1. Identify the manganese product in the reaction

Acidified potassium permanganate contains the permanganate ion, $\mathrm{MnO_4^-}$, where manganese is in the $+7$ oxidation state.

In acidic medium, permanganate oxidises oxalic acid to carbon dioxide and itself gets reduced as:

$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$

So, the manganese product formed is $\mathrm{Mn^{2+}}$.

2. Find the electronic configuration of $\mathrm{Mn^{2+}}$

Atomic number of Mn $= 25$.

Neutral manganese:

$\mathrm{Mn} : [Ar] \, 3d^5 4s^2$

For $\mathrm{Mn^{2+}}$, remove two electrons from $4s$ first:

$\mathrm{Mn^{2+}} : [Ar] \, 3d^5$

Thus, number of unpaired electrons,

$n = 5$

3. Use the spin-only magnetic moment formula

$\mu = \sqrt{n(n+2)} \text{ B.M.}$

Substituting $n=5$:

$\mu = \sqrt{5(5+2)} = \sqrt{35}$

$\mu \approx 5.92 \text{ B.M.}$

Nearest integer:

$6$

4. Compare with stored correct answer

Derived answer $= 6$

Stored correct answer $= 6$

They match.