A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. $\mathrm{H}_2 \mathrm{SO}_4$ and NaCl , chlorine gas is liberated and a compound (D) of manganese alongwith other products is formed. Identify compounds A to D and also explain the reactions involved.
Since, compound $(\mathrm{C})$ on treating with conc. $\mathrm{H}_2 \mathrm{SO}_4$ and NaCl gives $\mathrm{Cl}_2$ gas, so it is manganese dioxide $\left(\mathrm{MnO}_2\right)$. It is obtained alongwith $\mathrm{MnO}_4^{2-}$ when $\mathrm{KMnO}_4$ (violet) is heated.
Thus,
$$\begin{array}{ll} (A)=\mathrm{KMnO}_4 & (B)=\mathrm{K}_2 \mathrm{MnO}_4 \\ (C)=\mathrm{MnO}_2 & (D)=\mathrm{MnCl}_2 \end{array}$$
$\underset{[A]}{\mathrm{KMnO}_4} \xrightarrow{\Delta} \underset{[B]}{\mathrm{K}_2 \mathrm{MnO}_4}+\underset{[C]}{\mathrm{MnO}_2}+\mathrm{O}_2$
$2 \mathrm{MnO}_2+4 \mathrm{KOH}+\mathrm{O}_2 \longrightarrow 2 \mathrm{~K}_2 \mathrm{MnO}_4+2 \mathrm{H}_2 \mathrm{O}$
$$\mathrm{MnO}_2+4 \mathrm{NaCl}+4 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \underset{[D]}{\mathrm{MnCl}_2}+4 \mathrm{NaHSO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2$$