Match List-I with List-II.
| List-I (Coordination entity) | List-II (Hybridisation and nature) |
|---|---|
| (A) \( \ce{[Ni(CO)4]} \) | (I) sp3d2; outer orbital complex |
| (B) \( \ce{[NiCl4]^2-} \) | (II) sp3; diamagnetic |
| (C) \( \ce{[MnCl6]^3-} \) | (III) d2sp3; inner orbital complex |
| (D) \( \ce{[Fe(CN)6]^3-} \) | (IV) sp3; paramagnetic |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-II, B-IV, C-I, D-III
In the valence bond theory the metal uses hybrid orbitals to accept electron pairs from the ligands. Whether its d electrons are paired decides the magnetism, and whether inner (3d) or outer (4d) orbitals are used in an octahedral complex.
(A) \( \ce{[Ni(CO)4]} \) — This matches (II) sp3; diamagnetic. Nickel is in the zero oxidation state; the complex is tetrahedral and has no unpaired electron.
(B) \( \ce{[NiCl4]^2-} \) — This matches (IV) sp3; paramagnetic. Ni2+ is 3d8; one s and three p orbitals give a tetrahedral ion, and the 3d8 set keeps two unpaired electrons.
(C) \( \ce{[MnCl6]^3-} \) — This matches (I) sp3d2; outer orbital complex. Mn3+ (3d4) keeps four unpaired electrons with Cl−, so only one 3d orbital is left empty; two inner d orbitals are not available, and the outer 4d orbitals are used.
(D) \( \ce{[Fe(CN)6]^3-} \) — This matches (III) d2sp3; inner orbital complex. With CN− the five 3d electrons of Fe3+ pair up, leaving one unpaired electron and two empty 3d orbitals for hybridisation.
NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “In tetrahedral complexes one s and three p orbitals are hybridised to form four equivalent orbitals oriented tetrahedrally. This is illustrated below for [NiCl4]2-. … The compound is paramagnetic since it contains two unpaired electrons. Similarly, [Ni(CO)4] has tetrahedral geometry but is diamagnetic since nickel is in zero oxidation state and contains no unpaired electron.”
5.5.2 Magnetic Properties of Coordination Compounds – “…[Mn(CN)6]3–, [Fe(CN)6]3– and [Co(C2O4)3]3– are inner orbital complexes involving d2sp3 hybridisation … On the other hand, [MnCl6]3–, [FeF6]3– and [CoF6]3– are outer orbital complexes involving sp3d2 hybridisation and are paramagnetic corresponding to four, five and four unpaired electrons.”