The Language of Chemistry

Chloride of a metal X is XCl₂

By interchanging subscript and writing as superscript:

$$\begin{aligned}\underset{\phantom{1}{1}}{\text{X}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{Cl}} \Rightarrow \overset{\phantom{2}{2}}{\text{X}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{Cl}} \\[0.5em]\end{aligned}$$

Therefore, valency of metal X = 2.

Formula of the sulphate:

$$\begin{aligned}\text{X}^{2+} \phantom{\nearrow} \text{SO}_{4}^{2-} \\[0.5em] \overset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}}\text{O}_{4} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{S}}\text{O}_{4} \\[0.5em]\end{aligned}$$

As valency of both X and SO₄ is 2 so dividing by 2 we get 1 but 1 is never written so we get the formula as $\mathbf{XSO}_{\mathbf{4}}$

Formula of the hydroxide:

$$\begin{aligned}\text{X}^{2+} \phantom{\nearrow} \text{OH}^{1-} \\[0.5em] \overset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{O}}\text{H} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{O}}\text{H} \\[0.5em]\end{aligned}$$

Dropping 1 and enclosing OH in brackets, we get the formula as $\mathbf{X(OH)}_{\mathbf{2}}$

Therefore, we get

Formula of Sulphate : $\mathbf{XSO}_{\mathbf{4}}$

Formula of Hydroxide : $\mathbf{X(OH)}_{\mathbf{2}}$

Nitride of a metal X is XN.

Since valency of nitrogen is 3-
so valency of X is 3+

Formula of the sulphate:

$$\begin{aligned}\text{X}^{3+} \phantom{\nearrow} \text{SO}_{4}^{2-} \\[0.5em] \overset{\phantom{3}{3}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}}\text{O}_{4} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{S}}\text{O}_{4} \\[0.5em]\end{aligned}$$

So, we get the formula as $\mathbf{X}_2\mathbf{(SO}_{\mathbf{4}}\mathbf{)}_\mathbf{3}$

Formula of the hydroxide:

$$\begin{aligned}\text{X}^{3+} \phantom{\nearrow} \text{OH}^{1-} \\[0.5em] \overset{\phantom{3}{3}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{O}}\text{H} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{O}}\text{H} \\[0.5em]\end{aligned}$$

Dropping 1, we get the formula as $\mathbf{X(OH)}_{\mathbf{3}}$

Therefore, we get

Formula of Sulphate : $\mathbf{X}_2\mathbf{(SO}_{\mathbf{4}}\mathbf{)}_\mathbf{3}$

Formula of Hydroxide : $\mathbf{X(OH)}_{\mathbf{3}}$

$$\begin{aligned}\text{(i)} \space \text{NO} = \underset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]\end{aligned}$$

But valency of O is 2. Multiplying by 2, we get:

$$\begin{aligned}\overset{{2 \times 1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]\end{aligned}$$

Therefore, valency of Nitrogen is 2.

$$\begin{aligned}\text{(ii)} \space \text{N}_2\text{O} = \underset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]\end{aligned}$$

Therefore, valency of Nitrogen is 1.

$$\begin{aligned}\text{(iii)} \space \text{NO}_2 = \underset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]\end{aligned}$$

But valency of O is 2. Multiplying by 2, we get:

$$\begin{aligned}\overset{{2 \times 2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{4}{4}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]\end{aligned}$$

Therefore, valency of Nitrogen is 4.

Valency of X is 3+ and that of Oxygen is 2-

By interchanging the valency number and shifting it to the lower right side of the atom or radical, we get the formula : X₂O₃

Equation for the combustion of X in oxygen: 4X + 3O₂ ⟶ 2X₂O₃

(a) Elements whose symbols are derived from first letter of the name of the element :

• Carbon - C
• Sulphur - S
• Oxygen - O

(b) Elements whose symbols are derived from their Latin names :

• Kalium [Potassium] - K
• Ferrum [Iron] - Fe
• Natrium [Sodium] - Na

Eight metals which show variable valency are :

(a) Valency of sulphur in SO₂ is 4

By interchanging subscript and writing as superscript:

$$\begin{aligned}\underset{\phantom{1}{1}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]\end{aligned}$$

But valency of O is 2. Multiplying by 2, we get:

$$\begin{aligned}\overset{{2 \times 2}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{4}{4}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]\end{aligned}$$

Therefore, valency of Sulphur is 4.

