ICSE Questions Set in Previous Years — Question 45

Nitride of a metal X is XN.

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

Formula of the sulphate:

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

So, we get the formula as $\bold{X}_{\bold{2}}({\bold{SO}}_{\bold{4}})_{\bold{3}}$

Formula of the hydroxide:

$\text{X}^{3+} \phantom{\nearrow} \text{OH}^{1-} \\[0.5em] \overset{\phantom{2}{3}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{O}}\text{H} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{O}}\text{H} \\[0.5em]$

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

Therefore, we get

Formula of Sulphate : $\bold{X}_{\bold{2}}({\bold{SO}}_{\bold{4}})_{\bold{3}}$

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