1. Find oxidation state and unpaired electrons in $\mathrm{KMnO_4}$

In $\mathrm{KMnO_4}$:

• $\mathrm{K} = +1$
• $4\mathrm{O} = -8$

Let oxidation state of Mn be $x$: $x + 1 - 8 = 0 \Rightarrow x = +7$

So manganese is $\mathrm{Mn^{7+}}$.

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

For $\mathrm{Mn^{7+}}$, remove 7 electrons: $\mathrm{Mn^{7+}}: [Ar] \, 3d^0$

Thus number of unpaired electrons, $n = 0$

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0\,\mathrm{BM}$

2. Find the manganese product formed when $\mathrm{KMnO_4}$ titrates oxalic acid in acidic medium

In acidic medium, permanganate is reduced to $\mathrm{Mn^{2+}}$: $\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$

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

3. Find unpaired electrons in $\mathrm{Mn^{2+}}$

Mn atom: $[Ar] \, 3d^5 4s^2$

For $\mathrm{Mn^{2+}}$, remove two $4s$ electrons: $\mathrm{Mn^{2+}}: [Ar] \, 3d^5$

A $3d^5$ ion has 5 unpaired electrons.

So, $n = 5$

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\,\mathrm{BM}$

4. Difference in magnetic moments

$\Delta \mu = 5.92 - 0 = 5.92\,\mathrm{BM}$

Nearest integer: $\boxed{6}$