When Fe_(0.93)O is heated in presence of oxygen, it converts to Fe₂O₃. The number of correct statement/s from the following is ________
A. The equivalent weight of Fe_(0.93)O is (Molecular weight)/0.79
B. The number of moles of Fe²⁺ and Fe³⁺ in 1 mole of Fe_(0.93)O is 0.79 and 0.14 respectively
C. Fe_(0.93)O is metal deficient with lattice comprising of cubic closed packed arrangement of O²⁻ ions
D. The % composition of Fe²⁺ and Fe³⁺ in Fe_(0.93)O is 85% and 15% respectively

Solution

Chemistry ma’am · 2 steps

Find how many Fe are +2 and how many are +3 first; then every statement checks itself.

1. Balance the charge. In 1 mole of Fe_(0.93)O, let x moles of iron be Fe²⁺, so 0.93 - x moles are Fe³⁺. The total positive charge must cancel the -2 of the single O²⁻.

   2x + 3(0.93 - x) = 2 ⇒ x = 0.79, 0.93 - x = 0.14

   Statement	Check	Verdict
   A	Only Fe²⁺ is oxidised to Fe³⁺, losing 1 electron each, so n-factor = 0.79 and equivalent weight = M/0.79	Correct
   B	Charge balance gives Fe²⁺ = 0.79 mol and Fe³⁺ = 0.14 mol	Correct
   C	Iron is less than 1 per O, so the crystal is metal deficient. Like FeO it has a rock-salt structure with O²⁻ in ccp	Correct
   D	0.79/0.93 × 100 = 84.9% ≈ 85% and 0.14/0.93 × 100 = 15.1% ≈ 15%	Correct

2. So. All four statements are correct, so the answer is 4.

The common slip: 3. Marking D wrong gives 3. The 85% and 15% are shares of the 0.93 mol of total iron, so you divide 0.79 and 0.14 by 0.93, not by 1.

The idea to keep. In a non-stoichiometric oxide, charge balance splits the metal between its two oxidation states. Use that split for n-factors, and take percentages out of the total metal, not out of 1.

Answer from the NTA final answer key. Solution by PrepWiser.