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.

Answer 2

Step 1: Identify the reaction.
Chromite ore has the formula FeCr₂O₄.
When it is fused with Na₂CO₃ in the presence of air, oxidation takes place.
The reaction can be written as:
\[ 4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2 \]
Thus, the products formed are:
A = Na₂CrO₄ (sodium chromate)
B = Fe₂O₃ (ferric oxide)

Step 2: Determine oxidation states.
In Na₂CrO₄, Cr is in +6 oxidation state.
In Fe₂O₃, Fe is in +3 oxidation state.

Step 3: Calculate spin-only magnetic moments.
Use the formula:
\[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \]
where n = number of unpaired electrons.

For Cr⁶⁺: Electronic configuration = [Ar] 3d⁰ → n = 0
Hence, μ(A) = 0 B.M.

For Fe³⁺: Electronic configuration = [Ar] 3d⁵ → n = 5
\[ \mu(B) = \sqrt{5(5+2)} = \sqrt{35} ≈ 5.92 \, \text{B.M.} \]

Step 4: Sum of magnetic moments.
\[ \mu_{total} = \mu(A) + \mu(B) = 0 + 5.92 ≈ 6 \, \text{B.M.} \]

Step 5: Final Answer.
The sum of spin-only magnetic moments of A and B is 6 B.M.

Final Answer: 6 B.M.