Class 11 Chemistry

Some Basic Concepts Of Chemistry Multiple Choice Questions

Question 1. The number of significant figures in 0.0450 is

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

Answer: 1. 3

Question 2. The number of significant figures in 2.345 x 10⁴ is

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

Answer: 4. 4

Question 3. Which of the following is/are mixtures?

1. 20-carat gold
2. Iodised salt
3. Milk
4. Distilled water

Answer:

1. 20 carat gold
2. Iodised salt
3. Milk

Question 4. 1 mol of oxygen atoms represents

1. 16 g of oxygen
2. 6.022 x 10²³ atoms of oxygen
3. 22.7 L of oxygen at stp
4. 32 g of oxygen

Answer:

1. 16 g of oxygen
2. 6.022 x 10²³ atoms of oxygen
3. 22.7 L of oxygen at stp

Question 5. Which of the following is/are heteroatomic molecules?

1. CH₄
2. H₂O
3. O₂
4. Br₂

Answer:

1. CH₂
2. H₂O

Question 6. The sum of the atomic masses of two elements is 407.8 u. If one of them is mercury, the other element is

1. Silver
2. Tin
3. Lead
4. Bismuth

Answer: 3. Lead

Question 7. Which of the following represent/s 1 mol?

1. 11.2 L of a gas at stp
2. 18 g of water
3. 23 g of sodium
4. 100 g of calcium carbonate

Answer:

2. 18 g of water

3. 23 g of sodium

4. 100 g of calcium carbonate

Question 8. The molarity of a solution containing 2.8 g of KOH in 200 mL of water is

1. 2.8 M
2. 0.5 M
3. 0.28M
4. 0.25M

Answer: 4. 0.25M

Question 9. Which of the following contains the most molecules?

1. 1 g of H₂S
2. 1 g of H₂
3. 1 g of H₂O
4. 1 g of CH₄

Answer: 2. 1 g of H₂

Question 10. Which of the following has the least weight?

1. 1 mol of water
2. 3 gram-molecules of Cu₂
3. 0.2 mol of sucrose
4. 1 gram-atom of Na

Answer: 1. 1 mol of water

Question 11. Which of the following will have the least volume at stp?

1. 5g of HCl
2. 5g of HBr
3. 5g of HF
4. 5g of HI

Answer: 4. 5g of HI

Question 12. According to Dalton’s atomic theory’, which of the following happens during a reaction?

1. Atoms are destroyed.
2. Atoms are created.
3. Atoms are rearranged.
4. Atoms are converted into other kinds of atoms.

Answer: 3. Atoms are rearranged.

Question 13. 1 gram-atom of nitrogen weighs

1. 28g
2. 14 g
3. 7g
4. 56g

Answer: 1. 28g

Question 14. The volume occupied by 1 mol of hydrogen is

1. 2.24 L
2. 11.2 L
3. 224 L
4. 22.7 L

Answer: 4. 22.7 L

Question 15. The isotopes of an element differ in

1. The number of electrons
2. The number of protons
3. The number of neutrons
4. The number of electrons + protons

Answer: 3. The number of neutrons

Question 16. The mass of a silver atom is

1. 187.8g
2. 6.022 x 10²³ g/m
3. \(\frac{107.87}{6.022 \times 10^{23}} g\)
4. 3.01 x 10²³ g

Answer: 3. \(\frac{107.87}{6.022 \times 10^{23}} g\)

Question 17. The number of grams of oxygen in 0.10 mol of Na₂CO₄ -10H₂O is

1. 20.8
2. 16
3. 32
4. 8

Answer: 1. 20.8

Question 18. 1 mol each of ammonia and oxygen are made to react according to the following equation.

⇒ \(4 \mathrm{NH}_3+5 \mathrm{O}_2 \longrightarrow 4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O}\)

Which of the statements below is/are true?

1. 1 mol of H₂O is produced.
2. 1 mol of NO is produced.
3. All the ammonia is consumed.
4. All the oxygen is consumed.

Answer: 4. All the oxygen is consumed.

 

Chemical Equation

A chemical reaction can be represented by a chemical equation in which the reactants and products are represented by their molecular formulae. In other words, a chemical equation is a symbolic representation of a chemical change.

Essential of chemical equation

1. A chemical equation should represent a true chemical change. For example, the following equation represents a true chemical change.

⇒ \(2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{HCl}\)

On the other hand, silver and hydrogen do not react to form silver hydride, so the equation given below is not a chemical equation since it does not represent a true chemical reaction.

⇒ \(2 \mathrm{Ag}+\mathrm{H}_2 \longrightarrow 2 \mathrm{AgH}\)

2. A chemical equation must be balanced, i.e., the number of atoms of each element on both sides of the equation must be the same. This follows from the law of conservation of mass.

3. It should be molecular, i.e., every species involved in the reaction must be represented in its molecular form. The equation \(\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_2+2 \mathrm{H}\), for example, represents a true chemical change and is balanced, but it is still not a complete equation since hydrogen has not been expressed in its molecular form. The correct representation would be

⇒ \(\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_2+\mathrm{H}_2\)

Read and Learn More WBCHSE For Class11 Basic Chemistry Notes

Significance of a chemical equation: A chemical equation conveys a lot of information about the reaction, both qualitative and quantitative.

1. It gives the names of the reactants and products taking part in the chemical reaction.
2. It indicates the relative number of moles of the reactants and the products.
3. It shows the relative masses of the reactants and products.
4. It indicates the relative volumes of gaseous reactants and products.

Consider the information provided by the following equation.

⇒ \(\underset{\frac{40+12+3 \times 16}{100}}{\mathrm{CaCO}_3}+\underset{\frac{2(1+35.5)}{73}}{2 \mathrm{HCl}} \rightarrow \underset{\frac{40+2 \times 35.5}{111}}{\mathrm{CaCl}_2}+\underset{\frac{2 \times 1+16}{18}}{\mathrm{H}_2 \mathrm{O}}+\underset{\frac{12+2 \times 16}{44}}{\mathrm{CO}_2}\)

1. It shows that calcium carbonate reacts with hydrochloric acid to produce calcium chloride, carbon dioxide, and water.
2. It also shows that one molecule (one mole) of calcium carbonate reacts with two molecules (two moles) of hydrochloric acid to produce one molecule (one mole) each of calcium chloride, carbon dioxide, and water.
3. It indicates that 100 parts by weight of calcium carbonate reacts with 73 parts by weight of hydrochloric acid to give 111 parts by weight of calcium chloride, 44 parts by weight of carbon dioxide, and 18 parts by weight of water.

A chemical equation does provide a lot of valuable information about a chemical reaction, but it has a number of limitations. We can overcome these limitations by using certain appropriate additional symbols in the chemical equations. To begin with, the state of reactants and products can be indicated by (s) for solid, (l) for a liquid, (g) for gas, and (aq) for an aqueous solution. The arrow pointing upwards (↑) indicates the evolution of a gas whereas that pointing downward (↓) signifies precipitation.

⇒ \(\mathrm{AgNO}_3(\mathrm{aq})+\mathrm{NaCl}(\mathrm{aq}) \longrightarrow \mathrm{AgCl}(\mathrm{s}) \downarrow+\mathrm{NaNO}_3(\mathrm{aq})\)

The conditions (temperature, pressure, and catalyst) can be indicated on the arrow. The heat changes accompanying a chemical reaction can be indicated by change in enthalpy AH.

⇒ \(\mathrm{N}_3(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \underset{\text { Fe(anlyst) }}{\stackrel{773 \mathrm{~K}, 350 \mathrm{~atm}}{\longrightarrow}} 2 \mathrm{NH}_3(\mathrm{~g})\)

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \Delta H^{\oplus}=-890.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The superscript Θ denotes standard conditions of temperature and pressure.

The reversible nature of a reaction is indicated by a double-head arrow,

∴ \(2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3\)

Balancing chemical equation: According to the law of conservation of mass a chemical equation must be balanced, Le., the number of atoms of a particular kind on one side of the equation must he equal to the number of atoms of the same kind on the other side of the equation. Consider, for example, the reaction between hydrogen and oxygen to form water.

∴ H₂ +O₂ → H₂O

Writing the equation, as shown above, is not correct because the oxygen atoms on the two sides of the equation are not balanced. An equation of this kind is called skeleton equation. To balance an equation, use suitable coefficients such that the number of atoms of each species is the same on both sides of the equation. You can use one of the following methods to do this.

1. Trial and error method
2. Partial equation method

1. Trial and error method This method needs skill and practice. One has to keep trying until the equation is balanced. You can proceed in the following manner.

1. Write the skeleton equation, using symbols and formulae of the reactants and products.
2. Change elementary gases (like hydrogen, oxygen and nitrogen), if present, to their atomic states.
3. Start balancing the equation by selecting the formula containing the maximum number of atoms and balance the number of atoms of each of its constituents on both sides of the equation by multiplying with suitable numbers. Then proceed to balance the other atoms, if they are not balanced already.
4. Alternatively, first balance the atoms of the element which appear the least number of times on both sides of the equation. Then proceed to balance the other atoms.
5. Once the equation is balanced, change the elementary’ gases to their molecular form and multiply the other atoms by two.

Example: When steam is passed over iron, the products formed are magnetic oxide of iron (Fe₃O₄) and hydrogen. Write a balanced equation for the reaction.
Solution:

Let us write the skeleton equation first.

⇒ \(\mathrm{Fe}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{H}_2\)

Changing the elementary gas (hydrogen) to its atomic form:

⇒ \(\mathrm{Fe}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{H}\)

Fe₃O₄ has the largest number of atoms, so it should be balanced first. Let us start by multiplying Fe by 3 and H₂O by 4 on the left of the equation. This gives us 4 molecules of H₂O which contain 8 atoms of H on the left side. To balance hydrogen atoms on both sides, let us multiply H by 8 on the right side. The equation now becomes

⇒ \(3 \mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Fe}_3 \mathrm{O}_4+8 \mathrm{H}\)

The last step is to convert hydrogen to its molecular form and writing the balanced equation as shown below.

⇒ \(3 \mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{H}_2\)

This method is rather tedious and also difficult to use when the same element is repeated in a number of compounds in an equation. This method does not povide any information about the mechanism or pathway of a reaction either.

Partial equation method This method involves a number of steps which provide an insight into the mechanism of the reaction. The reaction is broken up into several steps and each step is represented by a partial equation. These equations are individually balanced by the trial and error method. If necessary, the partial equations are multiplied by suitable integers in order to cancel the intermediate products which do not occur in the final equation. The partial equations are then added to obtain the final balanced equation.

Example: Zinc reacts with dilute nitric acid to produce zinc nitrate, nitrous oxide and water. Write the balanced equation for this reaction.
Solution:

The skeleton equation for the reaction is given below.

⇒ \(\mathrm{Zn}+\mathrm{HNO}_3 \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\mathrm{N}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O}\)

The partial equations for the possible steps in the reaction are as follows

⇒ \(\mathrm{Zn}+\mathrm{HNO}_3 \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\mathrm{H}\)

⇒ \(\mathrm{HNO}_3+\mathrm{H} \longrightarrow \mathrm{N}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O}\)

Balancing the partial equations by the trial and error method

⇒ \(\begin{gathered}
{\left[\mathrm{Zn}+2 \mathrm{HNO}_3 \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+2 \mathrm{H}\right] \times 4} \\
2 \mathrm{HNO}_3+8 \mathrm{H} \longrightarrow \mathrm{N}_2 \mathrm{O}+5 \mathrm{H}_2 \mathrm{O} \\
\hline 4 \mathrm{Zn}+10 \mathrm{HNO}_3 \longrightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\mathrm{N}_2 \mathrm{O}+5 \mathrm{H}_2 \mathrm{O}
\end{gathered}\)

Stoichiometric calculations: A balanced chemical equation provides information about the quantitative relations between the various reactants and products or the mass and volume relations between them. This aspect of an equation, or the relative proportions in which the reactants react and the products are formed, is called stoichiometry (from the Greek word meaning ‘to measure an element’).

The quantitative information provided by an equation can be used to make the following calculations.

1. If the mass of one of the reactants or products is given, the mass of another reactant or product can be calculated.
2. If the mass or volume of one of the reactants or products is given, the mass or volume of another reactant or product can be calculated.
3. If the volume of one of the reactants or products is given, the volume of another reactant or product can be calculated.

While doing any such calculation, proceed step by step. First, write the balanced chemical equation. Then, write the number of moles, gram-atomic, or gram-molecular masses of the reactants and the products. If one of the reactants or products is gas, write its volume at stp instead of the number of moles, gram-atomic, or gram molecular mass.

Example: Methane bums in oxygen to form carbon dioxide and water, raising the amount of oxygen required to bum 4.0 g of methane. The balanced chemical equation for the reaction is \(\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}\).
Solution:

⇒ \(\underset{1 \mathrm{~mol}}{\mathrm{CH}_4}+\underset{2 \mathrm{~mol}}{2 \mathrm{O}_2} \longrightarrow \underset{1 \mathrm{~mol}}{\mathrm{CO}_2}+\underset{2 \mathrm{~mol}}{2 \mathrm{H}_2 \mathrm{O}}\)

⇒ \(\underset{16 \mathrm{~g}}{12+4} \underset{164 \mathrm{~g}}{2 \times 32}\)

16 g of methane needs 64 g of oxygen for complete burning.

∴ 4 g of methane will need 64/16 x 4 = 16 g of oxygen for complete burning.

Limiting reagent: A chemical reaction always takes place in the molar ratio given by the balanced chemical equation. If the reactants are present in this ratio, they will be consumed completely. But what if they are not? Then the reaction will stop as soon as one of the reactants is used up completely.

Suppose A and B react in the molar ratio 1 : 2 and that the amounts of the reactants taken in an experiment are in the ratio 2 : 2. Then the reaction will stop as soon as the reactant B is completely used up and half the amount of A will be left unused. In such cases B is called the Untiling reagent, because the amount of B present determines when the reaction will stop, or rather, the amount of the product formed. One could define the limiting reagent as the reactant which is completely consumed during a reaction. This aspect is also important in stoichiometric calculations. The following solved example will make it clear.

Example: How much water can be obtained from 2.00 g of hydrogen and 2.00 g of oxygen by the reaction \(2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}\)? Which is the limiting reagent in this case? Calculate the amount of the other reactant that will be left behind.
Solution:

2.00 g of H₂ = \(\frac{2.00}{2.00}\) = 1 mole of hydrogen.

2.0 g O₂ = \(\frac{2.00}{32.00}\) = 0.062 moles of oxygen.

From the equation, 1 mol of oxygen reacts with 2 mol of hydrogen to form 2 mol of water. Therefore, 0. 062 mol of oxygen will react with 2 x 0.062 = 0.124 mol of hydrogen to form 0.124 mol of water.

∴ Hydrogen used up = 0.124 mol

∴ hydrogen left = 1 – 0.124 = 0.876 mol.

∴ Mass of hydrogen left = 0.876 x 2 = 1.752 g

Mass of water formed = 0.124 x 18 = 2.132 g

The limiting reagent is obviously oxygen since all of it is consumed.

Reactions in solution: A solution is a homogeneous mixture of at least two non reacting substances. The component present in the smaller amount is called the solute, while the other component is called the solvent. The relative amounts of solvent and solute present in a solution is expressed in terms of concentration. The concentration is usually expressed in terms of mass percent, mole fraction, molality, or molarity.

Mass percent It is the grams of solute in 100 g of the solution and is obtained by the following relation:

∴ Mass percentage = \(\frac{\text { mass of solute }}{\text { mass of solution }} \times 100\)

Example 1. A solution is prepared by adding 5 g of cane sugar in 20 g of water. Calculate the mass per cent of the solute.
Solution:

Mass percent of solute = \(\frac{\text { mass of cane sugar }}{\text { mass of solution }} \times 100 \)

= \(\frac{5 \mathrm{~g}}{5 \mathrm{~g} \text { of cane sugar }+20 \mathrm{~g} \text { water }} \times 100 \)

= \(\frac{5 \mathrm{~g}}{25 \mathrm{~g}} \times 100=20 \% \).

Mole fraction The mole fraction (x) of any component in a solution is given by the number of moles of the component divided by the total number of moles making up the solution (including the solvent). For example, if nA moles of substance A is dissolved in nB moles of substance B then the mole fraction of A(x_a) and that of B (x_b), respectively, are

⇒ \(x_{\mathrm{A}}=\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}} \text { and } x_{\mathrm{B}}=\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}+n_{\mathrm{B}}} \text {. }\)

Note that a mole fraction has no units (it is a ratio) and has a value between 0 and 1 (since x_a + x_b = 1). Except in a few situations, mole fractions are not often used for liquid solutions but for calculations involving gaseous mixtures.

Example 2. 58,5 g of sodium chloride is dissolved in 180 g of water. Calculate the mole fraction of sodium chloride in the solution.
Solution:

58.5 g of sodium chloride = 1 mol.

∴ 180 gof H₂O = 10 moL

Mole fraction of NaCl = \(=\frac{\text { number of moles of } \mathrm{NaCl}}{\text { total number of moles of solution }}\)

= \(\frac{1}{1+10}=\frac{1}{11}=0.090\)

Molality It is defined as the number of moles of solute present in 1 kg of the solvent and is denoted by m.

