Question

Consider the following reaction: $P_4+3NaOH+3H_{2}O\to 3Na{H}_{2}P{O}_2+P{H}_3$. In the reaction:

$1$. Phosphorus is oxidised

$2$. Phosphorus is reduced

Which of the above is/are correct?

• A. $1$ only
• B. $2$ only
• C. Both $1$ and $2$
• D. Neither $1$ nor $2$

Answer 1

The correct option is C Both $1$ and $2$

Initial oxidation state of $P$ in $P_4=0$

(because it is in the elemental state)

Final oxidation state of $P$ in $Na{H}_{2}P{O}_2$ be $x$,

Applying formula, $\text{Sum of total oxidation state of all atoms}=\text{Overall charge on the compound}.$

$1+2\times \left(1\right)+x+2\times (-2)=0x=1$

Let oxidation state of $P$ in $P_4$ be $x$,

Applying above formula,

$x+3\times \left(1\right)=0x=-3$

$P$ is oxidized from ($0$ to $1$) and reduced from ($0$ to $-3$).