The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products $\mathrm{A}$ and $\mathrm{B}$ along with the evolution of $\mathrm{CO}_2$. The sum of spin-only magnetic moment values of A and B is _________ B.M. (Nearest integer)

[Given atomic number : $\mathrm{C}: 6, \mathrm{Na}: 11, \mathrm{O}: 8, \mathrm{Fe}: 26, \mathrm{Cr}: 24$]

<p>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.</p> <p>The reaction provided is:</p> <p>$4 \mathrm{FeCr}_2 \mathrm{O}_4 + 8 \mathrm{Na}_2 \mathrm{CO}_3 + 7 \mathrm{O}_2 \rightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4 (\mathrm{A}) + 2 \mathrm{Fe}_2 \mathrm{O}_3 (\mathrm{B}) + 8 \mathrm{CO}_2$</p> <strong>Identifying the Products:</strong> <br/><br/> <ul> <li>Product A is sodium chromate, $\mathrm{Na}_2 \mathrm{CrO}_4$.</li><br> <li>Product B is ferric oxide, $\mathrm{Fe}_2 \mathrm{O}_3$.</li> </ul><br/> <strong>Determining the Magnetic Moments:</strong> <br/><br/> <ol> <li><strong>For Product A ($\mathrm{Na}_2 \mathrm{CrO}_4$):</strong></li> </ol><br/> <ul> <li>In $\mathrm{Na}_2 \mathrm{CrO}_4$, the chromium is in the +6 oxidation state.</li><br> <li>The electronic configuration of $\mathrm{Cr}^{6+}$ is $[\mathrm{Ar}] 3d^0$.</li><br> <li>Since there are no unpaired electrons in the $3d$ orbitals, the spin-only magnetic moment is zero.</li> </ul> <p>$ \mu = \sqrt{n(n+2)} \text{ BM} = \sqrt{0(0+2)} \text{ BM} = 0 \text{ BM} $</p> <ol> <li><strong>For Product B ($\mathrm{Fe}_2 \mathrm{O}_3$):</strong></li> </ol> <ul> <li>In $\mathrm{Fe}_2 \mathrm{O}_3$, the iron is in the +3 oxidation state.</li><br> <li>The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3d^5$.</li><br> <li>$\mathrm{Fe}^{3+}$ has 5 unpaired electrons in the $3d$ orbitals.</li><br> <li>The spin-only magnetic moment can be calculated using the formula:</li> </ul> <p></p> <p>$ \mu = \sqrt{n(n+2)} \text{ BM} = \sqrt{5(5+2)} \text{ BM} = \sqrt{35} \text{ BM} \approx 5.9 \text{ BM} $</p> <strong>Sum of Magnetic Moments:</strong> <ul> <li>The spin-only magnetic moment of Product A is $0 \text{ BM}$.</li><br> <li>The spin-only magnetic moment of Product B is $5.9 \text{ BM}$.</li> </ul> <p>Therefore, the sum of the spin-only magnetic moments of A and B is:</p> <p>$ 0 + 5.9 \approx 6 \text{ BM}$</p> <strong>Conclusion:</strong> <p>The sum of the spin-only magnetic moment values of A and B is approximately 6 B.M. (to the nearest integer).</p>