The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products A and B along with the evolution of CO₂. The sum of spin-only magnetic moment values of A and B is _________ B.M. (Nearest integer)
[Given atomic number : C: 6, Na: 11, O: 8, Fe: 26, Cr: 24]

Solution

Chemistry ma’am · 4 steps

Write the fusion reaction first, then find the d-electron count of the metal ion in each product.

1. The reaction. Chromite ore is FeCr₂O₄. Fused with sodium carbonate in air, oxygen oxidises both metals: Cr goes from +3 to +6 and Fe goes from +2 to +3.

   4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

2. Product A. In Na₂CrO₄, Cr is +6. Cr is [Ar] 3d⁵ 4s¹, so Cr⁶⁺ is 3d⁰. With no unpaired electrons, μ = 0.
3. Product B. In Fe₂O₃, Fe is +3. Fe is [Ar] 3d⁶ 4s², so Fe³⁺ is 3d⁵ with n = 5 unpaired electrons. Then μ = √n(n+2) = √35 = 5.92 B.M.
4. So. The sum is 0 + 5.92 = 5.92 B.M., which rounds to 6.

The common slip: 5. Keeping iron as Fe²⁺ (as in the ore) gives d⁶ with 4 unpaired electrons, so μ = √24 = 4.90, which rounds to 5, not 6.

The idea to keep. In chromite fusion, air oxidises everything: chromium ends as CrO₄²⁻ (Cr +6, d⁰) and iron ends as Fe₂O₃ (Fe +3, d⁵). Fix the oxidation state before counting unpaired electrons.

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