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

Question

Philoid · Accepted 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