JEE Main 2019 — Chemistry Inorganic Chemistry

Potassium manganate is water soluble and so is leached out as green filtrate. MnO$_{2}$2​ is insoluble in water and gives black residue. The oxygen gas is released in this reaction.

In the first step potassium permanganate $({\mathrm{KMnO}}_{4})$(KMnO4​) the compound (X) decomposes on heating to give potassium manganate $({K}_{2}{\mathrm{MnO}}_{4})$(K2​MnO4​) which is the compound (Y), ${\mathrm{MnO}}_{2}$MnO2​and the gaseous product is oxygen.

In the second step ${\mathrm{MnO}}_{2}$MnO2​ further reacts with $\mathrm{NaCl}$NaCl and concentrated sulfuric acid to give a pungent-smelling gas that is Chlorine gas $({\mathrm{Cl}}_{2}),{\mathrm{MnSO}}_{4},{\mathrm{Na}}_{2}{\mathrm{SO}}_{4}$(Cl2​),MnSO4​,Na2​SO4​ and ${H}_{2}O$H2​O.

$$\underset{(X)}{2{\mathrm{KMnO}}_{4}}\rightarrow _{\Delta }^{513K}\Delta {K}_{2}\underset{(Y)}{\mathrm{Mn}}{O}_{4}+{\mathrm{MnO}}_{2}+{O}_{2(g)}$$(X)2KMnO4​​→Δ513K​ΔK2​(Y)Mn​O4​+MnO2​+O2(g)​

$${\mathrm{MnO}}_{2}+4\mathrm{NaCl}+4{H}_{2}{\mathrm{SO}}_{4}\rightarrow {\mathrm{MnCl}}_{2}+4{\mathrm{NaHSO}}_{4}+2{H}_{2}O+\underset{(Z)\mathrm{Pungent}\mathrm{gas}}{{\mathrm{Cl}}_{2(g)}}$$MnO2​+4NaCl+4H2​SO4​→MnCl2​+4NaHSO4​+2H2​O+(Z)PungentgasCl2(g)​​

The compound$X={\mathrm{KMnO}}_{4},Y={K}_{2}{\mathrm{MnO}}_{4},Z={\mathrm{Cl}}_{2}$X=KMnO4​,Y=K2​MnO4​,Z=Cl2​.