What is the oxidation number of underlined element?

$[N_{2} - - - H_{5})_{2} S O_{4}$

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Solution

Let $x$ be the oxidation number of $N$ in $[N_{2} - - - H_{5})_{2} S O_{4}$ .
Since the overall charge on the compound is $0$ , the sum of oxidation states of all elements in it should be equal to $0$ .
Therefore, $2 (2 x + 5 (+ 1)) + (- 2) = 0$
Hence, $x = - 2$
Hence, the oxidation number of $N$ in $[N_{2} - - - H_{5})_{2} S O_{4}$ is $- 2$ .

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