1. Use spin-only magnetic moment formula

For transition metal ions,

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

where $n$ is the number of unpaired electrons.

The given values are:

• $1.73 \Rightarrow n=1$
• $2.84 \Rightarrow n=2$
• $3.87 \Rightarrow n=3$
• $0.00 \Rightarrow n=0$

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2. Find electronic configuration and unpaired electrons

(A) $\mathrm{Ti}^{3+}$

Ti has $Z=22$. Neutral Ti: $[\mathrm{Ar}]\,3d^2 4s^2$

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

$$\mathrm{Ti}^{3+} = [\mathrm{Ar}]\,3d^1$$

So, number of unpaired electrons $n=1$. Hence,

$$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$$

Therefore,

$$(A) \to (III)$$

(B) $\mathrm{V}^{2+}$

V has $Z=23$. Neutral V: $[\mathrm{Ar}]\,3d^3 4s^2$

For $\mathrm{V}^{2+}$, remove two $4s$ electrons:

$$\mathrm{V}^{2+} = [\mathrm{Ar}]\,3d^3$$

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

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

Therefore,

$$(B) \to (I)$$

(C) $\mathrm{Ni}^{2+}$

Ni has $Z=28$. Neutral Ni: $[\mathrm{Ar}]\,3d^8 4s^2$

For $\mathrm{Ni}^{2+}$, remove two $4s$ electrons:

$$\mathrm{Ni}^{2+} = [\mathrm{Ar}]\,3d^8$$

A $3d^8$ configuration has $2$ unpaired electrons. Hence,

$$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84$$

Therefore,

$$(C) \to (IV)$$

(D) $\mathrm{Sc}^{3+}$

Sc has $Z=21$. Neutral Sc: $[\mathrm{Ar}]\,3d^1 4s^2$

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

$$\mathrm{Sc}^{3+} = [\mathrm{Ar}]$$

So, number of unpaired electrons $n=0$. Hence,

$$\mu = 0.00$$

Therefore,

$$(D) \to (II)$$

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3. Final matching

$$(A)-(III),\quad (B)-(I),\quad (C)-(IV),\quad (D)-(II)$$

This corresponds to Option A.