1. Find oxidation states of the metal ions

• In $\mathrm{Mn_2O_3}$: $2x + 3(-2)=0 \Rightarrow 2x-6=0 \Rightarrow x=+3$ So manganese is $\mathrm{Mn^{3+}}$.

• In $\mathrm{TiO}$: $x+(-2)=0 \Rightarrow x=+2$ So titanium is $\mathrm{Ti^{2+}}$.

• In $\mathrm{VO}$: $x+(-2)=0 \Rightarrow x=+2$ So vanadium is $\mathrm{V^{2+}}$.

2. Determine which compound has the strongest oxidising power

An oxidising agent gets reduced, so the species in the higher oxidation state generally shows greater oxidising power.

Here the metal oxidation states are:

• $\mathrm{Mn^{3+}}$
• $\mathrm{Ti^{2+}}$
• $\mathrm{V^{2+}}$

Thus, among these, $\mathrm{Mn_2O_3}$ containing $\mathrm{Mn^{3+}}$ has the strongest oxidising power.

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

Atomic number of Mn = 25.

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

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

So number of unpaired electrons, $n=4$.

4. Calculate spin-only magnetic moment

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

Substitute $n=4$: $\mu = \sqrt{4(4+2)}=\sqrt{24} \approx 4.90\;\text{B.M.}$

Nearest integer: $\boxed{5}$

5. Comparison with stored answer

Derived answer = $5$

Stored correct answer = $5$

Hence, they agree.