Which of the following examples does not represent disproportionation?

M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O

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2 H_{2} O_{2} \to 2 H_{2} O + O_{2}

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4 K C l O_{3} \to 3 K C l O_{4} + K C l

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3 C l_{2} + 6 N a O H \to 5 N a C l + N a C l O_{3} + 3 H_{2} O

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Solution

The correct option is A $M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O$
Disproportionation reaction is that reaction in which same substance undergoes oxidation as well as reduction to give two different products.
In A) reaction,oxidation state of chlorine changes from -1 in $H C l$ to 0 in $C l_{2}$ and -1 in $M n C l_{2}$ .Here chlorine undergoes only oxidation.So this is not disproportionation.
In B) reaction,oxidation state of oxygen changes from -1 in $H_{2} O_{2}$ to 0 in $O_{2}$ and -2 in $H_{2} O$ .Here oxygen undergoes only oxidation as well as reduction.So this is disproportionation reaction.
In C) reaction,oxidation state of chlorine changes from +5 in $K C l O_{3}$ to -1 in $K C l$ and +7 in $K C l O_{4}$ .Here chlorine undergoes only oxidation as well as reduction.So this is disproportionation reaction.
In D) reaction,oxidation state of chlorine changes from 0 in $C l_{2}$ to -1 in $N a C l$ and +5 in $N a C l O_{3}$ .Here chlorine undergoes only oxidation as well as reduction.So this is disproportionation reaction.

Suggest Corrections

Similar questions

4 H C l + M n O_{2} \to M n C l_{2} + 2 H_{2} O + C l_{2}

3 {C l}_{2} + 6 N a O H ⟶ 5 N a C l + N a C l O_{3} + 3 H_{2} O

M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}

{3 C l}_{2} + 6 N a O H ⟶ 5 N a C l + {N a C l O}_{3} + {3 H}_{2} O, {C l}_{2}

3 C l_{2} + 6 N a O H \to 5 N a C l + N a C l O_{3} + 3 H_{2} O

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