Molality (m) = \(\frac{\text { number of moles of solute }}{\text { mass of solvent in } \mathrm{kg}} \text {. }\)

Molarity The molarity of a solution is the number of moles of the solute per litre of the solution. The unit of molarity js mol L^-1. A more common unit is mol dm^-3.

⇒ \(\text { Molarity }(M)=\frac{\text { number of moles of solute }}{\text { volume of solution in litres }(V)}\)

⇒ \(\text { Number of moles }=\frac{\text { weight of solute }(W)}{\text { molecular weight }(M) \text { of solute }}\)

∴ \(\text { Molarity }=\frac{\text { weight of solute }}{\text { molecular weight of solute }} \times \frac{1000}{\text { volume of solution in } \mathrm{mL}}\)

⇒ \( M=\frac{W}{M} \times \frac{1000}{\mathrm{~V}(\mathrm{~mL})}\)

If some more solvent is added to a solution, it becomes more dilute, but the number of moles present in if does not change. Suppose the molarity of a solution is M₁ and its volume is V₁. Thus the number of moles of solute present in it is M₁, V₁. Now suppose this solution is diluted to volume V₂ and the molarity of the diluted solution is M₂. Then the number of moles present in the dilute solution, i.e., M₂V₂, must be the same as the number of moles in the original solution.

∴ M₁V₁=M₂V₂.

This equation is called the molarity equation. The molarity of a solution depends upon temperature because volume is temperature-dependent,

Example 3. Find the molality 0.02M NaCl solution.
Solution:

Since M = 0.02 mol L^-1,

0. 02 moles of NaCl are presept ip 1 L of water.

In case of water, for dilute solutions:

1000 mL = 1000 g (since the density of water Is 1 g mL^-1)

So the mass of the solvent (HO) is 1000 g = 1 kg

Molality(m) = \(\frac{\text { no. of moles of solute }}{\text { mass of solvent in } \mathrm{kg}}\)

= \(\frac{0.02 \mathrm{~mol}}{1 \mathrm{~kg}}=0.02 \mathrm{~mol} \mathrm{~kg}^{-1}\)

Example 4. A solution sodium hydroxide is prepared by dissolving 20.0 g of sodium hydroxide (NaOH) in distilled water to give 250mL of a solution. Calculate the molarity of the sodium hydroxide solution.
Solution:

Molar mass of NaOH = \(\frac{20.0 \mathrm{~g}}{40.0 \mathrm{~g} \mathrm{~mol}^{-1}}=0.50 \mathrm{~mol}\)

Volume of solution = 250 mL = 0.25 L.

Molarity = \(\frac{0.50 \mathrm{~mol}}{0.25 \mathrm{~L}}\) = 2.0 molL = 2.0 M.

Example 5. 250 cm³ of a solution of oxalic acid contains 1,26 g of axqlic acid. Calculate its molarity.
Solution:

Weight of solute = 1.26 g.

Volume of solution = 250 cm³ = 250 mL.

Molecular weight of oxalic acid [(COOH)₂-2H₂O] = 126.

M = \(\frac{1.26}{126} \times \frac{1000}{250}\) = 0.04 M

Example 6.

1. When 4.2 g of NaHCO₃ was added to a sample of acetic acid (CH₃COOH) weighing 10.0 g, 2.2 g of CO₂ was produced and the residue left weighed 12.0 g. Show that these observations are in agreement with the law of conservation of mass.
2. In another experiment, 6.3 g of NaHCO₃ Teas was added to 15.0 g of acetic acid. The residue weighed 18.0 g. Find the mass of CO₂ released in the reaction.

Solution:

1. The given reaction can be represented by the following chemical equation.

⇒ \(\mathrm{NaHCO}_3+\mathrm{CH}_3 \mathrm{COOH} \longrightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

Sum of the masses of the reactants = 4.2 + 10.0 = 14.2g.

Sum of the masses of the products = 12 + 2.2 = 14.2 g.

Thus, sum of the masses of the reactants is the same as that of the products. This is in accordance with the law of conservation or mass.

2. According to the law of conservation of mass, the mass of the reactants must be the same as the mass of the products.

⇒ \(\mathrm{NaHCO}_3+\mathrm{CH}_3 \mathrm{COOH} \longrightarrow \underbrace{\mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}}+\mathrm{CO}_2 \uparrow\)

Mass of NaHCO₃ = 6.3 g. (given)

Mass of CH₃COOH = 15.0 g.

Mass of CH₃COONa + H₂O = 18 g,

∴ Mass of CO₂ = (15.0 + 6.3) -18 = 3.3 g.

Example 7. Phosphorus trichloride contains 22.55% of phosphorus, phosphine contains 91.18% phosphorus, while HCl contains 97.26% of chlorine. Shore that this data illustrates the law of reciprocal proportions.
Solution:

In PCI_2, weight of phosphorus = 22.55 g and weight of chlorine = 100 – 22.55 = 77.45 g,

i.e., 22.55 g of P combines with 77.45 g Cl.

∴ 1 g of P will combine with \(\frac{77.45}{22.55}\) = 3.43 g of Cl.

In PH₃, weight of phosphorus = 91.18 g.

and weight of hydrogen = 100 – 9118 = 8.82 g,

i.e., 91.18 g of P combines with 8.82 g of hydrogen.

∴ 1 g of P combines with \(\frac{8.82}{91.18}\) = 0.096 g of H.

In HCl, the weight of chlorine = 97.26 g.

and weight of hydrogen = 100 – 97.26 = 2.74 g,

i.e. 2.74 g of hydrogen combines with 97.26 g of chlorine.

∴ 1 g of hydrogen combines with \(\frac{97.26}{2.74}\) = 35.49 g of chlorine.

The ratio of the weights of H and Cl which combine with a fixed weight of P (1 g) = 0.096 : 3.43 = 1:35.7.

The ratio of the weights of H and Cl which combine directly with each other = 2.74:97.26 = 1:35.49.

Thus, the two ratios are the same, in accordance with the law of reciprocal proportions.

Example 8. Carbon combines with hydrogen to form three compounds A, B, and C. The percentages of hydrogen in A, B, and C are 25,14.3, and 7.7 respectively. Which law of chemical combination does this example illustrate?
Solution:

In compound A, 25 parts of carbon combine with 75 parts of hydrogen.

∴ 1 part of carbon combines with \(\frac{75}{25}\) = 3 parts of H.

In compound B, 14.3 parts of carbon combine with 85.7 parts of hydrogen.

∴ 1 part of carbon combines with \(\frac{85.7}{14.3}\) = 6 parts of hydrogen.

In compound C, 7.7 parts of carbon combine with 92.3 parts of hydrogen.

∴ 1 part of carbon combines with \(\frac{92.3}{7.7}\) = 12 parts of hydrogen.

This example illustrates the law of multiple proportions because the weights of hydrogen that combine with a fixed weight of carbon bear a simple whole-number ratio to one another, viz., 3:6:12, or 1: 2:4.

Example 9. When 5.625 g of the higher oxide of a metal was heated it produced 5.325 g of the lower oxide. The sample of the lower oxide, on reductions yielded 5.025 g of the metal. Show that this illustrates the law of multiple proportions.
Solution:

Weight of lower oxide = 5.325 g.

Weight of higher oxide = 5.625 g.

Weight of oxygen in lower oxide = 5.325 – 5.025 = 0.300 g.

Weight of oxygen in higher oxide = 5.625 – 5.025 = 0.600 g (because the weight of the metal is the same in both the oxides).

Therefore, the ratio of the weight of oxygen that combines with a fixed weight of the metal in the two oxides = 0.300: 0.600 = 1:2.

This is a simple whole-number ratio, so the experiment illustrates the law of multiple proportions.

Example 10. Iron reacts zvith steam to produce iron oxide (Fe₃O₄). Calculate the zveight of iron zuhich will be converted into its oxide (Fe₃O₄) by 9 g of steam.
Solution:

The balanced chemical equation is 3 \(\mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{H}_2\)

(4 x 18) g of steam reacts with (3 x 56) g of iron

or 72 g of steam reacts with 168 g of iron.

∴ 9 g of steam will react with \(\frac{168}{72}\) x 9 = 21 g of iron.

Thus 9 g of steam can convert 21 g of iron into its oxide.

Example 11. How much potassium chlorate ivould be required to produce 1.12 L of oxygen at stp?
Solution:

The balanced chemical equation for the reaction is

⇒ \(\underset{2(39+35.5+3 \times 16) \mathrm{g} / \mathrm{L}}{2 \mathrm{KClO}_3} \longrightarrow 2 \mathrm{KCl}+\underset{3 \times 227}{3 \mathrm{O}_2}\)

From the equation, 22.7 x 3 L of oxygen is produced by 2(39 + 35.5 + 3 x 16)g of KCIO₃

∴ 1.12 L of oxygen can be obtained from \(\frac{245}{68.1}\) x 1.12 = 4.03 g of KCIO₃.

The required amount of KCIO₃ is 4.03 g.

Example 12. How much hydrochloric acid (HCl) is produced by 0.8 g of hydrogen and an excess of chlorine?
Solution:

The balanced chemical equation for the reaction is

∴ \(\mathrm{H}_2+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{HCl}\)

From the equation, 2 g of hydrogen will yield 2(1 + 35.5) = 73 g of HC1.

∴ 0. 8 g of hydrogen will yield \(\frac{73}{2}\) x 0.8 = 29.2 g of HCl.

The amount of HCl produced is 29.2 g.

Example 13. 2 g of an impure sample of magnesium sulfate was dissolved in water and treated with an excess of barium chloride. The weight of the dry precipitate of BaSO₄ produced was 3.0 g. Calculate the percentage purity of the sample.
Solution:

The chemical equation for tlie reaction is

⇒ \(\mathrm{MgSO}_4+\mathrm{BaCl}_2 \longrightarrow \mathrm{BaSO}_4+\mathrm{MgCl}_2\)

From the equation, (137 + 32 + 4 x 16) g of BaSO₄ is obtained from (24 + 32 + 4 x 16) g of MgSO₄ or 233 g of BaSO₄ is obtained from 120 g of MgSO₄.

∴ 3.0 g of BaSO₄, will be obtained from \(\frac{120}{233}\) x 3.0 = 1.54 g of MgSO₄.

This means that 2 g of the impure sample of MgSO₄ contains 1.54 g of pure MgSO₄.

∴ 100 g of the impure sample will contain \(\frac{154}{2}\) x 100 = 77.0 g of pure MgSO₄.

Therefore, the percentage purity of MgSO₄ is 77.0.

Example 14. What is the volume of oxygen that will be required to burn 1L of methane completely? How much carbon dioxide will be produced in the process?
Solution:

The chemical equation for the reaction is

⇒ \(\mathrm{CaCO}_3+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

According to the equation, 2 mol of CH₄ combines with 3 mol of O₂ to produce 2 mol of CO₂ and 2 mol of water vapour

or 2 volumes of CH₄ combine with 3 volumes of O₂ to produce 2 volumes of CO₂ and 2 volumes of water vapour

or 2 L of CH₄ combine with 3 L of O₂ to produce 2 L of CO₂ and 2 L of water vapor,

i. e., 2 L of CH₄ require 3 L of O₂.

∴ 1L of CH₄ requires — = 15 L of oxygen.

Also 2 L of CH₄ yield 2 L of CO₂

∴ 1 L of CH₄ yields 2/2 = 1L of CO₂.

Example 15. What is the volume of carbon dioxide measured at stp that would be obtained by the action of dilute hydrochloric acid on 15.0 g of calcium carbonate?
Solution:

The chemical equation representing the reaction is

⇒ \(\mathrm{CaCO}_3+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

From the equation (40 +12 + 3 x 16) g = 100 g of CaCO₃ reacts with HCl to produce 22.7 L of CO₂ at stp.

∴ 15 g of CaCO₃ will produce \(\frac{22.7}{100}\) x 15 L = 3.41 L of CO₂ at stp.

Example 16. How much oxygen will be required at stp for the complete combustion of 300 cm³, of acetylene? How much CO₂ will be produced in the process?
Solution:

The chemical equation for the reaction is

⇒ \(2 \mathrm{C}_2 \mathrm{H}_2+5 \mathrm{O}_2 \longrightarrow 4 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}\)

From the equation, 2 volumes of C₂H₂ bum in 5 volumes of qxygen to produce 4 volumes of CO₂, and 2 volumes of water vapour.

Using Gay-Lussac’s law of gaseous volumes, 300 cm³ of acetylene would require 5/2 x 300 = 750 cm3 of oxygen at stp for complete combustion.

Also, 2 volumes of acetylene produce 4 volumes of CO₂ at stp.

∴ 300 cm³ of acetylene will produce \(\frac{4}{2}\) x 300 = 600 cm³ of CO₂ at stp.

Example 17. How much hydrogen will be liberated if 6.5 g of Zn is added to 1 L of a hydrochloric acid solution containing 3.65 g of HCl per litre? Which reactant will be left behind when the reaction stops? How much of it will be left behind?
Solution:

Number of moles of Zil taken = \(\frac{6.5}{65}\) = 0.L

Number of moles of HCl in solution = \(\frac{3.65}{36.5}\) = 0.1

The equation for the reaction is

⇒ \(\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_2+\mathrm{H}_2\)

According to the equation, 1 mol of Zn reacts with 2 mol of HCl to produce 1 mol of H₂.

⇒ 0.1 mol of Zn Would require 0.2 mol Of HCl.

But the solution contains 0.1 mol of HCl.

Therefore, the limiting reagent is HCl. The reaction will stop when all the HCl is used up. Number of moles of H₂ liberated = 0.05.

∴ mass of H₂ liberated = 0.05 x 2 = 0.10 g.

∴ Number of moles of Zn used up = 0.05.

The number of moles of Zn left = 0.1 – 0.05 = 0.05.

∴ mass of Zn left = 0.05 x 65 = 3.25 g.

Example 18. Determine the number of moles of barium phosphate produced when 0.2 mol of sodium phosphate is treated with 0.1 mol of barium nitrate if the reaction proceeds according to the following equation.

⇒ \(2 \mathrm{Na}_3 \mathrm{PO}_4+3 \mathrm{Ba}\left(\mathrm{NO}_3\right)_2 \longrightarrow \mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2+6 \mathrm{NaNO}_3\)

Solution:

According to the equation, 2 mol of Na₃PO₄ reacts with 3 mol of Ba(NO₃)₂ to produce 1 mol of Ba₃(PO₄)₂.

Therefore, 0.1 mol of Ba(NO₃)₂ will react with \(\frac{2}{3}\) x 0.1 = 0.066 mol of Na₃PO₄. Thus, barium nitrate is the limiting reagent.

0.1 mol of Ba(NO₃)₂ will produce \(\frac{1}{3}\) x 0.1 = 0.033 mol of Ba₃(PO₄)₂.

Example 19. Find the number of moles o/KMn04 required to oxidise 1.35 g of oxalic acid completely in an acidic medium. How much 0.05-M KMnO₄ will be required for the above oxidation? The reaction takes place according to the following equation.

⇒ \(2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

Solution:

Molecular mass of oxalic acid = 2×1 + 2×12 + 4×16 = 90.0 u.

∴ number of moles of oxalic acid in 1.35 g of it = \(\frac{135}{90}\) = 0.015.

According to the equation given, 5 mol of oxalic acid is oxidised by 2 mol of KMnO₄.

∴0.015 mol of oxalic acid will require \(\frac{2 \times 0.015}{5}\) = 0.006 rnol of KMnO₄.

It is given that 0.05 mol of KMnO₄ is present in 1000 cm of the solution.

∴ 0.06 mol KMnO₄ will be contained in \(\frac{1000}{0.05}\) x 0.006 = 120 cm³ of the solution.

So the required volume of the 0.05-M KMnO₄ solution = 120 cm³

Example 20. 100 cm³ of a solution of NaOH contains 2.0 g of NaOH. Calculate its molarity.
Solution:

Weight of solute = 2.0 g.

Volume of solution = 100 cm³ = 0.1 L.

Molecular weight of NaOH = 23 + 16 + 1 = 40.

∴ Molarity = \(\frac{\text { weight of solute }}{\text { molecular weight }} \times \frac{1000}{\text { volume }\left(\mathrm{cm}^3\right)}\)

= \(\frac{2}{40} \times \frac{1000}{100}=0.5 \mathrm{M}\) .

Example 21. Calculate the amount of NaOH required to prepare 250 cm³ of a 0.1 M NaOH solution.
Solution:

Weight of solute = molarity x molecular weight x volume (L)

= 0.1 x 40 x \(\frac{250}{1000}\) = 1 g.

Example 22. How many grams 0f BaSO₄ will beformed when 500 cm³ of a 0.250-M Na₂SO₄ solution is added to an aqueous solution of 15.00 g o/BaCl₂?
Solution:

The balanced equation for the given reaction is

⇒ \(\mathrm{BaCl}_2+\mathrm{Na}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{NaCl}+\mathrm{BaSO}_4\)

According to the equation,1 mol of BaCl₂ reacts with 1 mol of N2SO₄ to form 1 mol of BaSO₄.

