Based on the Valence Bond Theory, which of the following complexes will exhibit either dsp2 or d2sp3 hybridisations?
A. [Ni(CN)4]2−
B. [NiCl4]2−
C. [Co(NH3)6]3+
D. [CoF6]3−
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A and C only
In valence bond theory, the metal ion uses a set of empty hybrid orbitals to accept the ligand lone pairs. Which orbitals are free depends on the ligand:
A strong field ligand (\( \ce{CN^{-}} \), \( \ce{NH3} \) with \( \ce{Co^{3+}} \)) forces the d electrons to pair up, freeing inner 3d orbitals: dsp2 (square planar, 4 ligands) or d2sp3 (inner orbital octahedral, 6 ligands).
A weak field ligand (\( \ce{Cl^{-}} \), \( \ce{F^{-}} \)) does not cause pairing, so outer orbitals are used: sp3 (tetrahedral) or sp3d2 (outer orbital octahedral).
(A) [Ni(CN)4]2− — Correct. x + 4(−1) = −2, so nickel is +2. Ni (Z = 28) is [Ar]3d84s2; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d8. \( \ce{CN^{-}} \) pairs the 8 electrons into four 3d orbitals, leaving one 3d orbital empty: dsp2, square planar, diamagnetic.
(B) [NiCl4]2− — Incorrect. x + 4(−1) = −2, so nickel is again \( \ce{Ni^{2+}} \), 3d8. \( \ce{Cl^{-}} \) is weak, so the two unpaired electrons stay: sp3, tetrahedral, paramagnetic.
(C) [Co(NH3)6]3+ — Correct. \( \ce{NH3} \) 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 6 electrons into three 3d orbitals, leaving two 3d orbitals empty: d2sp3, inner orbital octahedral, diamagnetic.
(D) [CoF6]3− — Incorrect. x + 6(−1) = −3, so cobalt is again \( \ce{Co^{3+}} \), 3d6. \( \ce{F^{-}} \) is weak, so four electrons stay unpaired and the outer 4d orbitals are used: sp3d2, outer orbital octahedral, paramagnetic.