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(B) In the reaction $Na_2S + H_2SO_4 \longrightarrow S + SO_2 + Na_2SO_4$,we analyze the oxidation states:$S$ in $Na_2S$ is $-2$.$S$ in $H_2SO_4$ is $

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$B$. for comproportionation reaction.

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(B) In the reaction $Na_2S + H_2SO_4 \longrightarrow S + SO_2 + Na_2SO_4$,we analyze the oxidation states:$S$ in $Na_2S$ is $-2$.$S$ in $H_2SO_4$ is $+6$.In the products,$S$ in $S\downarrow$ is $0$ and $S$ in $SO_2$ is $+4$.Here,the sulfur atom with oxidation state $-2$ is oxidized to $0$,and the sulfur atom with oxidation state $+6$ is reduced to $+4$.Since two different oxidation states of the same element ($S^{-2}$ and $S^{+6}$) combine to form intermediate oxidation states ($S^0$ and $S^{+4}$),this is a comproportionation reaction.Therefore,the correct assignment is $B$.

Assign $A, B, C, D$ from the given type of reaction.
$Na_2S + H_2SO_4$ $(Conc.)$ $\longrightarrow S\downarrow + SO_2 + Na_2SO_4$

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Assign $A, B, C, D$ from the given type of reaction.$Na_2S + H_2SO_4$ $(Conc.)$ $\longrightarrow S\downarrow + SO_2 + Na_2SO_4$