QJEEM24Ch0203: JEE Main 2024 Chemistry - Class 12 JEEM24Ch0203 Solution: The fusion of chromite ore with sodium carbonate in the…

To determine the spin-only magnetic moments of products A and B formed from the fusion of chromite ore with sodium carbonate in the presence of air, we first need to examine the given reaction and the properties of the products.

The reaction provided is:

$4 \mathrm{FeCr}_2 \mathrm{O}_4 + 8 \mathrm{Na}_2 \mathrm{CO}_3 + 7 \mathrm{O}_2 \rightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4 (\mathrm{A}) + 2 \mathrm{Fe}_2 \mathrm{O}_3 (\mathrm{B}) + 8 \mathrm{CO}_2$

• Product A is sodium chromate, $\mathrm{Na}_2 \mathrm{CrO}_4$ .


• Product B is ferric oxide, $\mathrm{Fe}_2 \mathrm{O}_3$ .

1. For Product A ( $\mathrm{Na}_2 \mathrm{CrO}_4$ ):

• In $\mathrm{Na}_2 \mathrm{CrO}_4$ , the chromium is in the +6 oxidation state.


• The electronic configuration of $\mathrm{Cr}^{6+}$ is $[\mathrm{Ar}] 3d^0$ .


• Since there are no unpaired electrons in the $3d$ orbitals, the spin-only magnetic moment is zero.

$\mu = \sqrt{n(n+2)} \text{ BM} = \sqrt{0(0+2)} \text{ BM} = 0 \text{ BM}$

1. For Product B ( $\mathrm{Fe}_2 \mathrm{O}_3$ ):

• In $\mathrm{Fe}_2 \mathrm{O}_3$ , the iron is in the +3 oxidation state.


• The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3d^5$ .


• $\mathrm{Fe}^{3+}$ has 5 unpaired electrons in the $3d$ orbitals.


• The spin-only magnetic moment can be calculated using the formula:

$\mu = \sqrt{n(n+2)} \text{ BM} = \sqrt{5(5+2)} \text{ BM} = \sqrt{35} \text{ BM} \approx 5.9 \text{ BM}$

• The spin-only magnetic moment of Product A is $0 \text{ BM}$ .


• The spin-only magnetic moment of Product B is $5.9 \text{ BM}$ .

Therefore, the sum of the spin-only magnetic moments of A and B is:

$0 + 5.9 \approx 6 \text{ BM}$

The sum of the spin-only magnetic moment values of A and B is approximately 6 B.M. (to the nearest integer).