The metal ion (in gaseous state) with lowest spin-only magnetic moment value is

We need to find the metal ion (in gaseous state) with the lowest spin-only magnetic moment.

$$V^{2+}$$: $$[Ar] 3d^3$$ — 3 unpaired electrons

$$Ni^{2+}$$: $$[Ar] 3d^8$$ — 2 unpaired electrons

$$Cr^{2+}$$: $$[Ar] 3d^4$$ — 4 unpaired electrons

$$Fe^{2+}$$: $$[Ar] 3d^6$$ — 4 unpaired electrons

$$V^{2+}$$: $$\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87$$ B.M.

$$Ni^{2+}$$: $$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83$$ B.M.

$$Cr^{2+}$$: $$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$$ B.M.

$$Fe^{2+}$$: $$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$$ B.M.

$$Ni^{2+}$$ has the lowest spin-only magnetic moment of $$2.83$$ B.M. with only 2 unpaired electrons.

The correct answer is Option B.