Question

What product is formed on the disproportionation of \[{H_3}P{O_3}\]?

Answer 1

Hint: In the disproportionate reaction two products are formed. The reactant undergoes oxidation and reduction simultaneously to form the product. One of the products has a higher oxidation state and the other has a lower oxidation state, whereas the reactant has an intermediate oxidation state.

Complete answer:
Disproportionation reaction is a type of redox reaction that simultaneous oxidation and reduction of atoms of same element from one oxidation state (OS) to two different oxidation states. The two compounds formed as products are one with the higher oxidation state and another one with the lower oxidation state.
\[{H_3}P{O_3}\] also undergoes a disproportionate reaction on heating, to give two products with different oxidation states. The reaction of disproportionation of \[{H_3}P{O_3}\]is:
\[4{H_3}P{O_3}\;\xrightarrow{\Delta }{\text{ }}3{H_3}P{O_4}\; + {\text{ }}P{H_3}\]
In this reaction, when \[{H_3}P{O_3}\] is heated it undergoes disproportionation and the products formed are ortho phosphoric acid $\left( {{H_3}P{O_4}} \right)$ and phosphine $\left( {P{H_3}} \right)$.
The oxidation state of phosphorus in the reactant is $ + 3$ whereas on the reactant side the oxidation state of phosphorus in phosphoric acid is $ + 5$ and in phosphine, it is $ - 3$. Hence we can see that in \[{H_3}P{O_3}\] phosphorus is undergoing oxidation and reduction simultaneously.
The disproportionate reaction is also called the dismutation reaction. The reverse of this reaction is called comproportionation or disproportionation. In this reaction, two reactants one of higher oxidation state and the other of lower oxidation state combine to form the product with an intermediate oxidation state.

Note:
Oxidation means that the electrons were lost during the reaction by the molecules. Here the oxidation state of the molecule increases. Reduction means that the electrons were gained during the reaction by the molecule. In this, the oxidation state of the molecule decreases.