1. Identify product A

Given reaction: $\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow A + \mathrm{H_2O}$

This is the well-known preparation of potassium manganate: $2\mathrm{MnO}_2 + 4\mathrm{KOH} + \mathrm{O}_2 \rightarrow 2\mathrm{K}_2\mathrm{MnO}_4 + 2\mathrm{H_2O}$

So, $A = \mathrm{K}_2\mathrm{MnO}_4$

Here Mn is in oxidation state $+6$.

2. Disproportionation of A in neutral/acidic medium

Manganate ion $\mathrm{MnO_4^{2-}}$ disproportionates in neutral or acidic medium: $3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO}_2 + 2\mathrm{H_2O}$

Thus,

• $B = \mathrm{KMnO_4}$ (permanganate, Mn in $+7$ state)
• $C = \mathrm{MnO_2}$ (Mn in $+4$ state)

3. Find unpaired electrons in B and C

Mn has atomic number $25$: $\mathrm{Mn}: [\mathrm{Ar}]\,3d^5 4s^2$

For $B = \mathrm{KMnO_4}$

Mn is $+7$: $\mathrm{Mn^{7+}}: 3d^0$ So number of unpaired electrons, $n=0$ Spin-only magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)}=0\,\text{BM}$

For $C = \mathrm{MnO_2}$

Mn is $+4$: $\mathrm{Mn^{4+}}: 3d^3$ So number of unpaired electrons, $n=3$ Spin-only magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\,\text{BM}$

4. Sum of magnetic moments

$\mu_B + \mu_C = 0 + 3.87 = 3.87\,\text{BM}$

Nearest integer: $\boxed{4}$

5. Comparison with stored answer

Stored correct answer = $4$

Our derived answer also = $4$, so they agree.