Question

Calculate the number of peroxy linkage(s) in $H_{3}P{O}_5$.

• A. $1$
• B. $2$
• C. $0$
• D. $\text{None of the above}$

Answer 1

The correct option is A $1$

$H_{3}P{O}_5$, number of central atom $=1$
$\text{Number of peroxy linkage}=\frac{\text{Calculated oxidation state - Maximum oxidation state}}{2}$
Consider the oxidation state of $P$ as '$x$' in $H_{3}P{O}_5$
$\therefore x+5\times (-2)+3=0$
$\Rightarrow x=+7$
Thus calculated oxidation state of $P=+7$
Maximum oxidation of phosphorous $=$ Number of valence electron $=+5$

$\therefore \text{Number of peroxy linkage}=\frac{7-5}{2}$
$\Rightarrow \text{Number of peroxy linkage}=1$