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(B) In the given reaction: $NH_4Cl + NaNO_2 \xrightarrow{\Delta} N_2 + NaCl + 2H_2O$.The oxidation state of $N$ in $NH_4^+$ is $-3$ and in $NO_2^-$ is

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$B$. for comproportionation reaction.

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(B) In the given reaction: $NH_4Cl + NaNO_2 \xrightarrow{\Delta} N_2 + NaCl + 2H_2O$.The oxidation state of $N$ in $NH_4^+$ is $-3$ and in $NO_2^-$ is $+3$.In the product $N_2$,the oxidation state of $N$ is $0$.Since the oxidation state of nitrogen changes from $-3$ to $0$ (oxidation) and from $+3$ to $0$ (reduction),this is a comproportionation reaction.Therefore,the correct classification is $B$.

Assign $A, B, C, D$ from the given type of reaction.
$NH_4Cl + NaNO_2 \xrightarrow{\Delta} N_2 \uparrow + NaCl + 2H_2O$

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Assign $A, B, C, D$ from the given type of reaction.$NH_4Cl + NaNO_2 \xrightarrow{\Delta} N_2 \uparrow + NaCl + 2H_2O$