Number of moles of BaCl₂ =\(\frac{\text { weight of } \mathrm{BaCl}_2}{\text { molecular weight of } \mathrm{BaCl}_2}=\frac{15}{208.4}=0.072\)

Number of moles of Na₂SO₄ = molarity x volume (L) = 0.250 x \(\frac{500}{1000}\) = 0.125

Thus the limiting reagent is BaCl₂.

0.072 mol of BaCl₂ will react with 0.072 mol of Na₂SO₄ to yield 0.072 mole of BaSO₄.

∴ mass of BaSO₄ = 0.072 x 233.4 =16.80 g.

Example 23. Commercially available concentrated sulphuric acid is 98% by mass. Its density is 1.84 g cm^-3.

1. Calculate the molarity of the solution.
2. What is the volume of concentrated H₂SO₄ that will be required to prepare 2 L of a 5-M sulphuric acid solution?

Solution:

1. The concentration of the given solution can be determined by two methods.

Density of solution = 1.84 g cm³.

∴ weight of 1000 cm³ of solution =184 x 1000 =1840 g.

Concentration of solution = 98% by mass

i.e., 100 g of solution contains = 98 g of H₂SO₄.

∴ 1840 g of solution contains =\(\frac{98 \times 1840}{100}\) g of H₂SO₄

Molarity = \(\frac{\text { weight of solute }}{\text { molecular weight }} \times \frac{1}{\text { volume in litre }}\)

= \(\frac{98 \times 1840}{98 \times 100}=18.4\)

Alternative method

Volume = \(\frac{\text { mass }}{\text { density }}\)

Volume 100 g of the solution = \(\frac{\text { mass percentage }}{\text { molecular weight }} \times \frac{1000}{\text { volume }\left(\mathrm{cm}^3\right)}\)

= \(\frac{98}{98} \times \frac{1000}{54.3}=18.4 \mathrm{M}\)

2. M₁V = M₂V₂

M₁ = 18.4 M; V₁ = ?

M₂ = 5 M; V₂ = 20 L66

∴ \(V_1=\frac{M_2 V_2}{M_1}=\frac{5 \times 20}{18.4}=5.43 \mathrm{~L}\)

The volume of the concentrated H₂SO₄ required is 5.43 L

The Mole

When we are dealing with elements and compounds and the reactions between them, it often becomes necessary (or more convenient) to know the number of atoms or molecules that a sample of an element or compound may contain. This is because atoms or molecules are the basic units taking part in a chemical reaction. Reactions take place between atoms of reactants and molecules of products are formed. Consider one example.

We know that one molecule of oxygen combines with two molecules of hydrogen to produce two molecules of water vapour. Suppose we wish to take molecules of oxygen and hydrogen in the ratio 1: 2, so that all the molecules of the reactants are used up in a reaction. How do we do so, since we cannot count molecules? The mole, a chemist’s unit for counting atoms and molecules, was introduced to solve such problems.

• Let us go back to the concept of (relative) atomic mass. The ratio of the masses of the hydrogen atom and the oxygen atom is 1.008:16, so the ratio of the masses of the hydrogen molecule and the oxygen molecule is 2.016: 32. Then the ratio of the mass of n molecules of hydrogen to that of n molecules of oxygen must also be 2.016: 32.
• In that case, if the ratio of the mass of a sample of hydrogen to the mass of a sample of oxygen is 2.016: 32, they must both contain the same number of molecules. In particular, a sample of hydrogen weighing 2.016 g and a sample of oxygen weighing 32 g must contain the same number of molecules. Let us say this number is N. by the same logic a gram-molecule of any compound (or element) must contain the same number of molecules as 2.016 g of hydrogen, i.e., N. This equals the number of molecules present in one gram-molecule of a compound (or element).
• As mentioned earlier, units like the gram-atom and gram-molecule have been replaced by the mole. A mole of a substance contains tire Avogadro number of elementary particles. These elementary particles may be atoms or molecules or any other discrete particles such as ions, electrons and protons.
• A mole is defined as the amount of a substance that contains as many entities (atoms, molecules, or other particles like ions) as there are atoms in exactly 0.012 kg (or 12 g) of the carbon-12 isotope. In SI units, we represent mole by the symbol mol.

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In order to determine the number of atoms in 12 g of ¹²C isotope, the mass of a carbon-12 atom was determined by a mass spectrometer, which came out to be 1.992648 x 10^-23 g. Knowing that 1 mol of carbon weighs 12 g, the number of atoms in it is equal to

⇒ \(\frac{12 \mathrm{~g}}{1.992648 \times 10^{-23} \mathrm{~g}}=6.022137 \times 10^{23}\) = 6.022137 x 10²³

Thus, we can say that a mole has 6.022137 x 10²³ atoms, molecules, ions, etc. This number is denoted by the symbol NA and is known as the Avogadro constant as it is a physical quantity and not a pure number. In most of the calculations, its value with four significant figures is used, i.e., 6.022 x10²³. Put simply, the mole is the chemical counting unit used by chemists for 6.022 x 10²³ particles of a substance.

• Although, while using the term mole, the kind of particles involved are specified, this may not always be the case. If we simply refer to one mole of a substance, it means we are talking about the naturally occurring form of that substance. For example, one mole of oxygen contains 6.022 x 10 ²³ oxygen molecules. Similarly, one mole of sodium contains 6.022 x 10²³ atoms of sodium, and one mole of sodium carbonate (Na₂CO₃) contains 6.022 x 10²³ molecules of sodium carbonate.
• According to Avogadro’s hypothesis, equal volumes of all gases under similar conditions of temperature and pressure contain equal numbers of molecules. This means 6.022 x 10²³ molecules of any gas at stp (0°C and 1 bar pressure) must have a particular volume. This volume has been found experimentally to be equal to 22.7 L and is called the molar volume. Put simply, 1 mole of any gas at stp has a volume of 22.7 L.
• It should be clear from the preceding discussion that a mole can be expressed in terms of mass, number, or volume.
• In terms of mass, a mole is defined as that amount of a substance whose mass in grams is numerically equal to its atomic mass (if the substance is atomic) or molecular mass (if the substance is molecular) in amu or u.

∴ 1 mole of oxygen (atomic) weighs 16 g.

∴ 1 mole of oxygen (molecular) weighs 32 g.

∴ 1 mole of carbon (atomic) weighs 12 g.

∴ 1 mole of H₂O (molecular) weighs 18 g.

Here again we must remember that one mole of hydrogen weighs 2.0 g if it has not been specified whether we are talking of atoms or molecules. Similarly one mole of sodium weighs 23.0 g and one mole of sodium carbonate weighs 106.0 g. Mass of one mole of any substance will be its molar mass.

∴ Molar mass of hydrogen (H₂) = 2.0 g/mol^-1

∴ Molar mass of sodium (Na) = 23.0 g mol^-1.

∴ Molar mass of sodium chloride (NaCl) = 585 g/mol^-1.

• Therefore, the molar mass is equal to the atomic mass or molecular mass expressed in grams, depending upon whether the substance contain atoms or molecules.
• In terms of volume (related to gases), a mole is the amount of a gas that occupies a volume of 22.7 L at stp. This means 1 mole of oxygen (gas) or 1 mole of CO₂ (gas) occupies 22.7 L at stp.
• Tire mole concept can also be applied to ionic compounds. Ionic compounds do not contain molecules. They are made up of positively and negatively charged ions. The formula of an ionic compound does not represent one molecule, but provides the ratio between the constituent ions.
• The formula of sodium chloride, for example, does not represent a molecule of sodium chloride—it merely expresses the ratio of sodium and chloride ions in sodium chloride. While referring to ionic compounds, we use the term formula unit, rather 6.022 x 10²³ formula units. The mass of one mole of formula units of an ionic compound is equal to its formula mass (rather than molecular mass) expressed in grams, or gram-formula mass.

Thus, a mole of NaCl weighs 58.5 g (1 gram-formula mass) and contains 6.022 x 10²³ formula units of NaCl or 6.022 x 10²³ Na⁺ ions and 6.022 X 10²³ Cl^– ions. Similarly, a mole of ferric chloride (FeCl₃)weighs 161.7 g and contains 6.022 x 10²³ FeCl₃ formula units or 6.022 x 10²³ Fe³⁺ ions and 3 x 6.022 x 10²³Cl^– ions.

As you know, the mole is a counting unit used for a huge number of particles (Avogadro constant). Thus, the mole is not a useful measure for things much larger than atoms or molecules. Further, being an SI unit, it can be used with prefixes.

For example, 1 m mol = 10^-3 mol,

∴ 1 μ mol = 10^-6 mol,

∴ 1 n mol =10^-9 mol.

We can conveniently convert a larger number of entities (atoms, molecules, etc.) into a number of moles and vice versa using the Avogadro constant.

⇒ Number of moles of entities/particles = \(\frac{\text { number of entities } / \text { particles }}{N_{\mathrm{A}}}\)

The following summary should help you remember what you have leamt about the mole and the Avogadro constant.

The mole concept and Avogadro constant are used in various types of chemical calculations.

• Molar masses are also used as conversion factors between masses in grams and number of moles. In order to find out the number of moles of an element we divide the mass of the element by its molar mass. To calculate the number of atoms in a given number of moles of a substance, we multiply the number of mole by the Avogadro constant. To calculate the mass of one atom we divide the atomic mass by the Avogadro constant.
• It is easier to calculate the molar volume of a liquid or solid if we know the molar mass and density at any given temperature and pressure as these do not change much with variation in temperature and pressure.

Example 1. Calculate the mass of

1. An atom of copper and
2. A molecule of sulphur dioxide.

Solution:

1. Atomic mass of copper = 64 u.

∴ mass of 1 mole of copper atoms = 64 g.

⇒ mass of 6.022 x 10²³ atoms of copper = 64 g.

∴ mass of 1 atom of Cu = \(\frac{64}{6.022 \times 10^{23}}\) = 10.63 x 10^-23 g.

2. Molecular mass of SO₂ =1×32 +2×16 = 64 u.

∴ mass of 1 mole of SO₂ = 64 g

⇒  mass of 6.022 x 10²³ molecules of SO₂ = 64 g.

∴ mass of 1 molecule of SO₂ = \(\frac{64}{6.022 \times 10^{23}}\) = 10.63 x 10^-23 g.

Example 2. Calculate the number of ozone molecules and that of oxygen atoms present in 64 g of ozone.
Solution:

Molecular mass of ozone (O₃) = 16 x 3 = 48 u.

∴ mass of 1 mole of ozone molecules = 48 g.

⇒ 48 g of ozone contains of 6.022 x 10²³ molecules.

∴ 64 g of ozone will contain \(\frac{6.022 \times 10^{23}}{48}\) x 64 = 830 x 10²³ molecules.

1 molecule of ozone contains 3 atoms of oxygen.

∴ 830 x 10²³ molecules of ozone will contain 3 x 830 x 10²³ = 24.9 x 10²³ oxygen atoms.

Example 3. Assuming that the density of water is 1 g cm^-3, calculate (a) the volume of one molecule of water and (b) the radius of a water molecule, taking it to be spherical.
Solution:

1. Volume of 1 mole of H₂O = 18 cm³ (because density of H₂O = 1g cm^-3 and mass of 1 mole of H₂O = 18 g)

⇒ volume of 6.022 x 10²³ molecules of H₂O = 18 cm³.

∴ volume of 1 molecule of H₂O = \(\frac{18}{6.022 \times 10^{23}} \mathrm{~cm}^3\) = 2.99 x 10^-23 cm³.

2. Let R be the radius of a water molecule. Then

4/3 πR³ = 2.99 x 10^-23 cm³ (volume of a sphere of radius r =4/3 πr³).

∴ R³ = 7.13 x 10^-24 = (7.13)^1/3 x 10^-8= 1.92 x 10^-8 cm.

Example 4. Calculate the number of moles in the following.

1. 460 g of H₂SO₄
2. 67,2 L of CO₂ at stp

Solution:

1. Mass of 1 mole of H₂SO₄ = 2 x 1 + 32 + 4 x 16 = 98 g.

∴ number of moles in 460 g of H₂SO₄ = 1/98 x 460 = 5.

2. 1 mole of CO₂ occupies 22.4 L at stp.

∴ 67.2 L of CO₂ at stp contains 1/22.4 x 67.2 = 3 moles of CO₂.

Laws Of Chemical Combination

The laws of chemical combination are the outcome of the dedicated work of Lavoisier. Son of a wealthy French lawyer, Lavoisier graduated in law but was far more fascinated by chemistry and spent his life studying chemical phenomena. He was perhaps the first chemist to realise how important quantitative measurements are to the study of chemical processes. Unfortunately, he was guillotined (beheaded) during the French Revolution.

It was Lavoisier who showed that weighing substances before and after a chemical change could be an important tool in understanding chemical phenomena. Through careful weighing he proved that mercury combines with the oxygen present in air to form a red solid, mercuric oxide, which when heated strongly gives the same amount of mercury as was used to prepare the mercuric oxide originally.

⇒ \(2 \mathrm{Hg}+\mathrm{O}_2 \longrightarrow 2 \mathrm{HgO}\)

⇒ \(2 \mathrm{HgO} \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{Hg}+\mathrm{O}_2\)

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This proved that mercuric oxide is not an element, but a compound. Lavoisier’s studies revealed the nature of combustion and proved that air consists of oxygen, which supports combustion, and nitrogen, which does not. Out of the work of Lavoisier and other scientists after him arose a few experimental laws on how elements combine with each other to form compounds. These laws were based entirely on observations related to the weight-weight, weight-volume, and volume-volume relationships between reacting substances and the products formed. These law’s are listed below.

1. Law of conservation of mass (Lavoisier, 1774)
2. Law of constant compositions (Proust, 1799)
3. Law of reciprocal proportions (Richter, 1792)
4. Law of multiple proportions (Dalton, 1803)
5. Gay-Lussac’s law of combining volumes (Gay-Lussac, 1808)

1. Law of conservation of mass: This law states that in a chemical change, the total mass of the products is equal to the total mass of the reactants. In other words, matter can neither be created nor destroyed. Landolt, a chemist, carried out an experiment to verify this law. The experiment has come to be known as Landolt’s experiment.

He took a solution of sodium chloride (NaCl) and a solution of silver nitrate (AgNO₃) in the two limbs of an H-shaped tube (called Landolt’s tube), sealed the tube and weighed it. Then he allowed the two solutions to mix and react by shaking the tube. After the AgNO₃ and NaCl solutions had reacted to form a white precipitate of AgCl he weighed the tube again.

⇒ \(\mathrm{AgNO}_3(\mathrm{aq})+\mathrm{NaCl}(\mathrm{aq}) \longrightarrow \mathrm{AgCl}(\mathrm{s})+\mathrm{NaNO}_3(\mathrm{aq})\)

He found that the weight of the tube and its contents after the reaction was the same as its weight before the reaction. This proved Lavoisier’s law of conservation of mass or indestructibility of matter.

Example: When 4.48 g of KClO₃ is heated, 1.73 g of oxygen is produced and the residue of KCI left behind weighs 2.75 g. Show that these observations illustrate the law of conservation of mass.
Solution:

The chemical equation is

⇒ \(\underset{4.48 \mathrm{~g}}{2 \mathrm{KClO}_3} \stackrel{\text { heat }}{\longrightarrow} \underset{2.75 \mathrm{~g}}{2 \mathrm{KCl}}+\underset{1.73 \mathrm{~g}}{3 \mathrm{O}_2}\)

The total mass of the products (2.75 + 1.73 = 4.48 g) formed is equal to the mass of the reactant (4.48 g).

This proves that the law of conservation of mass holds good for this reaction.

The discovery of nuclear reactions and radioactive disintegration has changed this view of the indestructibility of matter. The change in mass in such reactions is very significant. The mass that seems to be lost is actually converted into energy in accordance with Einstein’s equation, E = mc²,

∴ where m is the mass in kilograms, E is the energy in joules, and c is the velocity of light (3 x 10⁸ ms^-1).

In view of such reactions, the law of conservation of mass has now been modified and is known as the law of conservation of mass-energy. According to the modified law, mass and energy are interconvertible and the sum of the mass and energy of the system remains constant.

You may wonder why there is no decrease in the mass of the reactants in a chemical reaction in which heat is liberated, or why there is no gain in mass in reactions in which heat is absorbed. The truth is that the gain or loss in mass in such (chemical) reactions is too small to be detected by ordinary methods of chemical analysis. If, for example, the energy produced (heat liberated) in a reaction is of the order of 9 x 10⁷ kJ, the corresponding loss in mass is only 0.001 g. This is why the law of conservation of mass holds good for all chemical reactions.

2. Law of constant composition: This law, postulated by Joseph Louis Proust (1799), a French chemist, states that a sample of a pure chemical compound always consists of the same elements combined in the same definite proportion by mass. For example, carbon dioxide will always contain only carbon and oxygen combined in the ratio of 3 : 8 by mass, no matter what its source.

