Oxidation number of N in (NH_4)_2SO_4 is | vedclass

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Question

(D) In $(NH_4)_2SO_4$,the compound dissociates into $2NH_4^+$ and $SO_4^{2-}$ ions.We need to find the oxidation number of $N$ in the ammonium ion,$NH

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(D) In $(NH_4)_2SO_4$,the compound dissociates into $2NH_4^+$ and $SO_4^{2-}$ ions.We need to find the oxidation number of $N$ in the ammonium ion,$NH_4^+$.Let the oxidation number of $N$ be $x$.The oxidation number of $H$ is $+1$.For the $NH_4^+$ ion,the sum of oxidation numbers equals the charge on the ion:$x + 4(+1) = +1$$x + 4 = +1$$x = 1 - 4 = -3$Therefore,the oxidation number of $N$ is $-3$.

Oxidation number of $N$ in $(NH_4)_2SO_4$ is

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