(b) Valency of sulphur in SO₃ is 6

$$\begin{aligned}\underset{\phantom{1}{1}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{3}{\text{O}} \Rightarrow \overset{\phantom{3}{3}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]\end{aligned}$$

But valency of O is 2. Multiplying by 2, we get:

$$\begin{aligned}\overset{{2 \times 3}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{6}{6}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]\end{aligned}$$

Therefore, valency of Sulphur is 6.

The divalent elements or radicals are:

• Nickel
• Sulphite
• Dichromate

Symbol represents the short form of an element.

Qualitative meaning — A symbol represents specific element or one atom of an element. E.g., 'S' represents one atom of the element sulphur.

Quantitative meaning — A symbol also represents the weight of the element equal to its atomic weight i.e. it represents how many times an atom is heavier than one atomic mass unit [a.m.u.] which is defined as $\frac{1}{12}$th the mass of a carbon atom C¹²

Valency is the number of hydrogen atoms which can combine with or displace one atom of the element or radical so as to form a compound.

Valency of a metal is the number of electrons lost per atom of the metal. Valency of all metals and hydrogen is considered positive. Therefore, valency of potassium is 1+ as it has 1 electron in outermost shell which it loses and becomes K⁺.

Valency of non-metal is the number of electrons gained per atom of the non-metal. Valency of all non-metals/groups of non-metals is taken as negative. Therefore, valency of non-metal chlorine is -1, because chlorine [Cl] has 7 electrons in valence shell and gains 1 electron and becomes [Cl^–].

Certain elements exhibit more than one valency hence it is said that these elements have variable valency.

Reasons for exhibiting variable valency — An atom of an element can sometimes lose more electrons than are present in its valence shell. This happens when it loses electrons from the penultimate [i.e., last but one] shell and hence exhibit more than one or variable valency.

Variable valency of Silver — Atomic number of Silver (Ag) is 47. Its electronic configuration is [2, 8, 18, 18, 1]. The outermost shell has one electron and the penultimate shell has 18 electrons. However, the penultimate shell has not attained stability and one more electron sometimes jumps to the outermost shell thereby increasing valence electrons and the new configuration [2, 8, 18, 17, 2] loses two electrons and has valency [+2]. Therefore, Silver exhibits variable valency forming Ag¹⁺ and Ag²⁺.

Compound is a pure substance composed of two or more elements combined chemically in a fixed proportion. For example, water (H₂O) is composed of two elements — Hydrogen and Oxygen in a fixed proportion by weight.

Main characteristics - with special reference to the compound iron [II] sulphide:

1. Components are present in a definite proportion.
   • Elements iron and sulphur combine in a definite proportion.
2. Compound is always homogeneous.
3. Particles in a compound are of one kind.
   • Composition of iron [II] sulphide, cannot be varied, hence has a uniform composition.
   • Components - may not be seen separately.
4. Compound [iron [II] sulphide] has a definite set of properties.
5. Components in [Iron [II] Sulphide] do not retain their original properties and can be separated only by chemical means.
   • Particles in iron [II] sulphide are chemically combined and hence iron in iron [II] sulphide cannot be attracted by a magnet and does not give H₂ with dil. acid.
   • Sulphur present in iron [II] sulphide is insoluble in solvent CS₂.

A molecule of a substance i.e., element or compound could be represented by symbols. This representation is known as chemical formula.

As molecular formula also indicates the number of units of the radicals present in a compound with the proper subscript outside the unit of the radical and in the case of ZnCO₃, there are 1 atom of Zn and C whereas 3 atoms of O. Therefore, we get ZnCO₃.

Potassium

Sodium

Note on variable valency — Copper shows variable valencies of +1 and +2.

Note on variable valency — Iron shows variable valencies of +2 and +3.

Note on variable valency — Silver shows variable valencies of +1 and +2.

