A: White $P$ reacts with caustic soda, the products are ${P H}_{3}$ and ${N a H}_{2} {P O}_{2}$ . This reaction is an example of a disproportionation reaction.
R: In disproportionation reaction one and the same substance may act simultaneously as an oxidizing agent and as a reduced agent.

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Both A and R are true and R is the correct explanation

Both A and R are true but R is not the correct explanation of A

A is true but R is false

A is false but R is true

answer is A.

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Detailed Solution

$P_{4} + 3 N a O H + 3 H_{2} O \to {P H}_{3} + 3 {N a H}_{2} {P O}_{2}$

Let’s check the oxidation number of P in the reaction,

$P_{4}^{0} + 3 {N a}^{+ 1} O H + 3 H_{2} O \to P^{- 3} H_{3} + 3 {N a H}_{2} P^{+ 1} O_{2}$

Phosphorous, $P_{4}$ gets reduced to $P^{- 3}$ in ${P H}_{3}$ and $P_{4}$ also oxidized to $P^{+ 1}$ in ${N a H}_{2} {P O}_{2}$ . The reaction in which a compound/substance undergoes both oxidation and reduction is known as disproportionation reaction. Hence, both the statements are true and $R$ is the correct explanation of $A$ .