Ncert Exercises & Solutions

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. $H_{2} S O_{4}$ and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.

Thinking Process

This problem is based on preparation and properties of $K M n O_{4}, K_{2} M n O_{4} a n d M n O_{2}$

NEETprep Answer: Since, compound (C) on treating with conc.

H_{2} S O_{4} a n d N a C l

gives

C l_{2}

gas, so it is manganese dioxide

(M n O_{2})

. It is obtained along with

M n O_{4}^{2 -}

when

K M n O_{4}

(violet) is heated.

T h u s, (A) = K M n O_{4} (B) = K_{2} M n O_{4}

(C) = M n O_{2} (D) = M n C l_{2}

K M n O_{4} \overset{∆}{\to} K_{2} M n O_{4} + M n O_{2} + O_{2}

[A] [B] [C]

2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n O_{4} + 2 H_{2} O

M n O_{2} + 4 N a C L + 4 H_{2} S O_{4} \to M n C L_{2} + 4 N a H S O_{4} + 2 H_{2} O + C L_{2}

[D]

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