• Similarly, pure water, obtained from any source, will always contain hydrogen and oxygen combined in the ratio of 1 : 8 by mass. This may sound very obvious to you now, but when Proust came up with this conclusion, after painstaking observations, it was a major breakthrough in chemistry. The law of constant composition is also called the law of definite proportions.
• The law of constant composition is not valid if the same compound is obtained from isotopes of an element, since the mass of each isotope is different. For example, two samples of sodium chloride obtained from two isotopes of chlorine (Cl³⁵ and Cl³⁷) have different compositions. The ratio of the masses of the elements in the two samples would be 23:35 and 23:37 respectively.

Example: A sample of copper nitrate was obtained by dissolving 1.58 g of copper in nitric acid. This sample of copper nitrate yielded on ignition 1.97 g of copper oxide. In another experiment, 1.02 g of copper was dissolved in nitric acid and precipitated as copper hydroxide by adding sodium hydroxide to the solution. The sample of copper hydroxide (after being washed and dried) produced 1.26 g of copper oxide when it was ignited. Show that the results of the tivo experiments illustrate the law of constant composition.
Solution:

In the first experiment, 100 g of copper oxide contained \(\frac{1.58}{1.97}\) x 100 = 80.1g of copper.

In the second experiment, 100 g copper oxide contained \(\frac{1.02}{1.28}\) x 100 = 79.9 g of copper.

The percentage of copper in the two samples of copper oxide is nearly the same, so the results illustrate the law of constant composition.

3. Law of multiple proportions: John Dalton, a British chemist and physicist, formulated this law in 1803. In essence this law states that there is a simple relationship between the masses of elements which combine to form different compounds.

• A more precise way to say this is, when two elements A and B combine to form two or more compounds (example,  AB, A₂B, AB₂), the different masses of one of the elements (say A) that combine with a fixed mass of the other (say B) (or vice versa) are in a simple ratio. For example, nitrogen and oxygen combine to form 5 oxides, viz., nitrous oxide (N₂O), nitric oxide (NO), nitrogen trioxide (N₂O₃), nitrogen tetraoxide (N₂O₄) and nitrogen pentoxide (N₂O₅). Calculations show that the masses of oxygen which combine with a fixed mass of nitrogen to form these oxides are in the simple ratio 1:2:3 :4: 5.
• Similarly, if two pure samples of cuprous oxide (Cu₂O) and cupric oxide (CuO) are heated strongly in two crucibles, while a current of hydrogen is passed over them, it can be shown that the weights of copper which combined with the same weight of oxygen in the two samples are in the ratio 2:1. The weights of copper are obtained by first weighing the empty crucibles, then the crucibles with the samples of Cu₂O and CuO and finally, the crucibles after they have been heated strongly in a current of hydrogen.

⇒ \(\mathrm{CuO}+\mathrm{H}_2 \stackrel{\text { heat }}{\longrightarrow} \mathrm{Cu}+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{Cu}_2 \mathrm{O}+\mathrm{H}_2 \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{Cu}+\mathrm{H}_2 \mathrm{O}\)

Example: Three oxides of nitrogen contained 63.6%, 46.7%, and 30.4% nitrogen respectively. Show that these figures illustrate the law of multiple proportions.
Solution:

The first oxide of nitrogen contains 63.6% N

⇒ 63.6 g of N reacts with (100 – 63.6) g of O = 36.4 g of O.

∴ 1 g of N will react with \(\frac{36.4}{63.6}\) g of O = 0.57 g of O.

The second oxide of nitrogen contains 46.7% N

⇒ 46.7 g of N reacts with (100 – 46.7) g of O = 53.3 g of O.

∴ 1 g of N will react with \(\frac{53.3}{46.7}\) g of O = 1.14 g of O.

The third oxide of nitrogen contains 30.4% N

⇒ 30.4 g of N reacts with (100 – 30.4) g of O = 69.6 g of O.

∴ 1 g of N will react with \(\frac{69.6}{30.4}\) g of O = 2.26 g of O.

This means the ratio of the masses of oxygen which combine with 1 g of nitrogen is 0.57:1.14:2.26, i.e., 1:2:4. This is obviously in accordance with the law of multiple proportions.

4. Law of reciprocal proportions: This law, postulated by Richter in 1792, also deals with the relationship between the masses of elements that combine with each other. According to Richter, when two elements (say A and B) combine separately with the same weight of a third element (say C), the ratio in which they do so is the same or some simple multiple of the ratio in which they (A and B) combine with each other.

• Take sulphur, oxygen, and hydrogen, for example. Sulphur and oxygen combine with hydrogen to form hydrogen sulphide (H₂S) and water (H₂O) respectively. They also combine with each other to form sulfur dioxide (SO₂). According to this law, the ratio of the weights of S and O which combine with the same weight of H will either be the same or a simple multiple of the ratio in which S and O combine with each other.
• This can be easily verified. In H₂S, two parts by weight of hydrogen combine with 32 parts by weight of sulphur. In H₂O, two parts by weight of hydrogen combine with 16 parts by weight of oxygen. Therefore, the ratio of the weights of S and O which combine separately with a fixed weight (2 parts) of hydrogen is 32:16 or 2 :1. And the ratio of S and O in SO₂ is 32: 32, or 1 :1. The two ratios are related to each other as (2/1): (1/1)/ or 2 :1. Which means that the first ratio is a simple multiple of the second.

Example: Methane contains 75% carbon and 25% hydrogen, carbon dioxide contains 27.27% carbon and 72.73% oxygen and water contains 11.10% hydrogen and 88.96% oxygen. Can you use these to prove the law of reciprocal proportions?
Solution:

In methane, 1 g of H combines with \(\frac{75}{25}\) g of C = 3 g of C.

In water 1 g of H combines with \(\frac{88.96}{11.10}\) g of O = 8 g of O.

∴ ratio of weights of C and O which combine with 1 g of H = 3:8 = 1: 2.66.

In CO₂ ratio of the weights of C and O = 27.27: 72.73 = 1: 2.68.

∴ The two ratios are nearly the same, which proves the law of reciprocal proportions.

5. Gay-Lussac’s law of gaseous volumes: While studying reactions between gases, Joseph Louis Gay-Lussac, a French chemist and physicist, found that there is a simple relationship between the volumes of gaseous reactants and products. Gay-Lussac’s law of combining volumes (as his law is called) states that under similar conditions of temperature and pressure, gases react with each other in volumes which bear a simple ratio to one another and to the volumes of the products (if those too are gases). It can be shown experimentally, for example, that 1 volume of hydrogen reacts with 1 volume of chlorine to produce 2 volumes of hydrogen chloride.

⇒ \(\underset{\text { 1 volume }}{\mathrm{H}_2}+\underset{\text { 1 volume }}{\mathrm{Cl}_2} \longrightarrow \underset{2 \text { volumes }}{2 \mathrm{HCl}}\)

The ratio between the volumes of the reactants and the product in this reaction is 1:1:2, which is a simple ratio.

Berzelius Hypothesis

Dalton’s theory could not explain Gay-Lussac’s law of combining volumes, which made Dalton doubt Gay-Lussac’s observations. A Swedish scientist, Jons Jacob Berzelius, convinced of the correctness of Gay-Lussac’s observations, put forward a hypothesis to reconcile Dalton’s theory with Gay-Lussac’s experimental law. Berzelius stated that equal volumes of all gases under the same conditions of temperature and pressure contain the same number of atoms.

However, this hypothesis seemed to have a drawback. Let us apply it to the formation of hydrochloric acid gas from hydrogen and chlorine to see what the drawback was. It can be shown experimentally that one volume of hydrogen combines with one volume of chlorine to form two volumes of hydrochloric acid gas.

⇒ \(\underset{1 \mathrm{vol}}{\text { Hydrogen }}+\underset{1 \mathrm{vol}}{\text { Chlorine }} \longrightarrow \underset{2 \mathrm{vol}}{\text { Hydrogen chloride gas }}\)

Let one volume of a gas contain n atoms. Then

⇒ \(\text { Hydrogen }+ \text { Chlorine } \longrightarrow \text { Hydrogen chloride gas }\)

⇒ \(\begin{array}{lll}
n \text { atoms } & \text { watoms } & 2 \text { ri compound atoms } \\
1 \text { atom } & 1 \text { atom } & 2 \text { compound atoms } \\
\frac{1}{2} \text { atom } & \frac{1}{2} \text { atom } & 1 \text { compound atom }
\end{array}\)

The Berzelius hypothesis would imply that one compound atom of HC1 is formed by the combination of half an atom each of hydrogen and chlorine. However, the possibility that atoms may undergo division during a chemical reaction is against Dalton’s theory. This is why this hypothesis could not be accepted.

Avogadro’s Hypothesis

In 1811, an Italian physicist and lawyer, Amedeo Avogadro, finally resolved the problem that Berzelius had been unable to resolve. This was another flash of insight, a seemingly simple idea possible only for a great mind to conceive of.

• Avogadro suggested that matter consists of two kinds of particles—atoms and molecules. An atom is the smallest particle of an element which participates in chemical reactions, but it may or may not have independent existence.
• A molecule, according to Avogadro, is the smallest particle of an element or compound which is capable of independent existence. Based on this view of the smallest particles of matter, Avogadro put forward a hypothesis which was really a modification of the one proposed by Berzelius. He said that equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules. This hypothesis has been found to hold for all gaseous reactions.

Let us apply Avogadro’s hypothesis to the reaction we considered in the previous section.

⇒ \(\begin{gathered}
\text { Hydrogen } \\
1 \mathrm{vol}
\end{gathered}+\underset{1 \mathrm{vol}}{\text { Chlorine }} \longrightarrow \underset{2 \mathrm{vol}}{\text { Hydrogen chloride }}\)

Let us assume that 1 volume of a gas contains n molecules.

\(\text { Hydrogen }+ \text { Chlorine } \longrightarrow \text { Hydrogen chloride }\)

⇒ \(\begin{array}{lll}
n \text { molecules } & n \text { molecules } & 2 n \text { molecules } \\
\frac{1}{2} \text { molecule } & \frac{1}{2} \text { molecule } & 1 \text { molecule }
\end{array}\)

This implies that one molecule of hydrogen chloride contains half a molecule of hydrogen and half a molecule of chlorine. A molecule is the smallest particle of an element (or compound) which is capable of independent existence and may contain two or more atoms, so according to Avogadro’s hypothesis the existence of half a molecule is possible. In case of hydrogen and chlorine half a molecule contains one atom, as both are diatomic molecules. Therefore, a half-molecule will contain at least one atom, which is in agreement with Dalton’s atomic theory.

Atomicity of gases: The number of atoms present in one molecule of a gas is known as its atomicity. We know now that the number of hydrogen atoms in a hydrogen molecule is two, the number of oxygen atoms in an oxygen molecule is two and the number of oxygen atoms in a molecule of ozone is three, so the atomicity of hydrogen is two, the atomicity of oxygen is two, and that of ozone is three.

Avogadro’s hypothesis can be used to find the atomicity of a gas. Consider hydrogen. Two volumes of hydrogen combine with 1 volume of oxygen to form 2 volumes of water vapour.

⇒ \(\underset{2 \text { vol }}{\text { Hydrogen }}+\underset{1 \text { vol }}{\text { Oxygen }} \rightarrow \underset{2 \text { vol }}{\text { Water vapour }}\)

Assuming that 1 volume of a gas contains n molecules

\(\text { Hydrogen }+ \text { Oxygen } \longrightarrow \text { Water vapour }\)

⇒ \(\begin{array}{lll}
2 n \text { molecules } & n \text { molecules } & 2 n \text { molecules } \\
1 \text { molecule } & \frac{1}{2} \text { molecule } & 1 \text { molecule }
\end{array}\)

Thus, 1 molecule of water contains 1/2 molecule of oxygen. It has been found experimentally that 1 molecule of water contains 1 atom of oxygen.

Hence 1/2 molecule of oxygen = 1 atom of oxygen.

or 1 molecule of oxygen = 2 atoms of oxygen.

∴ the atomicity of oxygen = 2.

Atomic Mass

After Dalton said that the atoms of a substance had a particular mass, the problem that assailed scientists was how to find this mass. An atom is too small to be seen or isolated, so it was not possible to determine its mass by weighing. Theoretically, though the mass of an atom could be determined by weighing a large sample of an element and then dividing it by the number of atoms it contained. But no one could come up with a method to count the number of atoms in a sample of an element.

• Avogadro’s hypothesis at least provided a way to find the relative masses of elements. Since equal volumes of different gases under similar conditions of temperature and pressure contain equal numbers of molecules it stands to reason that if one weighs equal volumes of hydrogen and oxygen under the same temperature and pressure one would be able to know the weights of equal numbers of hydrogen and oxygen molecules.
• Comparing the two weights, one would be able to determine how much heavier (or lighter) an oxygen molecule is compared to a hydrogen molecule. This is exactly what scientists did and they found that a molecule of oxygen is 16 times heavier than a molecule of hydrogen. They knew by then that one molecule of hydrogen contains two atoms of hydrogen and that one molecule of oxygen contains two atoms of oxygen, so they concluded that an atom of oxygen is 16 times heavier than an atom of hydrogen.

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Similarly, the relative (compared to hydrogen) atomic masses of other gases were determined. Berzelius and the Italian chemist Stanislao Cannizzaro (separately) did some pioneering work in determining the relative atomic masses of various elements. The atomic mass of hydrogen was taken as one, and the relative atomic masses of other elements were determined with the hydrogen atom as the reference. The choice of this reference was made by Dalton because hydrogen is the lightest element.

Atomic Mass Of Elements

• The relative atomic mass of oxygen was actually found to be 15.88 on this scale. Later, the oxygen atom was chosen as the reference and it was assigned a mass of 16. This was done for convenience because with the oxygen atom as the reference the relative atomic masses of the other elements worked out to whole numbers or close to whole numbers. Also, oxygen was considered more reactive, forming a large number of compounds. On this scale, the mass of a hydrogen atom works out to 1.008.
• Much later, the exact mass of the hydrogen atom was determined by an indirect method and turned out to be L66 x 10^-24 g. However, to get back to the determination of relative atomic masses, in 1961, the International Union of Chemists selected a new reference for expressing them. They accepted ¹²C with the mass number 12 (the stable isotope of carbon) as the standard for comparing the atomic and molecular masses of elements and compounds. Today this scale depends upon measurements of atomic mass by mass speptrqmeters. A. mass spectroipeter is an instrument used for making accurate measurements of mass by comparing the mass of art atom with the mass of a particular atom chosen as the standard.
• Now, when we refer to the atomic mass of an element, we mean its relative atomic mass compared to 1/12 of the mass of an atom of ¹²C, which is taken as 12. In other words, 1/12 of the mass of a ¹²C atom is taken as the standard and is referred to as 1 atomic mass unit (amu). However, IUPAC has recommended a symbol (u) (unified mass) in place of amu. The average atomic masses of hydrogen, oxygen, nitrogen, and sulphur on this scale work out to 1.008,16,14 and 32 u respectively. The atomic masses of some common elements with the carbon-12 isotope as the reference are given in Table.

Average atomic mass: A large number of elements occur in nature as a mixture of several isotopes. The abundance of an isotope is defined ns the percentage of atoms of that isotope ih a sample of the element. The relative abundance of an element is fixed, thus independent of its state. For example, chlorine exists as two isotopes of masses Cl-35 and Cl-37. The relative abundance (per cent occurrence) of Cl-35 is 75.770 and that of Cl-i7 is 24.229. Also the precise f atomic mass of both the isotopes Cl-35 and Cl-37 are 34.9698 u and 36.9659 u respectively. The average atomic mass of chlorine can be calculated as follows.

Atomic Mass Of Common Elements

Average atomic mass of chlorine = \(=\frac{(34.9698 \times 75.770)+(36.9659 \times 24.229)}{100}=35.45 \mathrm{u}\)

The gram-atomic mass of an element is that quantity of it whose mass in grams is numerically equal to its atomic mass. In other words, it is the atomic mass of art element expressed in grams. It is also known as one gram-atom of the element. Atomic mass of oxygen = 16 u.

Gram-atomic mass of oxygen (or one gram-atom of oxygen) = 16 g.

Molecular mass: The average relative mass of a molecule compared to 1/12 of the mass of an atom of carbon (¹²C) is called its molecular mass. In other words, the molecular mass of a substance indicates how many times a molecule of it is heavier than l/12th the mass of an atom of carbon. The molecular mass of carbon dioxide, for example, is 44.0 u, which means a carbon dioxide molecule is 44 times heavier then 1/12th of a carbon atom.

The molecular mass of a compound or an element can be calculated by adding the atomic masses of all the atoms in one molecule of the compound or element. For example, the molecular formula of water is H₂O.

Atomic Weights Of Elements

Therefore, the molecular mass of water (H₂O) = (2 x 1008 u) for 2 hydrogen atoms + (1 x 16.0 u) for one oxygen atom = 18.02 u.

In case of ionic substances, we find out the relative formula mass instead of the molecular mass. The formula mass of a substance is the sum of the atomic masses of all the atoms in a formula unit of the compound. For example, the formula mass of NaCl is 58.5 u. The atomic masses of sodium and chlorine are 23.0 u and 35.5 u respectively.

