How many of the following metal ions have similar value of spin only magnetic moment in gaseous state?
(Given: Atomic number: V, 23; Cr, 24; Fe, 26; Ni, 28)
V$$^{3+}$$, Cr$$^{3+}$$, Fe$$^{2+}$$, Ni$$^{3+}$$

For V$$^{3+}$$, Cr$$^{3+}$$, Fe$$^{2+}$$, and Ni$$^{3+}$$ in the gaseous state, we determine how many have similar spin-only magnetic moments.

In the gaseous state (no ligand field), electrons occupy d-orbitals following Hund’s rule (maximum multiplicity).

For V$$^{3+}$$ (Z = 23), the configuration is [Ar] 3d$$^2$$, giving 2 unpaired electrons; for Cr$$^{3+}$$ (Z = 24), [Ar] 3d$$^3$$ gives 3 unpaired electrons; for Fe$$^{2+}$$ (Z = 26), [Ar] 3d$$^6$$ yields 4 unpaired electrons (↑↓ ↑ ↑ ↑ ↑); and for Ni$$^{3+}$$ (Z = 28), [Ar] 3d$$^7$$ yields 3 unpaired electrons (↑↓ ↑↓ ↑ ↑ ↑).

The spin-only magnetic moment formula is $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons.

Thus, for V$$^{3+}$$, $$\mu = \sqrt{2 \times 4} = \sqrt{8} = 2\sqrt{2}$$ BM; for Cr$$^{3+}$$, $$\mu = \sqrt{3 \times 5} = \sqrt{15}$$ BM; for Fe$$^{2+}$$, $$\mu = \sqrt{4 \times 6} = \sqrt{24} = 2\sqrt{6}$$ BM; and for Ni$$^{3+}$$, $$\mu = \sqrt{3 \times 5} = \sqrt{15}$$ BM.

Cr$$^{3+}$$ and Ni$$^{3+}$$ both have $$\mu = \sqrt{15}$$ BM (3 unpaired electrons each), so two ions share the same magnetic moment.

Answer: 2