Chemical equation is a shorthand form for a chemical change. It shows the result of a chemical change in which the reactants and the products are represented by:

• symbols [in the case of elements]
• formula [in the case of compounds]

The substances which take part in a chemical change and are written on the left side of the equation are called the reactants.

The substances formed as a result of a chemical change and are written on the right side of the equation are called the products.

E.g., NH₃ + H₂O ⟶ NH₄OH

Here, NH₃ and H₂O are reactants and NH₄OH is the product.

(a) NH₃ + H₂O ⟶ NH₄OH

(b) C₂H₅Br + NaOH ⟶ C₂H₅OH + NaBr

(c) 6CO₂ + 12H₂O ⟶ C₆H₁₂O₆ + 6H₂O + 6O₂

(d) 2KMnO₄ + 16HCl ⟶ 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂

(a) A balanced equation is one in which the number of atoms of each element is the same on the side of the reactants and on the side of the products.

(b) As atoms of each reactant [K, Cl, O] on the L.H.S. is equal to the number of the atoms of products on the R.H.S., hence the given equation is balanced.

(c) MnO₂ is a catalyst [simply increases the rate of reaction] and it does not take part in the reaction and so it is written on the top of the arrow.

(i) ⟶ : The direction of the reaction is irreversible

(ii) ⇌ : The direction of the reaction is reversible

(iii) (s) : The state of the matter is solid

(iv) (l) : The state of the matter is liquid

(v) (g) : The state of the matter is gas

(vi) (aq.) : The substance is in aqueous form.

(a) Information provided by the equation:

• One molecule of calcium carbonate reacts with two molecules of [dil.] hydrochloric acid to produce one molecule of calcium chloride, one molecule of water and one molecule of carbon dioxide.
• The reaction is irreversible.
• Hydrochloric acid is in dil. form.
• Carbon dioxide produced is in gaseous form.

(b) Information not provided by the equation:

• The concentration of the reactants and products.
• The speed of the reaction
• Change in colour occurring
• Completion of the reaction i.e., whether the reaction is complete or not
• Changes in evolution of light or sound energy during the occurrence of the reaction.

