JEE Main 2026 — Chemistry Physical Chemistry

First, we calculate the millimoles of the reactants:

Millimoles of $\text{MnO}_4^-$MnO4−​ = $500 \text{ mL} \times 0.2 \text{ M} = 100 \text{ mmol}$500 mL×0.2 M=100 mmol

Millimoles of $\text{I}^-$I− = $500 \text{ mL} \times 1.5 \text{ M} = 750 \text{ mmol}$500 mL×1.5 M=750 mmol

In a basic medium, $\text{MnO}_4^-$MnO4−​ is reduced to $\text{MnO}_2$MnO2​. The change in oxidation state of Mn is from $+7$+7 to $+4$+4, so its n-factor is $3$3.

The problem explicitly states that iodide ions are oxidized to molecular iodine ($\text{I}_2$I2​). The change in oxidation state for iodine is from $-1$−1 to $0$0, so the n-factor for $\text{I}^-$I− is $1$1.

Equivalents of $\text{MnO}_4^-$MnO4−​ = $100 \times 3 = 300 \text{ meq}$100×3=300 meq

Equivalents of $\text{I}^-$I− = $750 \times 1 = 750 \text{ meq}$750×1=750 meq

Since $\text{MnO}_4^-$MnO4−​ is the limiting reagent, the equivalents of $\text{I}_2$I2​ produced will be equal to the equivalents of $\text{MnO}_4^-$MnO4−​ consumed.

Equivalents of $\text{I}_2$I2​ formed = $300 \text{ meq}$300 meq

The liberated iodine is titrated with thiosulphate ($\text{S}_2\text{O}_3^{2-}$S2​O32−​):

$$\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$$I2​+2S2​O32−​→2I−+S4​O62−​

For thiosulphate, the n-factor is $1$1 (as $2$2 moles of $\text{S}_2\text{O}_3^{2-}$S2​O32−​ lose $2$2 moles of electrons).

From the law of equivalence:

Equivalents of $\text{S}_2\text{O}_3^{2-}$S2​O32−​ = Equivalents of $\text{I}_2$I2​

$$\text{Molarity} \times \text{Volume (in mL)} \times \text{n-factor} = 300$$Molarity×Volume (in mL)×n-factor=300

$$x \times 300 \times 1 = 300$$x×300×1=300

$$x = 1$$x=1

Answer: $1$1