$N a H_{2} P - - O_{4}$
What will be the oxidation state of underlined element?

+ 5

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+ 3

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- 3

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- 5

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Solution

The correct option is A $+ 5$
Let, oxidation state of $P$ be $x$ in $N a H_{2} P - - O_{4}$ .
Oxidation number of $N a = + 1$
Oxidation number of $H = + 1$
Oxidation number of $O = - 2$
$+ 1 + (+ 1 \times 2) + x + (4 \times - 2) = 0$
$+ 1 + 2 + x - 8 = 0$
$x + 3 - 8 = 0$
$x = + 5$

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