Question

Equivalent mass of ${{\text{H}}_3}{\text{P}}{{\text{O}}_2}$ when it undergoes disproportionation to ${\text{P}}{{\text{H}}_3}$ and ${{\text{H}}_3}{\text{P}}{{\text{O}}_3}$ will be:
A. $\dfrac{{{\text{M}}.{\text{w}}}}{2}$
B. $\dfrac{{{\text{M}}.{\text{w}}}}{4}$
C. $\dfrac{{{\text{M}}.{\text{w}}}}{{24}}$
D. $\dfrac{{{\text{3M}}.{\text{w}}}}{4}$

Answer 1

Hint:
Disproportionation reaction is a type of redox reaction. In this reaction, a single substance can undergo both oxidation and reduction. Loss of electrons is termed as oxidation. Gain of electrons is termed as reduction.

Complete step by step answer:
Redox reactions are the chemical reactions in which the atoms change their oxidation state, i.e., there is a transfer of electrons occurring between the species. When an atom loses its electrons or its oxidation state gets increased, it is termed as oxidation. When an atom gains electrons or its oxidation state gets reduced, it is termed as reduction.
Disproportionation reactions are the reaction in which a single atom simultaneously gets oxidized and reduced. Disproportionation reaction of ${{\text{H}}_3}{\text{P}}{{\text{O}}_2}$ is given below:
\[{\text{3}}{{\text{H}}_3}{\text{P}}{{\text{O}}_2} \to 2{{\text{H}}_3}{\text{P}}{{\text{O}}_3} + {\text{P}}{{\text{H}}_3}\]
${{\text{H}}_3}{\text{P}}{{\text{O}}_3}$ is phosphorus acid, ${{\text{H}}_3}{\text{P}}{{\text{O}}_2}$ is hypophosphorous acid and \[{\text{P}}{{\text{H}}_3}\] is phosphine.
We can explain this reaction in terms of giving oxygen. Oxidation means gaining oxygen, and reduction means losing oxygen.
Here ${{\text{H}}_3}{\text{P}}{{\text{O}}_2}$ has undergone both oxidation and reduction.
Oxidation: \[{{\text{H}}_3}{\text{P}}{{\text{O}}_2} \to {{\text{H}}_3}{\text{P}}{{\text{O}}_3}\]
Here, the oxidation number changes from $ + 1$ to $ + 3$, thereby $2{{\text{e}}^ - }$ is given.
Reduction: \[{{\text{H}}_3}{\text{P}}{{\text{O}}_2} \to {\text{P}}{{\text{H}}_3}\]
Here, the oxidation number changes from $ + 1$ to $ - 3$, thereby $4{{\text{e}}^ - }$ is used.
Thus ${{\text{H}}_3}{\text{P}}{{\text{O}}_2}$ gets oxidized and reduced.
Equivalent weight can be calculated by dividing molecular weight by the n-factor. n-factor can be calculated by dividing the number of electrons exchanged by the number of moles used.
Total number of electrons exchanged is $4$ and the number of moles used is $3$. Thus n-factor $ = \dfrac{4}{3}$
Equivalent weight $ = \dfrac{{{\text{M}}.{\text{w}}}}{{\dfrac{4}{3}}} = {\text{M}}.{\text{w}} \times \dfrac{3}{4} = \dfrac{{3{\text{M}}.{\text{w}}}}{4}$
Hence option D is correct.

Additional information:
Combination reaction is a reaction in which two elements are combined to form a compound. Displacement reaction is a reaction in which one element is reduced and replaces another element. In decomposition reaction, the molecule decomposes into its constituents.


Note:
Disproportionation reaction is a special case of decomposition reaction. But in decomposition reaction, different elements are being oxidized and reduced. But in a disproportionation reaction, a single element is being oxidized and reduced, i.e. there will be exchange of electrons in the reaction. Equivalent depends upon molecular weight and n-factor.