Q. 52

Match the species in Column I with the type of hybrid orbitals in Column II.


Answer :

(i)(c);


The hybridisation of the S is sp3d.


No. of Hybrid Orbitals = (V+M-C+A)


= (6+4)


= 5(As it has 5 hybridised orbitals).


So, it’s hybridisation is sp3d.


(ii) (a);


The hybridization of the I is sp3d2.


Iodine has valency = 7.


No. of Hybrid Orbitals = (7+5)


= 6


Thus, It’s hybridisation is sp3d2.


(e);


As, there is 1 (+) charge.


No. of Hybrid Orbitals = (5-1)


=2.


The hybridization of the N is sp


(iv) (d);


As, there is 1 cation.


No. of hybrid orbitals = (5+4-1)


= 4.


The hybridization of the N is sp3


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