1. 2C + O₂ ⟶ 2CO

2. N₂ + O₂ ⇌ 2NO

3. 2ZnS + 3O₂ ⟶ 2ZnO + 2SO₂

4. 4Al + 3O₂ ⟶ 2Al₂O₃

5. 3Mg + N₂ ⟶ Mg₃N₂

6. 2Al + N₂ ⟶ 2AlN

7. 2NO + O₂ ⟶ 2NO₂

8. 2SO₂ + O₂ ⇌ 2SO₃

9. H₂ + Cl₂ ⟶ 2HCl

10. 2Fe + 3Cl₂ ⟶ 2FeCl₃

11. H₂S + Cl₂ ⟶ S + 2HCl

12. 2FeCl₂ + Cl₂ ⟶ 2FeCl₃

13. CO₂ + C ⟶ 2CO

14. 2KHCO₃ ⟶ K₂CO₃ + H₂O + CO₂

15. 4K + CO₂ ⟶ 2K₂O + C

16. Ca(OH)₂ + 2HNO₃ ⟶ Ca(NO₃)₂ + 2H₂O

17. 2K + 2H₂O ⟶ 2KOH + H₂

18. Ca + 2H₂O ⟶ Ca(OH)₂ + H₂

19. 2Al + 3H₂O ⟶ Al₂O₃ + 3H₂

20. 3Fe + 4H₂O ⟶ Fe₃O₄ + 4H₂

21. Zn + 2NaOH ⟶ Na₂ZnO₂ + H₂

22. Zn + 2HCl ⟶ ZnCl₂ + H₂

23. 2Al + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 3H₂

24. 2H₂ + O₂ ⟶ 2H₂O

25. N₂ + 3H₂ ⟶ 2NH₃

26. Fe₂O₃ + 3H₂ ⟶ 2Fe + 3H₂O

27. 2KBr + Cl₂ ⟶ 2KCl + Br₂

28. 2NaOH + Cl₂ ⟶ NaCl + NaClO + H₂O

29. 2NaHCO₃ + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O + 2CO₂

30. 2Mg + CO₂ ⟶ 2MgO + C

31. Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

32. CaO + 2HCl ⟶ CaCl₂ + H₂O

1. 2KNO₃ ⟶ 2KNO₂ + O₂
2. Ca + 2H₂O ⟶ Ca(OH)₂ + H₂
3. Fe + 2HCl ⟶ FeCl₂ + H₂
4. 4NO₂ + 2H₂O + O₂ ⟶ 4HNO₃
5. 2PbO₂ ⟶ 2PbO + O₂
6. 4Al + 3O₂ ⟶ 2Al₂O₃
7. 2Fe + 3Cl₂ ⟶ 2FeCl₃
8. 2KBr + Cl₂ ⟶ 2KCl + Br₂
9. 2KHCO₃ ⟶ K₂CO₃ + H₂O + CO₂
10. Ca(OH)₂ + 2NH₄Cl ⟶ CaCl₂ + 2H₂O + 2NH₃

1. 2NaHCO₃ + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O + 2CO₂

2. 2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

3. Pb(NO₃)₂ + 2NaCl ⟶ 2NaNO₃ + PbCl₂

4. FeSO₄ + 2NaOH ⟶ Na₂SO₄ + Fe(OH)₂

5. FeCl₃ + 3NaOH ⟶ 3NaCl + Fe(OH)₃

6. CuSO₄ + 2NaOH ⟶ Na₂SO₄ + Cu(OH)₂

7. FeCl₃ + 3NH₄OH ⟶ 3NH₄Cl + Fe(OH)₃

8. ZnO + 2NaOH ⟶ Na₂ZnO₂ + H₂O

9. Pb(OH)₂ + 2NaOH ⟶ Na₂PbO₂ + 2H₂O

10. Al₂O₃.2H₂O + 2NaOH ⟶ 2NaAlO₂ + 3H₂O

11. NaAlO₂ + 2H₂O ⟶ NaOH + Al(OH)₃

12. 2Al(OH)₃ ⟶ Al₂O₃ + 3H₂O

13. 2ZnS + 3O₂ ⟶ 2ZnO + 2SO₂

14. Fe₂O₃ + 2Al ⟶ Al₂O₃ + 2Fe

15. 2Al + 3Cl₂ ⟶ 2AlCl₃

16. 2NaCl + H₂SO₄ ⟶ Na₂SO₄ + 2HCl

17. Fe + 2HCl ⟶ FeCl₂ + H₂

18. Na₂CO₃ + 2HCl ⟶ 2NaCl + H₂O + CO₂

19. Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ + 2HNO₃

20. AgCl + 2NH₄OH ⟶ Ag(NH₃)₂Cl + 2H₂O

21. MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

22. Pb₃O₄ + 8HCl ⟶ 3PbCl₂ + 4H₂O + Cl₂

23. 2KMnO₄ + 16HCl ⟶ 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂

24. K₂Cr₂O₇ + 14HCl ⟶ 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

25. 2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃

26. (NH₄)₂SO₄ + 2NaOH ⟶ Na₂SO₄ + 2H₂O + 2NH₃

27. Mg₃N₂ + 6H₂O ⟶ 3Mg(OH)₂ + 2NH₃

28. AlN + 3H₂O ⟶ Al(OH)₃ + NH₃

29. 4NH₃ + 3O₂ ⟶ 2N₂ + 6H₂O [burning of NH₃]

30. 4NH₃ + 5O₂ ⟶ 4NO + 6H₂O [Catalytic oxidation of NH₃]

31. 2NH₄OH + H₂SO₄ ⟶ (NH₄)₂SO₄ + 2H₂O

32. 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂

33. NH₃ + 3Cl₂ ⟶ 3HCl + NCl₃ [nitrogen trichloride]

34. 4HNO₃ ⟶ 2H₂O + 4NO₂ + O₂

35. Ca(HCO₃)₂ + 2HNO₃ ⟶ Ca(NO₃)₂ + 2H₂O + 2CO₂

36. C + 4HNO₃ [conc.] ⟶ CO₂ + 2H₂O + 4NO₂

37. S + 6HNO₃ [conc.] ⟶ H₂SO₄ + 2H₂O + 6NO₂

38. Cu + 4HNO₃ [conc.] ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

39. C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2H₂O + 2SO₂

40. S + 2H₂SO₄ [conc.] ⟶ 3SO₂ + 2H₂O

41. Cu + 2H₂SO₄ [conc.] ⟶ CuSO₄ + 2H₂O + SO₂

(a) The number of times one atom of an element is heavier than 1⁄12^th the mass of an atom of carbon [C¹²] is known as the Relative Atomic Mass [RAM] of the element.