The molecular mass of a substance (element or compound) expressed in grams is called its gram-molecular mass. The amount of the substance is referred to as one gram-molecule of the substance.

∴ Molecular mass of CO₂ = 44.0 u.

Gram-molecular mass of CO₂ (or one gram-molecule of CO₂) = 44.0 g.

Note: Units like the gram-atom, gram-molecule, and gram-equivalent are no longer used by scientists. The mole has replaced these units.

Dalton’s Atomic Theory

John Dalton, who had one of the finest minds of his time, was born to a poor weaver and started his career as a teacher in a village school. Later he taught mathematics, chemistry, and physics at a college in Manchester. His passion, however, was studying physical and chemical phenomena. On the basis of his own studies and the laws of chemical combination he came to the conclusion that elements are made of atoms. This was a great leap forward for science and the theory put forward by him is called Dalton’s atomic theory in his honor. The main postulates of his theory, which was published in 1808, are given below.

1. Matter is made up of small, individual particles called atoms which are indivisible.
2. The atoms of a particular element are identical in all respects (i.e., shape, size, mass and chemical properties) but differ from atoms of other elements.
3. An atom is the smallest unit that takes part in chemical combinations.
4. Atoms of two or more elements combine in a simple, whole-number ratio to form compound atoms (now called molecules).
5. Atoms can neither be created nor destroyed in a chemical reaction.

At the time when Dalton put forward his theory, most scientists were already of the view that matter consists of atoms. What Dalton did was to make this idea more concrete by assigning specific properties to the atoms of elements. In other words, he said atoms of a particular element have particular properties like shape, size, and most importantly, mass. He reasoned that only this concept of atoms could explain the laws of chemical combination.

• He said, for example, that the atoms of a particular element are identical but differ from atoms of other elements. This explains why different samples of the same element behave identically (have the same properties) but samples of different elements behave differently.
• According to Dalton, atoms are indestructible. This explains the law of conservation of mass. If atoms cannot be created or destroyed then the mass of the reactants must be the same as the mass of the products in a chemical reaction.
  That an atom of an element has a fixed mass can explain why a fixed quantity of an element (containing a particular number of atpm§) has a fixed mass.

Dalton also said that atoms combine (in simple ratios) to form compound atoms molecules were called compound atoms then). This explains both the law of definite proportions and the law of multiple proportions. Dalton actually proposed his theory on the basis of the laws of conservation of mass and definite proportions. It was he who put forward the law pf multiple proportions as a logical consequence of his theory.

• Dalton’s theory was a great milestone and provided an impetus to other scientists to investigate the structure of matter, but it had certain shortcomings. He did say that an atom has a definite mass but it was not possible at that time to determine this mass. Nor was it possible to determine the formulae of compounds (which would indicate how many atoms pf different elements combined to form a molecule of a particular compound).
• One had to start somewhere, and Dalton started by assuming that all elements are monoatomic and that these individual atoms combine in the simplest possible ways to form molecules. Accordingly, water and ammonia were assigned the formulae OH and NH respectively. Now we know that this is not true.
• Dalton’s theory could also not explain Gay-Lussac’s law, though it could explain the laws of chemical combination related to the masses of reactants and products. Nothing was known at that time about the relationship between the volume of a gas and the number of atoms it contains. Subsequent work by several scientists like Thomson, Rutherford and Niels Bohr has shown that Dalton’s theory had the following drawbacks.

1. Dalton did say that the atoms of different elements are different but he could not explain the reason. This came to be known only later when the constituents of the atom (protons, neutrons and electrons) were discovered.
2. Dalton could only say that atoms combine to form compound atoms (molecules); his theory said nothing about the nature of atoms or the kind of molecules they may form. Molecules were just thought of as a combination of atoms.
3. Dalton’s theory could not explain why and how atoms combine to form compound atoms (molecules).
4. Since nothing was known about the structure of an atom, his theory did not say anything about the nature of the forces that hold atoms together in a compound atom (molecule).
5. Dalton could explain the laws of chemical combination related to masses of reactants and products but he could not explain the law of gaseous volumes.

Dalton’s atomic theory has been modified in view of the picture we now have of the structure of the atom. This picture evolved over years, thanks to the painstaking work of several scientists and some brilliant flashes of insight. For the moment the salient features of the modem atomic theory have been listed below.

1. An atom is the smallest particle of matter which takes part in a chemical reaction.
2. An atom has the properties of the element from which it is derived.
3. An atom is composed of electrons, protons, and neutrons, it is no longer considered to be indivisible.
4. Atoms of the same element may have different atomic masses, that is to say, atoms of the same element need not be identical in every respect. When atoms of the same element have different atomic masses they are called isotopes. Isotopes have the same chemical properties.
5. Atoms of different elements may have the same atomic mass, for example, calcium and argon (40 amu). Such atoms are called isobars. This means atoms of different elements can be similar in some respects.
6. Now we know that atoms of one element can be changed into atoms of other elements (in nuclear reactors, for example). We can no longer think that they are indestructible since such changes always involve the conversion of some mass into energy. Such changes are called transmutation. However, in the course of chemical reactions, atoms remain unchanged, and we can think of them as indestructible.
7. Atoms do not always combine in simple ratios to form compounds, e.g., in proteins or carbohydrates, the ratios in which the elements combine are not simple, though they are fixed. For example, in a sugar molecule (C₁₂H₂₂C₁₁) the ratio of C, H, and O atoms is not simple, it is 12:22:11.

Classification of Elements and Periodicity of Properties

Man has been fascinated by the chemical properties of the natural substances in his environment from almost the dawn of civilization. Initially, his curiosity was fuelled entirely by his need to improve his living conditions. He learned to harden clay by baking it at a high-temperature thousands of years ago.

• Then, quite by accident, he discovered metals, which changed his life completely and led to the study of alchemy. Alchemy was not really a scientific study of metals.
• It was based on this belief in the existence of a magical substance (Philosopher’s Stone) that could transform base metals into gold.

However, modern chemistry owes a lot to alchemists, who in their search for the Philosopher’s Stone, tested all the known substances and increased man’s knowledge of the properties of these substances.

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• Early scientists, who inherited the knowledge passed on by the alchemists, could afford to study the properties of the few known elements individually. But as knowledge grew, and more and more elements came to be known, this approach proved to be cumbersome.
• Scientists started looking for a way to group together similar elements so that it would be possible to study their properties and the properties of their compounds more systematically.
• After years of effort and several attempts at classifying all the known elements, they succeeded in coming up with an arrangement in which similar elements, or elements which have similar properties, are grouped together.
• This arrangement, called the periodic table, has systematized the study of elements and their compounds. It has provided a framework for organizing the vast amount of information available on the chemical behavior of elements.

History Of The Periodic Table

All the attempts made to classify elements, initially, were based on their atomic weights. This was because Dalton had proposed, earlier, that different elements have different types of atoms, characterized by their atomic weight.

Dobereiner’s triads: J W Dobereiner, a German chemist, classified similar elements into groups of three, or triads, such that the atomic weight of the central element was approximately the arithmetic mean of the atomic weights of the other two. He also pointed out that the properties of the middle element were in between those of the other two members.

These groups were called Dobereiner’s triads, for example, lithium (7), sodium (23) and potassium (39), calcium (40), strontium (88) and barium (137), and chlorine (35), bromine (80) and iodine (127). The major drawback of this classification was that it could be applied to only a limited number of elements.

Newlands’ law of octaves: An English chemist, John Newlands, came up with this arrangement in 1865, based on the observation that when the lighter elements are arranged in order of increasing atomic weight, every element is similar to the element eight placed from it in the list.

Hus was called Newlnnds’ law of octaves because it was like a musical scale in which the first note is the same as the eighth. Thus, the properties of lithium are similar to those of the element eight places from it, i.e., sodium.

This arrangement too was discarded because it could not be applied to elements with atomic weights greater than 40 u. Also, the discovery of noble gases meant that the eighth element was no longer similar to the first. However, for his work Newlands was awarded the Davy Medal in 1887 by the Royal Society, London.

Lothar Meyer’s arrangement: Julius Lothar Meyer (1830-1895) was a German chemist who studied the atomic volumes, melting points, boiling points, and other physical properties of various elements.

• He found that if a graph is plotted between the atomic volumes and atomic weights of the various elements, similar elements occupy similar positions on the curve.
• The most strongly electropositive alkali metals (Li, Na, K, Rb, Cs) occupy the peaks on the curve, the less electropositive metals (Be, Mg, Ca, Sr, Ba) occupy descending positions on the curve and the most electronegative elements (F, Cl, Br, I) occupy ascending positions on the curve.

On the basis of his observations, Meyer proposed in 1869 that the physical properties of the elements are a periodic function of their atomic weights.

Mendeleev’s periodic law: Around the same time, DI Mendeleev (1834-1907), a Russian chemist, was busy trying to arrange the elements in some order. He found that if they are arranged in increasing order of atomic weights, their chemical properties vary in a regular pattern. He proposed that the chemical properties of the elements are a periodic function of their atomic weights.

• When Mendeleev came to know of Lothar Meyer’s conclusion, he combined his own proposal with Meyer’s, and the Jaw came to be known as Lothar Meyer-Mendeleev law or Mendeleev’s periodic law.
• According to Mendeleev: The properties of the elements, as well as the formulae and properties of their compounds depend on the atomic weights of the elements. This means that when the elements are arranged in increasing order of atomic weight, – elements with similar properties occur at regular intervals in the list.

Mendeleev’s Periodic Table

Mendeleev’s arrangement of elements in increasing order of atomic weights is called Mendeleev’s periodic table. He grouped together similar elements and was insightful and courageous enough to leave blank spaces (on his table) for elements he believed would be discovered later.

The elements predicted by him were discovered later, and remarkably enough, were found to possess the properties he had predicted they would have. For example, gallium and germanium, called eka-aluminum and eka-silicon by Mendeleev, were discovered after he proposed his table and are very similar to aluminum and silicon respectively, as he had foreseen.

Mendeleev’s periodic table had eight vertical columns (groups). Later, when the inert gases were discovered, a new group called the zero group was added to the table.

Vertical Columns There are nine vertical columns in Mendeleev’s periodic table called groups. These groups are designated 1, 2, 3, 4, 5, 6, 7, 8 and zero. With the exception of the 8th and zero groups, every group is divided into two subgroups A and B. Group 8 contains nine elements in three sets of three each, and group zero comprises the inert gases.

Horizontal rows The seven horizontal rows are called periods. The first period contains only two elements, the second and third periods contain eight elements each and are called short periods. The fourth, fifth, and sixth periods contain 18 elements each, while the seventh is incomplete and contains only three elements.

Merits of Mendeleev’s periodic table: Mendeleev’s periodic table was a giant step forward in the study of chemistry.

• Systematic study His classification of the elements made the study of their properties simpler and more systematic. Placing elements which have similar chemical properties in the same group is helpful in that if one knows the properties of one element of a group, one can predict those of the rest of the members of the group.
• Prediction of new elements While arranging the 56 elements known then, Mendeleev left blank spaces for elements which had not been discovered at that time. He predicted the properties of these unknown elements on the basis of their positions in his table. He was proved right later, when these elements were discovered.
• Determination of atomic weights Mendeleev’s periodic table has helped correct errors made in the calculation of tire atomic weights of some elements. For example, beryllium was initially assigned an atomic weight of 13.5 on the basis of its equivalent weight (4.5) and valency, which had been wrongly calculated as 3.

This would have placed it between C and N in the periodic table. But there was no such space in Mendeleev’s table. Besides, its properties showed that its correct position should be between lithium (7) and boron (11), so beryllium was assigned a valency of 2 (atomic weight = 2 x 4.5 = 9) and placed where it belonged.

Drawbacks of Mendeleev’s periodic table: So great was Mendeleev’s contribution to the systematic study of elements that he is usually given credit for the periodic table as we know it today. Nonetheless, the table had the following drawbacks.

• Position of isotopes Isotopes are atoms of the same elements that have different atomic weights. As Mendeleev’s. classification is based on atomic weight, isotopes would have to be placed in different positions. For example, the isotopes of hydrogen, \({ }_1^1 \mathrm{H},{ }_1^2 \mathrm{H},{ }_1^3 \mathrm{H}\) would occupy different positions.
• Position of hydrogen Though hydrogen is placed in Group IA (alkali metals), it resembles the elements of both Group IA and Group 7A (halogens). Thus, the position of hydrogen in the periodic table is controversial.
• Anomalous positioning of elements Some elements with higher atomic weight are placed before those with lower atomic weight. For example, argon (atomic weight = 40) is placed before potassium (atomic weight = 39).

Also, some dissimilar elements are grouped together, while some similar elements are placed in different groups. For example, alkali metals, such as Li, Na, and K (LA) are grouped together with the coinage metals, Cu, Ag, and Au (IB) though their properties are different. On the other hand, some chemically similar elements like Cu (LA) and Hg (2B) are placed in different groups.

Elements of Group 8 The nine elements of Group 8 have not been placed in a proper order. They have been arranged in three triads without any justification.

Modern Periodic Law

Mendeleev’s arrangement of elements, though rational and systematic, was empirical. Several decades later, after the electron had been discovered and the modem theory of atomic structure had been developed, H G J Moseley studied the spectral lines emitted by heavy elements in the X-ray region of the electromagnetic spectrum.

• He found that the X-rays radiated by each element have a characteristic frequency that differs according to a regular pattern. The frequencies were different not because of the change in atomic weight but due to the charge on the nucleus (atomic number).
• Thus, he proposed that the elements should be classified on the basis of atomic number (Z) and not atomic weight. He said that the properties of an element depend on its atomic number, rather than its periodic functions of their atomic numbers.
• This means that if the elements are arranged in order of increasing atomic number, elements that have similar properties will appear at regular intervals. This is known as periodicity.

Cause of periodicity: It is not difficult to understand periodicity if one considers atomic structures and what happens to an atom during a chemical reaction. Neither the nucleus, nor the electrons of the inner shells of an atom participate in a chemical reaction.

• Only the valence electrons participate in chemical reactions, so it is these electrons that govern the chemical properties of atoms. In that case, atoms which have the same kind of arrangement of electrons in their outermost shells should have similar chemical properties.
• Now consider the electronic configurations of any group of elements (for example, the alkali metals or Group 1 in the periodic table).

• They have the same number of valence-shell electrons (Group 1 f elements have one electron in their valence shell). This explains why they have similar properties.
• The atomic number of an element is equal to the number of protons it contains, which is again equal to the number of electrons it has. And an atom’s chemical behaviour is governed by the number of electrons it has.
• That is why it makes more sense to arrange the elements in order of increasing atomic number, rather than in order of increasing atomic weight.

Long Form Of Periodic Table

Several periodic tables have been proposed after it was realised that the atomic number, and not the atomic weight V of elements, should be the basis for the arrangement of elements. The most commonly used one is called the long form of the periodic table. It has eighteen columns and seven rows.

The rows are called periods and the columns contain elements with similar valence-shell configuration. These columns are called groups and elements belonging to the same group (obviously with similar properties) constitute a family, for example, the halogen family (Group 17). In this form of the periodic table, 14 elements each of the sixth and seventh periods (i.e., lanthanides and actinides) are placed separately at the bottom.

Electronic configurations of the elements and the periodic table: As a result of elucidation of the structure of the atom, it is now recognised that the periodic law is essentially the consequence of the periodic variation in electronic configurations.

The electronic configuration of an element indeed determines the physical and chemical properties of the element and its compounds. The electronic configuration of elements can be best studied in terms of variation in periods and groups of the periodic table.

Periods: You already know that an electron in an atom is characterised by a set of four quantum numbers. The principal quantum number (M) denotes the main energy level or shell.

Each period begins with the filling up of a new energy shell and all the elements of a period have the same number of electron shells, or the same principal quantum number (n) of the outermost shell. In fact, the number of a period is the same as the principal quantum number of the valence shell of the elements it contains. The number of elements in a period is equal to the number of electrons required to fill the orbitals of that shell.

First period This corresponds to the filling up of the first energy shell (n = 1). This shell has only one orbital, the Is orbital, which can accommodate only two electrons, so the first period has only ttvo elements.

Second period This is associated with filling up of the second energy shell (n = 2). This shell has one 2s and three
2p orbitals, which can accommodate eight electrons. The second period, thus, has eight elements.

Third period This corresponds to the filling up of the third shell (n = 3). This shell has one 3s, three 3p and five 3d orbitals, but the 3d orbitals have higher energy levels than the 4s orbitals. Consequently, the 3d orbitals are filled after the 4s orbitals and the third period involves the filling up of only four orbitals. Thus this period contains only eight elements, from sodium to argon.

Fourth period This corresponds to the filling up of the fourth energy level (n = 4). It starts with the filling up of the 4s orbital after which the five 3d orbitals are filled and then the three 4p orbitals. The 4d and 4f orbitals can be filled only after the 5s orbital. Therefore, this period involves the filling up of nine orbitals and contains eighteen elements, from potassium to krypton.

Fifth period The fifth period is associated with the filling up of the fifth energy level (n = 5). After the 5s orbital the five 4d and three 5p orbitals are filled. Neither the 4f, nor the 5d orbitals can be filled until the 6s orbital is filled. Thus, this period too has only eighteen elements, from rubidium to xenon.

