JEE Main 2019 — d and f Block Elements Question with Solution

Potassium manganate is water soluble and so is leached out as green filtrate. MnO₂ is insoluble in water and gives black residue. The oxygen gas is released in this reaction.

In the first step potassium permanganate $\left({\mathrm{KMnO}}_4\right)$ the compound (X) decomposes on heating to give potassium manganate $\left({\mathrm{K}}_2{\mathrm{MnO}}_4\right)$ which is the compound (Y), ${\mathrm{MnO}}_{2 }$and the gaseous product is oxygen.

In the second step ${\mathrm{MnO}}_2$ further reacts with $\mathrm{NaCl}$ and concentrated sulfuric acid to give a pungent-smelling gas that is Chlorine gas $({\mathrm{Cl}}_2 ), {\mathrm{MnSO}}_4 , {\mathrm{Na}}_2{\mathrm{SO}}_4$ and ${\mathrm{H}}_2\mathrm{O}$. 

$\underset{\left(\mathrm{X}\right)}{2{\mathrm{KMnO}}_4}\underset{\mathrm{\Delta }}{\overset{513 \mathrm{K}}{\to }}\mathrm{\Delta } {\mathrm{K}}_2\underset{\left(\mathrm{Y}\right)}{\mathrm{Mn}}{\mathrm{O}}_4+{\mathrm{MnO}}_2+{\mathrm{O}}_{2\left(\mathrm{g}\right)}$

${\mathrm{MnO}}_2+4\mathrm{NaCl}+4{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{MnCl}}_2+4{\mathrm{NaHSO}}_4+2{\mathrm{H}}_2\mathrm{O}+\underset{\left(\mathrm{Z}\right) \mathrm{Pungent} \mathrm{gas}}{{\mathrm{Cl}}_{2\left(\mathrm{g}\right)}}$

The compound$\mathrm{X}= {\mathrm{KMnO}}_4 , \mathrm{Y}={\mathrm{K}}_2{\mathrm{MnO}}_4, \mathrm{Z}={\mathrm{Cl}}_2$.