Identify chromite ore and the fusion reaction

Chromite ore is $\mathrm{FeCr_2O_4}$

On fusion with sodium carbonate in presence of air, chromite gives sodium chromate and ferric oxide, with evolution of carbon dioxide: $4\mathrm{FeCr_2O_4}+8\mathrm{Na_2CO_3}+7\mathrm{O_2}\rightarrow 8\mathrm{Na_2CrO_4}+2\mathrm{Fe_2O_3}+8\mathrm{CO_2}$

Hence, $A=\mathrm{Na_2CrO_4}, \qquad B=\mathrm{Fe_2O_3}$

Find oxidation states and electronic configurations

For $A = \mathrm{Na_2CrO_4}$

In chromate ion $\mathrm{CrO_4^{2-}}$, chromium is in oxidation state $+6$.

Neutral Cr: $\mathrm{Cr}: [Ar]3d^5 4s^1$

For $\mathrm{Cr^{6+}}$, all 6 valence electrons are removed: $\mathrm{Cr^{6+}}: 3d^0$

Number of unpaired electrons $n=0$.

Spin-only magnetic moment: $\mu=\sqrt{n(n+2)}=\sqrt{0(0+2)}=0\ \text{B.M.}$

For $B = \mathrm{Fe_2O_3}$

In $\mathrm{Fe_2O_3}$, iron is in oxidation state $+3$.

Neutral Fe: $\mathrm{Fe}: [Ar]3d^6 4s^2$

So, $\mathrm{Fe^{3+}}: [Ar]3d^5$

In oxide, $\mathrm{Fe^{3+}}$ is high spin, so number of unpaired electrons: $n=5$

Spin-only magnetic moment: $\mu=\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{35}\approx 5.92\ \text{B.M.}$

Sum of spin-only magnetic moments of $A$ and $B$

$\mu_{\text{total}}=0+5.92=5.92\ \text{B.M.}$

Nearest integer: $\boxed{6}$

Comparison with stored answer

Stored correct answer = $6$.

Our derived answer also = $6$.