$P_{4} + 3 N a O H + 3 H_{2} O △ ⟶ P H_{3} + 3 N a H_{2} P O_{2}$
The given reaction is a disproportionation reaction.
If true enter 1, else enter 0.

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Solution

In this reaction, $P_{4}$ (oxidation no. = 0) has been reduced to $P H_{3}$ (ox. no. = -3) as well as oxidised to $H_{2} P O_{2}^{-}$ (ox. no. = +1).
Hence, this is a disproportionation reaction.
Therefore, the correct answer is $1$ .

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