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. $H_{2}S{O}_4$ 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)	$KMn{O}_4$	$K_{2}Mn{O}_4$	$MnC{l}_2$	$Mn{O}_2$
(b)	$K_{2}Mn{O}_4$	$Mn{O}_2$	$KMn{O}_4$	$MnC{l}_2$
(c)	$KMn{O}_4$	$K_{2}Mn{O}_4$	$Mn{O}_2$	$MnC{l}_2$
(d)	$K_{2}Mn{O}_4$	$KMn{O}_4$	$Mn{O}_2$	$MnC{l}_2$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (C)

$P=KMn{O}_4,Q=K_{2}Mn{O}_4,R=Mn{O}_2,S=MnC{l}_2$
$\underset{(P)}{2KMn{O}_4}\stackrel{\Delta }{\to }\underset{(Q)}{K_{2}Mn{O}_4}+\underset{(R)}{Mn{O}_2}+O_2$
$2Mn{O}_2+4KOH+O_2⟶2K_{2}Mn{O}_4+2H_{2}O$
$Mn{O}_2+4NaCl+4H_{2}S{O}_4⟶\underset{(S)}{MnC{l}_2}+2NaHS{O}_4+2H_{2}O+C{l}_2$