Sixth period This corresponds to the filling up of the sixth energy level (n = 6). Starting with the 6s orbital, it involves the filling up of sixteen orbitals (6s, 4f, 5d and 6p), so it contains thirty-tivo elements, from caesium to radon. The fourteen elements from cerium to lutetium correspond to the filling up of the seven 4f orbitals. These are separated from the main table and placed below it to save space. They constitute the first f-transition series and are called lanthanides.

Seventh period This corresponds to the filling up of the seventh energy level (n = 7). Like the sixth period this should also have contained thirty-two elements, corresponding to the filling up of the 7s, 5f, 6d and 7p orbitals, but it is still incomplete and contains only twenty-five elements. It includes most of the man-made radioactive elements. The fourteen elements corresponding to the filling of the 5f orbitals are called actinides and constitute the second f-transition series. These too have been placed below the main table to save space and to allow similar elements to come under the same group.

The first three periods are called short periods and the next three are called long periods.

Division of elements into blocks: The long form of the periodic table has four blocks—s, p, d and f. Each block is named after the atomic orbital which receives the last electron during the filling up of orbitals in order of their increasing energies, s Block elements in which the last electron enters the s orbital of their respective valence shells are called s-block elements.

This subshell has only one orbital which can accommodate only two electrons, so there are only two groups of elements in this block—Groups 1 and 2. The general electronic configuration of the outermost shell of these elements is ns^1-2 where n represents the value of the principal quantum number of the valence shell. Group 1 comprises alkali metals, while Group 2 contains the alkaline earth metals. These elements have the following general characteristics.

1. They are soft metals.
2. They have low ionisation enthalpies and a re highly electropositive.
3. They are very reactive and form ionic compounds.
4. The metallic character and reactivity of the elements increase down the group.
5. They exhibit oxidation states of +1 and +2 as they have only one or two electrons in their valence shell.
6. They are good reducing agents because they lose electrons easily.

p block This block includes elements in which the last electron enters the p subshell of the valence shell, helium (1s²) being the exception. The general electronic configuration of the outermost shell of these elements is ns²np⁶.

This block contains the groups 13 to 18. Group 18 comprises noble gases with completely filled orbitals (ns²np⁶). Thus, each period ends in a noble gas with a closed-shell configuration. The other two important groups are chalcogens (Group 16) and halogens (Group 17).

The p-block elements have the following general characteristics.

1. Most of them are nonmetals and the character increases as we move from left to right across a period.
2. Most of them are highly electronegative.
3. They exhibit variable oxidation states.
4. They form ionic and covalent compounds.
5. Most of them form acidic oxides.

The elements of the s and p blocks together are called representative elements.

• Two exceptions are observed in this classification. The electronic configuration of helium is 1s² so that it should belong to the s block, but it has been placed in the p block with Group 18 elements (noble gases). Its position is justified because it has a completely filled valence shell and shows the characteristic properties of a noble gas. The other exception is hydrogen. The electronic configuration of hydrogen is 1s¹, hence it can be placed in Group 1 (s block).
• However, like halogens it is one electron short of noble-gas configuration. This controversy regarding position of hydrogen in the periodic table has been discussed in detail. It is placed at the top of the alkali metals in Period 1 of the periodic table.

d block Elements in which the last electron enters the d subshell of the penultimate (second last) energy level are called d-block elements. These include elements of groups 3 to 12. Their general valence shell configuration is (n – 1) d^1-10 ns^1-2 where n represents the outermost energy level. This block contains three rows of ten elements each. The fourth row is incomplete.

Tire rows which are complete are called the first, second and third transition series. They correspond to the filling of the 3d, 4d and 5d orbitals respectively. Members of this block are called transition elements because they represent a change in character from reactive metals on one side to nonmetals on the other. These elements have the following general characteristics.

1. They are hard metals with high melting points.
2. They exhibit variable valencies and oxidation states.
3. They form ionic as well as covalent compounds.
4. They form coloured complexes.
5. Most exhibit paramagnetism.
6. Most possess catalytic properties.

f block This block contains elements in which the last electron enters the f from the outermost) shell. The general electronic configuration of f-block elements is \((n-2) \mathrm{f}^{1-14}(n-1) \mathrm{d}^{0-1} n \mathrm{~s}^2\), where n is the principal quantum number of the valence shell. This block comprises two series of elements placed below the main body of the periodic table. Elements of the first series are called lanthanides, while elements of the second series are called actinides. Together they are called the inner transition elements.

These elements have the following general characteristics.

1. They show variable oxidation states.
2. They are metals with high densities and high melting points.
3. They form coloured compounds.
4. Most elements of the actinide series are radioactive.

Elements up to uranium (Z = 92) arc found in nature, except technetium (Z = 43) and promethium (Z = 61), which arc produced from the disintegration of radioactive elements. Elements beyond uranium are called synthetic or transuranic as they are produced synthetically. Many of these have been made only in nanogram quantities or even less by nuclear reactions and their chemistry is not fully known.

• In addition to the classification of elements into s, p, d and f blocks another classification is done, based on their properties. The elements can be broadly classified into metals and nonmetals.
• Elements on the left-hand side of the periodic table are metals. More than seventy-five per cent of all known elements are metals. Elements at the top right-hand side of the periodic table are nonmetals.
• Metals are generally solids at room temperature (except mercury) with high melting and boiling points. They are malleable and ductile, and good conductors of heat and electricity. Nonmetals are generally solids or gases at room temperature with low melting and boiling points. They are brittle and poor conductors of heat and electricity. The change from metallic to nonmetallic character is gradual. Some elements like germanium, arsenic, silicon and antimony show characteristics of both metals and nonmetals. These elements are called semimetals or metalloids.

Predicting period, group number and block of an element: If you have to predict the group, period and block of an element, proceed step by step.

1. Write the electronic configuration of the element according to the Aufbau principle. Do not change the configuration by considering the relative stability of half-filled and completely filled orbitals. Do not rearrange in order of increasing values of quantum number.
2. The period of the element is the same as the value of the principal quantum number of the valence shell. For example, consider aluminium, whose atomic number is 13. Its electronic configuration is 1s² 2s² 2p⁶ 3s² Sp¹. The value of the principal quantum number of its valence shell is 3, so it belongs to the third period.
3. The block of an element is predicted on the basis of the subshell which receives the last electron. Consider copper, whose electronic configuration should be 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹.
4. However, due to a slight difference in the energies of 4s and 3d subshells, and stability of the completely filled and half-filled orbitals, the outer electronic configuration of Cu is 3d¹⁰ 4s¹. The last electron is received by the d subshell, so it belongs to the d block.
5. The group number is predicted on the basis of the number of electrons in the outermost or the penultimate shell.

• For an s-block element, the group number is the same as the number of electrons in the valence shell. For example, the electronic configuration of lithium is 1s² 2s¹. It belongs to Group 1 as there is one electron in its valence shell.
• For d-block elements, the group number is obtained by adding 1 to the number of electrons in the d subshell of the penultimate shell. For example, the electronic configuration of copper is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰. Its group number is (10 +1) = 11.
• For p-block elements, the group number is obtained by adding 12 to the number of electrons in the p subshell of the valence shell. For example, the electronic configuration of gallium is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p¹. It belongs to Group 13.

Nomenclature of the elements with atomic number more than 100: Enrico Fermi and his co-workers in 1934 made attempts to prepare the elements beyond uranium, The new elements were prepared by the bombardment of uranium with slow neutrons. As of now, elements with numbers up to 112 and 114 have been discovered.

• Earlier the naming of new elements was done y discoverer(s) traditionally. In recent years, disputes have arisen over the original discoveries of some of the elements of atomic number 104 and above, when two scientists from two different countries have common criteria for discovering the same element.
• To avoid such discrepancies IUPAC has recommended that until the c a m or a newly discovered element is proved and its official name announced, a nomenclature is to be followed of name these new elements.
• The nomenclature of new elements and elements which are yet to be discovered is based on the Latin words for the atomic number of the elements. The names are derived using the numerical roots for 0 and numbers 1 to 9.
• The roots are combined together in the sequence of digits which make up the atomic number and (ium) is added at the end. The notation for IUPAC nomenclature of elements is shown in Table 3.7. For example, the IUPAC name and symbol for the element with atomic number 122 may be written from the table. The roots for 1 and 2 are un and bi respectively. Hence the name is un (1) + bi (2) + bi (2) + ium or unbibium and the symbol assigned is Ubb.

Example: Write the IUPAC names and symbols for the elements with atomic numbers 121 and 140. Also, identify the group where these elements would be placed (once discovered) by giving their electronic configuration.
Solution:

The IUPAC names for the elements with atomic numbers 121 and 140 are Unbiunium (Ubu) and Unquadnilium (Uqn) respectively.

• The electronic configuration of elements which are yet to be discovered can be written on the basis of the periodicity in filling of the valence shell orbitals of elements in the periodic table.
• Thus, the electronic configuration of the element with atomic number 121 is [Uuo]7d¹ 8s². Therefore, it will be placed in Group 3.
• The electronic configuration of the element with atomic number 140 is [Uuo]6f¹⁴ 7d⁶ 8s². Therefore, it will be placed in Group 8.

 

Periodic Trends In Chemical Properties

We have learnt about the periodicity shown by atomic properties like atomic size, ionisation enthalpy, electron gain enthalpy and electronegativity. The discussion involving trends in chemical properties can be quite elaborate.

Here we shall restrict ourselves to the periodicity of valency and anomalous properties of second-period elements only. We have already discussed the periodicity of valency earlier in the chapter.

Example: Refer to the periodic table and predict the formulae of compounds which might be formed by the following pairs k, of elements:

1. Carbon and chlorine, and
2. Boron and oxygen.

Solution:

1. Carbon is a Group 14 element with a valency of 4, while chlorine belongs to Group 17 with a valency of 1. Therefore, the formula of the compound formed will be CCl₄.
2. Boron is a Group 13 element with a valency of 3, while oxygen is a member of Group 16 and possesses a valency of 2. Hence, the formula of the compound formed would be B₂O₃.

Anomalous properties of second-period elements: The representative elements of the second period—lithium, beryllium, boron, etc., show anomalous characteristics, as compared to the other members in the respective groups.

• This is attributed to the high ionisation enthalpy, high electronegativity, small size and large charge-to-radius ratio of the first members of the said groups.
• Further these elements showing anomalous characteristics have a close chemical similarity to their diagonal neighbours in the next group of the third period. This is a relationship within the periodic table and is referred to as diagonal relationship. You will study about the anomalous characteristics and similarity in properties of the s block elements due to the diagonal relationship.
• The p-block elements of the second period have a tendency to enter into 7t bonding (or p_π-p_π multiple bonding) with themselves, as in C=C, C≡C and N=N, O=O. They also form π bonds with the other second-period elements. Such bonds exist in oxides of nitrogen, (N=O), cyanide and carboxyl groups (C=N, C=O).
• However, the subsequent members of the same group do not form σ bonds with other elements. For example, in silicon dioxide, there are only a bonds.

Periodic Trends And Chemical Reactivity

By now you are quite aware of the periodic trends in certain fundamental properties of elements. The ionisation enthalpy decreases down a group and increases across a period. The size (atomic and ionic radii) increases from J top to bottom in a group, and decreases from left to right in a period.

• The electronegativity and electron affinity both increase from left to right in a period and decrease from top to bottom in a group. All these properties are related to electronic configuration. But how does this affect the chemical reactivity of elements in the periodic table?
• The elements at the extreme left of the periodic table are very reactive due to their low ionisation enthalpies. The elements at the extreme right are also very reactive due to their high electronegativities or electron affinities.
• The elements in the middle of the periodic table are relatively unreactive. Thus, transition elements are only moderately reactive since their ionisation enthalpies are relatively large and electronegativities or electron affinities are relatively small.
• Tire elements at the extreme left of the periodic table are very reactive due to their tendency to lose an electron and form cations. Similarly, the elements at the extreme right of the periodic table readily accept an electron to form anions, and are, thus, very reactive. The metallic character decreases and nonmetallic character increases on moving from left to right across the periodic table.

The chemical reactivity of an element can be best shown by its reaction with oxygen. Elements at the two extremes of a period form oxides easily. The elements on extreme left form basic oxides (like Na₂O) while elements in the extreme right form acidic oxides (like Cl₂O₇).

The elements in the centre of the periodic table form amphoteric oxides (like Al₂O₃, and As₂O₃) or neutral oxides (like CO, NO, N₂O). Amphoteric oxides behave as acidic oxides with bases and as basic oxides with acids. However, neutral oxides have no acidic or basic properties.

Example: Show that Na₂O is a basic oxide and Cl₂O₇ is an acidic oxide.
Solution:

On reaction with water, Na₂O forms a strong base (NaOH) and so it is a basic oxide.

⇒ \(\mathrm{Na}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}\)

On reaction with water, Cl₂O₇ forms a strong acid (HClO₇) so it is an acidic oxide.

⇒ \(\mathrm{Cl}_2 \mathrm{O}_7+7 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{HClO}_7\)

The acidic or basic nature of the oxides can be easily tested by the action of litmus paper on their aqueous solutions.

Example: Predict the formulae of the stable compounds that ivottld be formed by the combination of the following pair of elements.

1. Lithium and oxygen
2. Magnesium and nitrogen
3. Aluminium and iodine
4. Silicon and oxygen
5. Phosphorus and fluorine
6. Element with atomic number 71 (Lu) and fluorine

Solution:

1. Li₂O
2. Mg₃N₂
3. All₃
4. SiO₂
5. PF₃ or PF₅
6. LuF₃

 

Periodic Properties

Most of the properties of atoms (elements), such as valency, ionisation enthalpy and electron affinity depend upon their electronic configuration. Elements of the same group have the same valence shell configuration, so they exhibit similar chemical properties. However, atomic size increases down a group, so the reactivity of elements belonging to the same group varies down the group.

Valency: The chemical properties of an element depend upon the number of electrons present in the outermost shell (valence shell) of its atoms. These electrons, called valence electrons, determine the valency of the atom.

The valency of an element is defined as its combining capacity. It is often expressed in terms of the number of hydrogen atoms, chlorine atoms, or double the number of oxygen atoms that one atom of the element combines with. For the representative elements, valency is either equal to the number of valence electrons or eight minus the number of valence electrons (if the number of valence electrons is five or more). The transition elements show variable valency.

Read and Learn More WBCHSE For Class11 Basic Chemistry Notes

Variation of valency in a period The number of valence electrons in representative elements increases from 1 to 8 while moving across a period. For s-block elements, the group number is the same as the number of valence electron (1 or 2), so valency is the same as the group number. For p-block elements, the number of valence electrons is the group number minus ten.

Variation of valency in a group The number of valence electrons is the same for all the elements of a group, so members of the same group exhibit the same valency.

Valencies of the elements show some periodic trends. Transition elements (Groups 3 to 12) exhibit variable valencies and oxidation states. The oxidation state is a measure of the electron control that an atom has in a compound compared to the atom in a pure element. Periodic trends in the oxidation states of the elements of Groups 1 and 2 (s block) and Groups 13 to 17 (p block) are shown in Table with the help of the formulae of their hydrides and oxides.

Atomic size: For the purpose of determining its size an atom is considered to be a sphere, and the atomic radius is the radius of this sphere. The atomic radius is defined as the distance from the centre of the nucleus of the atom to the outermost shell of electrons. It can also be defined as the distance from the centre of the nucleus to the point up to which the density of the electron cloud (or the probability of finding electrons) is the maximum.

However, it is not possible to determine the atomic radius precisely because of the following reasons.

1. It is not possible to isolate an atom.
2. According to the wave mechanical model of the atom, the probability of finding an electron is never zero, even at a very large distance from the nucleus. In other words, it is not possible to define the boundary of an atom.
3. The probability distribution of the electrons, of an atom is affected by the presence of other atoms in its neighbourhood.
4. The atomic radius changes from one bonding state to another. This is the reason why atomic radii may be given different operational names, like covalent radius, van der Waals radius and metallic radius.

Covalent radius The approximate radius of an atom can be determined by measuring the distance between two atoms in a covalent molecule. This is done by X-ray diffraction and other spectroscopic techniques. The covalent radius of an atom may be defined as one-half the distance between the centres of the nuclei of two similar atoms bonded by a single covalent bond.

For a homonuclear diatomic molecule, for example, chlorine (Cl₂), half the distance between the centres of the two chlorine atoms is the covalent or atomic radius of the chlorine atom, which is equal to 99 pm.

r_covalent = 1/2[intemuclear distance between two bonded atoms]

Vander Waals radius This is defined as half the distance between the centres of the nuclei of the two closest nonbonded atoms of adjacent molecules of an element in the solid state.

Obviously, the van der Waals radius of an atom depends upon the packing of the atoms when the element is in the solid state. The van der Waals radius of an atom is greater than its covalent radius because the formation of a covalent bond involves the overlapping of the orbitals of two atoms, which decreases the internuclear distance.

