Question

What are the hybridization and shapes of the following molecules $?$
(i) $C{H_3}F$
(ii) $HC \equiv N$
(A)(i) $s{p^2}$ ,trigonal planar; (ii) $s{p^3}$ ,tetrahedral
(B)(i) $s{p^3}$ ,tetrahedral; (ii) $sp$ ,linear
(C)(i) $sp$ ,linear; (ii) $s{p^2}$ ,trigonal planar
(D)(i) $s{p^2}$ ,trigonal planar; (ii) $s{p^2}$ ,trigonal planar

Answer 1

Hybridization can be defined as the mixing of the two atomic orbitals that have the same energy level to form new hybrid orbitals. To find the shape and hybridization of molecules, we must know the number of sigma bonds and number of lone pairs around the central atom. This gives the steric number. Steric number determines the hybridisation and thus the shape of the molecule.

Complete step By Step Answer:

To find the hybridization and shape of a molecule, we must find the steric number. It can be found out by drawing the Lewis structure of the molecule and calculating the number of sigma bonds and lone pairs of the central atom.

Steric number $=$ number of $\sigma$ bonds $+$ number of lone pairs

Each steric number denotes a specific hybridisation and the related shape of the molecule.

(i) $C{H_3}F$

Number of $\sigma$ bonds around C atom $=$ $4$

Number of lone pairs around C atom $=$ $0$

Therefore, steric number $=$ $4$

Hybridisation of $C{H_3}F$ is $s{p^3}$ and the shape is tetrahedral.

(ii) $HC \equiv N$

Number of $\sigma$ bonds around C atom $=$ $2$

Number of lone pairs around C atom $=$ $0$

Therefore, steric number $=$ $2$

Hybridisation of $HC \equiv N$ is sp and shape is linear.

The correct option is (B) (i) $s{p^3}$ ,tetrahedral; (ii) $sp$ ,linear.

Note:

In the above two cases, the steric number is the number of sigma bonds. The number of sigma bonds gives the number of atomic orbitals hybridised which determines the hybridisation. But when lone pairs are present, the shape of the molecules changes due to the strong lone pair-lone pair repulsion. The bond angles and shape changes in order to minimise repulsion and maximise stability of the molecule.