Question

Oxidation number of $P$ in $P_{4},PH_{3}\&NaH_{2}PO_{2}$ respectively are :-

• A. $0,-3,+1$
• B. $0,-3,-1$
• C. $0,+3,-1$
• D. $0,+3,+1$

Answer 1

Answer: (A)

$P_{4} \rightarrow O\cdot N=0$
$\left(PH\right)_{3} \rightarrow x+3\left(\right.+1\left.\right)=0$
$x=-3$
$\left(NaH\right)_{2}\left(PO\right)_{2} \rightarrow \left(\right.+1\left.\right)+2\left(\right.+1\left.\right)+x+2\left(\right.-2\left.\right)=0$
$x=+1$