1. Find oxidation states of Mn in each compound

• In $\mathrm{MnF_2}$: fluorine is $-1$, so $x + 2(-1) = 0 \Rightarrow x = +2$ Hence Mn is $\mathrm{Mn^{2+}}$.

• In $\mathrm{MnF_3}$: $x + 3(-1) = 0 \Rightarrow x = +3$ Hence Mn is $\mathrm{Mn^{3+}}$.

• In $\mathrm{MnF_4}$: $x + 4(-1) = 0 \Rightarrow x = +4$ Hence Mn is $\mathrm{Mn^{4+}}$.

2. Identify which has the strongest oxidizing ability

An oxidizing agent gets reduced. For manganese fluorides, the compound with the highest oxidation state of Mn will have the greatest tendency to get reduced, hence the strongest oxidizing ability.

Therefore, among $\mathrm{MnF_2}, \mathrm{MnF_3}, \mathrm{MnF_4}$, the strongest oxidizing agent is: $\mathrm{MnF_4}$

3. Find electronic configuration of Mn in $\mathrm{MnF_4}$

Atomic number of Mn is $25$.

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

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

So number of unpaired electrons, $n = 3$

4. Calculate spin-only magnetic moment

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

Substitute $n=3$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ B.M.}$

Nearest integer: $4$

5. Compare with stored answer

Stored correct answer is $5$, but the calculated value is $4$ (nearest integer).

So the stored answer does not agree with the correct spin-only value for the strongest oxidizing compound.