JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.) · numerical answer

JEE Main 2023 Chemistry question on Redox and Electrochemistry

When Fe0.93O\text{Fe}_{0.93}\text{O} is heated in presence of oxygen, it converts to Fe2O3\text{Fe}_2\text{O}_3. The number of correct statement/s from the following is __________

A. The equivalent weight of Fe0.93O\text{Fe}_{0.93}\text{O} is Molecular weight0.79\dfrac{\text{Molecular weight}}{0.79}

B. The number of moles of Fe2+\text{Fe}^{2+} and Fe3+\text{Fe}^{3+} in 1 mole of Fe0.93O\text{Fe}_{0.93}\text{O} is 0.79 and 0.14 respectively

C. Fe0.93O\text{Fe}_{0.93}\text{O} is metal deficient with lattice comprising of cubic closed packed arrangement of O2−\text{O}^{2-} ions

D. The % composition of Fe2+\text{Fe}^{2+} and Fe3+\text{Fe}^{3+} in Fe0.93O\text{Fe}_{0.93}\text{O} is 85% and 15% respectively
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Answer: 4

Solution

## Key Analysis of Fe₀.₉₃O

Finding Fe²⁺ and Fe³⁺ composition:

Let x = moles of Fe³⁺, then (0.93 - x) = moles of Fe²⁺ in 1 mole of Fe₀.₉₃O.

Charge balance: 2(0.93 - x) + 3x = 2(1) [since O²⁻ contributes 2⁻]

1.86 - 2x + 3x = 2 → x = 0.14

So Fe³⁺ = 0.14 mol, Fe²⁺ = 0.79 mol ✓ (Statement B is correct)

Checking Statement A (equivalent weight):

The n-factor of Fe₀.₉₃O when converted to Fe₂O₃: each Fe²⁺ loses 1 electron, so total electrons lost per formula unit = 0.79 × 1 = 0.79. Equivalent weight = Molecular weight/0.79 ✓ (Statement A is correct)

Checking Statement D (% composition):

% Fe²⁺ = (0.79/0.93) × 100 = 84.9% ≈ 85%, % Fe³⁺ = (0.14/0.93) × 100 = 15% ✓ (Statement D is correct)

Checking Statement C:

Wüstite (Fe₀.₉₃O) does have a rock-salt structure with **cubic close-packed O

[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]

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