ICSE Questions Set in Previous Years — Question 27

Chloride of a metal X is XCl₂

By interchanging subscript and writing as superscript:

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

Therefore, valency of metal X = 2.

Formula of the sulphate:

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

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 $\bold{XSO}_{\bold{4}}$

Formula of the hydroxide:

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

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

Therefore, we get

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

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