Which of the following is the electronic configuration of the central…

The correct answer is Option (3) → \(t_{2g}^5e_g^2\)

In an octahedral complex the d orbitals split into a lower t_2g set (three orbitals) and an upper e_g set (two orbitals), separated by Δ_o. With a weak field ligand (Δ_o < pairing energy P), electrons occupy the e_g orbitals singly before pairing in t_2g (high spin). With a strong field ligand they pair in t_2g first (low spin).

The oxidation state of cobalt: x + 6(0) = +2, so x = +2. Co (Z = 27) is [Ar]3d⁷4s²; losing the two 4s electrons gives \( \ce{Co^{2+}} \): 3d⁷.

\( \ce{H2O} \) is a weak field ligand, so the seven electrons are arranged high spin: five fill t_2g and e_g singly, and the last two pair up in t_2g:

    \( t_{2g}^{\,5}\,e_g^{\,2} \) (3 unpaired electrons)

Why the other options are incorrect

• Option (1) \(t_{2g}^6e_g^1\) — Incorrect. This is the low spin arrangement, found with a strong field ligand.

• Option (2) \(t_{2g}^4e_g^3\) — Incorrect. It puts more electrons in the higher e_g level than needed; this is never the ground state.

• Option (4) \(t_{2g}^6e_g^0\) — Incorrect. It has only six electrons, i.e. d⁶ (\( \ce{Co^{3+}} \)), not d⁷.