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

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.