Metallic radius The metallic radius of an atom in a metallic crystal is half the distance between the centres of the nuclei of two adjacent atoms in the metallic crystal. The metallic radius of an atom is greater than its covalent radius because the metallic bond is weaker than the covalent bond, so the internuclear distance between adjacent atoms in a metallic crystal is greater than the internuclear distance between two atoms held by a covalent bond.

Variation of atomic radius in a period The atomic radius decreases with increase in atomic number across a period. This is because the nuclear charge increases as the atomic number increases, say by one unit, but the extra electron is added to the same principal shell. The electron cloud is pulled closer to the nucleus by the increased nuclear charge so the atomic radius becomes smaller. Table shows the variation in atomic radius across the second period.

As you can see from the table, the atomic radius decreases across the period and then there is an abrupt increase in the atomic radius of neon. This is because while for the other elements the values represent the covalent radii, the value for Ne represents its van der Waals radius (Ne is an inert gas).

Variation of atomic radius in a group In general, the atomic radius increases as one moves down a group. This is because though the nuclear charge increases, the effect of the extra charge is not as pronounced as the effect of the additional shell of electrons.

With the addition of each electron shell (increase in principal quantum number), the distance of the outermost electrons from the nucleus increases, i.e., the atomic radius increases. This is observed in Table.

Ionic radius: A neutral atom becomes a cation when it loses one or more electrons and an anion when it gains one or more electrons. The ionic radius of an atom is the radius it exhibits in an ionic crystal like KCl in which the K⁺ and Cl^– ions are packed together with their outermost shells of electrons in contact with each other.

• A cation has the same number of protons as the neutral atom, but fewer electrons than it. So, the nuclear influence over the electrons is greater in the cation than in the neutral atom. Consequently, the size of the cation is smaller than that of the parent atom.
• Quite often the formation of a cation involves the removal of the valence shell altogether. Obviously this means a reduction in size, as in the formation of Na⁺ from Na. The metallic radius of Na is 186 pm, while the ionic radius of Na⁺ is 102 pm.
• An anion, on the other hand, has the same number of protons as the parent atom, but a greater number of electrons than it. Since the same nuclear charge acts on a greater number of electrons, the effective nuclear charge per electron decreases, and the electron cloud is held less tightly by the nucleus. Consequently, the anion is bigger than the parent atom.
• Take, for example, the case of the chloride ion which has 18 electrons, one electron more than the chlorine atom. Its size is 184 pm while the size of the parent atom is 99 pm.

• Isoelectronic ions Ions with the same number of electrons but different magnitudes of nuclear charge are called isoclectronic ions. These are really ions of different elements with the same electronic configuration, example, Cl^– and K⁺ both have 18 electrons but the former has 17 protons, while the latter has 19 protons.
• Within a series of isoelectronic ions, as the nuclear charge increases, the influence of the nucleus over the electrons increases, so the ionic radius decreases. That is to say, the size of isoelectronic ions decreases with increase in effective nuclear charge.
• We may also conclude that the cation with greater positive charge will have smaller radius. Anions with a greater negative charge will be larger in size as tire net repulsion of electrons will counteract the nuclear charge.

Trends in noble gas radii: Noble gases ordinarily do not form covalent bonds. In crystals of noble gases, therefore, no chemical forces operate between the atoms. The van der Waals forces are the only attractive forces between the atoms of these elements. Hence the atomic radii of noble gases are van der Waals radii, which represent the distance of closest approach of the two adjacent atoms in the solid state.

The van der Waals radii of noble gases are, therefore, much larger than the covalent radii of the corresponding adjacent elements.

Ionisation enthalpy: Energy is required to promote an electron from lower energy level to higher energy level, but if the amount of energy supplied is sufficiently large the electron may be completely removed. This energy is referred to as ionisation enthalpy. Defined in a more formal way, ionisation enthalpy is the enthalpy change accompanying the removal of an electron from a gas-phase atom.

⇒ \(A(g)+\text { Energy }(\text { IE }) \longrightarrow A^{+}(\mathrm{g})+\mathrm{e}^{-}\)

or, \(\mathrm{A}(\mathrm{g})\longrightarrow \mathrm{A}^{+}(\mathrm{g})+\mathrm{e}^{-} ; \Delta H\)

Ionisation enthalpy is also called ionisation potential because it is the potential required to remove the electron from the gaseous atom. It is expressed either in electronvolts per atom or kilojoules per mole of atoms.

∴ 1 eV = 96.3 kJ mol^-1

Energy is always required to remove electrons from an atom and hence ionisation enthalpies are always positive.

Successive ionisation enthalpies The amounts of energy required to remove the second, third and fourth electrons from the unipositive, dipositive and tripositive ions are called the second, third and fourth Ionisation enthalpy respectively. These are designated IE₂, IE₃ and IE₄, and are collectively known as successive ionisation enthalpies.

⇒ \(\mathrm{A}(\mathrm{g})+\mathrm{IE}_1 \longrightarrow \mathrm{A}(\mathrm{g})^{+}+\mathrm{e}^{-}\)

⇒ \(\mathrm{A}^{+}(\mathrm{g})+\mathrm{IE}_2 \longrightarrow \mathrm{A}^{2+}(\mathrm{g})+\mathrm{e}^{-}\)

⇒ \(\mathrm{A}^{2+}(\mathrm{g})+\mathrm{IE}_3 \longrightarrow \mathrm{A}^{3+}(\mathrm{g})+\mathrm{e}^{-}\)

After the removal of the first electron, the atom changes into a monopositive ion which has the same number of protons as the parent atom but fewer electrons than it. Consequently, the effective nuclear charge per electron increases and more energy is required to remove the second electron. Thus, the value of the second ionisation enthalpy (IE₂) is greater than that of the first (IE₁). Similarly, the value of IE₃ is greater than that of IE₂.

Like valency and atomic size, the ionisation enthalpy of elements exhibits certain periodicity, as is evident from figure.

The graph in the figure shows two striking features.

1. The noble gases have the highest ionisation enthalpies in their respective periods.
2. The alkali metals Li, Na, K and Rb have the lowest ionisation enthalpies in their respective periods.

To understand this periodicity, let us discuss the factors upon which the ionisation enthalpy of an element depends. The ionisation enthalpy depends upon the following factors.

• Atomic size
• Magnitude of nuclear charge
• Screening effect of inner electrons
• Penetration effect of electrons
• Electronic configuration

Atomic size The ionisation enthalpy depends upon the attractive force between the electron (to be removed) and the nucleus. And this force is inversely proportional to square of the distance between them \(F=q_1 q_2 / d^2\)

Consequently, as the size of the atom increases, and the hold of the nucleus over valence electrons decreases, the ionisation enthalpy decreases. In general, ionisation enthalpy decreases down the group.

Magnitude of nuclear charge The force of attraction between the valence electrons and the nucleus increases with , the increase in the nuclear charge if the principal energy shell remains the same because the attractive force is directly proportional to the product of the charge on the nucleus and that on the electrons (F = q). In other words, ionisation enthalpy increases with increase in nuclear charge. In general, the first ionisation enthalpy increases as we go across a period.

Screening effect of inner electrons The electrons of the valence shell of an atom are attracted to the nucleus because of the nuclear charge. How much of an impact the nuclear charge has on the outermost shell of electrons depends not only on the magnitude of the nuclear charge and the distance between the nucleus and the outermost shell but also on the inner shells of electrons.

• These inner shells exert a force of repulsion on the outer shell electrons, thus acting as a shield or screen which does not let the outer electrons experience the full impact of the nuclear charge. This is called the screening effect of the inner electrons. Naturally, the screening effect depends on the number of electrons present in the inner shells and as the screening effect increases, the ionisation enthalpy decreases. This is why ionisation enthalpy decreases down a group, with the successive addition of electron shells.
• Ionisation enthalpy increases across a period because though nuclear charge increases, electrons are added to the same shell and the screening effect of the inner electrons does not increase.

Penetration effect of electrons Among all the electrons of an atom, those of the s orbital have the maximum probability of being found near the nucleus.

The probability of finding p-orbital electrons near the nucleus is much less. Similar is the case with the d- and f-orbital electrons. In other words, electrons of the s orbital are more penetrating than electrons of the p orbital. In a given shell, the penetrating power decreases in the following order.

∴ s > p > d > f

If the penetration of an electron is greater, it is closer to the nucleus and held more firmly by it. Consequently, it is more difficult to remove such an electron from the atom than a more loosely held electron. Thus, for the same shell, it is easier to remove the p electrons than the s electrons.

Electronic configuration Certain electronic configurations are more stable than others. Obviously, atoms which have stable configurations do not tend to lose electrons easily, so they have higher ionisation enthalpies. What makes for a stable configuration? Either completely filled orbitals or exactly half-filled orbitals of the same subshell. An atom with three half-filled p orbitals in the outermost shell is more stable than an atom with one completely filled p orbital and two half-filled p orbitals.

1. The noble gases have the highest ionisation enthalpies in their respective periods because they have completely filled orbitals. The np² ns⁶ configuration of noble gases is very stable.
2. Take Be(1s² 2s²) and B(1s² 2s² 2p¹), for example. Be with a completely filled s orbital in the outermost shell is more stable than B, which has one half-filled and two vacant p orbitals in the outermost shell. Had all three p orbitals of B been half-filled, it would have been more stable. Similarly, \(M g\left(1 s^2 2 s^2 2 p^6 3 s^2\right)\) is more stable (higher ionisation enthalpy) than \(\mathrm{Al}\left(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}_x^1 3 \mathrm{p}_y 3 \mathrm{p}_z\right)\). However, if A1 had two half- filled p orbitals and one vacant p orbital, it would have been less stable than it is.
3. The fact that an atom in which all the orbitals of a particular subshell are half-filled is more stable than one in which some are half-filled, while others are vacant or completely filled, becomes clear if one considers \(\mathrm{N}\left(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}_x^1 2 \mathrm{p}_y^1 2 \mathrm{p}_z^1\right) \text { and } \mathrm{O}\left(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}_x^2 2 \mathrm{p}_y^1 2 \mathrm{p}_z^1\right)\). All the p orbitals in N are half-filled, so it is more stable (higher ionisation enthalpy) than O. Similarly, the ionisation enthalpy of \(P\left(1 s^2 2 s^2 2 p^6 3 s^2 3 p_x^1 3 p_y^1 3 p_z^1\right)\) is higher than that of \(\mathrm{S}\left(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}_x^2 3 \mathrm{p}_y^1 3 \mathrm{p}_z^1\right)\).

Variation in IE across a period In general, ionisation enthalpy increases across a period. This is because the nuclear charge increases but the electrons are added to the same principal shell, or the principal quantum number. remains the same. Consequently, the atomic size decreases and the valence electrons are held more tightly.

However, there are some exceptions to this rule. Let us consider the first ionisation enthalpies of the elements of the second period. They are in the following order.

₃Li < ₄Be > ₅B< ₆C< ₇N> ₈O < ₉F < ₁₀Ne

Irregularities in the curve occur at two points. The ionisation enthalpy of B is lower than that of Be and the ionisation enthalpy of O is lower than that of N. The ionisation enthalpy of B is lower than that of Be though B has a higher nuclear charge than Be. This is due to the following reasons.

1. The electronic configuration of \({ }_5 \mathrm{~B}\left(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^1\right)[latex] is less stable than that of [latex]4 { }_4 \mathrm{Be}\left(1 \mathrm{~s}^2 2 \mathrm{~s}^2\right)\), which has a completely filled s orbital in its outermost shell.
2. The outermost electron of B belongs to the 2p orbital while the outermost electrons of Be belong to the 2s orbital. The penetrating power of s electrons is more than that of p electrons, so it takes less energy to remove an electron from the p subshell than from the s subshell. Hence, the first ionisation enthalpy of Be is more than that of B.

N(1s² s² 2p³) has a more stable configuration than O(1s² 2s² 2p⁴), so there is a slight decrease in ionisation enthalpy from N to O.

Variation in IE down a group Ionisation enthalpy decreases regularly down a group. This can be explained in terms of increasing atomic size and the screening effect of inner shells.

• The atomic size increases down a group because of the addition of new principal shells, which means that the distance between the nucleus and the valence electrons increases, hence the attractive force exerted by the nucleus on the outermost electrons decreases. This leads to the lowering of ionisation enthalpy.
• The addition of new shells also means that the screening effect increases, or the force of attraction felt by the valence shell electrons decreases. This is the second factor which lowers the value of ionisation enthalpy.
• The nuclear charge increases as the atomic number increases. This should lead to an increase in ionisation enthalpy down a group.
• However, the screening effect more than compensates for the impact of the increased nuclear charge. In case of alkali metals, shielding is the most effective as the orbitals in the inner shells are completely filled.

Electron gain enthalpy (Δ_eg H): The electron gain enthalpy of an atom is a measure of its tendency to form an anion or a measure of the firmness with which an extra electron is bound to it. It is defined as the enthalpy change accompanying the process when an electron is added to a neutral gaseous atom to form a monovalent anion.

⇒ \(\mathrm{X}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{X}^{-}(\mathrm{g}) ; \Delta H=\Delta_{\mathrm{eg}} H\)

• The more negative the electron gain enthalpy of an atom, the its greater is tendency to accept to an electron gain enthalpy is expressed in eV per atom or kJ mol^-1.
• The process of electron gain can be endothermic or exothermic, depending on the nature of the element, and so the values of electron gain enthalpy may be positive or negative respectively. In case of electronegative elements, energy is released when one electron is added to a neutral gaseous atom.

After the addition of one electron, the atomf becomes negatively charged and the addition of a second electron is opposed by the electrostatic force of repulsion between the uninegative anion and the electron which is to be added. So, energy has to be supplied for the addition of the second electron. Thus, the second electron gain enthalpy is positive, and so is the third, and so on.

⇒ \(\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta_{\mathrm{eg}} H_1=-141 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta_{\mathrm{eg}} H_2=+844 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The electron gain enthalpy of an atom depends on the following factors.

Atomic size The distance between the nucleus and the outermost shell (which receives additional electrons) increases as atomic size increases. Consequently, the force of attraction between the nucleus and the additional electron decreases and the electron gain enthalpy becomes less negative.

Nuclear charge The force of attraction between the nucleus and the additional electron increases with increase in nuclear charge. With the increase in nuclear charge, electron gain enthalpy becomes more negative.

Electronic configuration Elements which have completely filled orbitals or exactly half-filled orbitals (of the same subshell) are more stable than other elements. Such elements do not accept additional electrons readily, so their electron gain enthalpies have large positive values.

The variation of electron gain enthalpy across a period or down a group is not absolutely regular. However, there is a general trend, despite some exceptions.

Variation of Δ_eg H in a group Both the atomic size and the nuclear charge increase down a group. But the effect of the increase in atomic size is more pronounced. Thus, the attractive force experienced by an additional electron decreases as atomic number increases and electron gain enthalpy is less negative down a group. For example, the Δ_eg H of bromine is greater than that of chlorine.

• However, the electron gain enthalpies of some elements of the second period, like O and F, are less negative than the electron gain enthalpies of the corresponding elements S and Cl of the third period. This is because the extremely small size of these atoms leads to the presence of considerable electron-electron repulsions within these atoms (of the elements of the second period).
• This electronic repulsion more than offsets the impact of the smaller atomic size and opposes the addition of an electron. In other words, though the smaller size should have led to greater nuclear attraction for an additional electron, the interelectronic repulsion actually makes the atom less ready to accept an electron. The AegH of oxygen (-141 kJ mol^-1) is less negative than that of sulphur (-200 kJ mol^-1).

Variation of Δ_eg H in a period In general, electron gain enthalpy becomes more negative across a period. This is because the atomic size decreases, while the magnitude of the nuclear charge increases. This increases the nuclear attraction experienced by an additional electron or makes the atom more ready to accept an electron.

• Halogens have the most negative electron gain enthalpies in their respective periods. This is because halogens (ns²np⁵) have seven electrons in the valence shell.
• They require only one electron to acquire the stable configuration of eight electrons in the outermost shell, so they have a strong tendency to accept an electron. They also have the smallest size in their respective periods. This, too, makes them more ready to accept an electron than other members of their period.
• The value of electron gain enthalpy becomes less negative from chlorine to iodine because of increasing atomic radius. However, as already explained, the electron gain enthalpy of F is unexpectedly less negative than that of Cl.

Δ_eg H of elements with completely filled or exactly half-filled orbitals

1. The alkaline earth metals, or elements of Group 2, have positive electron gain enthalpies. This is due to the fact that the s subshell of these elements is completely filled. This makes them very stable and they do not have a tendency to accept electrons.
2. Nitrogen and phosphorus have very low electron gain enthalpies because they have stable valence shell configurations with exactly half-filled p orbitals.
3. The electron gain enthalpy of noble gases is high and positive. Their valence-shell configuration is extremely stable with completely filled subshells (ns² np⁶). Any additional electron would have to be placed in an orbital of the next higher energy shell. The shielding effect of the inner electrons and the large distance from the nucleus makes the addition of an electron impossible.

