The magnetic moment of a transition metal compound has been calculated to be 3.87 BM. The metal ion is

The magnetic moment is given by the formula:

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

where $$n$$ is the number of unpaired electrons.

Given $$\mu = 3.87 \text{ BM}$$, substitute into the formula:

$$3.87 = \sqrt{n(n+2)}$$

Square both sides:

$$(3.87)^2 = n(n+2)$$

Calculate $$(3.87)^2$$:

$$3.87 \times 3.87 = 14.9769 \approx 15$$

So,

$$n(n+2) = 15$$

Solve the quadratic equation:

$$n^2 + 2n - 15 = 0$$

Factorize:

$$(n + 5)(n - 3) = 0$$

Thus, $$n = 3$$ or $$n = -5$$. Since the number of unpaired electrons cannot be negative, $$n = 3$$.

Now, determine which metal ion has 3 unpaired electrons by examining their electron configurations:

Option A: $$Cr^{2+}$$
Chromium (atomic number 24) has electron configuration $$[Ar] 4s^1 3d^5$$. $$Cr^{2+}$$ loses two electrons: first the 4s electron, then one from 3d, resulting in $$[Ar] 3d^4$$. For a free ion, the 3d orbitals are degenerate. Placing 4 electrons in 5 orbitals with maximum unpaired electrons gives 4 unpaired electrons.

Option B: $$Mn^{2+}$$
Manganese (atomic number 25) has configuration $$[Ar] 4s^2 3d^5$$. $$Mn^{2+}$$ loses two 4s electrons, resulting in $$[Ar] 3d^5$$. With 5 electrons in degenerate orbitals, all are unpaired, giving 5 unpaired electrons.

Option C: $$V^{2+}$$
Vanadium (atomic number 23) has configuration $$[Ar] 4s^2 3d^3$$. $$V^{2+}$$ loses two 4s electrons, resulting in $$[Ar] 3d^3$$. Placing 3 electrons in degenerate orbitals gives 3 unpaired electrons.

Option D: $$Ti^{2+}$$
Titanium (atomic number 22) has configuration $$[Ar] 4s^2 3d^2$$. $$Ti^{2+}$$ loses two 4s electrons, resulting in $$[Ar] 3d^2$$. Placing 2 electrons in degenerate orbitals gives 2 unpaired electrons.

Therefore, only $$V^{2+}$$ has 3 unpaired electrons, matching the calculated magnetic moment.

Verification of magnetic moments:

• $$Cr^{2+}$$: $$\mu = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90 \text{ BM}$$
• $$Mn^{2+}$$: $$\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92 \text{ BM}$$
• $$V^{2+}$$: $$\mu = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \text{ BM}$$
• $$Ti^{2+}$$: $$\mu = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83 \text{ BM}$$

Thus, the correct answer is option C: $$V^{2+}$$.