JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

In order to oxidise a mixture of $1$ mole each of $FeC_2O_4$ , $Fe_2(C_2O_4)_3$ , $FeSO_4$ and $Fe_2(SO_4)_3$ in acidic medium, the number of moles of $KMnO_4$ required is

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

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

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Solution

In acidic medium, $KMnO_4$ acts as an oxidising agent and gets reduced to $Mn^{2+}$ . The n-factor for $KMnO_4$ is $5$ .

The n-factors for the given reducing agents are calculated based on the total number of moles of electrons lost per mole of the compound:

For $FeC_2O_4$ :

$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$

$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^{-}$

Total electrons lost $= 3$ , so n-factor $= 3$ .

Equivalents $= 1 \times 3 = 3$ .

For $Fe_2(C_2O_4)_3$ :

$Fe^{3+}$ is not oxidised.

$3C_2O_4^{2-} \rightarrow 6CO_2 + 6e^{-}$

Total electrons lost $= 6$ , so n-factor $= 6$ .

Equivalents $= 1 \times 6 = 6$ .

For $FeSO_4$ :

$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$

$SO_4^{2-}$ is not oxidised.

Total electrons lost $= 1$ , so n-factor $= 1$ .

Equivalents $= 1 \times 1 = 1$ .

For $Fe_2(SO_4)_3$ :

Neither $Fe^{3+}$ nor $SO_4^{2-}$ can be oxidised further.

n-factor $= 0$ .

Equivalents $= 0$ .

Total equivalents of the reducing mixture $= 3 + 6 + 1 + 0 = 10$ .

Let the number of moles of $KMnO_4$ required be $x$ .

Equivalents of $KMnO_4 = x \times 5$ .

Equating the equivalents of the oxidising and reducing agents:

$5x = 10$

$x = 2$

Answer: $2$

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Solution

In acidic medium, $KMnO_4$ acts as an oxidising agent and gets reduced to $Mn^{2+}$ . The n-factor for $KMnO_4$ is $5$ .

The n-factors for the given reducing agents are calculated based on the total number of moles of electrons lost per mole of the compound:

For $FeC_2O_4$ :

$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$

$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^{-}$

Total electrons lost $= 3$ , so n-factor $= 3$ .

Equivalents $= 1 \times 3 = 3$ .

For $Fe_2(C_2O_4)_3$ :

$Fe^{3+}$ is not oxidised.

$3C_2O_4^{2-} \rightarrow 6CO_2 + 6e^{-}$

Total electrons lost $= 6$ , so n-factor $= 6$ .

Equivalents $= 1 \times 6 = 6$ .

For $FeSO_4$ :

$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$

$SO_4^{2-}$ is not oxidised.

Total electrons lost $= 1$ , so n-factor $= 1$ .

Equivalents $= 1 \times 1 = 1$ .

For $Fe_2(SO_4)_3$ :

Neither $Fe^{3+}$ nor $SO_4^{2-}$ can be oxidised further.

n-factor $= 0$ .

Equivalents $= 0$ .

Total equivalents of the reducing mixture $= 3 + 6 + 1 + 0 = 10$ .

Let the number of moles of $KMnO_4$ required be $x$ .

Equivalents of $KMnO_4 = x \times 5$ .

Equating the equivalents of the oxidising and reducing agents:

$5x = 10$

$x = 2$

Answer: $2$

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

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JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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