1. Use spin-only magnetic moment formula

For a gaseous transition metal ion, the spin-only magnetic moment is

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

where $n$ is the number of unpaired electrons.

So, we must find the number of unpaired electrons in each ion.

2. Find electronic configurations of the ions

Recall the atomic configurations:

• $V: [Ar]3d^34s^2$
• $Cr: [Ar]3d^54s^1$
• $Fe: [Ar]3d^64s^2$
• $Ni: [Ar]3d^84s^2$

For cations, electrons are removed first from $4s$ and then from $3d$.

(A) $V^{2+}$

$V^{2+}: [Ar]3d^3$ Unpaired electrons $n=3$

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

(B) $Ni^{2+}$

$Ni^{2+}: [Ar]3d^8$ In $3d^8$, the number of unpaired electrons is $n=2$.

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

(C) $Cr^{2+}$

$Cr^{2+}: [Ar]3d^4$ Unpaired electrons $n=4$

$\mu = \sqrt{4(4+2)} = \sqrt{24}$

(D) $Fe^{2+}$

$Fe^{2+}: [Ar]3d^6$ For free gaseous ion, $3d^6$ has $4$ unpaired electrons.

$\mu = \sqrt{4(4+2)} = \sqrt{24}$

3. Compare the values

• $V^{2+}: \sqrt{15}$
• $Ni^{2+}: \sqrt{8}$
• $Cr^{2+}: \sqrt{24}$
• $Fe^{2+}: \sqrt{24}$

Clearly,

$\sqrt{8} < \sqrt{15} < \sqrt{24}$

So the lowest spin-only magnetic moment is for

$\boxed{Ni^{2+}}$

4. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B ($Ni^{2+}$)

They match.