Question

The oxidation states of P atom in $POC{l}_3,H_{2}P{O}_3$ and $H_4{P}_2{O}_6$, respectively are:

• A. $+5,+4,+4$
• B. $+5,+5,+4$
• C. $+4,+4,+5$
• D. $+3,+4,+5$

Answer 1

The correct option is A $+5,+4,+4$

Oxygen has an oxidation state of $-2$ and hydrogen has $+1$.

$POC{l}_3=x+(-2)+3(-1)=0,x=+5$

$H_{2}P{O}_3=2(+1)+x+3(-2)=0,x=+4$

$H_4{P}_2{O}_6=4(+1)+2\left(x\right)+6(-2)=0,x=+4$

Hence, option A is correct.