Oxidation number of N in $(N H_{4})_{2} S O_{4}$ is

$\frac{- 1}{3}$

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Solution

The correct option is D
-3

$(N H_{4})_{2} S O_{4} = 2 N H_{4}^{+} + S O_{$}^{- -}$
$N H_{4}^{+}$
x+4=+1; x=1-4=-3

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