$$\begin{matrix} & \text{Mass of one atom} \\ \text{RAM} = & \text{of the element} \\ & \overline{(\frac{1}{12}) \space \text{Mass of one atom}} \\ & \text{of carbon [C}^{12}] \end{matrix}$$

(b) The number of times one molecule of the substance is heavier than 1⁄12^th the mass of an atom of carbon [C¹²] is known as Relative Molecular Mass [RMM] of the element.

$$\begin{matrix} & \text{Mass of one molecule} \\ \text{RMM} = & \text{of the substance} \\ & \overline{(\frac{1}{12}) \space \text{Mass of one atom}} \\ & \text{of carbon [C}^{12}] \end{matrix}$$

As atoms are extremely small and very light, hence cannot be weighed directly. So indirect methods are utilized to determine the absolute mass of an atom.

The relative mass of an atom or molecule is hence considered by considering a mass of a light atom and relating the mass of other atoms or molecules to it.

(a) Relative Molecular Mass of ZnCO₃:

Molecular wt. of ZnCO₃ = At. wt. of Zn + At. wt. of C + 3(At. wt. of O)
= 65 + 12 + 3(16)
= 65 + 12 + (48)
= 65 + 60
= 125

Hence, Relative Molecular Mass of ZnCO₃ = 125

(b) Relative Molecular Mass of CaSO₄:

Molecular wt. of CaSO₄ = At. wt. of Ca + At. wt. of S + 4(At. wt. of O)
= 40 + 32 + 4(16)
= 40 + 32 + 64
= 136

Hence, Relative Molecular Mass of CaSO₄ = 136

(a) Percentage composition of Calcium Chloride:

Chemical Formula of Calcium Chloride = CaCl₂

Mol. wt. of calcium chloride = At. wt. of Ca + 2(At. wt. of Cl)
= 40 + 2(35.5)
= 40 + 71
= 111

111 g of CaCl₂ contains 40 g of Calcium,

∴ 100 g of CaCl₂ will contain $\frac{40}{111}$ x 100 % = 36.04% of Calcium [percentage].

Similarly, 111 g of CaCl₂ will contain 71 g of Chlorine,

∴ 100 g of CaCl₂ will contain $\frac{71}{111}$ x 100 % = 63.96% of Chlorine.

Hence, Calcium Chloride contains 36.04% of Calcium and 63.96% of Chlorine.

(b) Percentage composition of Calcium Nitrate:

Chemical Formula of Calcium Nitrate = Ca(NO₃)₂

Mol. wt. of Calcium Nitrate = At. wt. of Ca + 2[At. wt. of N] + 6[At. wt. of O]
= 40 + 2(14) + 6(16)
= 40 + 28 + 96
= 164

164 g of Ca(NO₃)₂ contains 40 g of Calcium,

∴ 100 g of Ca(NO₃)₂ will contain $\frac{40}{164}$ x 100 % = 24.39% of Calcium.

Similarly, 164 g of Ca(NO₃)₂ will contain 28 g of Nitrogen,

∴ 100 g of Ca(NO₃)₂ will contain $\frac{28}{164}$ x 100 % = 17.07% of Nitrogen.

Similarly, 164 g of Ca(NO₃)₂ will contain 96 g of Oxygen,

∴ 100 g of Ca(NO₃)₂ will contain $\frac{96}{164}$ x 100 % = 58.54% of Oxygen.

Hence, Calcium Nitrate contains 24.39% of Calcium, 17.07% of Nitrogen and 58.54% of Oxygen.

