Question

The oxidation number of phosphorus in $Ba(H_{2}P{O}_2{)}_2$ is:

• A. $+3$
• B. $+2$
• C. $+1$
• D. $-1$

Answer 1

The correct option is C $+1$

$Ba{{(H}_2{PO}_2)}_2$ can be written as ${Ba}^{+2}{{(H}_2{PO}_2)}^{-1}$

Let the oxidation number of $P$ in ${{(H}_2{PO}_2)}^{-1}$ be $x$. The oxidation states of $H$ and $O$ are $+1$ and $-2$ respectively. Since the atom as whole is single negatively charged, following equation holds true:

$2\times (+1)+1\times \left(x\right)+2\times (-2)=-1⟹2+x-4=-1⟹x=4-2-1⟹x=4-3⟹x=+1$

Hence, an answer is $C$.