituents, the numbering must be done in such a way that the sum of the locants on the parent chain is the lowest possible.

4. When different substituents are present on the carbon chain, they are written in alphabetical order.

5. In case two substituents are present on the same carbon atom, the number indicating the position is written twice.

6. Identical substituents are indicated by using appropriate prefixes di, tri, tetra, etc. The numbers are separated by commas.

When two chains of equal length compete for selection as the parent chain, choose the chain with the greater number of substituents.

When two different substituents are in equivalent positions (from the two ends of the carbon chain), the groups are numbered in alphabetical order, while numbering the chain.

On removal of one hydrogen atom, both butane and isobutane give rise to two butyl groups.

The following examples illustrate the naming of compounds containing branched alkyl groups.

The prefixes iso, see, test and neo are approved by 1UPAC

Classification of carbon atoms

Being tetravalent, a carbon atom is attached to four other carbon or hydrogen atoms in hydrocarbons.

One carbon atom may be attached to one other carbon atom and three hydrogen atoms. Also, one carbon atom may be attached to three other carbon atoms and one hydrogen atom.

Depending upon these arrangements carbon atoms are classified into four types atom is attached to one other carbon—the primary atom.(1°), A secondary secondary carbon (2°), tertiary atom is(3°) attached to quaternary two other(4°). carbonA primary atoms, carbon and so on.

A hydrogen atom is also referred to as 1°, 2° or 3° depending on whether it is attached to a 1°, 2° or 3° carbon atom. In the following examples, 1°, 2°, 3° or 4° carbon atoms are shown.

Cycloalkanes

Cycloalkanes with only one ring are named by attaching the prefix ‘cydo’ to the names of the corresponding straight-chain alkanes. For example,

Substituted cycloalkanes are named applying the same rules as applied to branched-chain alkanes. If more than one substituent is present, its position is indicated by the appropriate number.

While naming the compound, the substituents are placed in alphabetical order. Also, the numbering is so done that the substituent(s) get the lowest possible number(s). The following examples illustrate the abovementioned points.

When a single ring is attached to a single chain with the same or a greater number of carbon atoms than in the ring, the compound is named as follows.

The carbon atom attached to the more-branched chain is assigned the lower number.

When the chain attached to the ring is longer than the ring, the ring becomes a substituent.

Alkenes

Alkenes are unsaturated hydrocarbons containing a double bond between two adjacent carbon atoms.

Ethene or ethylene is the first member of this homologous series. The rules for the IUPAC nomenclature of alkenes are similar in many respects to those of alkanes.

1. The longest carbon chain containing the double bond is chosen as the parent alkene. The suffix part of the corresponding alkane is changed into one. For example, the carbon chain with six carbon atoms and a double bond is hexene.

2. The numbering of the carbon chain is so done that the double bond has the lowest number. Designate the location of the double bond by writing the lower number as a prefix. The locant (number) is placed just before the functional group suffix.

3. The locations of the different substituents are indicated by the numbers of the carbon atoms to which they are attached. The lowest sum rule studied earlier in the case of alkanes also applies here.

4. Cycloalkenes are numbered in a way that gives the carbon atoms of the double bond the 1- and 2- positions and that also gives the substituent groups the lower numbers.

5. Compounds containing a double bond and an alcohol group are named by giving the alcohol carbon the lowest number.

The terminal e in the name of the alkene is dropped if it is followed by a suffix beginning with a, i,o, u or y. (The rule is the same for alkynes.)

6. The removal of the terminal hydrogen atom from ethene and propene gives the vinyl and the allyl group respectively. If a halogen atom is attached to these groups we obtain compounds with common names vinyl halide and allyl halide.

7. If two identical groups are on the same side of the double bond, the compound is designated as cis and if they are on opposite sides, it is designated as trans.

8. If there are two or three double bonds in a chain then the compound is called a diene or a triene respectively.

⇒ \(
\stackrel{1}{\mathrm{C}} \mathrm{H}_2=\stackrel{2}{\mathrm{C}} \mathrm{H}-\stackrel{3}{\mathrm{C}} \mathrm{H}=\stackrel{4}{\mathrm{C}} \mathrm{H}_2
Buta-1,3-diene\)

Alkynes

Alkynes are unsaturated hydrocarbons that contain a triple bond between two adjacent carbon atoms. The first member of the homologous series is acetylene or ethyne (HC=CH)

Alkynes are named in the same way as alkenes. The unbranched alkynes are named by replacing the suffix ane of the corresponding alkane with the suffix one. The first three members of the alkyne homologous scries ethyne, propyne and butyner.

The trivial name acetylene has been retained by the IUPAC system for HC=CH and is frequently used.

Where the double as well as the triple bond is present in the same molecule, the double bond is given the lowest number (order of priority of functional groups).

The locations of substituent groups of substituted alkynes are also indicated with the appropriate numbers. A hydroxyl (—OH) group is given priority over the triple bond while numbering the chain.

Here you should remember that the terminal e of meor yne is not dropped in case it is followed by di, tri or tetra.

Polyfunctional compounds

As stated earlier in the chapter, compounds containing the same functional group belong to a family. While naming the compound, the functional group is identified and the appropriate suffix is used.

The parent chain is so numbered that the carbon atom attached to the functional group gets the lowest possible number (as in the case of alkenes and alkynes).

But what if the compound has more than one functional group? In the case of polyfunctional compounds, one of the functional groups is chosen as the principal functional group according to the order of priority of functional groups.

The compound is named on the basis of that principal functional group and the remaining functional group(s) is named as substituents using appropriate prefixes.

The order of priority of some functional groups is COOH, —S03H, —COOR, —COG, —CONH2, —CN, =C—. —HC=0, —OH, -NH2, /C=CC

The alkyl (—R), phenyl (—C6H5), halogens (F, Cl, Br, I), nitro (—NOz), alkoxy (—OR), etc., are substituents and always prefixed.

Along with the different families of compounds, their respective prefixes and suffixes, Table 12.4 also contains examples of polyfunctional compounds and their IUPAC names.

Example Name the following compounds according to IUPAC rules

Solution:

1. The longest continuous chain is cyclic and of four carbon atoms. Hence the corresponding hydrocarbon is cyclobutane.
   1. Two functional groups (a double bond and an aldehyde) are present in the compound.
   2. According to the order of priority of functional groups, the aldehyde precedes the alkene.
   3. Hence the carbon attached to —CHO is numbered 1.
   4. Hence, the name of the compound is l-Formylcyclobut-2-ene.
2. The longest continuous chain (with two substituents) is of four carbon atoms, so the root word is but.
   1. Applying the lowest sum rule, the two substituents methyl (—CH3) are at carbons 2 and 3. The double bond is at position 1 of the carbon chain.
   2. Hence, the name of the compound is 2,3-Dimethyl-but-l-ene.

Aromatic Compounds

Benzene and substituted benzene compounds are the most important members of this class. Benzene contains six carbon atoms, numbered 1 to 6. The molecular formula of benzene is C6H6. It contains three double bonds.

The six hydrogen atoms in benzene are equivalent, each bonded to one carbon atom. Owing to this only one monosubstituted derivative is possible. But benzene can form three disubstituted derivatives with relative positions; 1,2;13 and 1,4.

The positions are indicated by trivial terms, ortho (o), meta (m) and para (p) respectively. While naming these derivatives according to IUPAC rules, we simply prefix the name of the substituent group to the word benzene.

Some derivatives of benzene have universally accepted, common names (written in brackets below) which have no resemblance to the names of the attached substituent.

Disubstituted or trisubstituted benzenes are named by numbering the positions of the substituents.

Dimethylbenzenes arc commonly called xylenes.

If different substituents are attached to the benzene ring then the following rules are observed.

1. The principal group or substituent of the base compound is assigned the number 1.
2. The numbering is so done that the substituents are located at the lowest numbers possible.
3. The substituents are written in alphabetical order.
4. In some cases, the common name of the die benzene derivative is considered the name of the parent compound.

Let us now look at a few categories of benzene derivatives.

The trivial names phenyl, benzyl, benzal and benzo have been accepted by IUPAC
Halogen derivatives.

Hydroxy derivatives

Amines

Ketones

Aromatic ethers

Aldehydes 

Carboxylic Acids 

Isomerism

When two or more compounds have the same molecular formula but differ in their physical and chemical properties, the phenomenon is called isomerism, and the compounds isomers (Greek, isos = equal; meros = parts).

Isomerism is of two types—structural isomerism and stereoisomerism.

Structural isomers are those which differ in the mode of linking of atoms whereas stereoisomers differ in the arrangement of atoms or groups in space. Stereoisomerism.

Structural isomerism is of different types, based on different functional groups, different carbon skeletons and different positions of the functional groups in the same carbon chain.

Chain isomerism This type of isomerism arises when the carbon skeletons of the isomers differ. For example, butane (C₄H₁₀) can have the following two structures.

In the same way the alcohol with the molecular formula, C₄H₉OH can be either n-butyl alcohol or is obutyl alcohol.

The alkene C₄H₈ can be either but -1-ene or isobutene.

Position isomerism When the isomers differ in the position of the substituent atom or group in the same carbon chain, the phenomenon is called position isomerism.

As you can see. the position of the —OH group is different in the same carbon chain of the isomers with the molecular formula C₃H₇OH.

Functional group isomerism When two or more compounds have the same molecular formula but different functional groups, they are called functional group isomers and the phenomenon is called functional group isomerism.

A compound with the molecular formula C₄H₁₀ can be an ether or an alcohol.

Similarly, C₃H₆₀ can be a ketone or an aldehyde.

C₂H₄0 can be an add or an ester.

Metamerism When the isomers differ in the alkyl groups attached to the same functional group, they are called metamers and the phenomenon is called metamerism.

An ether, C₄H₁₀, can be either methyl propyl ether or diethyl ether.

A ketone, C₅H₁₀, can have two metamers.

Tautomerism This is a type of isomerism in which the two isomers are in equilibrium. The most common land is keto-enol tautomerism.

In this form of tautomerism compound containing a —CH₂—CO— group (the keto form of the molecule) is in equilibrium with one containing the —CH=C(OH)— group (the enol).

This happens when a hydrogen atom migrates from a carbon atom to an oxygen atom bonded to an adjacent carbon.

Example What is the relationship between the following molecules and CH3CH(CH3)CH, CH3?

Solution:

1. The bond line formula for CH₃CH(CH₃ )CH₂CH₃ is which is the same as and Thus, all three molecules are the same.
2. Molecule
   • Is neopentane, i.e.,
3. Molecule is CH₃CH(CH₃)CH₂CH,CH₃ which is the next homologue of CH₃CH(CH₃)CH₂CH₃.

CH₃CH(CH₃)CH₂CH₃ and neopentane are chain isomers.

Reaction Mechanisms

Many reactions take place via intermediates, i.e., they are multi-step. When the products in a reaction are formed directly without any intermediate, the reaction is called a single-step reaction. In general, an organic reaction may be represented as follows.

⇒ \(\text { Substrate } \stackrel{\text { reagent }}{\longrightarrow} \text { [Intermediate] } \longrightarrow \text { Products }\)

The steps leading to the cleavage of bonds (covalent bonds in the substrate molecule) and the formation of new bonds in the products constitute the reaction mechanism.

A clear understanding of reaction mechanisms helps us identify patterns in complex organic reactions.

And recognising patterns helps us apply our knowledge about one reaction to many similar reactions.

The reaction mechanism of a complex organic reaction may depend on many factors such as the following.

1. The reactivity of the substrate molecule may be known by an understanding of the electron displacement effects caused due to the interaction of substrate and reagent molecules and the difference in electronegativities of the atoms in a molecule.
2. The nature of bond fission decides the type of the reaction intermediate formed. Carbocations, carbanions and free radicals are types of reaction intermediates.
3. The nature of the reagent, whether electrophilic or nucleophilic.
4. The type of reaction which occurs. Substitution, addition, elimination and rearrangement are some common types of organic reactions. You will study these reactions.

Let us now discuss these factors in greater detail.

Electron Displacement In A Molecule

Generally, unlike inorganic compounds, organic compounds do not react with each other on their own.

A reaction between organic compounds requires the presence of a reagent. Some examples of reagents are H2S04, HC1 and NaOH.

Although a substrate molecule is electrically neutral, it develops some polarity due to the difference in electronegativities of the atoms of the molecule or under the influence of an attacking reagent.

The polarity develops by the displacement of electrons in the substrate. This is referred to as the electron displacement effect.

We normally come across the following four types of such effects in organic molecules.

1. Inductive effect
2. Electromeric effect
3. Mesomeric effect
4. Hyperconjugation effect

Inductive effect When a covalent bond is formed between two dissimilar atoms, the shared electron pair will be displaced towards the atom which is more electronegative.

This causes the more electronegative atom to acquire a partial negative charge (5-) whereas a partial positive charge (8+) develops on the other atom.

For example, if we consider the covalent bond C—X, where X is more electronegative than carbon, then the bond is polarised so that C gains a partial positive charge (5+) and X gains a partial negative charge (8-).

Alternatively, in a covalent bond C—Y, where Y is less electronegative than C, the shared electron pair is displaced towards carbon, so that C attains a 8- charge and y attains a &+ charge.

The polarization induced in a substrate molecule is of a permanent nature.

Thus, an inductive effect involves the permanent shift of electrons towards the more electronegative element in a covalent bond.

When the electronegative group linked to a carbon atom withdraws electrons and gains an 8- charge it is said to be an electron-withdrawing group exerting a -I effect.

Some common electron-withdrawing groups (in decreasing order of -I effect) arc:

⇒ \(\mathrm{O}^{-}>\mathrm{CO}_2^{-}>\left(\mathrm{CH}_3\right)_3 \mathrm{C}>\left(\mathrm{CH}_3\right)_2 \mathrm{CH}>\mathrm{CH}_3 \mathrm{CH}_2>\mathrm{CH}_3\)

⇒ \(\mathrm{NO}_2>\mathrm{F}>\mathrm{COOH}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}>\mathrm{OH}\)

In contrast, if the substituent bonded to the carbon atom is electron donating (being less electronegative titan carbon) it acquires an 8+ charge and produces a +1 effect.

Some electron-donating groups in decreasing order of +1 effect are:

⇒ \(\mathrm{O}^{-}>\mathrm{CO}_2^{-}>\left(\mathrm{CH}_3\right)_3 \mathrm{C}>\left(\mathrm{CH}_3\right)_2 \mathrm{CH}>\mathrm{CH}_3 \mathrm{CH}_2>\mathrm{CH}_3\)

The inductive effect is transmitted along the chain of carbon atoms but the charge developed on all the carbon atoms in the chain decreases as the distance from the source increases.

In propyl chloride, the partial positive charge developed on C1 induces some positive charge on C2 and C3 gains still less positive charge.

The inductive effect is represented by an arrow pointing towards the electron-withdrawing atom or group, Electromeric effect When a reagent attacks a compound containing a multiple bond (double or a triple bond), n electrons are transferred completely from one atom to the other in the bond.

The atom that gains the electron pair becomes negatively charged and becomes positively charged. This is the electromagnetic effect.

The electromagnetic effect is temporary. The influence of an attacking reagent causes a polarisation in the substrate molecule.

As soon as the reagent is removed, the polarised molecule reverts to its original state.

The electromeric effect is represented by the symbol E and is said to be +E when the displacement of electrons is towards the atom to which the attacking reagent gets attached.

The effect is -E when the electrons are transferred to that atom to which the attacking reagent does not get attached. The curved arrow shows the displacement of the shared electron pair.

Mesomeric or resonance effect Like the inductive effect, the mesomeric effect also causes permanent polarisation in a molecule.

However, unlike the inductive effect which operates in molecules having single bonds (sigma bonds), the mesomeric or conjugative effect operates only in those molecules which have unsaturated conjugated systems.

A conjugated system in a molecule can be described as a double or triple bond separated by one single bond. In such molecules the delocalisation of some n.

orbital electrons occur between carbon atoms linked by a single bond. Consider the case of a carbonyl group. Its structural formulae I and II do not satisfactorily explain all its properties.

This led to the idea that it exists in a state which is some combination of two or more electronic structures. This is the resonance hybrid (HI) in which electrons are drawn partially towards oxygen.

The mesomeric effect is more pronounced if the carboxyl group is conjugated with an unsaturated system such as CH₃=CH—CH=CH—CH=0. The TI electrons of the C—O bond get displaced towards oxygen (which is more electronegative) giving rise to the resonance structures.

The resonance structure which has more dispersal of charge (negative charge on the more electronegative atom and positive charge on the more electropositive atom) is more stable.

As explained above, polarization in a molecule is transmitted via the n electrons. The mesomeric effect (denoted by M) can be +M or -M. A group of atoms causes a -M effect when the direction of displacement of electrons is towards it. Examples of groups that cause a -M effect are

The -M effect of the NO₂ group is shown as follows.

This effect can also be seen in a cyclic system with alternate single and double bonds, for example, nitrobenzene.

A group of atoms has a +M effect if the electron pair is displaced away from it. Some examples of electron-repelling groups are as follows.

These groups have one or more lone pairs of electrons which they furnish for conjugation with the attached unsaturated system. For example, the NH2 group, which has a lone pair of electrons, shows a +M effect.

This kind of electron transfer is called mesomeric or resonance effect. Recall that when the structure of a molecule can be represented by two or more conventional formulae, the phenomenon is known as resonance.

Hyperconjugation When an H—C bond is attached to an unsaturated n system (double or triple bond) which has an unshared p orbital, there is a conjugation.

In other words, the electrons of the H —C bond become less localised by entering into partial conjugation with the attached unsaturated system.

