Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

[Ni(CO) 4 ]° why in this complex hybridization is sp3 and not dsp2?

[Ni(CO)4]° why in this complex hybridization is sp3 and not dsp2?

Grade:12

1 Answers

SAI DEEPAK
17 Points
10 months ago

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 upaired electrons. Hence, the 4s electrons goes to the 3 d orbital. Now, 3 d orbital is completely filled, but 4s and 4p are still available. These 4 orbitals form a degenerate set of orbitals, that means hybridisation is sp3 hybridised.

now let us compair with 

In case of [Ni(CN)4 ]2-, oxidation state of Nickel is +2. So, Ni2+ : 3d8 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.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free