QJEEAdv17Ch57d4: JEE Advanced 2017 Chemistry - Class 12 JEEAdv17Ch57d4 Solution: W and X are, respectively

(i) Potassium chlorate $\left(\mathrm{KClO}_3\right)$ is decomposed in presence of $\mathrm{MnO}_2$ as catalyst to form potassium chloride and oxygen gas.

$2 \mathrm{KClO}_3(s) \underset{\mathrm{MnO}_2}{\stackrel{\Delta}{\longrightarrow}} 2 \mathrm{KCl}+\underset{(w)}{3 \mathrm{O}_2(g)}$

The gas $(w)$ is oxygen.

(ii) Reaction of white phosphorous with excess of gas $w$ (i.e., $\mathrm{O}_2$ ) gives $\mathrm{P}_4 \mathrm{O}_{10}$ (a dimer of phosphorous pentaoxide).

$$\underset{\text{White}}{{\mathrm{P}}_4(s)}+\underset{\text{(excess)}}{5{\mathrm{O}}_2(g)}\to \underset{(X)}{{\mathrm{P}}_4{\mathrm{O}}_{10}(s)}$$

phosphorous oxygen

The compound $(X)$ is $\mathrm{P}_4 \mathrm{O}_{10}$ .