Match the species in Column I with the type of hybrid orbitals in Column II.
The hybridisation of the S is sp3d.
No. of Hybrid Orbitals = (V+M-C+A)
= 5(As it has 5 hybridised orbitals).
So, it’s hybridisation is sp3d.
The hybridization of the I is sp3d2.
Iodine has valency = 7.
No. of Hybrid Orbitals = (7+5)
Thus, It’s hybridisation is sp3d2.
As, there is 1 (+) charge.
No. of Hybrid Orbitals = (5-1)
The hybridization of the N is sp
As, there is 1 cation.
No. of hybrid orbitals = (5+4-1)
The hybridization of the N is sp3
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