Match the complex given in List-I with their hybridization and magnet…

The correct answer is Option (1) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)

According to valence bond theory, an octahedral complex uses six hybrid orbitals. If two inner 3d orbitals are free (or can be made free by pairing electrons with a strong field ligand), the hybridisation is d²sp³ (inner orbital complex). If the ligand is weak or the 3d orbitals are full, the outer 4d orbitals are used and the hybridisation is sp³d² (outer orbital complex). The complex is paramagnetic if any electron is unpaired.

• (A) \( \ce{[Zn(H2O)6]^{2+}} \) — This matches (III) sp³d², diamagnetic. Zn (Z = 30) is [Ar]3d¹⁰4s²; water is neutral, so zinc is +2, and losing the two 4s electrons gives \( \ce{Zn^{2+}} \): 3d¹⁰, so no inner d orbital is free and the outer 4d orbitals are used. All electrons are paired.

• (B) \( \ce{[Co(NH3)6]^{3+}} \) — This matches (IV) d²sp³, diamagnetic. \( \ce{NH3} \) is neutral, so cobalt is +3. Co (Z = 27) is [Ar]3d⁷4s²; losing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): 3d⁶. \( \ce{NH3} \) pairs the six electrons into three 3d orbitals, leaving two 3d orbitals free, and no electron remains unpaired.

• (C) \( \ce{[Mn(CN)6]^{3-}} \) — This matches (II) d²sp³, paramagnetic. Charge balance: x + 6(−1) = −3, so x = +3. Mn (Z = 25) is [Ar]3d⁵4s²; losing the two 4s electrons and one 3d electron gives \( \ce{Mn^{3+}} \): 3d⁴. The strong field \( \ce{CN^{-}} \) ligand pairs the electrons into three 3d orbitals, leaving two free for hybridisation, but two electrons stay unpaired.

• (D) \( \ce{[CoF6]^{3-}} \) — This matches (I) sp³d², paramagnetic. Charge balance: x + 6(−1) = −3, so x = +3, and \( \ce{Co^{3+}} \) is 3d⁶ as worked out in (B); the weak field \( \ce{F^{-}} \) ligand does not pair the electrons. The outer 4d orbitals are used, and four electrons stay unpaired.

Shortcut: The pair (C)–(II) appears only in Option (1).