The interaction of and n electrons is made possible by the partial overlap of orbitals: sp³—s (C—H bond) with the empty p orbital of an adjacent positively charged carbon atom.

Hyperconjugation stabilises a molecule, for example, propene is more stable than expected.

Here the electrons of the C —H bond become partially delocalised through conjugation so that the hydrogen atoms are not free —only the ionic character of the C—H bond is increased. This is one way of looking at the hyperconjugative effect. Hyperconjugation may also be termed “no bond resonance”. It is a permanent effect.

Bond Fission

You already know from your previous classes that when a chemical reaction occurs the bonds in the reactant molecules are broken to form new ones in the products.

The whole mechanism of the reaction depends on the way these bonds are broken. The fission of a covalent bond can take place in two ways—homolytic and heterolytic.

The way in which the bond is broken depends more on the nature of the atoms constituting it and to some extent on the conditions prevailing during the reaction.

Bond fission leads to the formation of reaction intermediates, which are of transitory existence. Most organic reactions take place via the formation of reaction intermediates.

Being very reactive, reaction intermediates may or may not be isolated under normal conditions.

However, their structures have been established by indirect means either chemically or spectroscopically.

The most common reaction intermediates that we come across usually at the present level are carbocations, carbanions and free radicals.

Homolytic bond fission

When, during the fission of a bond, the separating atoms take away one electron each of the shared pair, the fission is symmetrical and called homolytic fission or homolysis.

Curved arrow notations are used to represent the shift of electrons. A single electron shift is represented by a half arrowhead (fish arrow).

The two fragments, which may be atoms or groups of atoms, obtained as a result of homolytic bond fission carry an unpaired electron each and are called free radicals.

These are highly reactive reaction intermediates and tend to pair up or react with other radicals or molecules to restore stability by being a part of the bonding pair.

Free radicals can be produced by photolysis or pyrolysis in which a bond is broken and ions are not formed.

For example, the cleavage of a chlorine or an alkyl halide molecule takes place in the presence of ultraviolet light or heat giving rise to two chlorine and one alkyl and one halide free radical respectively.

Alkyl free radicals can be classified as primary, secondary or tertiary, depending upon which hydrogen (1°, 2°, or 3°) is abstracted from the alkyl. Since the ease of abstraction of hydrogen atoms follows the sequence 3° > 2° > 1° > CH 4, the ease of formation (and stability) of free radicals follows the sequence:

Titus, the stability of alkyl free radicals increases from primary to tertiary. Organic reactions which have free radicals as intermediates are called free-radical or nonpolar reactions.

Heterolytic bond fission

In heterolytic cleavage, when the bond between two atoms breaks, both the electrons of the shared pair are taken over by one of the atoms.

Heterolytic cleavage is represented by a curved arrow that starts from the bonded pair of electrons and points at the atom which retains the shared pair of electrons.

Thus a heterolytic bond fission gives rise to one positive species, which lacks a pair of electrons in its valence shell and one negative species with a valence octet.

The more electronegative atom retains the shared pair of electrons. Heterolytic bond fission in organic molecules leads to the formation of either carbocations or carbanions.

In a carbocation, the carbon atom carries a positive charge, i.e., lacks a pair of electrons in its valence shell. In contrast, the carbanion, as the name suggests, is a negative group in which the carbon atom carries a negative charge, i.e., has an unshared pair of electrons.

The positively charged carbon atom with a sextet of electrons is sp² hybridised and it uses all its three hybridised orbitals to form bonds with other atoms. The remaining p_z orbital is empty and is perpendicular to the plane of the molecule.

The shape of a carbocation may be described as planar (trigonal coplanar) with a bond angle of 120°

Carbocations are also classified as primary (1°), secondary (2°) or tertiary (3°) depending on whether one, two or three carbon atoms respectively are attached to the carbon bearing the positive charge.

Since a carbocation has a positively charged (electron deficient) carbon, its relative stability is determined chiefly by how well it accommodates that charge.

Dispersal of the charge leads to the stability of the charged system. (This applies equally well to both positively and negatively charged systems.)

The greater the number of alkyl groups attached to a positively charged carbon atom, the greater the charge dispersion due to hyperconjugation lending to a greater stability of the carbocation.

In the hyperconjugative effect, electrons are provided by o bonds to the empty p orbital of the electron-deficient carbon. Thus, the order of stability of carbocations is

The arrows pointing towards the electron-deficient carbon indicate electron release from the alkyl groups. Benzyl and allyl carbocations, though primary in nature, are stable due to the resonance effect.

The greater the number of canonical structures for a carbocation, the more stable it is. In view of this, the order of stability of the carbocations is

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2>\mathrm{CH}_2=\mathrm{CH}-\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2>\mathrm{CH}_3 \mathrm{CH}_2 \stackrel{\stackrel{\circ}{\mathrm{C}} \mathrm{H}_2}{ }\)

A carbanion is a species containing a carbon atom with one pair of unshared electrons and thus carrying a negative charge.

The carbanion is pyramidal and the central carbon atom is sp³ hybridised with the unshared or lone pair of electrons occupying one sp³ hybrid orbital.

The order of stability of carbanions is just the reverse of that in carbocations. This can be explained on the basis of the +1 (inductive) effect of the alkyl group.

Being electron-releasing, the alkyl group tends to increase the electron density on the negatively charged carbon atom, lowering its stability. Thus, the order of stability of carbanions is

Types Of Reagents

As already stated, generally an organic reaction proceeds when a reagent attacks a substrate molecule.

Under suitable reaction conditions and the influence of an attacking reagent, the substrate molecule undergoes bond fission to form reaction intermediates, which combine with the reagent to give the final product.

Attacking reagents may be of two types—electrophilic and nucleophilic, depending on the electron density at the attacking centre.

Electrophilic reagents

These are electron-loving species. Being electron-deficient, they attack the electron-rich centre in the substrate molecule—that specific atom in the substrate molecule which is electron-rich. Electrophiles (E+) include positively charged species and neutral molecules.

For example, H+, H30+ and carbocations (CH₃ ,RH₂C+). Neutral molecules like A1C1₃, BF₃ and SO₃ act as electrophiles as they are electron-deficient.

Molecules having polarisable functional groups like the carbonyl group (ÿC=0) and alkyl halides (halogen atom is polarised) also act as electrophiles.

In alkyl halides, due to the polarity of the C—X bond, a partial positive charge develops on the carbon atom, which becomes an electrophilic centre

Nucleophile reagents

These reagents are nucleus-loving and attack electron-deficient centres in the substrate molecule. Nucleophiles (Nu:) may be negatively charged ions or neutral molecules with a lone pair of electrons. For example, Cl~,OH_,CN+ and carbanions (R3C:).

Neutral molecules containing a lone pair of electrons such as H₂O, R₃N, NH₃, ROH and ROR act as nucleophiles.

As stated earlier, the mechanism of organic reactions, apart from the reactivity of the substrate, the nature of bond fission and reagents, also depends on the type of reaction that occurs.

You will come across different types of reactions while studying the preparation and properties of hydrocarbons.

Purification Of Organic Compounds

To study the structure and properties of an organic compound, it is essential to obtain the compound in its purest form.

An organic compound isolated from various sources contains many types of impurities.

Many methods are used to purify organic compounds. The choice of the purification method employed depends on the type of impurities present and the nature of the organic compound.

The common methods of purification are as follows.

1. Crystallisation
2. Distillation
3. Differential extraction
4. Sublimation
5. Chromatography

Crystallisation

Crystallisation is the process of the formation of a crystalline solid from a solution, generally by the evaporation of the solvent.

The process is most commonly used for the purification of solid organic compounds. The method is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.

The choice of the solvent is very important. A solvent which dissolves less of the organic compound at room temperature and an appreciable quantity at a higher temperature is suitable for preparing the solution.

Also, the solvent should not dissolve the impurities and should not react with the compound. A hot saturated solution is filtered and then allowed to cool down slowly to room temperature when the compound separates in the form of crystals.

In case any coloured impurities are present in the organic compound, crystallisation is carried out by heating the hot saturated solution with a small amount of activated charcoal, which adsorbs the impurities.

Repeated crystallisation of a compound from a solvent to obtain the increasingly pure form is called recrystallisation.

It is generally done for the purification of compounds containing impurities of comparable solubilities. A mixture of two or more compounds can be separated by fractional crystallisation.

The method is based on the different solubilities of compounds in the same solvent. The less soluble compound separates out first while the more soluble one remains in the solution.

A mixture of benzoic acid and cane sugar can be separated by this method. Benzoic acid is less soluble in water than cane sugar.

Distillation

The process is used to separate liquids with appreciable differences in their boiling points. Also, a volatile liquid can be separated from nonvolatile impurities.

The process involves the conversion of the liquid into its vapours by heating followed by condensation of the vapours to obtain the pure liquid.

Distillation is of the following types:

1. Simple distillation
2. Fractional distillation
3. Distillation under reduced pressure, or vacuum distillation
4. Steam distillation

The choice of the distillation process depends on the nature of the liquids to be separated and the kind of impurities present in them.

Simple distillation

This method is used for the purification of liquids which boil without decomposition and contain nonvolatile impurities. The apparatus consists of a distillation flask with a side tube, connected to a Liebig condenser, through which water is passed.

The liquid to be purified is taken in the distillation flask. A few pieces of glass beads or unglazed porcelain are put in the flask.

This helps to avoid bumping (violent boiling of a liquid caused by super-heating) during distillation. For a liquid with a low boiling point, say 353-363 K, the flask is heated on a water bath and for a liquid with a high boiling point, say more than 373 K, an oil bath is used.

In place of the oil bath, the distillation flask can also be heated with a small flame over a wire gauze.

On heating, the liquid changes to vapours, which are passed through a condenser where they condense into liquid.

The pure distilled liquid is collected and the impurities are left behind in the distillation flask. The constant temperature at which most of the liquid distils is known as the boiling point of that liquid.

Fractional Distillation

Simple distillation cannot be used for the separation and purification of liquids with comparable boiling points. This is because both the liquids will be distilled simultaneously.

The separation of such liquid mixtures can be achieved by employing fractional distillation. This involves the use of a suitable fractionating column. Different types of fractionating columns can be used.

Basically, a fractionating column is a long tube filled with glass beads or a tube in which the walls have many inward-pointing indentations.

The fractionating column obstructs the passage of the vapours moving upwards and the liquid moving downwards. In fact, the indentations increase the cooling surface.

The fractionating column is fitted onto the mouth of a round bottom flask, which contains the liquid mixture.

When the flask is heated, the vapours of the more volatile liquid rise up in the fractionating column and the less volatile fraction condenses back into the flask (higher up the column, the temperature being lower).

As the vapour mixture rises up, it becomes rich in the low-boiling (volatile) component.

Thus, by the time the low-boiling vapours reach the top of the column, they become concentrated and pure. These vapours then pass through the condenser and pure liquid is collected.

After many successive distillations, the remaining liquid in the flask gets enriched in the high-boiling component.

Each successive condensation and vaporisation unit in the fractionating column is called a theoretical plate.

The temperature at the top of the fractionating column indicates the fraction being distilled.

Industrial applications of fractionating columns are in petroleum refineries, petrochemical plants and natural gas processing plants. This method is also used to separate the components of air.

Distillation under reduced pressure (vacuum distillation)

This method is used for the separation and purification of liquids which have high boiling points and those which decompose on heating. The distillation is carried out under reduced pressure.

The depression in the boiling point of the substance distilled means that the temperature is lower, which prevents the substance from decomposing. Low pressure can be achieved by using a vacuum pump.

The apparatus used in this method is more or less of the same type as that used in simple distillation except that instead of a round bottom flask, a two-necked flask, called the Claisenflask, is used.

In one of the necks, a capillary tube is fitted, which allows air bubbles to pass through (to prevent bumping) and in the second neck, a thermometer is fitted.

The pressure under which the liquid distils is measured with the help of a manometer.

A high-boiling liquid like glycerol (b.p. 563 K) can be easily separated from spent lye in the soap manufacturing process by this method.

Spent lye is a mixture of brine and glycerine, which is separated from crude soap during soap manufacturing. Another application of vacuum distillation is in the concentration of sugar cane juice.

Steam distillation

This method is used for the purification of those compounds which are volatile in steam, immiscible with water, and which contain nonvolatile impurities and possess a high vapour pressure.

In this process, the liquid to be purified is taken in a distillation flask and steam is continuously bubbled through it. The steam is generated using a steam generator.

The distillation flask is also heated so as to maintain the temperature at about 373 K. Initially, the steam heats the liquid but itself gets condensed.

But after some time, the mixture begins to boil. You already know that a liquid boils when its vapour pressure is equal to the atmospheric pressure.

In this case, water and the organic substance vaporise together and the total vapour pressure of the two becomes equal to the atmospheric pressure.

This implies that in steam distillation, the organic substance (being steam volatile) vapourises and gets distilled at a lower temperature than the boiling point, thus avoiding decomposition.

Thus, steam distillation serves the same purpose as distillation under reduced pressure.

The organic liquid is separated from water using a separating funnel. A mixture of water and aniline can be separated by this method. The process is also employed in the manufacture of essential oils.

Differential Extraction

This method is used to isolate or recover organic compounds (liquids or solids) present in an aqueous solution.

In this process, an aqueous solution of the organic compound is taken in a separating funnel, and mixed with a suitable organic solvent. The mixture is shaken.

The solubility of the organic compound should be higher in the organic solvent than in water. Also, the organic solvent should be immiscible with water.

Ether, benzene, chloroform, ethyl acetate, etc., are some of the common organic solvents used for this purpose.

When the contents are shaken thoroughly in the stoppered separating funnel, the organic phase dissolves the compound present in the aqueous phase.

The separating funnel is then allowed to stand for some time when the aqueous and the organic phase (solvent) separate as distinct layers. The two layers are extracted one after the other by opening the stop code, thus the phenomenon is known as differential extraction.

The process is repeated 2-3 times till all the organic compound originally dissolved in the aqueous layer is extracted with the organic solvent.

The organic compound is separated from the solvent by distillation. If the solubility of the organic compound is less in the solvent, then a large quantity of the solvent is used to extract a small quantity of the compound.

In that case, the same solvent is employed repeatedly. This technique is known as continuous extraction.

The method of differential extraction is useful for the isolation of nonvolatile compounds.

For example, benzoic add, which is soluble in water, can be extracted by the extraction of the aqueous solution using ether as the solvent.

Sublimation

On heating, some organic compounds change from the solid to the vapour state directly without passing through the intervening liquid state.

This process is called sublimation. Only those substances sublime whose vapour pressures equal the atmospheric pressure much before their melting points.

This process is useful for the purification of such solids which sublime on heating and contain nonvolatile impurities.

In this process, the impure solid is heated in a china dish which is covered with an inverted funnel.

A perforated filter paper is folded in the shape of the funnel and adjusted on the inner surface of the funnel. The stem of the funnel is plugged with cotton.

On heating, the substance volatilises and the vapours rise, solidify and get deposited on the inner, less hot surface of the inverted funnel. The non-volatile impurities are left in the china dish.

The perforations in the filter paper allow only the vapours to pass through it. The compounds which can be purified by sublimation are camphor, naphthalene and anthracene.

Chromatography

Chromatography is a method of separation in which the compounds to be separated are in two phases, one of which is stationary (stationary phase) while the other (mobile phase) moves in a definite direction.

The method was first used to separate coloured pigments found in plants, hence the name (Greek, Chroma-colour).

Generally speaking, the different chromatographic techniques work on either of the two principles— the difference between adsorption affinities and solubilities of the various components of the mixture.

Based on this, we shall study the two categories of chromatography—adsorption chromatography and partition chromatography.

Adsorption chromatography

Chromatography in which separation is based mainly on the differences between the adsorption affinities of the components in a sample for the surface of an active solid is called adsorption chromatography.

A porous solid with adsorptive properties which help in chromatographic separations is known as an active solid.

Here the stationary phase is the adsorbent (active solid, e.g., alumina). Thus the principle involved is differential adsorption.

The two main types of chromatography based on the principle of differential adsorption are column chromatography and thin-layer chromatography. Column chromatography The stationary bed, in this technique, is within a tube.

The adsorbent or solid stationary phase may fill the inside of the tube or be concentrated along the inside tube wall leaving an unrestricted path for the mobile phase in the middle part of the tube. The most commonly used adsorbents are alumina and silica gel.

Once the adsorbent is packed in the glass tube, the sample is poured into the column and continuously washed with a solvent a process known as elution. The solvent is called an elu

Different components of the sample are adsorbed to different extents and move down the column at different rates. Thus the most readily adsorbed components are at the top of the column.

The solution is the mobile phase in this case. The usual method is to collect the solution as it passes out from the column in fractions.

Thin-layer chromatography (TLC) Chromatography is carried out on a layer of adsorbent spread on a glass plate.

The layer is thin (about 0.2 mm) and known as a thin-layer chromatography plate or chrome plate. A small spot of the solution of the mixture is put on the plate at a distance of about 2 cm from the base.

This is done with the help of a capillary tube. The plate is then dried and placed in a glass jar containing the eluent at the bottom. It is ascertained that the spot on the TLC plate does not touch or dip in the eluent. The jar is covered with a glass plate.

The eluent moves up the plate due to capillary action and meets the sample mixture. The components of the mixture get dissolved in the solvent (eluent) and are carried up the plate.

The components of the mixture get separated into a number of spots due to their different adsorption affinities. Each component corresponds to a component of the mixture.

The plate is taken out, and the solvent front is marked and dried. Each component is finally eluted with a suitable solvent to get the pure component.

The relative adsorption of each component of the mixture is denoted by its retention factor (Rÿ).

