Number of complexes from the following with even number of unpaired "d" electrons is: $$[V(H_2O)_6]^{3+},\ [Cr(H_2O)_6]^{2+},\ [Fe(H_2O)_6]^{3+},\ [Ni(H_2O)_6]^{3+},\ [Cu(H_2O)_6]^{2+}$$. [Given atomic numbers: V=23, Cr=24, Fe=26, Ni=28, Cu=29]

We need to count how many of the given aqua complexes have an even number of unpaired d-electrons.

Key point: $$H_2O$$ is a weak field ligand, so all these complexes are high-spin (octahedral).

1. $$[V(H_2O)_6]^{3+}$$: V³⁺ has configuration $$[Ar]3d^2$$. High-spin octahedral: $$t_{2g}^2 e_g^0$$. Unpaired electrons = 2 (even).

2. $$[Cr(H_2O)_6]^{2+}$$: Cr²⁺ has configuration $$[Ar]3d^4$$. High-spin octahedral: $$t_{2g}^3 e_g^1$$. Unpaired electrons = 4 (even).

3. $$[Fe(H_2O)_6]^{3+}$$: Fe³⁺ has configuration $$[Ar]3d^5$$. High-spin octahedral: $$t_{2g}^3 e_g^2$$. Unpaired electrons = 5 (odd).

4. $$[Ni(H_2O)_6]^{3+}$$: Ni³⁺ has configuration $$[Ar]3d^7$$. High-spin octahedral: $$t_{2g}^5 e_g^2$$. Unpaired electrons = 3 (odd).

5. $$[Cu(H_2O)_6]^{2+}$$: Cu²⁺ has configuration $$[Ar]3d^9$$. Octahedral: $$t_{2g}^6 e_g^3$$. Unpaired electrons = 1 (odd).

Complexes with even number of unpaired d-electrons: $$[V(H_2O)_6]^{3+}$$ (2) and $$[Cr(H_2O)_6]^{2+}$$ (4). Total = 2.

The correct answer is Option (1): 2.