Match the LIST-I with LIST-II.LIST-I(Complex)LIST-II(Number of t2g or…

The correct answer is Option (2) → A-III, B-IV, C-I, D-II

In an octahedral complex, the five d orbitals split into a lower set, t_2g (three orbitals), and a higher set, e_g (two orbitals). The gap between them is the crystal field splitting, Δ_o. Electrons beyond the third one can go two ways:

• Weak field ligands (such as F⁻): Δ_o is smaller than the pairing energy P, so the electrons enter e_g unpaired, giving a high spin complex.

• Strong field ligands (such as CN⁻, and NH₃ with Co³⁺): Δ_o is larger than P, so the electrons pair up in t_2g first, giving a low spin complex.

With 1–3 d electrons, all of them go into t_2g whatever the ligand.

• A. \( \ce{[CoF6]^{3-}} \) — This matches III. \( t_{2g}^{4} \). Charge balance x + 6(−1) = −3 gives Co(+3). Co (Z = 27) is [Ar]3d⁷4s²; removing two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): 3d⁶. \( \ce{F^{-}} \) is a weak field ligand, so the complex is high spin: t_2g⁴ e_g².

• B. \( \ce{[Fe(CN)6]^{3-}} \) — This matches IV. \( t_{2g}^{5} \). Charge balance x + 6(−1) = −3 gives Fe(+3). Fe (Z = 26) is [Ar]3d⁶4s², so \( \ce{Fe^{3+}} \) is 3d⁵. \( \ce{CN^{-}} \) is a strong field ligand, so all five electrons pair up in t_2g: t_2g⁵ e_g⁰.

• C. \( \ce{[V(CN)6]^{4-}} \) — This matches I. \( t_{2g}^{3} \). Charge balance x + 6(−1) = −4 gives V(+2). V (Z = 23) is [Ar]3d³4s², so \( \ce{V^{2+}} \) is 3d³. Three electrons occupy the three t_2g orbitals singly, whatever the ligand: t_2g³.

• D. \( \ce{[Co(NH3)6]^{3+}} \) — This matches II. \( t_{2g}^{6} \). Charge balance x + 6(0) = +3 gives Co(+3), which is 3d⁶. With \( \ce{Co^{3+}} \), \( \ce{NH3} \) acts as a strong field ligand, so all six electrons pair up in t_2g: t_2g⁶ e_g⁰, and the complex is diamagnetic.

Shortcut: The pair C–I appears only in Option (2).