Match List-I (complex) with List-II (hybridisation of the central atom).
| List-I | List-II |
| (A) [Ni(CN)4]2− | (I) sp3 |
| (B) [Fe(CN)6]4− | (II) sp3d2 |
| (C) [FeF6]4− | (III) d2sp3 |
| (D) [Ni(CO)4] | (IV) dsp2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-IV, B-III, C-II, D-I
In valence bond theory the metal uses empty orbitals to hold the ligands’ electron pairs. A strong field ligand (\( \ce{CN^-} \), CO) makes the d electrons pair up and can free inner 3d orbitals; a weak field ligand (\( \ce{F^-} \)) leaves them unpaired, so outer 4d orbitals are used.
(A) [Ni(CN)4]2− — This matches (IV) dsp2. x + 4(−1) = −2, so Ni is +2: Ni (Z = 28) [Ar]3d84s2 → \( \ce{Ni^2+} \) 3d8. The \( \ce{CN^-} \) ligands pair the eight electrons into four 3d orbitals; one 3d, the 4s and two 4p orbitals give dsp2 (square planar).
(B) [Fe(CN)6]4− — This matches (III) d2sp3. x + 6(−1) = −4, so Fe is +2: Fe (Z = 26) [Ar]3d64s2 → \( \ce{Fe^2+} \) 3d6. The strong field \( \ce{CN^-} \) pairs the six electrons into three 3d orbitals; the two free 3d, the 4s and three 4p orbitals give d2sp3 (inner orbital complex, as \( \ce{CN^-} \) also gives with iron in [Fe(CN)6]3−).
(C) [FeF6]4− — This matches (II) sp3d2. Fe is again +2 (3d6), but the weak field \( \ce{F^-} \) does not force pairing: the six 3d electrons occupy all five 3d orbitals (4 unpaired), so the 4s, three 4p and two outer 4d orbitals give sp3d2 (outer orbital complex, as \( \ce{F^-} \) also gives with iron in [FeF6]3−).
(D) [Ni(CO)4] — This matches (I) sp3. CO is neutral, so Ni is in the zero oxidation state; the complex is tetrahedral and diamagnetic, using one 4s and three 4p orbitals: sp3.
NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “In the square planar complexes, the hybridisation involved is dsp2. An example is [Ni(CN)4]2–. Here nickel is in +2 oxidation state and has the electronic configuration 3d8.”
“In tetrahedral complexes one s and three p orbitals are hybridised to form four equivalent orbitals oriented tetrahedrally. … 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, the former two complexes are paramagnetic and the latter diamagnetic. 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.”