Question

Equivalent weight of $H_{3}P{O}_2$ in the given dispropotionation reaction is:$2H_{3}P{O}_2\to H_{3}P{O}_4+P{H}_3$

• A. one fourth of its molecular weight
• B. half of its molecular weight
• C. equal to its molecular weight
• D. one fifth of its molecular weight

Answer 1

The correct option is B half of its molecular weight

Option B is the correct option.

Explanation:

The reaction is given as $2H_{3}P{O}_2\to H_{3}P{O}_4+P{H}_3$

Equivalent weight is given by the formula $EqWt=\frac{M}{nfactor}$

Let the molar mass of $$H_{3}P{O}_2$$ be $M$ .

For the first half of reaction:

$H_{3}P{O}_2\to H_{3}P{O}_4$

The change in oxidation state $5-1=4$

Therefore $E{q}_1=\frac{M}{4}$

For the second half of the reaction:

$$H_{3}P{O}_2\to HP{H}_3$$

The change in oxidation no. is $-3-1=4$

Therefore $E{q}_2=\frac{M}{4}$

Adding both the equivalent masses we get:

$Eq=\frac{M}{4}+\frac{M}{4}=\frac{M}{2}$

Which is half of the molar mass of $H_{3}P{O}_2$

Hence option B is correct.