Question

When P reacts with caustic soda, the products are $P{{H}_{3}}$ and $Na{{H}_{2}}P{{O}_{2}}$. The reaction is an example of:
A. oxidation
B. reduction
C. both oxidation and reduction
D. neutralisation

Answer 1

In order to find the reaction type, we will first find the oxidation number of phosphorus in all the reactants and products containing it. As we know that oxidation is a reaction in which removal of an electron from a substance, reduction is a reaction in which addition of an electron to a substance takes place.

Complete Solution :
- We can write the reaction when Phosphorus reacts with caustic soda as:
${{P}_{4}}+3NaOH+3{{H}_{2}}O\to P{{H}_{3}}+Na{{H}_{2}}P{{O}_{2}}$

- Firstly, let’s determine the oxidation numbers:
As we know that the oxidation number of Phosphorus in white Phosphorus $\left( {{P}_{4}} \right)$ is 0.
Now, let us consider the oxidation number of Phosphorus in phosphine $\left( P{{H}_{3}} \right)$ be x.
As there are 3 hydrogens and the oxidation number of H is +1. So,
$\begin{aligned} & x+1(3)=0 \\\ & \Rightarrow x=-3 \\\ \end{aligned}$

- Now, let y be the oxidation number of Phosphorus in sodium hypophosphite($Na{{H}_{2}}P{{O}_{2}}$).
There is one Na and oxidation number of Na is +1.
There are two H and oxidation numbers of H is +1.
There are two O and the oxidation number of O is -2.
So, we can calculate y as:
$\begin{aligned} & +1+1(2)+y-2(2)=0 \\\ & \Rightarrow y=+1 \\\ \end{aligned}$

- As we know that when oxidation number increases, the element is said to be oxidised, and when oxidation number decreases, the element is said to be reduced.
- We can see that phosphorus is getting both oxidised and reduced, therefore this reaction is a disproportionation reaction. Basically, it is the reaction in which the same species get oxidised and reduced.
So, the correct answer is “Option C”.

Note: - We can solve problems by finding the oxidation number of the species in different reactants and products. We should know the concept to find the oxidation number and must have the knowledge about different reactions.