The Rj values of a spot can be determined by dividing the distance travelled by that spot (component) by the total distance travelled by the solvent (solvent front). The distances are measured from the baseline.

The spots of coloured compounds are visible on the TLC plate but analysing colourless compounds becomes difficult. There are several methods to make the colourless spots visible.

If a small amount of fluorescent dye is added to the adsorbent, ultraviolet radiation absorbing, colourless spots can be seen. Iodine vapours are also used as the colour reagent.

Specific colour reagents are also sprayed on the TLC plate. For instance, ninhydrin gives a yellow colour when it reacts with the amino add proline and a blue colour with other amino acids.

Partition Chromatography

In this type of chromatography, the separation of the components of the mixture is based mainly on the differences between their solubilities in the stationary and mobile phases.

Paper chromatography is a type of partition chromatography. In this case, a strip of chromatographic paper acts as the adsorbent. The molecules of water trapped in the chromatographic paper serve as the stationary phase. The mobile phase is the suitable solvent or a mixture of solvents.

A spot of the mixture to be separated is placed near one edge of the paper and the sheet is suspended vertically in a solvent, which rises through the paper by capillary action, carrying the components with it. The components of the mixture move at different rates and separate.

The adsorption of components to different extents on the paper and the partition between the solvent and the moisture in the paper are the two factors responsible for the separation of the components of the mixture.

The paper is removed, the solvent front is marked, and the paper is dried. The dried chromatographic paper with a line of spots of different components of the mixture is called a chromatogram.

The components can be identified by the distance they move in a given time. Colourless substances are detected by using ultraviolet radiations or by spraying with a substance which reacts to give a coloured spot (substance).

Rf can be calculated for each component. If for a given system at a known temperature is a characteristic of the component and can be used to identify compounds.

Qualitative Analysis

Once the compound has been purified, its structure can be determined. To do so, it is necessary to identify the elements constituting the compound. This is known as qualitative analysis.

You already know that organic compounds invariably contain carbon and hydrogen. Besides, they may also contain nitrogen, sulphur, a halogen and phosphorus. The presence or the absence of these elements is ascertained as follows.

Detection Of Carbon And Hydrogen

The presence of carbon and hydrogen is detected by strongly heating the compound with cupric oxide.

The carbon in the compound is oxidised to carbon dioxide (tested by passing the gas through limewater) and hydrogen is converted into water (tested by using anhydrous copper sulphate, which turns blue).

⇒ \(\begin{aligned}
& \mathrm{C}+2 \mathrm{CuO} \longrightarrow \mathrm{CO}_2+2 \mathrm{Cu} \\
& 2 \mathrm{H}+\mathrm{CuO} \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{Cu}
\end{aligned}\)

⇒ \(\mathrm{CO}_2+\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \mathrm{CaCO}_3 \downarrow+\mathrm{H}_2 \mathrm{O}\)

⇒ \(
\begin{aligned}
& \mathrm{C}+2 \mathrm{CuO} \longrightarrow \mathrm{CO}_2+2 \mathrm{Cu} \\
& 2 \mathrm{H}+\mathrm{CuO} \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{Cu} \\
& \mathrm{CO}_2+\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \mathrm{CaCO}_3 \downarrow+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{H}_2 \mathrm{O}+\underset{\text { (White) }}{\mathrm{CuSO}_4} \longrightarrow \underset{\text { (Blue) }}{\mathrm{CuSO}_2 \cdot 5 \mathrm{H}_2 \mathrm{O}}
\end{aligned}
\)

Detection Of Other Elements

Apart from carbon and hydrogen, the other elements generally present are detected by Lassaigne’s test.

The compound to be analysed is fused with sodium metal. The elements covalently bonded to the carbon or the hydrogen atoms in the molecule of the compound get converted into water-soluble, ionic salts. The conversion to ionic forms makes the detection of elements easier.

⇒ \(Organic compound +\mathrm{Na} \longrightarrow \mathrm{NaCN}+\mathrm{Na}_2 \mathrm{~S}+\mathrm{NaX}
(C, H, O, N, S, X)\)

In case both nitrogen and sulphur are present in the same molecule, sodium thiocyanate (NaSCN) is obtained on fusion with sodium. The ionic salts arc extracted from the fused mass by boiling it with distilled water.

The solution obtained is the sodium extract or Lassaignc’s extract, which is then used to detect the elements.

Test for nitrogen

The sodium extract is heated to boiling with a few crystals of ferrous sulphate and then acidified with dilute sulphuric acid, and 2-3 drops of ferric chloride solution are added.

If nitrogen is present, a Prussian blue precipitate of ferric ferrocyanide is obtained. The blue colour is due to the cyanide ions.

Hydrazine (H2N—NH2) cannot be detected by Lassaigne’s test due to the presence of the N—N bond.

⇒ \(2 \mathrm{NaCN}+\mathrm{FeSO}_4 \longrightarrow \mathrm{Fe}(\mathrm{CN})_2+\mathrm{Na}_2 \mathrm{SO}_4\)

⇒ \(
\begin{gathered}
\mathrm{Fe}(\mathrm{CN})_2+4 \mathrm{NaCN} \longrightarrow \underset{\text { Sodium hexacyanoferrate(II) }}{\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]} \\
3 \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+4 \mathrm{FeCl}_3 \longrightarrow \underset{\text { Ferriferrocyanide }}{\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3+12 \mathrm{NaCl}}
\end{gathered}\)

The presence of nitrogen in a compound can also be detected by the soda lime test. The compound is heated with soda lime —the evolution of ammonia confirms the presence of nitrogen.

⇒ \(\underset{\text { Actamide }}{\mathrm{CH}_3 \mathrm{CONH}_2+\mathrm{NaOH}} \underset{\text { theat }}{\stackrel{\mathrm{CnO}}{\longrightarrow}} \mathrm{CH}_3 \mathrm{COONa}+\mathrm{NH}_3\)

The evolved ammonia can be tested by bringing a glass rod dipped in concentrated HC1 to the mouth of the test tube. White fumes of NH4C1 are seen.

Test for sulphur

Acetic acid followed by a lead acetate solution is added to the sodium extract. A black precipitate indicates the presence of sulphur.

⇒ \(\left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \mathrm{~Pb}+\mathrm{Na}_2 \mathrm{~S} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \underset{\text { (black) }}{\mathrm{PbS} \downarrow}+2 \mathrm{CH}_3 \mathrm{COONa}\)

Alternatively, when sodium nitroprusside solution is added to the sodium extract, a violet colour indicates the presence of sulphur.

⇒ \(\underset{\text { Sodium nitroprusside }}{\mathrm{Na}_2 \mathrm{~S}}+\underset{\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]}{\longrightarrow} \underset{\text { violet }}{\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]}\)

In case sulphur is present along with nitrogen, then the sodium extract will contain sodium thiocyanate, which instead of Prussian blue gives a blood red colour with ferric chloride due to the presence of the thiocyanate.

⇒ \(3 \mathrm{NaSCN}+\mathrm{FeCl}_3 \longrightarrow \underset{\substack{\text { Ferrithiocyanate } \\ \text { (blood red) }}}{\mathrm{Fe}(\mathrm{SCN})_3}+3 \mathrm{NaCl}\)

However, with excess sodium, during sodium fusion, the thiocyanate decomposes to yield cyanide and sulphide, which respond positively to their usual tests.

⇒ \(\mathrm{NaSCN}+2 \mathrm{Na} \longrightarrow \mathrm{NaCN}+\mathrm{Na}_2 \mathrm{~S}\)

Test for halogens

The sodium extract is boiled with dilute nitric acid to convert sodium cyanide or sodium sulphide (if present) to hydrogen cyanide and hydrogen sulphide respectively.

The cyanide or sulphide ions otherwise interfere with further tests. The cooled solution is treated with silver nitrate.

A white precipitate soluble in ammonium hydroxide confirms the presence of chlorine.

A yellowish precipitate sparingly soluble in ammonium hydroxide indicates bromine while a yellow precipitate insoluble in ammonium hydroxide confirms the presence of iodine.

The presence of bromine and iodine can also be detected by using carbon disulphide or carbon tetrachloride.

The sodium extract is boiled with dilute H2S04, cooled, and chlorine water followed by CS2 or CC14 is added.

The solution is shaken and allowed to stand. The layer of CS2 or CC14 (organic layer) turns yellow or orange in the presence of bromine and violet in case iodine is present.

Test for phosphorus

To detect phosphorus, the organic compound is fused with sodium peroxide, which is an oxidising agent.

The phosphorus present is oxidised to phosphate (Na3P04). The aqueous extract containing the phosphate is boiled with concentrated HN03 and a few drops of ammonium molybdate solution are added to it.

A yellow solution or precipitate shows the presence of phosphorus. This yellow precipitate is due to the formation of ammonium phosphomolybdate.

⇒ \(\mathrm{Na}_3 \mathrm{PO}_4+3 \mathrm{HNO}_3 \longrightarrow \mathrm{H}_3 \mathrm{PO}_4+3 \mathrm{NaNO}_3\)

⇒ \(\mathrm{H}_3 \mathrm{PO}_4+12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4+21 \mathrm{HNO}_3 \longrightarrow\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3+21 \mathrm{NH}_4 \mathrm{NO}_3+12 \mathrm{H}_2 \mathrm{O}\)

There is no direct test for the detection of oxygen in an organic compound. Its presence can, however, be inferred by testing for the presence of an oxygen-containing functional group like —OH, —CHO, —COOH and NO₂.

Quantitative Analysis

Once the various elements present in an organic compound are detected by qualitative analysis, in order to derive its empirical formula, it is necessary to determine quantitatively the percentages of these elements in the compound.

Estimation Of Carbon And Hydrogen

Liebig’s method In this method, a known mass of the organic compound is heated in a current of pure dry oxygen in the presence of cupric oxide till all the carbon and hydrogen are oxidised to carbon dioxide and water respectively.

⇒ \(\text { Organic compound }+\mathrm{O}_2+\underset{\text { excess }}{\mathrm{CuO}} \stackrel{\text { heat }}{\longrightarrow} \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{Cu}\)

After combustion, the gases are passed first through a U-tube containing anhydrous calcium chloride, which absorbs water and then through a U-tube containing potassium hydroxide solution, which absorbs carbon dioxide.

Both the U-tubes are weighed before and after the combustion so that the amount of HzO and CO₂ evolved from a known amount of the sample can be calculated.

Let the mass of the organic compound be m g.

• Mass of water formed = m1 g.
• Mass of carbon dioxide formed = m2 g.
• Since 44 g of CO₂ =12 g of carbon,
• m2 g of CO₂ \(\equiv \frac{12}{44} \times m_2\) ni2 g of carbon.

This is the mass of carbon present in m g of the organic compound.

∴ Percentage of C in the compound \(=\frac{12}{44} \times m_2 \frac{100}{m} \text {. }\)

Similarly, the percentage of H in the compound \(=\frac{2}{18} \times m_1 \times \frac{100}{m}.\)

Estimation Of Nitrogen

Nitrogen can be estimated either by the Dumas method or by the Kjeldahl method. Dumas method In this method, a weighed sample of the organic compound is heated with cupric oxide in an atmosphere of carbon dioxide.

On heating, the carbon and hydrogen are oxidised to CO₂ and H₂O respectively, while N2 is set free.

Any oxides of nitrogen produced during combustion are reduced back to nitrogen by passing the sample over a heated copper gauze. The gaseous mixture is collected in a nitrometer, which contains a solution of potassium hydroxide.

The mercury at the bottom of the nitrometer does not allow the potassium hydroxide solution to move from the nitrometer.

All gases except nitrogen are absorbed in the KOH solution. The volume of nitrogen gas collected at the top of the tube is measured.

The percentage of N in the given organic compound is calculated as follows.

• Let the mass of the organic compound = mg.
• The volume of nitrogen collected in the nitrometer = V mL.
• Room temperature = TK
• Pressure of dry nitrogen = (atmospheric pressure- aqueous tension at t° C)
  = p bar.

From the relation \(\frac{p_1 V_1}{t_1}=\frac{p_2 V_2}{t_2}\) We can find the volume of N2 gas at stp.

Experimental values

p₁ =pbar
V₁= VmL
t₁ =TK

At stp
p₂ = 1bar
V₂=?
T₂ = 273 K

Thus, \(V_2=\frac{p_1 \times V_1 \times 273}{T \times 1}\)

We know that 22,700 mL of N2 at stp weighs 28 g.

∴ \(V_2 \mathrm{~mL} \text { of } \mathrm{N}_2 \text { at stp will weigh } \frac{28 \times V_2}{22,700} \mathrm{~g} \text {. }\)

Percentage of N = \(\frac{\text { mass of nitrogen }}{\text { mass of organic compound }} \times 100\)

⇒ \(=\frac{28 \times V_2}{22700} \times \frac{100}{m} .\)

Kjeldahl method In this method a known mass of the organic compound is heated with concentrated sulphuric acid in the presence of copper sulphate, which acts as a catalyst.

The nitrogen of the organic compound gets converted into ammonium sulphate. The solution is then heated with an excess of sodium hydroxide.

⇒ \(\text { Organic compound }+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\)

⇒ \(\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{NH}_3+2 \mathrm{H}_2 \mathrm{O}\)

The liberated ammonia gas is absorbed in a known volume (taken in excess) of a standard solution of sulphuric add.

⇒ \(2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\)

The add that remains unused is estimated by titrating the solution with a standard solution of alkali.

The percentage of nitrogen in the organic compound can be calculated as follows.

• Let the mass of the organic compound = m g.
• Let the volume of the standard solution of added taking molarity M=V mL.
• Let the volume of unreacted acid (with ammonia) = V1 mL.

Let the volume of the standard solution of alkali of molarity M used for neutralising the unreacted acid = V₂ mL.

The chemical reaction involved in the titration is

⇒ \(2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}\)

Thus at the endpoint,

⇒ \(n_1 M_1 V_1=n_2 M_2 V_2 \text {, }\)

where n1, M₁ and V₁ are the basicity, molarity and volume respectively of the acid used and;n2, M₂ and V₂ are the acidity, molarity and volume respectively of the base used. All these quantities are relevant only to the titration.

‍∴ \(\quad V_1=\frac{n_2 M_2 V_2}{n_1 M_1}=\frac{1 \times M \times V_2}{2 \times M}=\frac{V_2}{2}\)

or the volume of unreacted acid (with ammonia) \((V)=\frac{V_2}{2},\)

∴ The volume of acid reacted with NH₃ = (V- V₂/2) mL.

∴ Millimoles of NH3 that react with add- 2(V- V₂/2)M

or moles of NH3 \(=\frac{2\left(V-V_2 / 2\right) M}{1000} .\)

Now moles of N in NH3 =1 x mole of NH3 \(=\frac{1 \times 2\left(V-V_2 / 2\right)}{1000} .\)

Mass of N \(=\frac{2\left(V-V_2 / 2\right)}{1000} \times 14 \mathrm{~g}\)

∴ Percentage of N in the organic compound

⇒ \(\begin{aligned}
& =\frac{2\left(V-V_2 / 2\right) M \times 14 \times 100}{1000 \times m} \\
& =\frac{14 \times M \times 2\left(V-V_2 / 2\right)}{m} .
\end{aligned}\)

Estimation Of Halogen

The quantity of a halogen in an organic compound is estimated by the Carius method. In this method, a known mass of the organic compound is taken in a hard glass-sealed tube (Carius tube) with fuming nitric acid in the presence of silver nitrate.

The Carius tube is then heated in a furnace. On heating carbon and hydrogen are oxidised to CO₂ and H₂O respectively while halogens form a precipitate of silver halide (AgX). The silver halide precipitate is filtered and washed with water and alcohol. It is then dried and weighed.

The percentage of the halogen is calculated as follows.

• Let the mass of the organic compound = m g.
• Mass of silver halide (AgX) obtained = m2 g.

108 + x parts by mass of AgX contains x parts by mass of halogen (x is the atomic mass of the halogen atom and 108 is the atomic mass of Ag).

(108 + x) g of AgX contains x g of X.

Therefore, mt g of AgX contains \(\frac{x}{108+x} \times m_1 g \text { of } \mathrm{X} .\)

Percentage of halogen X \(\mathrm{X}=\frac{x \times m_1}{(108+x)} \times \frac{100}{m} .\)

Estimation Of Sulphur

Sulphur is also estimated by the Carius method. A known mass of the organic compound is heated with fuming nitric acid in the Carius tube.

Sulphur is converted to sulphuric acid, which is precipitated by barium chloride as barium sulphate.

The precipitate of barium sulphate is filtered, washed, dried and weighed. From the mass of barium sulphate obtained, the percentage of sulphur can be calculated.

⇒ \(\begin{aligned}
& \mathrm{S}+\mathrm{H}_2 \mathrm{O}+3 \mathrm{O} \stackrel{\mathrm{HNO}_3}{\longrightarrow} \mathrm{H}_2 \mathrm{SO}_4 \\
& \mathrm{H}_2 \mathrm{SO}_4+\mathrm{BaCl}_2 \longrightarrow \mathrm{BaSO}_4 \downarrow+2 \mathrm{HCl}
\end{aligned}\)

Let the mass of the organic compound = m g.

Let the mass of the precipitate = m1 g.

233 g of BaS04 contains 32 g of S.

Therefore, m1 g of BaSQ4 contains \(\frac{32}{233} \times m_1 g \text { of } S\)

Percentage of S \(=\frac{32}{233} \times m_1 \times \frac{100}{m} .\)

Estimation Of Phosphorus

A known mass of the organic compound is heated with fuming nitric acid. The phosphorus in the organic compound is oxidised to phosphoric acid.

It is then treated with magnesia mixture (a solution containing magnesium chloride, ammonium chloride and a little ammonia).

