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Equivalent weight of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ when it disproportionate into ${\text{P}}{{\text{H}}_{\text{3}}}$ and ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$ (mol. wt. = M) is:

A. M
B. $\dfrac{{\text{M}}}{{\text{2}}}$
C. $\dfrac{{\text{M}}}{4}$
D. $\dfrac{{{\text{3M}}}}{4}$

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Last updated date: 17th Jun 2024
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Answer
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Hint: As we know in a chemical reaction, equivalence of a substance is explained as the amount of which combines with 1 mole of hydrogen atoms or replaces the same number of hydrogen atoms. Thus equivalent weight in grams is weight in grams of 1 equivalent.

Complete step by step answer:
In this question given that ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ disproportionate and form ${\text{P}}{{\text{H}}_{\text{3}}}$ . We can write this equation is as follows:
     $
  {{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}} \to {\text{P}}{{\text{H}}_{\text{3}}} \\
  {{\text{P}}^ + } + 4{e^ - } \to {{\text{P}}^{3 - }} \\
 $
Thus equivalent weight of $\left( {{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}} \right)$ in ${\text{P}}{{\text{H}}_{\text{3}}}$ is,
$
  {\text{ = }}\dfrac{{{\text{Molecular}}\,{\text{weight}}}}{{{\text{Valance}}\,{\text{factor}}}} \\
  {\text{ = }}\dfrac{{\text{M}}}{{\text{4}}} \\
 $
(Here valance factor is 4)
Again, ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ disproportionate and form ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}$ . We can write this equation is as follows:
$
  {{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}} \to {{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3} \\
  {{\text{P}}^ + } - 2{e^ - } \to {{\text{P}}^{3 + }} \\
 $
Thus equivalent weight of \[\left( {{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}} \right)\] in ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}$ is,
$
  {\text{ = }}\dfrac{{{\text{Molecular}}\,{\text{weight}}}}{{{\text{Valance}}\,{\text{factor}}}} \\
  {\text{ = }}\dfrac{{\text{M}}}{2} \\
 $
(Here valance factor is 2)
Now we add equivalence weight of ${\text{P}}{{\text{H}}_{\text{3}}}$ and ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}$ .

$\dfrac{{\text{M}}}{{\text{4}}}{\text{ + }}\dfrac{{\text{M}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{3M}}}}{{\text{4}}}$

Thus option D is correct.

Additional Information: We know that the law of equivalence is one equivalence of an element combined with one equivalent of others. Equivalent weight of an acid in an acid base neutralization reaction is the portion of weight of 1 mole of the acid that can furnish 1 mole of hydrogen ion and equivalent weight of a base is part of weight of one mole of base that can furnish 1 mole of hydroxide ion or accept 1 mole of hydrogen ion.

Note:
As we know a redox reaction that reactant is transformed into product by simultaneous reduction reaction as well as oxidation reaction is termed as disproportionation reaction. For a reaction, say hydrogen peroxide transformed into water and oxygen. In this reaction hydrogen peroxide oxidizes and forms oxygen and it reduces and forms water.