Question

The given reaction is a disproportional reaction. If true enter 1, else enter 0.
\[{{P}_{4}}+3NaOH+3{{H}_{2}}O\xrightarrow{\Delta }P{{H}_{3}}+3Na{{H}_{2}}P{{O}_{2}}\]

Answer 1

Hint: the disproportional reaction is defined as in which we observe that one reactant is being oxidised and the same reactant gets reduced in the same reaction. The other name for disproportional reaction is dismutation reaction. The reverse of a disproportional reaction is called a comproportionation reaction.

Complete answer:
- In this disproportional reaction we see both reduction and oxidation reactions of the same reactant but it is slightly different from the redox reaction.
- In the given equation the reactant P that is phosphorus has its oxidation state equal to 0 while the same element P is present with oxidation state of -3 in the product \[P{{H}_{3}}\]. The same element in the product \[Na{{H}_{2}}P{{O}_{2}}\] is present with the oxidation state of +1.
- So we see that the reactant \[{{P}_{4}}\] is undergoing the process of reduction and oxidation in the same reaction. So the \[Na{{H}_{2}}P{{O}_{2}}\] is the oxidised product as in this the oxidation state of P increases from ) to +1. While the product \[P{{H}_{3}}\] is the reduced product as in this the oxidation state of P decreases from 0 to -3.

So the given reaction is a disproportional reaction so we will enter 1.

Note: Each disproportional reaction is the redox reaction but not every redox reaction is the disproportional reaction. The compound undergoes the process of disproportional reaction to gain their stability which occurs through oxidation or reduction. Always remember that the gain of electrons is reduction while the loss of electrons is oxidation.