A precipitate of magnesium ammonium phosphate (MgNH4P04) is obtained. It is filtered, washed, dried and then ignited to give magnesium pyrophosphate (Mg2P₂O7), which is then weighed.

⇒ \(\mathrm{H}_3 \mathrm{PO}_4+\left(\mathrm{MgCl}_2+\mathrm{NH}_4 \mathrm{Cl}\right) \longrightarrow \mathrm{MgNH}_4 \mathrm{PO}_4
Magnesia mixture\)

Therefore, m1 g of Mg₂P₂O₇ \(\frac{62}{222} \times m_1\) contains of phosphorus

Percentage of phosphorus \(=\frac{62}{222} \times \frac{m_1}{m} \times 100\)

Alternatively, phosphoric acid may be precipitated as ammonium phosphomolybdate, (NH4)3P04 -12Mo03, by adding ammonia and ammonium molybdate.

⇒ \(\mathrm{H}_3 \mathrm{PO}_4+\underset{\substack{\text { Ammonium } \\ \text { molybdate }}}{12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4}+21 \mathrm{HNO}_3 \longrightarrow \underset{\substack{\text { Ammonium } \\ \text { phosphomolybdate }}}{\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 \downarrow}+21 \mathrm{NH}_4 \mathrm{NO}_3+12 \mathrm{H}_2 \mathrm{O}\)

Then the percentage of phosphorus will be \(=\frac{31 \times m_1 \times 100}{1877 \times m},\) where m mass 0f ammonium phosphomolybdate.

Estimation Of Oxygen

The percentage of oxygen is generally determined by subtracting the sum of percentages of all other elements from 100. However, it can be estimated directly by the following method.

A known amount of the organic compound is decomposed by heating in nitrogen gas.

The mixture of the gaseous product formed is passed over red hot coke when all the oxygen (present in the mixture) is converted into carbon monoxide.

The mixture is then passed through warm iodine pentoxide (I₂O₅) when carbon monoxide is oxidised to carbon dioxide and iodine is reduced.

⇒ \(\begin{gathered}
2 \mathrm{C}+\mathrm{O}_2 \stackrel{1373 \mathrm{~K}}{\longrightarrow} 2 \mathrm{CO} \\
\mathrm{I}_2 \mathrm{O}_5+5 \mathrm{CO} \longrightarrow 5 \mathrm{CO}_2+\mathrm{I}_2
\end{gathered}\)

The percentage of oxygen can be calculated from the amount of carbon dioxide or iodine produced.

Let the mass of the organic compound = mg.

Let the mass of carbon dioxide produced = m g.

44 g of CO₂ contains 32 g of O.

Therefore, m1 g of CO₂ contains \(\frac{32}{44} \times m_1 g \text { of } \mathrm{O}\)

Percentage of O \(=\frac{32 \times m_2 \times 100}{44 \times m} .\)

Earlier, for quantitative analysis large amounts of the pure compound (about 1 g) were required.

However, with the advances in the recent past, it is now possible to carry out the analysis with 3-4 mg of the substance. The accuracy in these estimations is ±0.03%. Automatic CHN elemental analysers are also available, which give results in a few minutes.

Example 1. Calculate the percentage of carbon, hydrogen and oxygen, if 02722g of an organic compound on complete combustion gives 05545g of carbon dioxide and 02227 g of water.
Solution: Percentage of C =

⇒ \(\begin{aligned} & =\frac{12}{44} \times \frac{\text { mass of } \mathrm{CO}_2}{\text { mass of compound }} \times 100 \\
& =\frac{12}{44} \times \frac{0.5545}{0.2722} \times 100=54.5 .
\end{aligned}\)

Percentage of H=

⇒ \(\begin{aligned} & =\frac{2}{18} \times \frac{\text { mass of } \mathrm{H}_2 \mathrm{O}}{\text { mass of compound }} \times 100 \\ & =\frac{2}{18} \times \frac{0.2227}{0.2722} \times 100=9 \end{aligned}\)

Thus, the percentage ofO =100- (54.4 + 9) = 36.6.

Example 2. During the estimation of nitrogen by the Dumas method, 0.45 g of an organic compound gave 75mL of nitrogen, which was collected at 300 K and at 715 mmHg pressure. Calculate the percentage of nitrogen in the compound. (The vapour pressure of water at 300K is 15mmHg.)
Solution: Vapour pressure of gas = (715- 15) mmHg = 700 mmHg.

Using the gas equation \(\frac{p_1 V_1}{t_1}=\frac{p_2 V_2}{t_2}\) the volume of nitrogen at 760 mmHg pressure and 0°C can be calculated as

⇒ \(V_2=\frac{700 \times 75 \times 273}{760 \times 300} \mathrm{~mL} .\)

22,400 mL of nitrogen at 760 mmHg pressure and (PC weighs 28 g.

Therefore, \(\frac{700 \times 75 \times 273}{760 \times 300}\)
mL of nitrogen at the same pressure and temperature weighs

⇒ \(\frac{28}{22,400} \times \frac{760 \times 300}{700 \times 75 \times 273}=0.0786 \mathrm{~g} .\)

Percentage of nitrogen \(=\frac{0.786}{0.45} \times 100=17.46\)

Example 3. During the estimation of nitrogen by the Kjeldahl method 0.5 g of an organic compound evolved ammonia, which was absorbed in 100 mL of \(\frac{M}{10} \mathrm{H}_2 \mathrm{SO}_4 \text {. }\)

The unused acid required 150 mL \(\frac{M}{10}\) NaOH solution. Calculate the percentage composition of nitrogen in the compound.

Solution: The chemical reaction taking place in the titration is

⇒ \(2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}\)

i.e., 2 moles of NaOH react with 1 mole of H₂SO₄.

M₁V₁=M₂V₂.

If the volume of the unused H2S04 is V1 then,

⇒ \(\begin{array}{rlrl}
& & V_1 \times \frac{1}{10} & =150 \times \frac{1}{10} \times \frac{1}{2} \\
\text { or } & V_1 & =75 \mathrm{~mL} .
\end{array}\)

Therefore, the volume of \(\frac{1}{10} \mathrm{MH}_2 \mathrm{SO}_4\) used by NH3 = 100- 75 = 25 mL.

Millimoles of H2S04 used by NH3 \(=25 \times \frac{1}{10}=2.5 .\)

Since \(2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4,\)

millimoles of NH3 = 2 x millimoles of H2S04

\(=2 \times 2.5=5.0 \text {. }\)

Mass of NH3 formed = moles x molar mass

\(=5.0 \times 10^{-3} \times 17 \text {. }\)

Mass of N \(=\frac{5.0 \times 10^{-3} \times 17 \times 14}{17} \mathrm{~g}\)

\(=5.0 \times 10^{-10} \times 14 \mathrm{~g}\) \(\text { Percentage of } \mathrm{N}=\frac{1 .}{1000} \times 5.0 \times \frac{100}{0.5}=14 .\)

Example 4. During bromine estimation by the Carius method, 0.165 g of a compound gave 0.132 g of AgBr. Find the percentage of bromine in the compound.
Solution: Molecular mass of AgBr = 188

188 g of AgBr contains 80 g of bromine.

0.132 g of AgBr contains \(\frac{80}{188} \times 0.132 \mathrm{~g} \text { of bromine. }\)

Percentage of bromine in the compound \(=\frac{80 \times 0.132}{188} \times \frac{100}{0.165}=34 .\)

Example 5. During sulphur estimation by the Carius method, 0.434 g of an organic compound gave 0640 g of BaSO, Calculate the percentage of sulphur in the compound.
Solution: 233 g of BaS04 contains 32 g of sulphur.

0.640 g of BaS04 contains \(\frac{32}{233} \times 0.640 \mathrm{~g}\) of sulphur.

Percentage of S \(=\frac{32}{233} \times \frac{0.640 \times 100}{0.434}=20.27\)

Organic Chemistry—Some Basic Principles And Techniques Multiple Choice Questions

Question 1. Which of the following has the least C—C bond length?

1. Ethane
2. Ethene
3. Ethyne
4. All are equal

Answer: 3. Ethyne

Question 2. How many tertiary hydrogen atoms are present in

1. One
2. Two
3. Three
4. Four

Answer: 1. One

Question 3. The compounds butene-1 and isobutene exhibit

1. Chain isomerism
2. Position isomerism
3. Functional isomerism
4. Metamerism

Answer: 1. Chain isomerism

Question 4. Which of the following has maximum -I effect?

1. OH
2. COOH
3. Br
4. NQ₂

Answer: 4. NQ₂

Question 5. Which of the carbon atoms numbered will have a maximum 6+ charge in the given compound?

⇒ \(\stackrel{3}{\mathrm{C}} \mathrm{H}_3-\stackrel{2}{\mathrm{C}} \mathrm{H}_2-\stackrel{1}{\mathrm{C}} \mathrm{H}_2-\mathrm{Cl}\)

1. Carbon 1
2. Carbon 2
3. Carbon 3
4. All will have equal 5+ charges.

Answer: 1. Carbon 1

Question 6. Which of the following carbocations is the most stable?

1. \(\mathrm{CH}_3 \mathrm{CH}_2 \stackrel{+}{\mathrm{C}} \mathrm{H}_2\)
2. \(\mathrm{CH}_2=\mathrm{CH}-\stackrel{+}{\mathrm{C}} \mathrm{H}_2\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{H}_2\)
4. All are Equally Stable

Answer: 3. \(\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{H}_2\)

Question 7. Which of the following free radicals is the most stable?

1. Methyl
2. Ethyl
3. Isopropyl
4. Tertiary butyl

Answer: 4. Tertiary butyl

Question 8. Which of the following is/are electrophilic?

1. NO_2⁺
2. 
3. C₂H₅Cl
4. All the above

Answer: 4. All the above

Question 9. In which of the following reactions has the hybridised state of a carbon atom changed?

Answer: 3 and 4

Question 10. The IUPAC name of

1. Butanal
2. Butan-l-al
3. Butane-l-carboxaldehyde
4. Pentan-l-al

Answer: 4. Pentan-l-al

Question 11. The IUPAC name of CH₃CH=CH—C=CH is

1. 4-Pentyn-2-ene
2. l-Pentyn-3-ene
3. 3-Penten-l-yne
4. 2-Penten-4-yne

Answer: 3. 3-Penten-l-yne

Question 12. Identify the molecule/ion in which the inductive, resonance and hyperconjugative effects are operative.

1. (CH₃)3C
2. CH₂=CH—CH=0
3. (CH₃)3C—CH₂
4. CH₃CH=CH—CH=0

Answer: 4. CH₃CH=CH—CH=0

Qualitative 13. The smallest alkane with a 3° carbon is

1. Neopentane
2. Isopentane
3. Isohexane
4. Isobutane

Answer: 4. Isobutane

Question 14. Sublimation can be used for the purification of

1. Glucose
2. Acetamide
3. Naphthalene
4. Formic acid

Answer: 3. Naphthalene

Question 15. Glycerol boils at 536 K and decomposes at its bp. It can be purified by

1. Fractional distillation
2. Steam distillation
3. Sublimation
4. Distillation under reduced pressure

Answer: 4. Distillation under reduced pressure

Question 16. A mixture of benzene and toluene can be separated by

1. Steam distillation
2. Fractional distillation
3. Vacuum distillation
4. Simple distillation

Answer: 2. Fractional distillation

Question 17. Which of the following will respond positively to Lassaigne’s test for nitrogen?

1. Nitrobenzene
2. Acetamide
3. Urea
4. All the Above

Answer: 4. All the Above

Question 18. In Lassaigne’s test, when both N and S are present, a blood red colour is obtained. This is due to the formation of

1. Ferriferrocyanide
2. Ferrithionate
3. Sodium thiocyanate
4. Ferricyanide

Answer: 2. Ferrithionate

Question 19. A mixture of acetone (b.p. 329 K) and methyl alcohol (b.p. 338 K) can be separated by

1. Simple distillation
2. Distillation under reduced pressure
3. Steam distillation
4. Fractional distillation

Answer: 4. Fractional distillation

Question 20. In column chromatography, the separation of a mixture of two substances depends on their

1. Different solubilities
2. Different melting points
3. Different specific gravities
4. Differential adsorption

Answer: 4. Differential adsorption

Question 21. A substance which is insoluble in water and possesses a vapour pressure of 10-15 mmHg at 373 K can be purified by

1. Crystallisation
2. Distillation
3. Steam distillation
4. Sublimation

Answer: 3. Steam distillation

Question 22. In steam distillation, the total pressure of the organic substance and water vapour becomes

1. Equal to the atmospheric pressure
2. More than the atmospheric pressure
3. Less than the atmospheric pressure
4. There is no fixed relationship

Answer: 1. Equal to the atmospheric pressure

Question 23. 0.4 g of an organic compound on complete combustion gives 0.18 g of water. What will be the percentage of hydrogen in the compound?

1. 10
2. 15
3. 5
4. 2.5

Answer: 3. 5

Question 24. Which of the following compounds will give a blood-red colouration when nitrogen is present?

1. M-nitrobenzoic add
2. P-toluidine
3. Urea
4. M-nitrobenzene sulphonic add

Answer: 4. M-nitrobenzene sulphonic add

Question 25. In the Kjeldahl method, the nitrogen present in an organic compound is estimated by measuring the amount of

1. N₂
2. NO₂
3. NH₃
4. (NH₄)2SO₄

Answer: 3. NH₃

Question 26. Which of the following reagents is added to the sodium extract of an organic compound to detect die presence of sulphur in the compound?

1. Sodium nitroprusside
2. Ammonium molybdate
3. Ammonium phosphomolybdate
4. Sodium hexacyanoferrate

Answer: 1. Sodium nitroprusside

Question 27. In the sodium fusion test of organic compounds, the nitrogen of an organic compound is converted to

1. Sodium nitrate
2. Sodium nitrite
3. Sodium amide
4. Sodium cyanide

Answer: 4. Sodium cyanide

Question 28. Carbon and hydrogen present in an organic compound are estimated by

1. The dumas method
2. The various method
3. The Liebig method
4. None of these

Answer: 3. The lie big method

The P-Block Elements

The elements in which the last electron enters the p orbital constitute the p-block elements. There are three degenerate p orbitals in a shell and each can accommodate a maximum of two electrons.

When tire energy levels are filled in the sequence to build up elements, the number of electrons present in the p orbitals varies from one to six, and hence there are six groups of p-block elements.

General Trends

The valence-shell electronic configuration of p-block elements is Ns²np^1-6 except for that of He, which is Is2. The size and metallic character of the elements increase on moving down the group and decrease on moving from left to right across a period.

Unlike s-block 1s² elements, which are all metallic in nature, the p-block elements comprise metals, metalloids and nonmetals; the lighter members are nonmetals while the heavier ones are predominantly metallic.

The chemistry of p-block elements is complex as both metals and nonmetals are to be dealt with in contrast to s-block elements, which are all metals. In addition, there are some elements whose properties are intermediate between those of the metals and the nonmetals.

These are referred to as metalloids, e.g., arsenic and antimony. It may be emphasised that except in the p block, nonmetals and metalloids do not occur anywhere in the periodic table.

Metals generally have low ionisation enthalpies and low electronegativities and form cations whereas nonmetals form anions. The p-block elements form both ionic and covalent compounds.

The ionisation enthalpy, electronegativity and oxidising power decrease down the group and increase across a period in the p block. A striking behaviour of the p-block elements is the display of two oxidation states—higher and lower.

The higher oxidation state is equal to the group number minus 10, i.e., the total number of s and p electrons in the valence shell and the lower one is two units less than the higher oxidation state, i.e., equal to the number of p electrons of the valence shell.

The stability of the lower oxidation state increases on moving down the group.

The p-block elements exhibit a higher oxidation state when both the ns and up electrons of the valence shell are used in bond formation. On the other hand, the lower oxidation state is observed only when the tup electron(s) of the valence shell participate in bonding.

This reluctance of the outermost s-orbital electron pair to participate in bond formation is called the inert-pair effect. This can be explained in terms of energy.

On the one hand, energy is needed to uncouple the valence-shell-paired s electrons and on the other, it is released when the electrons participate in bond formation.

If the bond energy is high enough to compensate for the amount needed to uncouple, the s electrons participate in bond formation.

On moving down the groups, the bond energy decreases and the valence-shells electrons remain paired and the lower oxidation state becomes more stable.

In contrast to s-block elements, which display only positive oxidation states, p-block elements display both positive and negative oxidation states.

As the electronegativity of the elements increases tire negative oxidation states become more stable.

This will become clearer to you when we discuss specific examples while studying the different groups separately.

A regular gradation in properties is not noted in the p-block elements (unlike the s-block elements) as we come across both metals and nonmetals.

The display of multiple oxidation states adds to the complexity. The first member behaves differently from the other members of the group.

We have observed this in s-block elements also —lithium and beryllium, due to their small size, show anomalous behaviour.

Another important characteristic of the p-block elements is their ability, from the third period onwards, to utilise the vacant d orbitals and hence expand their covalence.

Boron (second period) has a maximum covalence of four (one 2s and three 2p orbitals), e.g., BF^4-, but aluminium (third period) can utilise 3d orbitals in addition to 3s and 3p and form species like [AlF₆]^3- where the covalence is six.

In the third period elements of p-block, the general valence-shell electronic configuration is 3s² 3pⁿ.

The atoms of the elements have vacant 3D orbitals lying between the 3p and 4s energy levels. Using these orbitals the elements expand their covalence above four.

Another unique feature of p-block elements is the ability of the first member of each group to form pπ-pπ multiple bonds with itself or with the other elements in the same row.

