Among [Ni(CO)4], [Ni(CN)4]2-, [NiCl4]2- species, the hybridisation states at the Ni atom are, respectively [Atomic number of Ni = 28]
- sp3, dsp2, dsp2
- sp3, dsp2, sp3
- sp3, sp3, dsp2
- dsp2, sp3, sp3
Answer
(2) sp3, dsp2, sp3
Solution
[Ni(CO)4]: Ni(0) is 3d84s2. CO is a strong field ligand, so the 4s electrons are paired into 3d, giving 3d10, and the empty 4s and 4p orbitals accept the four CO pairs: sp3, tetrahedral.
[Ni(CN)4]2-: Ni2+ is 3d8. Strong field CN- pairs the two unpaired electrons, leaving one 3d orbital empty, so one 3d, one 4s and two 4p orbitals hybridise: dsp2, square planar (diamagnetic).
[NiCl4]2-: Cl- is a weak field ligand, so no pairing occurs in 3d8 (two unpaired electrons) and the 4s and three 4p orbitals are used: sp3, tetrahedral (paramagnetic).
Answer: (2)