Question

Match incorrect equivalent mass of reactant in following column ? (M = molar mass of reactant):-

Column-I	Column-II
(1) $FeS_2 \rightarrow Fe^{3+} + SO_3$	$\frac{M}{15}$
(2) $P_2H_4 \rightarrow P_4H_2 + PH_3$	$\frac{5M}{6}$
(3) $2Mn^{+7} \rightarrow Mn^{+2} + Mn^{+6}$	$\frac{M}{3}$
(4) $Ba(MnO_4)_2 \xrightarrow{H^+} Mn^{2+}$	$\frac{M}{5}$

• A. (1) and (2)
• B. (2) and (3)
• C. (1) and (4)
• D. (3) and (4)

Answer 1

Answer: (B)

(2) and (3)

Answer 2

The equivalent mass of a reactant in a redox reaction is calculated using the formula: $\text{Equivalent Mass} = \frac{\text{Molar Mass}}{\text{n-factor}}$ The n-factor represents the total change in oxidation state per molecule or ion. We need to identify the incorrect matches.

(1) $FeS_2 \rightarrow Fe^{3+} + SO_3$ In $FeS_2$, the oxidation state of Fe is +2 and each S is -1. In the products, Fe is +3 and in $SO_3$, S is +6. Change in oxidation state for Fe: $3 - 2 = +1$. Change in oxidation state for each S atom: $6 - (-1) = +7$. Since there are two S atoms in $FeS_2$, the total change for sulfur is $2 \times 7 = 14$. The total n-factor for $FeS_2$ is the sum of changes for all atoms: $1 + 14 = 15$. Equivalent Mass = $\frac{M}{15}$. This match is correct.

(2) $P_2H_4 \rightarrow P_4H_2 + PH_3$ In $P_2H_4$, the oxidation state of P is -2. In $P_4H_2$, the oxidation state of P is -1/2. In $PH_3$, the oxidation state of P is -3. This is a disproportionation reaction. Balancing the reaction gives: $5 P_2H_4 \rightarrow P_4H_2 + 6 PH_3$. The total change in oxidation state for 5 moles of $P_2H_4$ (containing 10 moles of P atoms) is: Oxidation: $4 \times (-1/2 - (-2)) = 4 \times (3/2) = 6$. Reduction: $6 \times (-3 - (-2)) = 6 \times (-1) = -6$. The total n-factor for 5 moles of $P_2H_4$ is $|+6| + |-6| = 12$. The n-factor per mole of $P_2H_4$ is $12/5$. Equivalent Mass = $\frac{M}{12/5} = \frac{5M}{12}$. The given match is $\frac{5M}{6}$. This match is incorrect.

(3) $2Mn^{+7} \rightarrow Mn^{+2} + Mn^{+6}$ This reaction as written is problematic because $Mn^{+7}$ cannot disproportionate. However, if we interpret it as a process where two $Mn^{+7}$ ions are involved, one is reduced to $Mn^{+2}$ and another to $Mn^{+6}$, then: Change for one $Mn^{+7}$ to $Mn^{+2}$: $7 - 2 = 5$. Change for one $Mn^{+7}$ to $Mn^{+6}$: $7 - 6 = 1$. The total change for $2Mn^{+7}$ is $5 + 1 = 6$. If M is the molar mass of the species containing $Mn^{+7}$, the n-factor is 6. Equivalent Mass = $\frac{M}{6}$. The given match is $\frac{M}{3}$. This match is incorrect.

(4) $Ba(MnO_4)_2 \xrightarrow{H^+} Mn^{2+}$ In $Ba(MnO_4)_2$, the oxidation state of Mn is +7. In the product, Mn is +2. The change in oxidation state for Mn is $7 - 2 = 5$. The n-factor for $Ba(MnO_4)_2$ is 5. Equivalent Mass = $\frac{M}{5}$. This match is correct.

Therefore, the incorrect matches are (2) and (3).