A first row transition metal in its +2 oxidation state has a spin-only magnetic moment value of $$3.86$$ BM. The atomic number of the metal is:

A first row transition metal in +2 oxidation state has a spin-only magnetic moment of 3.86 BM. Find the atomic number.

To determine the number of unpaired electrons, we use the spin-only magnetic moment formula: $$\mu = \sqrt{n(n+2)}$$ BM. Substituting $$\mu = 3.86$$ gives $$3.86 = \sqrt{n(n+2)}$$ and thus $$14.9 \approx n(n+2)$$. For $$n = 3$$, $$\sqrt{3 \times 5} = \sqrt{15} \approx 3.87$$ BM, which matches the observed value.

Next, we consider first row transition metals in the +2 oxidation state, which typically lose two 4s electrons, leaving only the 3d electrons. We look for a 3d configuration with three unpaired electrons:

V$$^{2+}$$ (Z=23): [Ar] 3d$$^3$$ - 3 unpaired electrons. Matches.

Cr$$^{2+}$$ (Z=24): [Ar] 3d$$^4$$ - 4 unpaired electrons (high spin). Does not match.

Mn$$^{2+}$$ (Z=25): [Ar] 3d$$^5$$ - 5 unpaired electrons. Does not match.

Fe$$^{2+}$$ (Z=26): [Ar] 3d$$^6$$ - 4 unpaired electrons (high spin). Does not match.

Co$$^{2+}$$ (Z=27): [Ar] 3d$$^7$$ - 3 unpaired electrons, but among the given options only Z=23 is listed.

The correct answer is Option 3: 23 (Vanadium).