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  4. Oxidation number of P in P4,PH3 &NaH2PO2 respectively are :-

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

NTA AbhyasNTA Abhyas 2020

Solution:

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