Which of the following paramagnetic ions would exhibit a magnetic moment (spin only) of the order of $$5$$BM? (At. Nos. Mn $$= 25$$, Cr $$= 24$$, V $$= 23$$, Ti $$= 22$$ )

The spin-only magnetic moment of a transition-metal ion is given by the formula
$$\mu_{\text{so}} = \sqrt{n(n+2)}\ \text{BM}$$
where $$n$$ is the number of unpaired d-electrons.

To obtain a moment of the order of $$5$$ BM we solve
$$\sqrt{n(n+2)} \approx 5$$
$$\Rightarrow n(n+2) \approx 25$$

Checking integral values of $$n$$:
  • $$n = 4 : \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ \text{BM}$$
  • $$n = 5 : \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{BM}$$
A value close to $$5$$ BM is obtained for $$n = 4$$ (≈$$4.9$$ BM).

Hence we need the ion that possesses a $$d^{4}$$ electronic configuration (four unpaired electrons). Calculate the d-electron count for each ion:

Case 1: Mn$$^{2+}$$
Mn: [Ar] $$3d^{5}\,4s^{2}$$; remove two 4s electrons ⇒ $$3d^{5}$$ (five unpaired).
$$\mu_{\text{so}} \approx \sqrt{5(5+2)} = 5.92\ \text{BM}$$ (too high).

Case 2: Ti$$^{2+}$$
Ti: [Ar] $$3d^{2}\,4s^{2}$$; remove two 4s electrons ⇒ $$3d^{2}$$ (two unpaired).
$$\mu_{\text{so}} \approx \sqrt{2(2+2)} = 2.83\ \text{BM}$$ (too low).

Case 3: V$$^{2+}$$
V: [Ar] $$3d^{3}\,4s^{2}$$; remove two 4s electrons ⇒ $$3d^{3}$$ (three unpaired).
$$\mu_{\text{so}} \approx \sqrt{3(3+2)} = 3.87\ \text{BM}$$ (below 5 BM).

Case 4: Cr$$^{2+}$$
Cr: [Ar] $$3d^{5}\,4s^{1}$$; remove one 4s and one 3d electron ⇒ $$3d^{4}$$ (four unpaired).
$$\mu_{\text{so}} \approx \sqrt{4(4+2)} = 4.90\ \text{BM}$$ (very close to 5 BM).

Therefore Cr$$^{2+}$$ is the only ion whose spin-only magnetic moment is of the order of $$5$$ BM.

Option D which is: Cr$$^{2+}$$