The term electron gain enthalpy is closely related to electron affinity. Electron affinity, E_ea, is the energy released when an electron is added to a neutral gaseous atom. It is taken as positive when energy is released and negative when energy is absorbed, in contrast to the thermodynamic convention. Thus, electron affinity and electron gain enthalpy have very similar numerical values but differ in sign.

Electronegativity: The electronegativity of an element is a qualitative measure of the ability of its atom to attract the pair of electrons shared by bonded atoms in a molecule. Linus Pauling formulated a numerical scale of electronegativity based on considerations of bond dissociation energies. Table lists values for the main-group elements.

An atom that has a high electronegativity is likely to be one having a high ionisation enthalpy (so that it is unlikely to lose electrons to another atom in the molecule). Also, such an atom will have a high negative electron gain enthalpy so that it is energetically favourable for an electron to move towards it.

Electronegativities show a periodicity and the elements with the highest electronegativities are those close to fluorine in the periodic table. Fluorine is the most electronegative element while caesium is the least electronegative element.

Variation of electronegativity in a period In a period, in general, electronegativity increases from left to right. This is because the attraction between the valence electrons and the nucleus increases as the atomic radius decreases across a period.

The increase in electronegativity across a period is directly related to the increase in nonmetallic s’- properties of elements when we move from left to right across a period. You already know that nonmetals have the tendency to accept electrons.

Variation of electronegativity in a group The electronegativity values decrease with the increase in atomic size down a group. For example, the electronegativity of fluorine is the most and that of the iodine is the least in Group 17.

Classification Of Elements And Periodicity Of Properties Multiple Choice Questions

Question 1. All the elements in a group in the periodic table have the same

1. Number of electrons
2. Number of valence electrons
3. Atomic weight
4. Atomic number

Answer: 2. Number of valence electrons

Question 2. On moving down a group, the electropositive character of elements

1. Increases
2. Decreases
3. Remains the same
4. None of the above

Answer: 1. Increases

Question 3. The 3d transition series contains elements with atomic numbers from

1. 22 to 30
2. 21 to 30
3. 21 to 31
4. 21 to 29

Answer: 2. 21 to 30

Question 4. The electronegativity of the following elements increases in the order

1. C, N, Si, P
2. N, Si, C, P
3. Si, P, C, N
4. P, Si, N, C

Answer: 3. Si, P, C, N

Question 5. A sudden jump in the difference between the second and third ionisation enthalpies is associated with the electronic configuration

1. \(1 s^2 2 s^2 2 p^6 3 s^1\)
2. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^1\)
3. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6\)
4. \(1 s^2 2 s^2 2 p^6 3 s^2\)

Answer: 4. \(1 s^2 2 s^2 2 p^6 3 s^2\)

Question 6. Which of the following pairs belongs to the same group?

1. Ba, Ca
2. Mg, Na
3. Na, Cl
4. Mg, Cu

Answer: 1. Ba, Ca

Question 7. Choose the correct order of ionic radii from the following.

1. \(\mathrm{Ca}^{2+}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{O}^{2-}\)
2. \(\mathrm{Ca}^{2+}>\mathrm{O}^{2-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}\)
3. \(\mathrm{Na}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{O}^{2-}\)
4. \(\mathrm{O}^{2-}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Na}^{+}\)

Answer: 2.\(\mathrm{Ca}^{2+}>\mathrm{O}^{2-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}\)

Question 8. The valence shell electronic configuration of the most electronegative element is

1. ns²np¹
2. ns²np⁵
3. ns²np⁴
4. ns²np³

Answer: 2. ns²np⁵

Question 9. Which of the following represents the electronic configuration of d-block elements?

1. \((n-1) \mathrm{d}^{1-10} n \mathrm{~s}^{1-2}\)
2. \((n-1) \mathrm{s}^{1-2} n \mathrm{~d}^{1-10}\)
3. \((n-1) \mathrm{d}^{1-10} n \mathrm{~s}^2 n \mathrm{p}^4\)
4. \((n-1) \mathrm{p}^4 n \mathrm{~s}^2\)

Answer: 1. \((n-1) \mathrm{d}^{1-10} n \mathrm{~s}^{1-2}\)

Question 10. An element with the atomic number 24 belongs to the

1. s block
2. p block
3. d block
4. f block

Answer: 3. d block

Question 11. Which of the following transitions involves the maximum amount of energy?

1. \(\mathrm{M}^{-}(\mathrm{g}) \rightarrow \mathrm{M}(\mathrm{g})\)
2. \(\mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+}(g)\)
3. \(\mathrm{M}^{2+}(\mathrm{g}) \rightarrow \mathrm{M}^{3+}(\mathrm{g})\)
4. \(\mathrm{M}^{+}(\mathrm{g}) \rightarrow \mathrm{M}^{2+}(\mathrm{g})\)

Answer: 3. \(\mathrm{M}^{2+}(\mathrm{g}) \rightarrow \mathrm{M}^{3+}(\mathrm{g})\)

Question 12. Which of the following species has the highest electron affinity?

1. F^–
2. O
3. O^–
4. Na⁺

Answer: 2. O

Question 13. Which of the following increases from iodine to fluorine?

1. Bond length
2. Electronegativity
3. The first ionisation enthalpy
4. Oxidising power

Answer:

2. Electronegativity

3. The first ionisation enthalpy

4. Oxidising power

Question 14. Elements of the same group are characterised by the same

1. Ionisation enthalpy
2. Electronegativity
3. Electron gain enthalpy
4. Number of valence electrons

Answer: 4. Number of valence electrons

Question 15. Which of the following have the highest ionisation enthalpies?

1. Halogens
2. Noble gases
3. Alkali metals
4. Alkaline Earth metals

Answer: 2. Noble gases

Question 16. The elements in which 4f orbitals are progressively filled in are called

1. Transition elements
2. Actinides
3. Lanthanides
4. Coinage metals

Answer: 3. Lanthanides

Question 17. The correct order of the first ionisation enthalpies among the elements Be, B, C, N and O is

1. Be < B< O < C < N
2. B < Be < C < O < N
3. N < C < O < B < Be
4. N < O < C < Be < B

Answer: 2. B < Be < C < O < N

 

 

Atomic Structure

The atom as pictured by Dalton, Berzelius, Cannizzaro, and Avogadro was an indestructible particle, which could neither be created nor destroyed. It was thought to be indivisible, and the smallest building block of matter. This concept of the atom, as you have read in the previous chapter, could not explain how or why atoms combine, or how exactly atoms of different elements differ from each other.

• The explanation had to wait until the end of the nineteenth century when some path-breaking experiments proved that the atom is made of still smaller particles—electrons, protons, and neutrons. These discoveries led to a clearer understanding of the chemical behaviour of atoms.
• They also led to the formulation of quantum mechanics, since Newtonian mechanics could not describe the motion of the electron. The laws of quantum mechanics helped to explain the stability of the atom in terms of the arrangement of electrons in the atom and the characteristic radiation emitted by atoms.

Faraday’s Contribution

The earliest indication that scientists had regarding the existence of a relation between matter and electricity came from the observation that static electricity is generated when substances like glass (or ebonite) are rubbed against silk (or fur). Michael Faraday, an English physicist, performed some experiments in the 1830s, which made this connection more evident.

Read and Learn More WBCHSE For Class11 Basic Chemistry Notes

Discovery Of Subatomic Particles

Faraday showed that when electricity is passed through an electrolyte, the electrolyte undergoes chemical changes. The phenomenon is called electrolysis and the quantitative results obtained by Faraday are expressed in Faraday’s laws of electrolysis.

According to these laws,

1. The weights of the elementary substances produced (liberated) during electrolysis are proportional to the quantity of charge (current x time) passed through the electrolyte and
2. The number of moles of various substances liberated by a fixed quantity of charge bears a simple, whole-number ratio with each other.

• These results showed clearly that there is a relation between atoms and electricity. They were also very similar to the laws of chemical combination. Only years later, a scientist called Stoney claimed that Faraday’s laws on the interaction between matter and electricity were like the laws of chemical combination, which dealt with the interaction between different materials.
• He claimed further that since the laws of chemical combination led to the formulation of the atomic theory of matter, Faraday’s laws should imply the discrete nature of electricity. In other words, that electricity must be made of discrete particles and he called these particles of electricity electrons.

Discovery Of The Electron

The electron was discovered during experiments involving the passage of electric current through a gas. Under ordinary conditions, gases do not conduct electricity.

Discovery Of Subatomic Particles

However, if a gas is filled in a sealed tube fitted with electrodes and the pressure inside the lube is lowered to about 10^-2 atm, the gas starts conducting electricity when a high voltage (5000-10,000 V) is applied across the electrodes. Such a tube is called a gas-discharge tube. The neon signs we are so familiar with are actually gas-discharge tubes filled with neon.

Discovery Of The Electron Cathode rays: When the pressure inside the discharge tube is 1 atm, no electric current flows through the tube even on applying high voltage. The pressure of the gas inside the tube can be decreased by pumping out the gas with the help of a vacuum pump. As the pressure is reduced to about 10^-2 atm, a purple light is emitted if the gas being used is air. The whole tube appears purple.

Discovery Of Subatomic Particles Class 11 Notes

• The colour of the light depends on the gas being used. If the pressure is reduced much further to about 10^-4 atm, the purple light disappears and the part of the glass tube directly opposite to the cathode begins to glow with a greenish light.
• A British physicist, Sir William Crookes, who conducted this experiment in 1879, showed that the green glow (fluorescence) is caused by the bombardment of invisible rays emitted by the cathode. An object placed inside the tube, between the cathode and the opposite end of the tube, casts a sharp shadow on the wall of the tube.
• Besides, the glow can then be seen only around the shadow. This proved that the glow was due to the emission of rays from the cathode and that the rays moved in straight lines. These rays are called cathode rays.

Discovery Of The Electron Properties of cathode rays: A series of experiments conducted by several scientists ultimately proved that cathode rays are actually a stream of electrons. Becquerel’s discovery of radioactivity and subsequent experiments conducted by other scientists to study the phenomenon of radioactivity acted as secondary proof to establish that the particles emitted by the cathode are electrons.

Discovery Of Subatomic Particles Class 11 Notes

• But first it had to be proved that the emission from the cathode consisted of a stream of material particles. This was done by placing a light paddle wheel in the path of the cathode rays. It was seen that the rays rotate the paddle wheel, which could happen only if they comprised particles with mass and energy.
• A metal foil placed in the path of the cathode rays gets heated if the rays are focussed on it. This, too, shows that emission from the cathode consists of fast-moving particles. Of course, light rays could also cause the foil to get heated. In fact, later experiments did show that cathode rays consist of electrons and light rays, but we will not concern ourselves with that at the moment.

Discovery Of Subatomic Particles Class 11 Notes

• The next thing that needed to be investigated was whether the particles emitted by the cathode were charged. An electric field was applied to the beam of cathode rays as shown in Figure and it was seen that the rays get deflected towards the positive plate. This proved that the particles of the cathode rays are negatively charged.
• When a magnetic field was applied to the beam of cathode rays, the rays were deflected towards the south pole, as negatively charged particles should.

Other experiments conducted to investigate the properties of cathode rays showed that they ionise the gas through which they pass (as charged particles should) and that they produce X-rays when they are made to fall on metals, such as tungsten.

Discovery Of The Electron Charge To Mass Ratio (e/m): Sir J J Thomson, a British physicist, observed the deflection of cathode rays under the influence of an electric field and a magnetic field applied simultaneously and at right angles to the electric field in a specially designed discharge tube.

• He then adjusted the two fields so that the cathode rays would strike a fluorescent screen placed at the end of the tube opposite the cathode at the original position as did before the two fields were applied. From his observations, he calculated the charge (e) to mass (m) ratio of the electron.
• It turned out to be the same (1.76 x 10⁸ C g^-1) irrespective of the nature of the cathode or the gas taken in the discharge tube. This proved that electrons, the negatively charged subatomic particles, are universal constituents of matter.

Discovery Of The Electron Charge : In 1909, Robert Millikan, an American physicist, measured the charge carried by an electron by doing an experiment which has come to be known as the Millikan oil-drop experiment. He allowed a spray of oil droplets (produced by an atomiser) to enter the apparatus through a small hole and fall between two charged plates.

• He observed the motion of the droplets of oil as they fell under the influence of the force of gravity through a telescope. He then passed X-rays into the chamber to ionise the air inside. The electrons thus produced imparted a negative charge to the oil droplets which then came under the influence of the electric field between the charged plates.
• He adjusted the electric field until it balanced the gravitational field. (In a balanced field a droplet would either be stationary or move with constant velocity in accordance with the first law of motion.) The magnitude of the electric field required to balance the two fields would be a measure of the charge on the droplet of oil. The charge on a droplet, according to Millikan’s observations, was always a whole-number multiple of 16 x 10^-19 C. He, therefore, concluded that the charge on an electron is 16 x 10^-19 C.

Discovery Of Mass Of The Electron: The mass of an electron can be calculated quite simply from the results of Thomson’s and Millikan’s experiments.

According to Thomson,

⇒ \(\frac{e}{m}=1.76 \times 10^8 \mathrm{Cg}^{-1}\)

According to Millikan, e =1.6 x 10^-19 C.

∴ mass of electron (m) = \(\frac{e}{e / m}\)

= \(\frac{16 \times 10^{-19}}{1.76 \times 10^8}=9.1 \times 10^{-28} \mathrm{~g}\)

= \(9.1 \times 10^{-31} \mathrm{~kg}\)

Anode Rays Or Canal Rays

After the discovery of the electron began the search for its positive counterpart. Since the atom as a whole is electrically neutral, scientists began to look for positively charged subatomic particles. In 1886, Eugen Goldstein, a German physicist, performed an experiment that led to the discovery of the proton.

• He used a perforated cathode instead of the normal one used in the gas discharge tube.
• When he passed an electric discharge through the tube, he observed a glow on the wall of the tube behind the perforated cathode. This obviously meant that a stream of particles (or light) was moving towards the cathode, passing through it and causing the tube to glow.
• The rays were named anode rays or canal rays. These rays traveled in a direction opposite to that of the cathode rays.

Investigations followed, as in the case of cathode rays, and the following conclusions were reached.

1. Anode rays travel in straight lines.
2. They consist of material particles.
3. They are positively charged since they are deflected towards the negatively charged plate in an electric field.
4. They are deflected by a magnetic field.
5. Their charge-to-mass (e/m) ratio depends upon the gas used in the discharge tube.
6. The mass of the positively charged particles in the anode rays is equal to the atomic mass of the gas used.

Origin Of Cathode And Anode Rays: When a high voltage is applied across the gas-discharge tube, the atoms or molecules of the gas get ionised, resulting in the formation of positively charged ions and negatively charged particles. The negatively charged particles, we know, are electrons.

⇒ \(\underset{substack{\text { a neutral } \\ \text { gaseous atom }}}{\mathrm{A}} \stackrel{\text { ionised }}{\longrightarrow} \mathrm{A}^{+}+\mathrm{e}^{-}\)

Under the influence of the high voltage, the electrons move at a high velocity towards the anode. In the process, they ionize more neutral gas atoms, or produce more electrons and positive ions, The negatively charged particles or electrons constitute cathode rays, while the positively charged ions constitute anode rays.

Obviously, the e/m ratio of the cathode rays is independent of the gas used because these rays comprise electrons. The e/m ratio of the anode rays, on the other hand, depends on the gas used because the mass of the ion depends on the mass of the original atom.

Cathode And Anode Rays Discovery Of Proton: Observations showed that the positive particles (ions) produced by hydrogen had the highest e/m ratio. In other words, hydrogen produced the lightest positive particle. This particle was called proton.

Its charge-to-mass ratio was found to be 9.58 x 10⁴ C g^-1. The charge on the proton is equal and opposite to that on the electron, i.e., 16 x 10^-19 C and its mass is 167 x 10^-24 g. The proton is about 1837 times heavier than the electron and its mass is almost the same as that of a hydrogen atom.

Radioactivity

As mentioned earlier, Becquerel’s investigations into the phenomenon of radioactivity further confirmed that the atom is divisible and is made up of charged particles. The spontaneous emission of radiation by some substances like uranium is called radioactivity. Elements that have this property are called radioactive elements.

• When the radiation emitted by a radioactive substance is allowed to pass through a strong electric field or magnetic field, it splits into three parts which are deflected in different directions.
• The rays deflected towards the negative plate are called α (alpha) rays. These consist of positively charged He²⁺ ions.
• The rays deflected towards the positive plate are called β (beta) rays and consist of negatively charged β particles or electrons.
• The undeflected rays are high-energy electromagnetic radiations called γ (gamma) rays. You will learn more about this phenomenon later.

Thomson’s Model Of The Atom

Once it was discovered that the atom consisted of negatively and positively charged particles (this was before Chadwick’s discovery of the neutron), scientists wished to know how these particles were arranged inside the atom.

In 1898 Thomson proposed a model of the atom in which the positive charge was supposed to be spread over a sphere of radius 10’8 cm. He assumed that electrons were embedded in this sphere, more or less evenly, as plums in a pudding.

Consequently, Thomson’s model of the atom came to be known as the plum pudding model. Thomson’s model was a pioneering effort to explain the atomic structure and he received the Nobel prize in physics in 1906.