Find the oxidation number of P in $N a H_{2} P O_{4}$ .

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Solution

Let X be the oxidation number of P in $N a H_{2} P O_{4}$ . The oxidation numbers of Na, H and O are $+ 1$ , $+ 1$ and $- 2$ respectively.
$+ 1 + 2 (+ 1) + X + 4 (- 2) = 0$
$3 + X + - 8 = 0$
$X + - 5 = 0$
$X = + 5$
The oxidation number of P in $N a H_{2} P O_{4}$ is $+ 5$ .

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