Thus we have innumerable species containing C=C and C=C bonds, but hardly any displaying Si=Si bonds. Elemental nitrogen and oxygen contain triple and double bonds respectively (N=N, 0=0) but for the lower members such linkages are not observed.

There are also many compounds which have C=0, C=N, and N=0 bonds. A pπ-pπ bond is formed by the sideways overlap of p orbitals.

For maximum overlap to occur, the orbitals should be small and of comparable size. For the heavier members, the orbitals become large and diffuse and the effective overlap does not occur.

This does not mean that the heavier (or lower) members do not form rr-bonds.

In fact, in SO₄²– and PO₄³– there is multiple bonding between sulphur and oxygen, and phosphorus and oxygen respectively. These bonds are formed involving p orbitals of oxygen and d orbitals of sulphur and phosphorus and are referred to as pπ-dπ bonds.

The d orbitals are higher in energy than the p orbitals, so they contribute less to the stability of the molecule than the pn-pn bond. You will study more about this in higher classes.

We will now discuss in detail Groups 13 and 14 with special emphasis on boron, aluminium, carbon and some important compounds.

Group 13 Elements

The elements in this group are boron, aluminium, gallium, indium and thallium. Boron is a nonmetal while the others are fairly reactive metals. The metallic character increases down the group. The general valence-shell electronic configuration is ns² np¹.

Occurrence

Boron is a rare element and occurs to the extent of 0.0001% by mass in the earth’s crust. There are two isotopes of boron — ¹⁰ B and 11 B —abundant in the ratio 1:4.

Boron occurs principally in the earth’s crust as boric acid (H3B03) and as borates, such as borax (Na₂B₄O₇.10H₂O), kemite (Na₂B₄O₇.4H₂O) and colemanite, (Ca₂B₆O₁₁ -5H₂O).

Aluminium is the most abundant metal (8.13%) in the earth’s crust and the third-most abundant element (after oxygen and silicon).

The most important ore of aluminium is bauxite, Al₂O₃ .xH₂O (varies between and 3). Bauxite is a mixture of aluminium oxide and hydroxide.

The other ores of aluminium are cryolite ( Na₃ A1F₆ ) and corundum (A1₂O₃). Aluminium is widely present in aluminosilicate rocks like feldspars and micas. These rocks form clay minerals on weathering.

Aluminium, however, cannot be extracted from feldspars, micas and clay minerals. Gallium, indium and thallium are much less abundant and mainly occur as sulphide ores.

The main ore of gallium is germanite (a mixed sulphide of zinc, copper, gallium, germanium and arsenic).

Indium and thallium are present in traces in the sulphide ores of zinc and lead respectively.

The important physical properties of Group 13 elements

Electronic Configuration

The general outer electronic configuration of the elements is ns2npx. The inner cores of boron and aluminium have noble-gas configuration while for the other members, there are ten d electrons present in addition to the noble-gas configuration.

The d electrons affect size, ionisation enthalpy, electronegativity and chemical properties of the elements.

Atomic and ionic radii

The atomic radii of Group 13 elements are less than those of the corresponding s-block elements.

The atomic size does not increase in a regular manner, as expected, on descending the group. The atomic radius of A1 (143 pm) is greater than that of B (85 pm) which is the expected trend.

However, the atomic radius of Ga (135 pm) is less than that of Al. This is because B and A1 follow the s-block elements whereas Ga follows immediately after the d-block elements.

Therefore, the inner core of Ga contains ten d electrons, which do not shield the nuclear charge efficiently.

(Shielding of electrons is in the order: s > p > d > f.) Therefore, the effective nuclear charge of Ga is more than that of Al so the outer electrons are attracted towards the nucleus and the size is smaller than expected.

Similarly, the inclusion of fourteen, poorly shielding 4f electrons affects the size of Tl.

Ionisation enthalpy

There is a decrease in ionisation enthalpies of the elements on moving down the group, but the decrease is not regular. The ionisation enthalpies of Ga and Tl are higher than expected due to the higher effective nuclear charge as a result of poor screening by d and f electrons.

The second and third ionisation enthalpies of Group 13 elements are very high as compared to their first ionisation enthalpies, which are even lower than those of the corresponding Group 2 elements.

This is because the first electron to be removed from Group 13 elements is an unpaired p electron whereas that in Group 2 elements is a paired s electron.

And we already know that the s orbital is more penetrating than the p orbital. Since the sum of the first three ionisation enthalpies is immense, it follows that a lot of energy is needed to form tri positive cations (M³⁺) B forms covalent compounds, as the sum of its first three ionisation enthalpies is very high.

Many simple compounds of Al and Ga are covalent when anhydrous. However, in solution, the large amount of hydration enthalpy evolved compensates for the high ionisation enthalpy and the compounds formed by Group 13 elements are ionic in nature.

Electronegativity

Due to irregularities in the atomic size of the elements, a regular variation in electronegativity is not noted.

Electronegativity first decreases from B to Al and then increases to a small extent. This is due to the differences in the atomic structures of the elements.

Oxidation states

The common oxidation states are +3 and +1. As already discussed, the lower oxidation state arises due to the inert pair effect.

The stability of the +1 oxidation state increases down the group because ns electrons are prevented from participating in bond formation due to the increased effective nuclear charge.

Thus for B, Al and Ga, the +3 oxidation state is stable (the +1 oxidation state for Ga is reducing), for both are comparable instability, while for Tl the predominant oxidation state is +1 (+3 state is oxidising). Ga appears to be divalent in GaCl₂, but actually, GaCl₂ is represented as Ga⁺[GaCl₄ ]^– which contains Ga(I) and Ga(III).

Physical properties

• The precise determination of the physical properties of elemental boron is hampered due to two difficulties— the existence of a large number of allotropic forms and contamination by irremovable impurities.
• It is a black, extremely hard, refractory solid with a high melting point, low density and low electrical conductivity.
• The other elements are low-melting, rather soft metals and are good conductors of electricity.
• Gallium has an unusually low melting point and contracts on melting.

Chemical reactivity

Tire elements of Group 13 are less reactive than those of the s block.

They generally form covalent compounds due to their small size, a large sum of the first three ionisation enthalpies and higher electronegativity values than those of the corresponding Group 1 and Group 2 elements.

Reaction with air

All elements of Group 13 react with oxygen at high temperatures to form oxides—M₂O₃

⇒ \(4 \mathrm{M}(\mathrm{s})+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{M}_2 \mathrm{O}_3(\mathrm{~s})\)

Thallium also forms T1₂O. A very thin layer of oxide forms on aluminium which prevents further reaction.

The basic character of oxides increases down the group—B₂O₃ is acidic, A1₂O₃ and Ga₂O₃ are amphoteric whereas ln₂O₃ and T1₂O₃ are basic.

T120 dissolves in water to form TlOH, which is as strong a base as KOH. Only boron and aluminium react with nitrogen at a very high temperature to form nitrides.

⇒ \(2 \mathrm{Al}+\mathrm{N}_2 \rightarrow 2 \mathrm{AlN}\)

Reaction with halogens

Trihalides of all elements are known. In addition, thallium also forms monohalides. The trihalides are electron-deficient compounds as the central atom has only six electrons (3 valence electrons and 3 from halogens, e.g., BC1₃).

Such compounds have a strong tendency to accept a pair of electrons and complete the octet, and thus act as Lewis acids.

On moving down the group, the tendency to behave as Lewis acid decreases. BC1₃ or BF₃ readily accepts a pair of electrons from Lewis bases like ammonia to form BF₃-NH₃.

Aluminium chloride remedies the electron deficiency by forming a dimer (a halogen-bridged molecule) in which each aluminium atom completes its octet by accepting an electron pair from chlorine.

The halides are covalent and get hydrolysed in water, forming four coordinate species like [M(OH)₄]^–.

These are tetrahedral in structure where the central element, M, is sp3 hybridised. The halides also have a tendency to form octahedral hydrated species like [A1(H₂O)₆]³⁺, which is formed when aluminium chloride is dissolved in acidified water. In such species, the central metal is sp³d² hybridised. Since boron does not have vacant d orbitals, a similar species containing boron is not known.

Reaction with acids and alkalis

Boron is unreactive towards acids and alkalis. Aluminium, however, dissolves in both acids and alkalis, liberating hydrogen.

⇒ \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \rightarrow 2 \mathrm{AlCl}_3(\mathrm{aq})+3 \mathrm{H}_2(\mathrm{~g})\)

⇒ \(2 \mathrm{Al}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \underset{\text { sodium tetrahydroxoaluminate (III) }}{2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right](\mathrm{aq})+3 \mathrm{H}_2(\mathrm{~g})}\)

Boron

Boron is a hard solid and exists in different allotropic forms. It is dark brown to black in colour and has low electrical conductivity.

Anomalous behaviour

• As with Groups 1 and 2, the tilt member of Group 13 is different from the other members in certain aspects.
• The main reasons for the anomalous behaviour are the small size of the boron atom and the nonavailability of d orbitals.
• Boron, apart from the differences with Al( Ga, In and Tl, also exhibits a diagonal relationship with silicon.

Some of the main differences in the characteristics of boron from the rest of the members of the group are as follows.

1. B is a nonmetal whereas Al, Ga, In and Tl are metals.
2. B₂O₃ (like SiO₂) is an acidic oxide, in contrast to Al₂O₃, which is amphoteric.
3. Boric acid, H₃BO₃, which is represented as B(OH)₃, is acidic whereas Al(OH)₃ is amphoteric.
4. Borates and silicates polymerise to form chains, rings and sheet structures. Aluminium forms no such compounds.
5. The hydrides of B and Si are gaseous, readily hydrolysed and inflammable whereas aluminium hydride is a nonvolatile polymeric solid.
6. BC1₃ is monomeric whereas the analogous aluminium compound (Al₂Cl₆) is dimeric in the solid state.

Chemical Properties

Boron has three outer electrons but it does not form B³⁺ ions (B is always covalent) due to its small size and high ionisation enthalpy. Pure crystalline boron is quite unreactive.

However, amorphous boron is more reactive. It reacts with dioxygen at a high temperature to form boron trioxide.

⇒ \(4 \mathrm{~B}+3 \mathrm{O}_2 \stackrel{973 \mathrm{~K}}{\longrightarrow} 2 \mathrm{~B}_2 \mathrm{O}_3\)

Boron combines with halogens at elevated temperatures to form the respective trihalides.

⇒ \(2 \mathrm{~B}+3 \mathrm{X}_2 \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{BX}_3\)

As already stated, the trihalides formed by Group 13 elements are electron-deficient compounds.

The bonded boron atom is surrounded by six electrons (three from B and one each from the three halogen atoms).

Boron halides act as Lewis acids and readily accept an electron pair from Lewis bases to complete the octet.

The boron trihalides, despite being electron deficient, are stable and do not dimerise.

They remedy the electron deficiency by forming a pπ-pπ bond, where tine empty 2p orbital on boron accepts nonbonding electrons of the halogen, as shown.

The molecule is a resonance hybrid of three canonical forms.

Thus the boron-fluorine bond length in BF3 is shorter than that of a single bond.

On heating to a very high temperature boron reacts with dinitrogen to form boron nitride, which is a slippery, white solid having a layered structure like that of graphite.

⇒ \(2 \mathrm{~B}+\mathrm{N}_2 \rightarrow 2 \mathrm{BN}\)

When fused with sodium hydroxide boron forms borates, and liberates hydrogen.

⇒ \(2 \mathrm{~B}+6 \mathrm{NaOH} \rightarrow 2 \mathrm{Na}_3 \mathrm{BO}_3+3 \mathrm{H}_2\)

Boron combines with many metals (not Group 1) to form borides, which are hard, high-melting solids.

⇒ \(\begin{aligned}
3 \mathrm{Mg}+2 \mathrm{~B} & \rightarrow \mathrm{Mg}_3 \mathrm{~B}_2 \\
\mathrm{Cr}+\mathrm{B} & \rightarrow \mathrm{CrB}
\end{aligned}\)

Boron does not react with nonoxidising acids like hydrochloric acid. However, it forms boric acid with hot concentrated oxidising acids.

⇒ \(\begin{aligned}
2 \mathrm{~B}+3 \mathrm{H}_2 \mathrm{SO}_4 & \rightarrow 2 \mathrm{H}_3 \mathrm{BO}_3+3 \mathrm{SO}_2 \\
\mathrm{~B}+3 \mathrm{HNO}_3 & \rightarrow \mathrm{H}_3 \mathrm{BO}_3+3 \mathrm{NO}_2
\end{aligned}\)

Boron acts as a reducing agent, reducing carbon dioxide and silica to carbon and silicon respectively.

Uses

• Boron is a hard, refractory solid having low density and low electrical conductivity.
• It is used to increase the hardness of steel, and crystalline boron is used in transistors.
• The isotope 10 B is a good neutron absorber and is used to make shields and control rods in nuclear reactors.
• Boron fibres are used in aircraft and for making bullet¬ proof vests. Borax, orthoboric acid and diboranes are some of the important compounds of boron which we will discuss here.

Borax

Borax, or sodium tetraborate decahydrate, is the most important compound of boron.

Initially formulated as Na 2B4O7-10H2O, it is now correctly represented as Na2[B405 (0H4)]-8H20. It naturally occurs in certain lakes in India, Tibet and the USA.

Preparation

Borax can be prepared from the mineral colemanite by boiling it with sodium carbonate solution.

⇒ \(\underset{\text { colemanite }}{\mathrm{Ca}_2 \mathrm{~B}_6 \mathrm{O}_{11}(\mathrm{~s})}+\underset{\mathrm{Na}_2 \mathrm{CO}_3(\mathrm{aq})}{2} \underset{\text { borax }}{\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7(\mathrm{aq})}+\underset{\text { sodium metaborate }}{2 \mathrm{NaBO}_2(\mathrm{aq})}+2 \mathrm{CaCO}_3(\mathrm{~s})\)

Calcium Carbonate Is Filtered Off. The Filtrate, On Concentration, Gives Crystals Of Borax. Carbon Dioxide Is Passed Through The Mother Liquor To Convert Sodium Metaborate To Borax.

⇒ \(4 \mathrm{NaBO}_2+\mathrm{CO}_2 \rightarrow \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+\mathrm{Na}_2 \mathrm{CO}_3\)

Properties and reactions

Borax is a white, crystalline solid slightly soluble in cold water but more soluble in hot water. It hydrolyses to give an alkaline solution

⇒ \(
\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+7 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 2 \mathrm{NaOH}+4 \mathrm{H}_3 \mathrm{BO}_3
orthoboric acid\)

On heating, borax loses its water of crystallisation and swells to give a white, opaque, puffy mass of the anhydrous salt, which on further heating gives a glassy mass comprising sodium metaborate and boron trioxide.

⇒ \(\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O} \stackrel{-10 \mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}_3\)

Boron trioxide is acidic in nature and reacts with metallic (basic) oxides, forming salts called metaborates.

The metaborates have characteristic colours and this reaction is the basis of the borax bead test for the detection of metal ions.

⇒ \(\mathrm{MO}+\mathrm{B}_2 \mathrm{O}_3 \stackrel{\text { heat }}{\longrightarrow} \mathrm{M}\left(\mathrm{BO}_2\right)_2\)

The test is performed on a metal salt by heating it with borax in the loop of a platinum wire.

On heating, the metal salt forms an oxide. This oxide fuses with the boron trioxide obtained from borax to form a metaborate.

When borax is treated with a calculated quantity of sodium hydroxide, sodium metaborate is formed.

⇒ \(\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+2 \mathrm{NaOH} \rightarrow 4 \mathrm{NaBO}_2+\mathrm{H}_2 \mathrm{O}\)

Borax, on treatment with a calculated quantity of concentrated sulphuric acid, gives boric acid.

⇒ \(\begin{array}{r}
\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{~B}_4 \mathrm{O}_7 \\
\downarrow \\
\downarrow \mathrm{H}_2 \mathrm{O} \\
4 \mathrm{H}_3 \mathrm{BO}_3
\end{array}\)

Uses

Borax is used

• In the manufacture of enamels and glazes for the manufacture of tiles
  and in pottery,
• In making optical glass and heat-resistant glass (borosil, pyrex, etc.),
• In medicinal soaps and eye washes due to its antiseptic properties, and
• In the borax bead test for the detection of metals.

Structure

Borax (Na2[B405(0H)4]-8H20) is made up of two triangular units and two tetrahedral units. The ion is actually [B405(0H)4]2_. The other water molecules are associated with the metal ions.

Boric acids

Orthoboric acid, H3 B03, commonly known as boric acid, and metaboric acid, HBO 2, are two common oxoacids of boron.

Orthoboric acid occurs naturally in the condensate from volcanic steam vents called stuffing.

Preparation

Boric acid can be prepared by acidifying an aqueous solution of borax with sulphuric acid or hydrochloric add.

On cooling, the crystals of boric add separate out.

⇒ \(\begin{gathered}
\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+4 \mathrm{H}_3 \mathrm{BO}_3 \\
\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+2 \mathrm{HCl}+5 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaCl}+4 \mathrm{H}_3 \mathrm{BO}_3
\end{gathered}\)

It can also be obtained when sulphur dioxide is passed through a hot solution of colemanite in water. The resulting solution, on cooling, gives crystals of boric acid.

⇒ \(\mathrm{Ca}_2 \mathrm{~B}_6 \mathrm{O}_{11}+4 \mathrm{SO}_2+11 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Ca}\left(\mathrm{HSO}_3\right)_2+6 \mathrm{H}_3 \mathrm{BO}_3\)

The hydrolysis of some compounds of boron (viz., halide, nitride, hydride) gives boric acid.

