A violet compound of manganese (F) decomposes on heating to liberate oxygen and compounds (Q) and (R) of manganese is formed. Compound (R) reacts with KOH in the presence of potassium nitrate to give compound (Q). On heating compound (R) with conc. $H_{2} S O_{4}$ and NaCl, chlorine gas is liberated and a compound (S) of manganese along with other products is formed. Compounds P to S are:

P - K M n O_{4}, Q - K_{2} M n O_{4}, R - M n C l_{2}, S - M n O_{2}

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P - K_{2} M n O_{4}, Q - M n O_{2}, R - K M n O_{4}, S - M n C l_{2}

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P - K M n O_{4}, Q - K_{2} M n O_{4}, R - M n O_{2}, S - M n C l_{2}

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P - K_{2} M n O_{4}, Q - K M n O_{4}, R - M n O_{2}, S - M n C l_{2}

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Solution

The correct option is D $P - K M n O_{4}, Q - K_{2} M n O_{4}, R - M n O_{2}, S - M n C l_{2}$
$P v i o l e t △ - \to O_{2} + Q + R$
$R K O H - --- \to K N O_{3} Q$
conc. $H_{2} S O_{4}$ & $N a C l$ $↓$ $△$
$S + C l_{2}$
$P = K M n O_{4}$ (violet)
$Q = K_{2} M n O_{4}$
$R = M n O_{2}$
$S = M n C l_{2}$
So the reaction goes this way,
$K M n O_{4} - \to △ K_{2} M n O_{4} + M n O_{2} + O_{2}$
P Q R
(Green) $↓ K O H / K N O_{3}$
$K_{2} M n O_{4}$
Q
$M n O_{2} c o n c . H_{2} S O_{4}, N a C l - ---------- \to △ M n C l_{2} + C l_{2}$

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Similar questions

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When an oxide of manganese (P) is fused with KOH in the presence of an oxidizing agent and dissolved in water, it gives a dark green solution of compound (Q). Compound (Q) disproportionates in neutral or acidic solution to give purple compound (R). An alkaline solution of compound (R) oxidizes potassium iodide solution to a compound (S) and compound (P) is also formed. Compounds P to S are:

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The compound R is:

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M n O_{2}

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