1. Find the electronic configuration of each ion

For first-row transition metal ions:

• $Sc: [Ar]3d^1 4s^2$
• $Ti: [Ar]3d^2 4s^2$
• $V: [Ar]3d^3 4s^2$

On ionization, $4s$ electrons are removed first.

So,

• $Sc^{3+} = [Ar]3d^0$
• $Ti^{3+} = [Ar]3d^1$
• $Ti^{2+} = [Ar]3d^2$
• $V^{2+} = [Ar]3d^3$

Since these are hydrated ions (weak-field $H_2O$ ligands), for these early $3d$ ions the number of unpaired electrons remains the same as the free-ion $d^n$ count.

2. Determine the number of unpaired electrons

• $Sc^{3+} : d^0 \Rightarrow n = 0$
• $Ti^{3+} : d^1 \Rightarrow n = 1$
• $Ti^{2+} : d^2 \Rightarrow n = 2$
• $V^{2+} : d^3 \Rightarrow n = 3$

3. Use spin-only magnetic moment formula

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

where $n$ is the number of unpaired electrons.

Thus,

• $Sc^{3+}: \mu = \sqrt{0(0+2)} = 0$
• $Ti^{3+}: \mu = \sqrt{1(1+2)} = \sqrt{3}$
• $Ti^{2+}: \mu = \sqrt{2(2+2)} = \sqrt{8}$
• $V^{2+}: \mu = \sqrt{3(3+2)} = \sqrt{15}$

Clearly,

$0 < \sqrt{3} < \sqrt{8} < \sqrt{15}$

So the correct order is

$Sc^{3+} < Ti^{3+} < Ti^{2+} < V^{2+}$

4. Match with the options

This corresponds to Option B.

5. Comparison with stored correct answer

Stored correct answer = B

Our derived answer = B

So they agree.