1. Identify the species and oxidation state

The given ion is manganate, $\mathrm{MnO_4^{2-}}$.

Let oxidation state of Mn be $x$:

$x + 4(-2) = -2$ $x - 8 = -2$ $x = +6$

So, Mn is in the $+6$ oxidation state in $\mathrm{MnO_4^{2-}}$.

2. Disproportionation in acidic medium

A species in an intermediate oxidation state can undergo disproportionation to form one species with higher oxidation state and one with lower oxidation state.

For $\mathrm{MnO_4^{2-}}$ in acidic medium, the two Mn compounds formed are:

• $\mathrm{MnO_4^-}$, where Mn is $+7$
• $\mathrm{MnO_2}$, where Mn is $+4$

Thus:

• $A$ has higher oxidation state: $\mathrm{MnO_4^-}$
• $B$ has lower oxidation state: $\mathrm{MnO_2}$

So, in $B$, Mn is in the $+4$ state.

3. Find the electronic configuration of $\mathrm{Mn^{4+}}$

Atomic number of Mn = 25.

Ground-state configuration of Mn:

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

For $\mathrm{Mn^{4+}}$, remove 2 electrons from $4s$ and 2 from $3d$:

$\mathrm{Mn^{4+}}: [Ar] \, 3d^3$

So number of unpaired electrons, $n = 3$.

4. Calculate spin-only magnetic moment

Formula:

$\mu = \sqrt{n(n+2)} \text{ BM}$

Substitute $n = 3$:

$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$

Nearest integer:

$\boxed{4}$

5. Comparison with stored answer

Derived answer = $4$

Stored correct answer = $4$

They agree.