Zinc

Reason — Symbol "Zn" is derived directly from Zinc, not from a Latin name.
Symbol "Na" comes from Latin name Natrium. Symbol "Hg" comes from Latin name Hydrargyrum. Symbol "Fe" comes from Latin name Ferrum

Lead & tin

Reason — Lead and tin, both show +2 and +4 valency.
Silver commonly show +1 and rarely show +2 valency. Gold show +1 and +3 valency. Mercury show +1 and +2 valency. Iron shows +2 and +3 valency.

Ammonium

Reason — A positive radical is a cation and Ammonium is a cation. Permanganate, Plumbite, Dichromate are anions or negative radicals.

Calcium hydroxide

Reason — Calcium hydroxide has two hydroxide groups which means two oxygen atoms.
Sodium hydroxide has 1 oxygen atom. Lead nitrate has 6 oxygen atoms. Nitric acid has 3 oxygen atoms.

Iron [III] oxide

Al₂(SO₄)₃

Reason — Aluminium generally shows only +3 valency.
Iron shows +2 and +3 valency. Lead shows +2 and +4 valency. Copper shows +1 and +2 valency.

Glauber's salt

Reason — Glauber's salt has 14 oxygen atoms.
Washing soda has 13 oxygen atoms. Green vitriol has 11 oxygen atoms. Gypsum has 6 oxygen atoms.

2NH₃ + 3CuO ⟶ 3Cu + 2H₂O + N₂

Reason — Hydrogen and oxygen atoms are not balanced in this equation. The left side has 6 hydrogen atoms and 3 oxygen atoms, while the right side has 4 hydrogen atoms and 2 oxygen atoms.

Hydroxyl

Reason — Hydroxyl is the negative atmospheric radical. It is present in the troposphere.
Plumbite and Zincate are negative radicals but they are not present in the atmosphere. Ammonium is a positive radical.

Reaction of excess ammonia with chlorine — ammonia as a reducing agent:

$$\begin{array}{l} 2\text{NH}_3 + 3\text{Cl}_2 \longrightarrow \cancel{6\text{HCl}} + \text{N}_2 \\ 6\text{NH}_3 + \cancel{6\text{HCl}} \longrightarrow 6\text{NH}_4\text{Cl} \\ \hline 8\text{NH}_3 + 3\text{Cl}_2 \longrightarrow 6\text{NH}_4\text{Cl} + \text{N}_2 \end{array}$$

(a) The acid having two oxygen atoms in its molecule is acetic acid.

(b) The compound having a trivalent negative ion is calcium nitride.

(c) The valency of nitrogen in nitrogen dioxide is +4.

1. A symbol represents a short form of an element.
2. Compounds are always homogeneous in nature.
3. Variable valency is exhibited, since electrons are lost from an element from the penultimate shell.
4. A chemical equation is a shorthand form for a chemical change.
5. Relative molecular mass of an element/compound is the number of times one molecule of the substance is heavier than 1⁄12^th the mass of an atom of carbon [C¹²].

1. Na₂SO₃ + HCl ⟶ 2NaCl + H₂O + SO₂
   Corrected Balancing — 2HCl
2. CaC₂ + N₂ ⟶ 2CaCN₂ + C
   Corrected Balancing — CaCN₂
3. Fe₂O₃ + 2H₂ ⟶ 2Fe + 3H₂O
   Corrected Balancing — 3H₂
4. Cl₂ + 2H₂O + SO₂ ⟶ 4HCl + H₂SO₄
   Corrected Balancing — 2HCl
5. 6NaOH + 3Cl₂ ⟶ 6NaCl + NaClO₃ + 3H₂O
   Corrected Balancing — 5NaCl
6. C₂H₅OH + 3O₂ ⟶ 2CO₂ + 2H₂O
   Corrected Balancing — 3H₂O
7. NaOH + CO₂ ⟶ Na₂CO₃ + H₂O
   Corrected Balancing — 2NaOH
8. 2H₂O + 2Cl₂ ⟶ 2HCl + O₂
   Corrected Balancing — 4HCl
9. 3CuO + NH₃ ⟶ 3Cu + 3H₂O + N₂
   Corrected Balancing — 2NH₃
10. PbO₂ + 4HCl ⟶ PbCl₂ + H₂O + Cl₂
    Corrected Balancing — 2H₂O

1. A
2. B
3. A
4. B
5. A