Question

Find the equivalent weight of $P_4$ in following reaction:
$P_4\to {PH}_3+H_3{PO}_3$

• A. $40$
• B. $20.66$
• C. $60$
• D. $31$

Answer 1

Answer: (B)

$20.66$

$P_4⟶P{H}_3+H_{3}P{O}_3$

Oxidation state of $P$ in $P_4=O$

Oxidation state of $P$ in $P{H}_3$ & $H_{3}P{O}_3=+3$

$\therefore$ Change of oxidation state of $P=3$

$\therefore$ Change of oxidation state of $P_4=\frac{3\times 4}{2}=\frac{12}{2}=6$

$\therefore$ Equivalent weight of $P_4=\frac{Molecularwt.}{12/2}=\frac{31\times 4}{12/2}=20.66$