Question

A violet compound of manganese $(P)$ decomposes on heating to liberate oxygen and compounds $(Q)$ and $(P)$ of manganese are formed. Compound $(R)$ reacts with $KOH$ in the presence of potassium nitrate to give compound $(Q)$. On heating compound $(P)$ with cone. $H_2SO_4$ and $NaCl$, chlorine gas is liberated and a compound $(S)$ of manganese along with other products is formed. Compounds $P$ to $S$ are

	P	Q	R	S
(a)	$KMnO_4$	$K_2MnO_4$	$MnCl_2$	$MnO_2$
(b)	$K_2MnO_4$	$MnO_2$	$KMnO_4$	$MnCl_2$
(c)	$KMnO_4$	$K_2MnO_4$	$MnO_2$	$MnCl_2$
(d)	$K_2MnO_4$	$KMnO_4$	$MnO_2$	$MnCl_2$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (C)

$P=K M n O_{4}, Q=K_{2} M n O_{4}, R=M n O_{2}, S=M n C l_{2}$
$\underset{(P)}{2 KMnO _{4}} \xrightarrow{\Delta} \underset{(Q)}{K _{2} MnO _{4}}+ \underset{(R)}{MnO _{2}}+O_{2}$
$2 MnO _{2}+4 KOH + O _{2} \longrightarrow 2 K _{2} MnO _{4}+2 H _{2} O$
$MnO _{2}+4 NaCl +4 H _{2} SO _{4} \longrightarrow \underset{(S)}{MnCl _{2}}+2 NaHSO _{4}+2 H _{2} O + Cl _{2}$