Number of Complexes with even number of electrons in $$t_{2g}$$ orbitals is - $$[Fe(H_2O)_6]^{2+}, [Co(H_2O)_6]^{2+}, [Co(H_2O)_6]^{3+}, [Cu(H_2O)_6]^{2+}, [Cr(H_2O)_6]^{2+}$$

Count complexes with an even number of electrons in $$t_{2g}$$ orbitals (weak field / high spin octahedral, since all use $$H_2O$$).

Analysis (all are weak-field/high-spin octahedral):

- $$[Fe(H_2O)_6]^{2+}$$: Fe²⁺ = d⁶. High spin: $$t_{2g}^4 e_g^2$$. $$t_{2g}$$ electrons = 4 (even).

- $$[Co(H_2O)_6]^{2+}$$: Co²⁺ = d⁷. High spin: $$t_{2g}^5 e_g^2$$. $$t_{2g}$$ electrons = 5 (odd).

- $$[Co(H_2O)_6]^{3+}$$: Co³⁺ = d⁶. Low spin (Co³⁺ is strong enough): $$t_{2g}^6 e_g^0$$. $$t_{2g}$$ electrons = 6 (even).

- $$[Cu(H_2O)_6]^{2+}$$: Cu²⁺ = d⁹. $$t_{2g}^6 e_g^3$$. $$t_{2g}$$ electrons = 6 (even).

- $$[Cr(H_2O)_6]^{2+}$$: Cr²⁺ = d⁴. High spin: $$t_{2g}^3 e_g^1$$. $$t_{2g}$$ electrons = 3 (odd).

Even $$t_{2g}$$ count: Fe²⁺ (4), Co³⁺ (6), Cu²⁺ (6) → 3 complexes.

The correct answer is Option (2): 3.