1. Use the spin-only magnetic moment formula

For a free ion, the spin-only magnetic moment is

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

where $n$ is the number of unpaired electrons.

2. Find electronic configurations of the ions

(i) $Ti^{3+}$

Atomic number of Ti = 22

Neutral Ti: $Ti = [Ar]3d^2 4s^2$

For $Ti^{3+}$, remove first two electrons from $4s$ and one from $3d$: $Ti^{3+} = [Ar]3d^1$

So, number of unpaired electrons: $n=1$

Hence, $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\ \text{BM}$

(ii) $V^{2+}$

Atomic number of V = 23

Neutral V: $V = [Ar]3d^3 4s^2$

For $V^{2+}$, remove two electrons from $4s$: $V^{2+} = [Ar]3d^3$

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

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

(iii) $Sc^{3+}$

Atomic number of Sc = 21

Neutral Sc: $Sc = [Ar]3d^1 4s^2$

For $Sc^{3+}$, remove two electrons from $4s$ and one from $3d$: $Sc^{3+} = [Ar]$

So, number of unpaired electrons: $n=0$

Hence, $\mu = 0\ \text{BM}$

3. Collect the values in the required order

For $Ti^{3+}, V^{2+}, Sc^{3+}$ respectively:

$1.73,\ 3.87,\ 0$

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.