How would you account for the following:
[Ni(CO)4] has tetrahedral geometry while [Ni(CO)4]2- has a square planar geometry?
Asked by Topperlearning User
| 7th Apr, 2014,
09:56: AM
Expert Answer:
Nickel in [Ni(CO)4] is in the zero oxidation state and has electronic configuration of [Ar]3d84s2 or 3d10.
sp3 hybrid orbitals accommodate four pair of electrons from four CO groups and the resulting tetrahedral complex is diamagnetic due to the absence of unpaired electrons.
Nickel in [Ni(CO)4]2- is in the +2 oxidation state and has electronic configuration of [Ar]3d104s2.
dsp2 hybrid orbitals accommodate four pair of electrons from four CO groups and the resulting square planer complex is diamagnetic due to the absence of unpaired electrons.
Answered by
| 7th Apr, 2014,
11:56: AM
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