Give reason The shape of [Ni(CO)4] is not the same as [Ni(CN)4]2-

Dear Student,

‚Äč
In [Ni(CO)4 ] , nickel is present in zero oxidation state {Ni = 3d8 4s2}. CO is a very strong ligand, so it can pair the unpaired electrons. Hence, the 4s electrons go to the 3 d orbital. Now, 3 d orbitals are completely filled, but 4s and 4p are still available. These 4 orbitals form a degenerate set of orbitals, that means hybridization is sp3 hybridized.

In case of [Ni(CN)4 ]2-, the oxidation state of Nickel is +2. So, Ni+2 3d 8 4s0. Now, cyanide also causes pairing of unpaired electrons, in 3d orbital, all the 8 electrons will get paired, so now, 1 more orbital is left.... and there are 4 ligands to bond with. Hence, the hybridization will be dsp2 so hence, it is a square planar complex because all
dsp 2 complexes are square planar. The singly unpaired electron will pair up only if the ligand field is very strong and that too only in the lower energy orbitals.

In 
dsp 2 hybridization, one d-orbital [which is d(x2–y2)] is involved in hybridization with one s and two p-orbital. This gives rise to the square planar geometry.


 
Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

 
Regards

 
 

  • 1
What are you looking for?