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(D) In the reaction $H_2O_2 \xrightarrow{R.T} H_2O + \frac{1}{2} O_2$,the oxidation state of oxygen in $H_2O_2$ is $-1$.In the products,the oxidation

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Both $(A)$ and $(C)$

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(D) In the reaction $H_2O_2 \xrightarrow{R.T} H_2O + \frac{1}{2} O_2$,the oxidation state of oxygen in $H_2O_2$ is $-1$.In the products,the oxidation state of oxygen in $H_2O$ is $-2$ and in $O_2$ is $0$.Since the oxidation state of oxygen decreases from $-1$ to $-2$ (reduction) and increases from $-1$ to $0$ (oxidation) in the same reaction,it is a disproportionation reaction.Also,since $H_2O_2$ decomposes into simpler substances upon heating or at room temperature,it is a thermal decomposition redox reaction.Therefore,both $(A)$ and $(C)$ are correct.

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$H_2O_2 \xrightarrow{R.T} H_2O + \frac{1}{2} O_2$

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Assign $A$,$B$,$C$,$D$ from the given type of reaction.$H_2O_2 \xrightarrow{R.T} H_2O + \frac{1}{2} O_2$