⇒ \(\begin{aligned}
\mathrm{BCl}_3+3 \mathrm{H}_2 \mathrm{O} & \rightarrow \mathrm{H}_3 \mathrm{BO}_3+3 \mathrm{HCl} \\
\mathrm{BN}+3 \mathrm{H}_2 \mathrm{O} & \rightarrow \mathrm{H}_3 \mathrm{BO}_3+\mathrm{NH}_3 \\
\mathrm{~B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} & \rightarrow 2 \mathrm{H}_3 \mathrm{BO}_3+6 \mathrm{H}_2
\end{aligned}\)

Properties And Reactions

Orthoboric acid is a flaky, white, crystalline solid, moderately soluble in water.

It is a very weak acid. Unlike acids, which donate protons, they accept a pair of electrons from OH ions and thus behave as a Lewis acid.

⇒ \(\mathrm{H}_3 \mathrm{BO}_3+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { metaborate ion }}{\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}}+\mathrm{H}_3 \mathrm{O}^{+}\)

On heating orthoboric acid loses water to form metaboric add, which on further heating yields tetraboric acid and finally boron trioxide.

⇒ \(\underset{\text { orthoboric acid }}{\mathrm{H}_3 \mathrm{BO}_3} \frac{-\mathrm{H}_2 \mathrm{O}}{375 \mathrm{~K}} \underset{\text { metaboric acid }}{\mathrm{HBO}_2}\)

Uses

Boric acid is used

• As an antiseptic as well as a preservative, and
• In the manufacture of heat-resistant glass, enamels and glazes in pottery.

Structure

Orthoboric add contains triangular B033” units, which are hydrogen bonded to give a layered structure.

The layers are quite a distance apart (318 pm) and the crystal breaks easily, thus making the fr’ compound soft and flaky.

Boron hydrides

Boron forms a series of volatile hydrides collectively called boranes. In view of its trivalency, boron is expected to form a simple hydride BH3.

However, no such compound is known, the simplest hydride is diborane (B2H6). Two series of boranes are known corresponding to the stoichiometries—BnHn+4 and BnHn+6.

The most important hydride is diborane. Its preparation, properties and structure are discussed as follows.

Preparation

1. Diborane can be prepared by the reduction of boron trifluoride etherate (Et20-BF3) with lithium aluminium hydride (LiAlH4) in ether or with sodium borohydride (NaBH4) in diglyme (CH3OCH2CH2OCH2CH2OCH3) as a solvent.

⇒ \(4 \mathrm{Et}_2 \mathrm{O} \cdot \mathrm{BF}_3+3 \mathrm{LiAlH}_4 \stackrel{\text { ether }}{\longrightarrow} 2 \mathrm{~B}_2 \mathrm{H}_6+3 \mathrm{LiF}+3 \mathrm{AlF}_3+4 \mathrm{Et}_2 \mathrm{O}\)

⇒ \(4 \mathrm{Et}_2 \mathrm{O} \cdot \mathrm{BF}_3+3 \mathrm{NaBH}_4 \stackrel{\text { diglyme }}{\longrightarrow} 2 \mathrm{~B}_2 \mathrm{H}_6+3 \mathrm{Na}\left[\mathrm{BF}_4\right]+4 \mathrm{Et}_2 \mathrm{O}\)

In the laboratory, diborane is prepared by heating sodium borohydride and iodine in the solvent diglyme.

⇒ \(2 \mathrm{NaBH}_4+\mathrm{I}_2 \stackrel{\text { diglyme }}{\longrightarrow} \mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaI}+\mathrm{H}_2\)

Diborane is industrially prepared by the reduction of boron trifluoride with sodium hydride.

⇒ \(2 \mathrm{BF}_3+6 \mathrm{NaH} \stackrel{450 \mathrm{~K}}{\longrightarrow} \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{NaF}\)

Properties and reactions

Diborane is a colourless, highly reactive gas. It catches fire spontaneously in the air and explodes in oxygen. The reaction is highly exothermic.

⇒ \(\mathrm{B}_2 \mathrm{H}_6+3 \mathrm{O}_2 \rightarrow \mathrm{B}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O} \quad \Delta_c H^{\ominus}=-2165 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

It has a higher heat of combustion per unit weight than most other fuels and was earlier used as a rocket fuel.

However, this was discontinued as the combustion was incomplete and the exhaust nozzles became blocked with I, a nonvolatile polymer.

Diborane is rapidly hydrolysed to yield boric acid and hydrogen.

⇒ \(\mathrm{B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{BO}_3+6 \mathrm{H}_2\)

When heated in a sealed tube a complex reaction occurs and the borane is converted to various higher boranes, depending on the prevailing experimental conditions.

⇒ \(\begin{aligned}
& \mathrm{B}_2 \mathrm{H}_6 \stackrel{373-523 \mathrm{~K}}{\longrightarrow} \mathrm{B}_4 \mathrm{H}_{10}+\mathrm{B}_5 \mathrm{H}_{11}+\mathrm{B}_6 \mathrm{H}_{12}+\mathrm{B}_{10} \mathrm{H}_{14} \\
& \mathrm{~B}_2 \mathrm{H}_6 \stackrel{353-523 \mathrm{~K}}{\underset{\text { high pressure }}{\longrightarrow}} \mathrm{B}_4 \mathrm{H}_{10} \\
& \mathrm{~B}_2 \mathrm{H}_6 \stackrel{423 \mathrm{~K}}{\longrightarrow} \mathrm{B}_{10} \mathrm{H}_{14}
\end{aligned}\)

When diborane, which is a Lewis acid, is treated with Lewis bases, like amines [(CH3)2NH or (CH3)3N], it undergoes cleavage to give BH3, which then forms an adduct with the amine. (Here Me represents the methyl I group.

⇒ \(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NMe}_3 \rightarrow 2 \mathrm{BH}_3 \cdot \mathrm{NMe}_3\)

⇒ \(\mathrm{B}_2 \mathrm{H}_6+\mathrm{NH}_3 \longrightarrow \underset{\substack{\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6+\mathrm{H}_2 \\ \text { borazine }}}{\left[\mathrm{BH}_2\left(\mathrm{NH}_3\right)_2\right]^{+}\left[\mathrm{BH}_4\right]}\)

On the reaction of diborane with ammonia, initially, an addition V product B2H6 2NH3 (better formulated as [BH2(NH3)21+[(BH4))-) is formed, which decomposes on heating to form a volatile compound, borazine. Borazine is isoelectronic and isostructural with benzene.

Diborane reacts with alkali metal hydrides to form tetrahydroborates or borohydrides, which contain the tetrahedral BH4-

⇒ latex]2 \mathrm{NaH}+\mathrm{B}_2 \mathrm{H}_6 \underset{\text { ether }}{\stackrel{\text { diethyl }}{\longrightarrow}} 2 \mathrm{NaBH}_4[/latex]

Lithium borohydride and sodium borohydride are reducing agents used in organic synthesis.

Diborane undergoes addition reactions with alkenes and alkynes in ether at room temperature to form organoboranes.

⇒ \(6 \mathrm{RCH}=\mathrm{CH}_2+\mathrm{B}_2 \mathrm{H}_6 \longrightarrow 2 \mathrm{~B}\left(\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{R}\right)_3\)

This is referred to as a hydroboration reaction. Organoboranes are useful starting materials for the synthesis of various organic compounds.

Structure

The structure of diborane is quite different from the one usually expected, as it cannot be explained on the basis of simple theories of bonding.

The electronic configuration of boron ([He]2s2 2p1) suggests that there are three valence electrons which implies that a boron atom can form three covalent bonds.

But, if each boron atom is bonded to three hydrogen atoms, then there will be no electron left to hold the two BH3 units together. The structure of diborane has been confirmed by electron diffraction studies.

Diborane is an electron-deficient compound as the total number of valence electrons (12) is not sufficient to form eight bonds.

The two hydrogen atoms which hold the boron atoms together are called the bridging hydrogen atoms whereas the other four hydrogen atoms are called the terminal hydrogen atoms.

The bridging hydrogen atoms, being in the plane perpendicular to that of the rest of the atoms, prevent rotation between the two boron atoms. The terminal B —H distance is the same as in a non-electron-deficient compound (119 pm).

The terminal B—H bonds are formed by sharing an electron pair between boron and hydrogen and are referred to as two-centre two-electron (2c-2e) bonds. The bridge bonds, B—H—B, are different from the normal covalent bonds.

Since the four terminal B —H bonds use eight out of twelve valence electrons, only four valence electrons are present to form the four bridge bonds. The formation of the bridge bond can be understood from the concept of hybridisation.

Each B atom which is sp3 hybridised contains four hybrid orbitals, one of which is vacant and the others singly filled by the three valence electrons.

Two of these orbitals of each atom overlap with the Is orbital of hydrogen to form the terminal B—H bonds.

One singly filled sp3 hybrid orbital on one boron atom and a vacant sp3 hybrid orbital on another boron atom overlap with the singly filled Is orbital of hydrogen to form a bonding orbital, B —H—B, shaped like a banana, covering all the three atoms.

Thus two electrons hold three atoms together and this is called a three-centre two-electron bond (3c-2e)

Aluminium

Aluminium is a moderately soft, light, nontoxic, corrosion-resistant metal. It is the most abundant metal. It has high thermal and electrical conductivity and tensile strength. Many of its mechanical properties are improved by alloying it with copper, manganese, magnesium, silicon and zinc.

Aluminium and its alloys are used in making utensils and as a structural material for aircraft, ships and the transportation industry. The metal is also used for making electric cables and in the packaging industry mainly as metal-foil cans and toothpaste tubes.

Aluminium is used as a reducing agent in the aluminothermic process for the extraction of chromium and manganese from their ores.

Alums are double salts comprising aluminium sulphate and an alkali metal sulphate. They may be A represented by the general formula: M2S04′ A12(S04)3-24H20 or 2MA1(S04)2-12H20 (M = Na or K).

Alums are in water purification and as a mordant in the dyeing and printing of textiles. (Mordants are inorganic oxides salts, which are absorbed on the fabric.)

Aluminium oxide exists in various crystalline modifications, some of which are used as the adsorbent in thin-layer chromatography and as catalyst support. Some others like corundum are used as refractory materials and abrasives.

Group 14 Elements

The elements in this group are carbon, silicon, germanium, tin and lead. A marked increase in metallic character is noted on descending the group. The first two elements are nonmetals while the others are metals.

Occurrence

Carbon is the seventeenth-most abundant element in the earth’s crust. In the free state, it is present as coal, diamond and graphite, while in the combined state it is present as carbon dioxide, carbonate rocks and hydrocarbons (petroleum, natural gas).

Carbon is an essential constituent of all living systems. Naturally occurring carbon has two stable isotopes—12C and 13C. In addition, a radioactive isotope, 14 C, of half-life 5770 years is also found.

Silicon is the second-most abundant element in the earth’s crust, next only to oxygen. It is present as silica (sand) and silicates (rocks). Germanium is a rare element occurring in traces in coal, lead and zinc ores. Tin occurs mainly as cassiterite (SnO₂) and lead as galena (PbS).

Electronic configuration

The valence-shell electronic configuration of Group 14 elements is n2np2.

The inner core of carbon and silicon contains only s and p electrons whereas that of germanium, tin and lead contains d electrons also.

In lead, fourteen f electrons are also present. The presence of d and f electrons in some members of the group results in the variation of properties among the Group 14 elements.

Covalent radii

The covalent radii increase down the group. There is a significant increase in covalent radius from carbon to silicon.

The covalent radius of germanium is only slightly more than that of silicon. This is due to the presence of ten d electrons in germanium, which do not shield the nuclear charge effectively.

A small increase in radius is noted on descending the group, i.e., between tin and lead. This is due to the presence of d and f electrons in the inner core. If you recall, such a behaviour was noted in Group 13 also.

Ionisation enthalpy

The ionisation enthalpies of the Group 14 elements are higher than those of the corresponding Group 13 elements.

This is expected as ionisation enthalpy increases across a period due to an increase in nuclear charge and a decrease in the size of the element, making the electrons tightly held.

The variation in ionisation enthalpy is not very regular though a decrease is noted in the first and second ionisation enthalpies on moving from carbon to tin. The irregularities occur due to the poor shielding effect of d and f orbitals.

Electronegativity

The elements of this group are more electronegative than the corresponding members of Group 13 due to their smaller size.

The electronegativity of carbon is more than that of the other members of the group.

Owing to this it can gain electrons to form anions like C2~ (carbide), C22 (acetylide) and C4~ (methanide). The electronegativity values of the other elements are almost the same.

Oxidation states

The common oxidation states are +4 and +2, the latter arising due to the inert-pair effect, which is common in the heavier members of the group. The sum of the first four ionisation enthalpies is immense.

Therefore, the elements in the +4 oxidation state generally form covalent compounds. On descending the group, the +4 oxidation state becomes less stable, and the +2 oxidation state is shown increasingly.

This is also reflected in the oxidising power of the tetravalent compounds of the heavier elements of the group.

If we consider the oxides CO₂, SiO₂and GeO₂, they are not oxidising; SnO₂ is a mild oxidising agent while PbO₂ is a strong oxidising agent.

The tendency to form ionic compounds in the +2 oxidation state increases on moving down the group.

PbF2 and PbCl 2 are ionic compounds. To sum up, both carbon and silicon exhibit mostly the +4 oxidation state; the +4 state is more stable for germanium than the +2 state.

Tin shows both the oxidation states whereas the +2 state is stable for lead.

Chemical Reactivity

The elements of Group 14 are, in general, relatively unreactive. Among the members of the group, the heavier ones are more reactive than the lighter ones.

Reaction With Air

All the elements of the group form oxides when heated with oxygen. Generally, monoxides (MO) and dioxides (MOz) are formed.

⇒ \(\begin{gathered}
2 \mathrm{M}+\mathrm{O}_2 \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{MO} \\
\mathrm{M}+\mathrm{O}_2 \stackrel{\text { heat }}{\longrightarrow} \mathrm{MO}_2
\end{gathered}\)

The dioxides are all thermally stable except PbO₂. This is because of the inert-pair effect, which enhances the stability of the +2 oxidation state in lead. PbO₂ is thus a good oxidising agent.

The acidic nature of dioxides decreases down the group as the nonmetallic character of the elements decreases—CO₂, SiO₂ and GeO₂ are acidic whereas SnO₂ and PbO₂ are amphoteric.

CO₂ is stable. SiO₂ however exists only at a high temperature. The monoxides of Ge, Sn and lead are known. GeO is reducing in nature.

Reaction with water

Most of the Group 14 elements are unaffected by water. Tin reacts with steam.

⇒ \(\mathrm{Sn}+2 \mathrm{H}_2 \mathrm{O} \stackrel{\text { heat }}{\longrightarrow} \mathrm{SnO}_2+2 \mathrm{H}_2\)

Lead does not undergo this reaction as it gets superficially oxidised and a protective oxide film is formed on the surface.

Reaction with halogens

The elements in this group form tetrahalides (MX2) as well as dihalides (MX2) (X = F, C1, Br, I), All tetrahalides are known except Pbl 4.

The bond enthalpy of Pb—I is not enough to compensate for the energy required to unpair the 6s2 electrons of lead and get four impaired electrons.

The stability of the tetrahalides decreases down the group and that of dihalides increases (inert-pair effect).

The fluorides, by virtue of the high electronegativity of fluorine, have appreciable ionic character.

SnF4 and PbF4 exist as ionic solids. The other tetrahalides are covalent. The central metal atom in tetrahalides is sp3 hybridised and the shape of the molecule is tetrahedral.

The Group 14 elements other than carbon can expand their covalence from four due to the presence of d orbitals.

Since complex formation by elements is favoured by the availability of empty orbitals, carbon does not form complexes. A complex is a compound in which molecules or ions form coordinate bonds with a metal atom or ion.

Tetrahalides of carbon are inert towards water (hydrolysis takes place by complex formation), but tetrahalides of silicon undergo hydrolysis readily.

⇒ \(\mathrm{SiCl}_4+4 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Si}(\mathrm{OH})_4+4 \mathrm{HCl} sillicic acid\)

The hydrolysis proceeds via the formation of an intermediate, in which a water molecule gets attached to the Si atom before a Si— —CI bond is broken.

Thus, the Si atom shows a coordination number of five and the lone pair of electrons present on the oxygen of the water molecule is accommodated in a vacant 3d orbital of the silicon atom.

(The coordination number is the number of species surrounding a given atom or ion in a complex.)

The process continues till all Si—Cl bonds are replaced by Si—OH bonds. The other trihalides of the group also have a tendency to hydrolyse but the hydrolysis can be suppressed by adding the appropriate halogen acid.

The tetrahalides are all examples of electron-precise molecules, i.e., the central metal is surrounded by an octet of electrons (4 from the Group 14 element and 1 each from the four halogens).

They are neither electron-deficient nor electron-rich species and thus do not behave as Lewis acids or bases. Carbon cannot extend its covalence beyond four due to the absence of d orbitals.

The tetrahalides of Si, Ge, Sn and Ph can increase their coordination number to 6 by forming complexes like [SiF₆]^-2 and [SnCl-6]^2-.

In these species, the central atom accepts electron pairs from donors and is spM1 hybridised. Germanium, tin and lend form dihalides. There is an increase in the stability of dihalides on moving down the tire group.

Carbon

Anomalous behaviour

Like in other groups, carbon, being the first member of Group 14, differs from the other members in many aspects.

The main reasons for these differences are

1. The small size of the carbon atom,
2. High electronegativity,
3. Unavailability of d orbitals and
4. High ionisation enthalpy.

Some of the specific anomalies are that carbon has a maximum covalence of four and is the only element to form stable pπ-pπ bonds.

