JEE Main 2024 — D And F Block Elements Question with Solution

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:

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

Identifying the Products:

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


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

Determining the Magnetic Moments:

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:

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

Sum of Magnetic Moments:

• 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:

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

Conclusion:

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