The oxidation number of N in NH_4Cl is | vedclass

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(D) The compound $NH_4Cl$ dissociates as: $NH_4Cl \rightarrow NH_4^+ + Cl^-$In the ammonium ion $(NH_4^+)$,let the oxidation number of $N$ be $x$.The

Vedclass · Accepted Answer

$-3$

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(D) The compound $NH_4Cl$ dissociates as: $NH_4Cl ightarrow NH_4^+ + Cl^-$In the ammonium ion $(NH_4^+)$,let the oxidation number of $N$ be $x$.The sum of oxidation numbers of all atoms in the ion is equal to its charge:$x + 4 	imes (+1) = +1$$x + 4 = +1$$x = 1 - 4 = -3$Therefore,the oxidation number of $N$ in $NH_4Cl$ is $-3$.

The oxidation number of $N$ in $NH_4Cl$ is

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