Assign oxidation number to the underlined element in $N a H S - - O_{4}$ .

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Solution

Let X be the oxidation state of S in $N a H S O_{4}$ .
We know that oxidation no. for $O = - 2, H = + 1, N a = + 1,$
$+ 1 + 1 + X + 4 (- 2) = 0$
$\Rightarrow X = + 6$
Hence, the oxidation state of S in $N a H S O_{4}$ is +6.

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