Question

Complete the following reaction given below by choosing products:
$NaOH + {P_4} + {H_2}O \to$
A. $P{H_3} + Na{H_2}P{O_2}$
B. $P{H_3} + N{a_2}P{O_4}$
C. $P{H_3} + N{a_2}HP{O_2}$
D. ${H_3}P{O_4} + NaO$

Answer 1

As we can see presence of sodium hydroxide, so the reaction is happening in the basic medium. So the phosphorus will undergo a disproportionation reaction.
Now to complete the reaction we need knowledge of two more keywords. They are oxidation states or oxidation numbers and how it affects the redox reactions, oxidation and reduction reactions.
Finally we can conclude by writing the balanced chemical reaction.

Complete step by step answer:
Now, to start the completion of reaction we need to understand a few terms. Let us start with a disproportionation reaction, as phosphorus will undergo disproportionation in presence of basic medium due to presence of Sodium hydroxide.
Disproportionation reaction: the reaction in which the same substance undergoes simultaneously oxidation and reduction.
Now to see whether the substance is oxidised or reduced, we need the concept of oxidation states or oxidation number. When oxidation number increases then substance is oxidised, and when oxidation number is reduced then the substance is undergoing reduction.
But, as we already started our discussion stating, that the Phosphorus will undergo reduction and oxidation simultaneously. Thus option (D) is surely incorrect, as it does not have disproportionation products.
So, we can write chemical reactions completely, based on our knowledge of properties of Phosphorus.
$3NaOH + {P_4} + 3{H_2}O \to P{H_3} + 3Na{H_2}P{O_2}$
Above here is a balanced chemical reaction.

Thus the correct option is (A) $P{H_3} + Na{H_2}P{O_2}$ .

Note:
In pure molecules of Phosphorus, the oxidation state is Zero (0), and in P{H_3}\ & Na{H_2}P{O_2} the oxidation state of P is -3 and +1 respectively. Thus the one which is less than Zero (0) is negative and that is undergoing reduction, while the other product which has positive oxidation state is undergoing oxidation. Thus the same phosphorus is undergoing oxidation and reduction, so it is called disproportionation reaction.