Identify the metal ions among $$Co^{2+}, Ni^{2+}, Fe^{2+}, V^{3+} \text{ and } Ti^{2+}$$ having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is __________.

Metal ions: Co²⁺(d⁷), Ni²⁺(d⁸), Fe²⁺(d⁶), V³⁺(d²), Ti²⁺(d²).

Spin-only magnetic moment: $$\mu = \sqrt{n(n+2)}$$ BM where n = unpaired electrons.

For $$\mu > 3.0$$ BM: $$n(n+2) > 9 \Rightarrow n \geq 3$$.

High spin octahedral:

Co²⁺(d⁷): $$t_{2g}^5 e_g^2$$, 3 unpaired ✓ ($$\mu = 3.87$$)

Ni²⁺(d⁸): $$t_{2g}^6 e_g^2$$, 2 unpaired ✗ ($$\mu = 2.83$$)

Fe²⁺(d⁶): $$t_{2g}^4 e_g^2$$, 4 unpaired ✓ ($$\mu = 4.90$$)

V³⁺(d²): $$t_{2g}^2$$, 2 unpaired ✗ ($$\mu = 2.83$$)

Ti²⁺(d²): $$t_{2g}^2$$, 2 unpaired ✗ ($$\mu = 2.83$$)

Sum of unpaired electrons for qualifying ions: 3 (Co²⁺) + 4 (Fe²⁺) = 7.

The answer is 7.