This apart, carbon has a remarkable property of forming long chains and rings of its atoms.

This is called catenation and is attributed to the strength of the C—C bond. The tendency to catenate decreases down the group with the decrease in the strength of the M —M bond.

The bond enthalpy decreases with the decrease in electronegativity and increase in the atomic size of elements.

Carbon forms pz-pir multiple bonds with itself and other small electronegative elements, e.g., C=C, C=C, C=0, C=S and C=N.

The pn-pz bonds are formed by the sideways overlap of a p orbital of carbon with a p orbital of another carbon, oxygen or nitrogen atom.

The heavier elements do not form pπ-pπ bonds as the orbitals are too large and diffuse to allow effective overlap. However, they can form p5-d5 bonds.

Allotropes

Due to its unique property of catenation and pn-pz bond formation, carbon exists in different allotropic modifications.

Until 1985, only two crystalline allotropes of carbon were known—diamond and graphite. In 1985, another allotrope of carbon known as fullerene was discovered by HW Krato, E Smalley and RF Curl.

Diamond

Diamond is the hardest substance known and has a very rigid structure. Each carbon atom in the crystalline lattice is sp3 hybridised and linked covalently to the tour other carbon atoms. Tire C—C distance is 154 pm.

The rigid three-dimensional structure extends throughout the lattice, which is difficult to break, making diamond t extremely hard.

Graphite

Unlike diamond, graphite has a layered structure.

Individual layers of the fused hexagonal rings of carbon atoms are held together by weak van der Waals forces.

Each carbon atom shared by three hexagons is sp2 hybridised. Out of the four valence electrons, the toe is involved in bond formation and the fourth electron is involved in delocalised re-bonding.

The C—C bond length within a layer is 141.5 pm while the inter-layer distance is 335.4 pm.

Due to weak van der Waals forces the layers can slide over one another and the jt electrons move within each layer, making graphite a good conductor of electricity and conferring lubricating properties.

Graphite cleaves readily between the layers and is soft and slippery. Some physical properties of diamond and graphite axes are summarised.

Fullerenes

A fascinating discovery was the synthesis of spheroidal carbon-cage molecules called fullerenes.

The discoverers of fullerene were awarded the Nobel Prize in Chemistry (1996). Fullerenes were first prepared by the evaporation of graphite using a laser.

A more practical method for the preparation is to heat graphite in an electric arc in an inert atmosphere (helium or argon).

A sooty material so formed consists of C with small amounts of C70 and other fullerenes containing an even number of carbon atoms upto 350.

Fullerenes have a smooth structure and, unlike diamond and graphite, dissolve in organic solvents like toluene.

C60 is the most stable fullerene. It has the shape of a football and is called buckminsterfullerene.

consists of fused five- and six-membered carbon rings. Each six-membered ring is surrounded alternately by hexagons and pentagons of carbon while the five-membered rings are fused to five hexagons.

The carbon atoms are sp2 hybridised, each carbon atom forms three bonds with the other three carbon atoms and the fourth electron is delocalised. Although all atoms are equivalent, the bonds are not.

In the structure, C —C bonds of two different bond lengths occur—at the fusion of two six-membered rings (C—C bond length 135.5 pm) and at the fusion of five- and six-membered rings (C—C bond length 146.7 pm).

The smallest known fullerene is C₂O, which is obtained from the hydrocarbon C₂OH₂O by a two-step reaction. First, the hydrogens are replaced by bromine. This is then followed by denomination.

Thermodynamically, the most stable allotrope of carbon is considered to be graphite. It is considered to be the standard state for carbon and its enthalpy of formation is taken as zero.

The enthalpies of the formation of diamond and fullerene are 1.90 and 38.1 kJ mol-1 respectively. In addition to these forms, there are a few other elemental forms of carbon like coke, lampblack, soot and charcoal. These are impure forms of graphite.

Uses

1. The isotope 12C has been assigned a mass of 12.0000 units and is the international standard for atomic mass measurements.
2. The isotope 14C is radioactive (half-life 5770 years). It is, therefore, used in radiocarbon dating, a method used to determine the age of samples of organic origin.
3. Diamond is used as a gemstone. Inferior quality diamonds are used as abrasives in cutting, drilling, and V also in sharpening and polishing tools. Diamond is also used for making dies for drawing thin wires from metals.
4. Graphite conducts electricity. It is, therefore, used for making electrodes in batteries and electrolytic cells. It is used in steel making, for making acid- and alkali-resistant crucibles and as a moderator in atomic reactors. It is also used in the lead of pencils and as a lubricant (being soft and slippery). Graphite fibres embedded in, plastic are used in tennis rackets, aircraft and fishing rods.
5. Activated charcoal is an excellent adsorbent. It is, therefore, used in gas masks (to adsorb toxic gases) and in air-conditioning systems to control odour. It is also used in the manufacture of sugar, refining of oils and in water filters. (When charcoal is activated for adsorption by steaming or heating in a vacuum, it is known as activated charcoal.)
6. Coke is a fuel and also finds use as a reducing agent in metallurgy.
7. Carbon black is used as a pigment in black ink and as a filler in tyres.

Oxides Of Carbon And Silicon

Among all compounds of carbon and silicon, oxides are the most important and abundant.

The oxides of carbon—carbon monoxide and carbon dioxide differ from the others as they contain pn-pn multiple bonds between carbon and oxygen.

Silicon forms two oxides—silicon monoxide and silicon dioxide, also called silica.

Carbon monoxide

It is obtained when carbon is burnt in a limited supply of air.

⇒ \(2 \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{CO}(\mathrm{g})\)

It is also obtained during the incomplete combustion of fuels like petrol and diesel and is thus always present in automobile exhausts. Pure carbon monoxide is obtained by the dehydration of formic acid with concentrated sulphuric acid at 373 K.

⇒ \(\mathrm{HCOOH} \stackrel{\text { heat }}{\stackrel{\text { canc. } \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}} \mathrm{CO}+\mathrm{H}_2 \mathrm{O}\)

It is commercially obtained by passing steam over red hot coke. The reaction also produces hydrogen along with carbon monoxide, and this mixture is called water gas or synthetic gas. Water gas is used as a fuel.

⇒ \(\mathrm{C}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \stackrel{473-1273 \mathrm{~K}}{\longrightarrow} \underbrace{\mathrm{CO}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g})}_{\text {water gas }}\)

If air is listed instead of steam, a mixture of carbon monoxide and nitrogen is produced. This is called producer I gas and is an important fuel.

⇒ \(2 \mathrm{C}(\mathrm{s})+\underbrace{\mathrm{O}_2(\mathrm{~g})+4 \mathrm{~N}_2(\mathrm{~g})}_{\text {air }} \longrightarrow \underbrace{2 \mathrm{CO}(\mathrm{g})+4 \mathrm{~N}_2(\mathrm{~g})}_{\text {producer gas }}\)

Properties and reactions

Carbon monoxide is a colourless and odourless gas. It is a neutral oxide insoluble in water.

On combustion, carbon monoxide gives carbon dioxide. The reaction is highly exothermic and this explains why producer gas and water gas, which contain carbon monoxide, are important fuels.

Among the two, water gas is more efficient than producer gas because both carbon monoxide and hydrogen bum and evolve heat.

⇒ \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) \quad \Delta_c H^{\ominus}=-566 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Due to its strong affinity for oxygen, carbon monoxide is a powerful reducing agent and is used to extract many metals (except alkali or alkaline earth metals) by the reduction of their oxides.

⇒ \(\begin{gathered}
\mathrm{ZnO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Zn}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \\
\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_2(\mathrm{~g}) \\
\mathrm{CuO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})
\end{gathered}\)

Carbon monoxide contains a triple bond between the carbon atom and the oxygen atom—one CT and two n bonds.

The lone pair of electrons on carbon (C=O) may be donated to a transition metal in its low oxidation state to form a compound called metal carbonyl. You will learn more about these in higher classes.

⇒ \(
\mathrm{Ni}+4 \mathrm{CO} \rightarrow \mathrm{Ni}(\mathrm{CO})_4
nickel carbonyl\)

Carbon monoxide is very toxic. The toxicity is due to its ability to form a very stable complex (carboxyhaemoglobin) with the haemoglobin of blood.

Haemoglobin acts as an oxygen carrier as it binds itself to oxygen in the lungs and transports it.

it to the tissues. The formation of carboxyhaemoglobin prevents haemoglobin from binding itself to oxygen.

If a person is exposed to large amounts of carbon monoxide, he or she may eventually die (due to lack of oxygen).

Carbon dioxide

Preparation

It is obtained when carbon or fossil fuels are burnt in an excess of air. In other words, carbon dioxide is produced by the complete combustion of carbon and carbon-containing fuels.

⇒ \(\begin{gathered}
2 \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \stackrel{\text { heat }}{\longrightarrow} \mathrm{CO}_2(\mathrm{~g}) \\
\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\
\mathrm{C}_5 \mathrm{H}_{12}(\mathrm{~g})+8 \mathrm{O}_2(\mathrm{~g}) \stackrel{\text { heat }}{\longrightarrow} 5 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})
\end{gathered}\)

Carbon dioxide can be prepared in the laboratory by the action of a dilute add-on carbonate. Generally, calcium carbonate in the form of marble chips is used.

⇒ \(\mathrm{CaCO}_3(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CaCl}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{CO}_2(\mathrm{~g})\)

The large-scale production of carbon dioxide involves the thermal decomposition of limestone in lime kilns.

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{1000 \mathrm{~K}}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

The main industrial source of carbon dioxide is as a by-product in the manufacture of hydrogen used in the Haber process to make ammonia.

⇒ \(\mathrm{CH}_4+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CO}_2+4 \mathrm{H}_2\)

Properties and reactions

Carbon dioxide is a colourless and odourless gas. It is slightly soluble in water. The solubility increases with an increase in pressure.

Soda water and aerated soft drinks contain carbon dioxide dissolved in water. Carbon dioxide dissolves in water to form carbonic acid, which is a weak, dibasic acid.

⇒ \(\begin{gathered}
\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \underset{\text { carbonic acid }}{\mathrm{H}_2 \mathrm{CO}_3(\mathrm{aq})} \\
\mathrm{H}_2 \mathrm{CO}_3(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\underset{\text { hydrogen carbonate }}{\mathrm{HCO}_3^{-}(\mathrm{aq})} \\
\mathrm{HCO}_3^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\underset{\text { carbonate }}{\mathrm{CO}_3{ }^{2-}(\mathrm{aq})}
\end{gathered}\)

Thus, a solution of carbon dioxide in water is a mixture of CO₂, H₂CO₃, HCO₃– and CO_3^2-. The dissolution in water is important biochemically as the H₂CO₃ /HCO^3- buffer helps to maintain the pH of blood between 7.2 and 7.4.

Tire presence of carbon dioxide in the atmosphere (0.03%) is balanced by two biological processes— photosynthesis and respiration.

Carbon dioxide is used by green plants in photosynthesis to make sugar, which on combustion during respiration releases carbon dioxide.

⇒ \(6 \mathrm{CO}_2+12 \mathrm{H}_2 \mathrm{O} \stackrel{\text { sunlight }}{\longrightarrow} \underset{\text { glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+6 \mathrm{O}_2+6 \mathrm{H}_2 \mathrm{O}\)

\(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}+\text { energy }\)

Though carbon dioxide is not toxic, an excess of it in the atmosphere causes climatic changes which may have far-reaching consequences.

Increased burning of fuels and indiscriminate reduction of forest cover is responsible for the increase of the carbon dioxide level in the atmosphere. This has caused global warming (greenhouse effect).

When carbon dioxide is cooled under pressure (50-60 atm), it solidifies. The solid formed is called dry ice, which sublimes at -78°C under atmospheric pressure.

It is used as a refrigerant. Carbon dioxide, being a nonsupporter of combustion, is used in fire extinguishers, especially those meant for electrical fires.

They contain liquid carbon dioxide under pressure. Rapid vaporisation of liquid carbon dioxide is used for blasting in coal mines.

Structure

Carbon dioxide has a linear structure with zero dipole moment. The carbon atom in the molecule is sp hybridised.

The carbon atom has two sp hybridised orbitals which form a bond with two oxygen atoms. The other two unpaired electrons on carbon form pπ-pπ bonds with oxygen. So the expected structure is:O—C=O:

However, the C —O bond length of 115 pm is less than the C—O double bond length of 122 pm. This can be explained by considering carbon dioxide to be a resonance hybrid of the following canonical forms.

⇒ \(: \overline{\mathrm{O}}-\mathrm{C} \equiv \stackrel{+}{\mathrm{O}}: \longrightarrow: \mathrm{O}=\mathrm{C}=\ddot{\mathrm{O}}: \longrightarrow:_{\mathrm{O}}^{+} \equiv \mathrm{C}-\overline{\mathrm{o}}:\)

Silicon dioxide

Silicon dioxide (SiO₂), commonly known as silica, along with silicates, make up a major portion of the earth’s crust. Silica exists in both crystalline and amorphous forms.

Flint is the amorphous form of silica whereas the common crystalline forms are quartz, tridymite and cristobalite, which can be interconverted with the change in temperature.

Unlike carbon dioxide, which consists of discrete molecules (due to pπ-pπ bonding) and is a gas, silicon dioxide is a high-melting solid.

Unlike carbon, silicon cannot form a pπ-pπ bond with oxygen; in silica, each silicon atom is tetrahedrally surrounded by four covalently bonded oxygen atoms and each oxygen atom is bonded to two silicon atoms.

This structure extends in three dimensions to form a macromolecule. A lot of energy is needed to break the rigid structure and thus silica has a very high melting point Since the Si—O bond is very strong, silica is not so reactive. Being an acidic oxide, it reacts with hot alkalis and does not add.

⇒ \(\mathrm{SiO}_2+2 \mathrm{NaOH} \stackrel{\text { heat }}{\longrightarrow} \mathrm{Na}_2 \mathrm{SiO}_3+\mathrm{H}_2 \mathrm{O}\)

However, it is attacked by hydrogen fluoride

⇒ \(\mathrm{SiO}_2+4 \mathrm{HF} \longrightarrow \mathrm{SiF}_4+2 \mathrm{H}_2 \mathrm{O}\)

Quartz is used as a piezoelectric material (having the ability to generate voltage when mechanical stress is applied or dee versa) for crystals in gramophone pickups and gas lighters, and for making crystal oscillators in radios, televisions and computers.

It is also used in optical components such as lenses and prisms. Silica gel, an amorphous form of silica, is used as a catalyst in the petroleum industry, in chromatography and as a drying agent Kieselguhr, another amorphous form of silica, is used in filtration plants and as a filtration aid (e.g., dynamite contains glyceryl trinitrate and glyceryl dinitrate with Kieselguhr as adsorbent).

Silicates

Any 3 substance containing negative ions composed of silicon and oxygen is a silicate. For example, phenate (Be₂SiO₄) contains beryllium and beryl (Be₃Al₂Si₁₆O₁₈) contains beryllium and aluminium along with silicate ions.

A large number of silicate minerals exist in nature. The basic unit in all silicates is an sp3 hybridised silicon atom linked to four oxygen atoms, i.e., the SiO_4^4- unit. This unit can be represented in either of the two ways.

These tetrahedra may exist as discrete units or may be linked to each other by sharing comers.

This gives rise to a wide variety of silicates, which comprise rings, chains, sheets and three-dimensional structures.

Cations linked to the negatively charged silicate units help maintain the electrical neutrality of the species.

If some silicon atoms are replaced by aluminium, we get aluminosilicates. Silicates are classified on the basis of the number of oxygen atoms shared per tetrahedron. A few representative types of silicates.

Some important two-dimensional silicates are beryl, mica and asbestos. Cement and glass are man-made silicates.

In addition to the two-dimensional silicates, three-dimensional silicates are also known. In these all four oxygen atoms of a tetrahedron are shared and some silicon atoms are replaced by aluminium, and some cations like Na⁺, K⁺ and Ca²⁺ are introduced to maintain electrical neutrality.

Two important three-dimensional silicates are feldspars and zeolites. Feldspars are the most important rock-forming minerals. Zeolites have an open honeycomb-like structure.

The cavities present may account for more than 50 per cent of the space by volume. The mineral is highly porous and contains water molecules in the channels.

The structure of the mineral is similar to that of a cage. Some examples of zeolites are natrolite, chabazite and sodalite. Zeolites act as molecular sieves as they allow small molecules to pass through them and retain the larger ones.

They are capable of separating straight-chain hydrocarbons from branched-chain ones and are, therefore, used in petroleum refining. They are also used as catalysts, particularly in petrochemical industries for the cracking of hydrocarbons and isomerisation.

A particular type of zeolite, ZSM-5, acts as a catalyst in the conversion of alcohol to gasoline. Zeolites have exchangeable cations, and so they are used in water-softening.

Silicones

These are a group of synthetic organo-silicon polymers. The basic unit is formed by alternating silicon and oxygen atoms with alkyl or aryl groups attached to the silicon. They are represented by the general formula (R₂2SiO)n or

where R is an alkyl or aryl group. The compounds contain Si—C bonds, which are almost as strong as C—C bonds.

Since the empirical formula R₃SiO is similar to that of a ketone (R₂CO), these materials are named silicones. They may be linear, cyclic or cross-linked.

The starting materials for the manufacture of silicones are alkyl- or arylsubstituted silicon chlorides (RnSiCl₄-n), where R is the alkyl or aryl group.

When methyl chloride is treated with silicon in the presence of copper powder (catalyst) at 570 K, a mixture of methyl-substituted chlorosilanes — CH₃SiCl₃, (CH₃)2SiCl₂, (CH₃)3SiCl, along with, a small amount of (CH₃)4Si, is formed. The products can be separated by fractional distillation.

