Question

The oxidation state of the central atom in the anion of the compound $Na{H}_{2}P{O}_2$ will be:

• A. $-3$
• B. $+1$
• C. $+3$
• D. $+5$

Answer 1

The correct option is B $+1$

In $Na{H}_{2}P{O}_2$, the cation is $N{a}^{+}$, and the anion is $(H_{2}P{O}_2{)}^{-}$
Calculating the oxidation state of the central atom:
In $(H_{2}P{O}_2{)}^{-}P$, is the central atom.
Let the oxidation state of $P$ be ${}^{\prime }{x}^{\prime }.$

The sum of oxidation numbers of all the atoms is equal to the charge on that species
$\therefore 2\text{oxidation number of H atom}+x+2\text{oxidation number of O atom}=-1$

$=2(+1)+x+2(-2)=-1$

$=2+x-4=-1$

$=x-2=-1$

$=x=+1$

Thus, the oxidation state of the central atom in the anion of the compound $Na{H}_{2}P{O}_2$ is $+1$.

So, option (C) is the correct answer.