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(D) In the reaction $NaH + H_2O \longrightarrow NaOH + H_2$,the oxidation states are:$Na$ changes from $+1$ to $+1$ (no change).$H$ in $NaH$ is $-1$ a

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Both $(B)$ and $(C)$.

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(D) In the reaction $NaH + H_2O \longrightarrow NaOH + H_2$,the oxidation states are:$Na$ changes from $+1$ to $+1$ (no change).$H$ in $NaH$ is $-1$ and in $H_2O$ is $+1$.In the product $NaOH$,$H$ is $+1$ and in $H_2$,$H$ is $0$.Here,$H$ in $NaH$ $(-1)$ is oxidized to $H_2$ $(0)$ and $H$ in $H_2O$ $(+1)$ is reduced to $H_2$ $(0)$.Since two different species (or atoms in different oxidation states) react to form a product with a common oxidation state,this is a comproportionation reaction.Additionally,it can be classified as a displacement reaction where $H^-$ displaces $H^+$ from water.Therefore,both $(B)$ and $(C)$ are correct.

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$NaH_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(aq)} + H_{2(g)} \uparrow$

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Assign $A$,$B$,$C$,$D$ from the given type of reaction.$NaH_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(aq)} + H_{2(g)} \uparrow$