JEE Main 2026 — Apr 4, Shift 1 — Chemistry solution

JEE Main2026Apr 4, Shift 1Chemistry

In order to oxidise a mixture of $1$ mole each of $Fe C_{2} O_{4}$ , $F e_{2} (C_{2} O_{4})_{3}$ , $FeS O_{4}$ and $F e_{2} (S O_{4})_{3}$ in acidic medium, the number of moles of $KMn O_{4}$ required is :

A
$3$
B
$2$
C
$5$
D
$7$

Solution

$KMn O_{4}$ in acidic medium acts with $n$ -factor $5$ (Mn $^{+ 7} \to$ Mn $^{+ 2}$ ). Count the $n$ -factors of each reducing species:

$∙ Fe C_{2} O_{4}$ : Fe $^{+ 2} \to$ Fe $^{+ 3}$ ( $1$ electron) plus $C_{2} O_{4}^{2 -} \to 2 C O_{2}$ ( $2$ electrons). Total $n = 3$ per mole.

$∙ F e_{2} (C_{2} O_{4})_{3}$ : Fe is already +3 (no Fe contribution), three $C_{2} O_{4}^{2 -}$ ions give $3 \times 2 = 6$ electrons.

$∙ FeS O_{4}$ : Fe $^{+ 2} \to$ Fe $^{+ 3}$ , $n = 1$ .

$∙ F e_{2} (S O_{4})_{3}$ : Fe is +3, no oxidation possible. $n = 0$ .

Total electrons released by $1$ mole of each $= 3 + 6 + 1 + 0 = 10$ .

Moles of $KMn O_{4}$ required $= \frac{10}{5} = 2$ .

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