The hydrolysis of dimethyl dichlorosilane yields a straight-chain polymer. During hydrolysis, the chlorine atoms are replaced by —OH groups to give silanols (similar to alcohols).

The silanols undergo intermolecular condensation to give a polymer.

The chain length of the polymer can be controlled by adding a little (CH₃)3SiCl. This on hydrolysis gives (CH₃)3Si—OH. This has only one —OH group, which attaches itself to one end of the chain, thereby limiting the chain size.

Similarly, the addition of CH₃SiCl₃ to (CH₃)2SiCl₂ during hydrolysis gives a cross-linked polymer. Commercially useful polymers are generally methyl or phenyl derivatives. Silicones may be used in the form of oils, rubbery elastomers or resins.

They have unique properties like chemical inertness, thermal stability and electrical insulation.

Owing to these properties silicones are used for making electrical insulators and nonstick utensils.

Since they are surrounded by nonpolar, hydrophobic organic groups which repel water, silicones are used for the waterproofing of textiles. They are also used as greases, varnishes and antifoam agents in breweries and sewage disposal plants.

Other uses of silicones include applications in surgical and cosmetic implants due to their biocompatible nature.

The p- Block Elements Multiple Choice Questions

Question 1. Which among the following shows the most stable +2 oxidation state?

1. Al
2. Tl
3. Si
4. Pb

Answer: 4. Pb

Question 2. The hybridisation of boron in BF3 is

1. Sp
2. Sp₂
3. Sp₃
4. Dsp₂

Answer: 2. Sp₂

Question 3. Which of these is not a Lewis acid?

1. BF₃
2. AlCl₃
3. CCl₄
4. Bcl₃

Answer: 3. CCl₄

Question 4. [SiF6]^2- is known but [CF6]^2- is not known because carbon

1. Is very electronegative
2. Has small size
3. Does not have d orbitaLs
4. None of these

Answer: 3. Does not have d orbitals

Question 5. A1C1₃ ionises in an aqueous solution because

1. It is a Lewis acid
2. It is an ionic compound
3. It is unstable
4. Its hydration enthalpy exceeds its lattice enthalpy

Answer: 4. Its hydration enthalpy exceeds its lattice enthalpy

Question 6. Which gas is evolved when aluminium is treated with NaOFI?

1. H₂
2. O₂
3. Both O₂
4. and H₂
5. Water vapour

Answer: 1. H₂

Question 7. Thermodynamically, the most stable form of carbon is

1. Diamond
2. Graphite
3. Fullerene
4. Coke

Answer: 2. Graphite

Question 8. Pb(IV) compounds are

1. Stable
2. Oxidising
3. Reducing
4. Not Known

Answer: 2. Oxidising

Question 9. Which of these is an acidic oxide?

1. SiO₂
2. pbO₂
3. SnO₂
4. PbO

Answer: 1. SiO₂

Question 10. Aluminium chloride exists as a

1. Dimer
2. Monomer
3. Trimer
4. Tetramer

Answer: 1. Dimmer

Question 11. B(OH)3 is a

1. Base
2. Monobasic Acid
3. Dibasic acid
4. Tribasic acid

Answer: 2. Monabasic Acid

Question 12. Which of these is electron-deficient?

1. SnCl₂
2. SnCl₄
3. CC1₄
4. AlCl₃

Answer: 4. AlCl₃

Question 13. Which of these is the most stable?

1. PbCl₂
2. SnCl₂
3. SiCl₂
4. CC1₂

Answer: 1. PbCl₂

Question 14. Carbon forms several compounds because of its

1. Variable valency
2. Property of catenation
3. High electronegativity
4. High ionisation enthalpy

Answer: 2. Property of catenation

Question 15. Which of these is the least stable?

1. Pbl₄
2. PbI₄
3. SnCl₂
4. SnCl₄

Answer: 2. PbI₄

Question 16. Which of these is an electron-precise molecule?

1. BF₃
2. CCL₄
3. PbCl₂
4. AlCl₃

Answer: 2. CCL₄

Question 17. Which among the following is expected to be oxidising?

1. CC1₄
2. SiCl₄
3. SnCl₄
4. PbCl₄

Answer: 4. PbCl₄

Question 18. Which among the following is a neutral oxide?

1. Pbo
2. A1₂O₂
3. CO
4. CO₂

Answer: 3. CO

Question 19. Which of these is an important constituent of many fuels?

1. CO
2. CO₂
3. SiO₂
4. None Of These

Answer: 1. CO

Hydrogen

Hydrogen Is The Lightest And The Simplest Atom Known In The Family Of Chemical Elements.

It Was Discovered By The English Chemist Henry Cavendish In 1766. Although On Earth Hydrogen In Combined Form Is The Ninth Most Abundant Element, It Is The Most Abundant Element In The Universe.

Some Of The Planets Appear To Have Significant Amounts Of It Including The Atmospheres Of Jupiter, Saturn And Uranus.

Collected By Gravitational Forces In Stars, Hydrogen Is Converted Into Helium By Nuclear Fusion.

This Process Leads To The Release Of Enormous Energy, Which Is Supplied To Stars Including The Sun.

Being a light gas, hydrogen continually escapes into space and is, therefore, found in low concentration in the Earth’s atmosphere. In 1781, Cavendish confirmed that water was produced by the burning of hydrogen.

This led the French chemist Antoine Lavoisier to coin the name ‘hydrogen’ (Greek: maker of water). The two heavier isotopes of hydrogen—deuterium and tritium—were discovered in 1931 and 1935, respectively.

The nuclear fusion of each of these isotopes occurs at a very high temperature and gives an enormous amount of energy. The hydrogen bomb makes use of this phenomenon.

Hydrogen Position In The Periodic Table

The atomic mass of hydrogen is 1.0079 u and the atomic number is 1. It has the simplest electronic configuration (Is¹), resembling those of alkali metals, which have one electron in their outermost orbits.

Alkali metals tend to lose the outermost electron and form univalent positive ions. Hydrogen can also lose an electron to form H+.

Therefore, hydrogen can be placed above alkali metals in Group 1 of the periodic table.

On the other hand, a closer look at the electronic configuration of hydrogen shows that it has one electron less than helium (Is²), a noble gas, and resembles halogens (F, Cl, Br, I), which too have one electron less in their outermost shells than their nearest noble gases.

Halogens gain an electron to form univalent negative ions. H- ions are rarely formed, and a hydrogen atom gains an electron from highly electropositive metals only.

On the basis of its unique properties, hydrogen can be placed either in Group 1 (along with the alkali metals) or in Group 17 (along with halogens).

Conventionally, hydrogen is said to be the first member of Group 1 of the periodic table. According to many, however, hydrogen should be treated as a group in itself.

In the present form of the periodic table hydrogen along with helium is placed in the first period.

Hydrogen shows similarities in chemical behaviour with alkali metals as well as halogens. However, it also differs from them in many ways.

Let us now compare hydrogen with alkali metals and halogens in greater detail.

Hydrogen Comparison With Alkali Metals Similarities

1. Electronic configuration As already stated, the hydrogen atom has one electron in its only s orbital, and alkali metals have only one electron in their outermost orbital.

• H Is¹
• Li 1S²2S¹
• Na lS²2s²2p3s¹

2. Electropositive character Hydrogen as well as alkali metals are electropositive.

They lose one electron to form the corresponding ion. Alkali metals are more electropositive than hydrogen. The hydrogen ion exists only in a discharge tube but not in solution.

In an aqueous solution, H+ combines with water and is represented as H+ (aq). For example, when hydrogen chloride is dissolved in water, it produces hydronium ions (H3O+).

The electropositive nature of hydrogen and alkali metals is also reflected by the fact that during the electrolysis of their salts, both are liberated at the cathode.

⇒ \(2 \mathrm{NaCl} \stackrel{\text { electrolyals }}{\longrightarrow} \frac{\text { At cathode }}{2 \mathrm{Na}}+\frac{\text { At anode }}{\mathrm{Cl}_2}\)

⇒ \(2 \mathrm{HCl} \stackrel{\text { electrolygis }}{\longrightarrow} \mathrm{H}_2+\mathrm{Cl}_2\)

3. Reducing nature Like alkali metals, hydrogen acts as a strong reducing agent.

⇒ \(\begin{aligned}
2 \mathrm{H}_2+\mathrm{O}_2 & \longrightarrow 2 \mathrm{H}_2 \mathrm{O} \\
2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} & \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_2
\end{aligned}\)

4. Oxidation state Alkali metals as well as hydrogen show an oxidation state of +1 in their compounds, e.g., hydrogen chloride (H+C1-) and sodium chloride (Na+Cl-).

Hydrogen also shows an oxidation state of 1 in hydrides.

5. Combination with nonmetals Hydrogen and alkali metals readily form compounds with nonmetals like oxygen, sulphur and halogens.

Differences

1. Ionisation enthalpy The ionisation enthalpy of hydrogen (1312 kJ Mol^-1) is much higher than the ionisation enthalpies of alkali metals, e.g., Li =520 kJ Mol^-1 and Cs = 375.6 kJ Mol^-1. This implies that hydrogen has less tendency to form positive ions than alkali metals do.
2. Halides Hydrogen halides and alkali-metal halides differ in character. For instance, HC1 is a covalent compound and NaCl is ionic; HC1 is gaseous and NaCl is a solid at room temperature.
3. Nonmetallic character Hydrogen is a nonmetal whereas all the Group1 elements are metals.
4. Atomicity The hydrogen molecule exists as H₂, i.e., it is diatomic. In contrast, alkali metals are monoatomic. Atomic hydrogen exists only at high temperatures.
5. Oxides Alkali-metal oxides, e.g., Li;0 and K20, are basic while the oxide of hydrogen—HzO—is neutral.
6. Ionic size H+ ions (~ L5 x 10_3pm) are much smaller than alkali-metal ions (50-220 pm).

Hydrogen Comparison With Halogens Similarities

1. Electronic configuration Halogens have seven electrons in their outermost shells, one electron less than the stable configuration of the nearest noble gas. Having one electron in its outermost shell, hydrogen is one electron short of the stable configuration of the corresponding noble gas (helium)
   • H(ls¹)
   • F(ls² 2s² 2p5)
   • Cl(ls² 2s²2ph 3s² 3p5)
   • Hc(ls²)
   • Ne(ls² 2s² 2pf )
   • Ar(ls² 2s² 2p6 3s² 3p6)
2. Atomicity Hydrogen as well as halogens exist as diatomic molecules (H₂, CI₂, F₂, llr₂, etc.),
3. Nonmetallic character Hydrogen as well as halogens arc nonmetals.
4. Formation of covalent compounds Hydrogen as well as halogens combine with nonmetals like carbon, and nitrogen and metalloids like germanium to form covalent compounds.
5. Ionisation enthalpy The ionisation enthalpies of hydrogen (1312 kJ Mol^-1) and halogens are comparable. Chlorine, for instance, has an ionisation enthalpy of 1255 kJ Mol^-1.
6. Electronegative nature Halogens readily gain one electron to form a halide ion. Hydrogen gains one electron to form a hydride ion (H-) only with a few extremely electropositive metals, e.g., LiH and CaH₂.

Differences

1. Nature of Oxides The oxide of hydrogen (HzO) is neutral, whereas some oxides of halogens, e.g., HI03, are acidic.
2. Absence of unshared electrons The hydrogen molecule has no unshared pairs of electrons while halogen molecules have six such pairs.

Degree of electronegativity Hydrogen is less electronegative than halogens. This is borne out by the fact that halogens form compounds with several metals while hydrogen docs so with very few highly electropositive metals such as sodium, lithium and calcium.

⇒ \(\mathrm{H}: \mathrm{H} \quad \text { : } \ddot{\mathrm{F}}-\ddot{\mathrm{F}}:\)

Hydrogen Occurrence

Hydrogen constitutes about 92% of the universe. However, its abundance in the earth’s atmosphere is very low. This is because the earth’s gravitational field is not strong enough to hold such a light element.

Some H₂, however, is found in volcanic gases. It occurs in water, organic compounds, fossil fuels (like coal, petroleum and natural gas), clay, ammonia and acids.

All animal and vegetable matter also contains hydrogen. In fact, hydrogen forms more compounds than any other element.

Although it is often said that there are more known compounds of carbon than of any other element, the fact is that almost all carbon compounds contain hydrogen.

Hydrogen Multiple-Choice Questions

Question 1. At 925 K, palladium adsorb

1. 50 volumes of hydrogen
2. 100 volumes of hydrogen
3. 800 volumes of hydrogen
4. 1000 volumes of hydrogen

Answer: 3. 800 volumes of hydrogen

Question 2. Interstitial hydrides are obtained using

1. Elements Of The D Block
2. Actinide elements of the f block
3. lanthanide elements in the f block
4. Ca⁺⁺, Na⁺ and K⁺

Answer: 1. Elements Of The D Block

Question 3. In the reaction the product obtained is

1. Nascent Hydrogen
2. Atomic hydrogen
3. Ortho hydrogen
4. Para hydrogen

Answer: 2. Atomic hydrogen

Question 4. Which of the following reactions is explosive in sunlight?

1. \(\mathrm{N}_2+3 \mathrm{H}_2 \longrightarrow 2 \mathrm{NH}_3\)
2. \(\mathrm{S}+\mathrm{O}_2 \longrightarrow \mathrm{SO}_2\)
3. \(\mathrm{H}_2+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{HCl}\)
4. \(\mathrm{Na}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NaOH}+\mathrm{H}_2[latex]\)

Answer: 3. \(\mathrm{H}_2+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{HCl}\)

Question 5. Deionised water can be obtained by

1. Distillation
2. The Calgon Process
3. The Permutit Process
4. The ion-exchange method

Answer: 1. Distillation

Question 6. In which of the following reactions does hydrogen form compounds by gaining electrons?

1. \(2 \mathrm{Na}+\mathrm{H}_2 \longrightarrow 2 \mathrm{NaH}\)
2. \(\mathrm{CuO}+\mathrm{H}_2 \longrightarrow \mathrm{Cu}+\mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{H}_2+\mathrm{Cl}_2 \stackrel{\text { darkness }}{\longrightarrow} 2 \mathrm{H}-\mathrm{Cl}\)
4. \(\mathrm{H}_2+\mathrm{F}_2 \underset{\text { light }}{\stackrel{\text { diffused }}{\longrightarrow}} 2 \mathrm{H}-\mathrm{F}\)

Answer: 1. \(2 \mathrm{Na}+\mathrm{H}_2 \longrightarrow 2 \mathrm{NaH}\)

Question 7. Hydrogen gas does not reduce heat

1. A1₂O₃
2. CuO
3. SnO₂
4. Fe₂O₃

Answer: 4. Fe₂O₃

Question 8. Which of the following is used as a moderator in nuclear reactions?

1. Heavy water
2. Hard water
3. Deionised water
4. Soft water

Answer: 1. Heavy water

Question 9. In H₂O₂ the O—O—H bond angle is

1. 94.8º
2. 106º
3. 109º 28′
4. 111º

Answer: 1. 94.8º

Question 10. Which of the following metals does not liberate hydrogen by reacting with dilute HC1?

1. Cu
2. Fe
3. Zn
4. Mg

Answer: 1. Cu

Question 11. The decomposition of H₂O₂ is prevented by

1. Urea
2. Acetanilide
3. Oxalic Acid
4. KOH

Answer: 1. Urea

Question 12. Usually, hydrogen combines with other elements by

1. Gaining an electron
2. Sharing an electron
3. Losing an electron
4. None of these

Answer: 2. Sharing an electron

Question 13. The permanent hardness of water is not removed by boiling because

1. It is caused due to the bicarbonates of calcium and magnesium
2. It is caused due to the sulphates of calcium and magnesium
3. It cannot drive away SO^2-₄
4. It cannot drive away CO₂

Answer: 2. It is caused due to the sulphates of calcium and magnesium

Question 14. H₂O₂ has a higher boiling point than water because it

1. Is More Dense Than Water
2. Has A Higher Dielectric Constant Than That Of Water
3. Is More Hydrogen Bonded Than Is Water
4. None Of These

Answer: 3. Is More Hydrogen Bonded Than Is Water

Question 15. Which of the following statement(s) is/are correct?

1. Both hydrogen and halogens have the tendency to accept electrons.
2. Hydrogen and halogens form hydrides and halide ions respectively with equal ease.
3. Both hydrogen and halogens combine with nonmetals to form covalent compounds.
4. Both hydrogen and halogens are very reactive.

Answer: 3. Both hydrogen and halogens combine with nonmetals to form covalent compounds.

Question 16. The exhausted permit is generally regenerated by percolating through it a solution of

1. Polyphosphates
2. Calcium Chloride
3. Aluminium Oxide
4. Polyphosphates

Answer: 3. Aluminium Oxide

Question 17. Hydrogen(c) sodium is chloride-commercially prepared from 4

1. Water Gas
2. Natural Gas
3. Coal Gas
4. Producer Gas

Answer: 1. Water Gas

Question 18. The oxide that gives hydrogen peroxide on treatment with a dilute acid is

1. PbO₂
2. MnO₂
3. BaO₂
4. TiO₂

Answer: 3. BaO₂

Question 19. The aggregates of water molecules, in the liquid state, are held due to

1. Intermolecular hydrogen bonds
2. Van der waals forces
3. Intramolecular hydrogen bonds
4. Intermolecular hydrogen bonds
5. Covalent bonds

Answer: 1. Intermolecular hydrogen bonds

Question 20. Which among the following is an electron-deficient covalent hydride?

1. CH₄
2. NH₃
3. B₂H₆
4.  H₂O

Answer: 3. B₂H₆