Question

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 CO$_2$. 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)

Answer 1

Correct Answer: 6

The reaction for the fusion of chromite ore is:
\[4\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2.\]
Here: \(A = \text{Na}_2\text{CrO}_4\), \(B = \text{Fe}_2\text{O}_3\).
Step 1: Calculate the spin-only magnetic moment
The spin-only magnetic moment is given by:
\[\mu_s = \sqrt{n(n+2)} \, \mu_B,\]
where \(n\) is the number of unpaired electrons.
For \(\text{Na}_2\text{CrO}_4\):
Chromium in \(\text{Na}_2\text{CrO}_4\) is in the \(+6\) oxidation state (\(\text{Cr}^{6+}\)).
The electronic configuration of \(\text{Cr}^{6+}\) is \([\text{Ar}]3d^0\).
\(n = 0\), so \(\mu_s = 0 \, \mu_B\).
For \(\text{Fe}_2\text{O}_3\):
Iron in \(\text{Fe}_2\text{O}_3\) is in the \(+3\) oxidation state (\(\text{Fe}^{3+}\)).
The electronic configuration of \(\text{Fe}^{3+}\) is \([\text{Ar}]3d^5\).
\(n = 5\), so:
\[\mu_s = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \, \mu_B.\]
Step 2: Sum of magnetic moments
The sum of spin-only magnetic moments of \(A\) and \(B\) is:
\[\mu_{\text{total}} = 0 + 5.9 = 5.9 \, \mu_B.\]
Rounding to the nearest integer:
\[\mu_{\text{total}} = 6 \, \mu_B.\]
Final Answer: 6 B.M.