n-factor of H3PO2 during its diproportionation is 3H3PO2→PH3+2H3PO3

Solution:

$n - f a c t o r$ measures the change in electrons taking place per molecule. In the given reaction, oxidation state of P in each compound is:

$3 H_{3} \overset{+ 1}{P} O_{2} \to \overset{- 3}{P} H_{3} + 2 H_{3} \overset{+ 3}{P} O_{3}$

Half cell reaction:

$H_{3} {PO}_{2} \to H_{3} {PO}_{3} + 2 e^{-} (oxidation) 2 H_{3} {PO}_{2} \to 2 H_{3} {PO}_{3} + 4 e^{-} (reduction) 3 H_{3} {PO}_{2} + 4 e^{-} \to {PH}_{3} + 2 H_{3} {PO}_{3} + 4 e^{-} (overall reaction)$

$n - f a c t o r$ is calculated as:

$n - factor = \frac{number of electrons involved}{number of molecules involved} = \frac{4}{3} = 1.33$

n-factor of $H_{3} P O_{2}$ during its diproportionation is $3 H_{3} P O_{2} \to P H_{3} + 2 H_{3} P O_{3}$

Solution:

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