JEE Main 2024 — 5 April, Shift 2 — Question 80

On fusion of chromite ore with sodium carbonate in the presence of air, chromite is oxidised to chromate and iron(III) oxide is formed with the evolution of carbon dioxide.

$$\mathrm{4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \xrightarrow{\Delta} 8Na_2CrO_4\ (A) + 2Fe_2O_3\ (B) + 8CO_2}$$

In $A = \mathrm{Na_2CrO_4}$, chromium is in the $+6$ oxidation state ($d^0$); hence its spin-only magnetic moment is $0\ \text{B.M.}$.

In $B = \mathrm{Fe_2O_3}$, iron is in the $+3$ oxidation state ($d^5$, high spin), having five unpaired electrons; therefore,

$$\mu = \sqrt{n(n+2)} = \sqrt{5(7)} \approx 5.92 \approx 6\ \text\;{B.M.}$$