Question

Hich one of the following statements about orbital hybridization is incorrect?

Answer 1

Answer : The incorrect statement is, (C) The nitrogen atom in NH₃ is sp² hybridized.

Explanation :

Formula used  :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(a) The given molecule is, $CH_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(b) The given molecule is, $CO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

(c) The given molecule is, $NH_3$

$\text{Number of electrons}=\frac{1}{2}\times [5+3]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are 3 atoms around the central nitrogen atom, the fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be trigonal pyramidal.

(d) sp² hybrid orbitals are coplanar, and at 120° to each other.

The sp² hybrid orbitals has trigonal planar geometry and are in the same plane that means coplanar. Thus, the bond angle is 120° to each other.

(e) sp hybrid orbitals lie at 180° to each other.

The sp hybrid orbitals has linear geometry. Thus, the bond angle is 180° to each other.

Hence, the incorrect statement is, (C) The nitrogen atom in NH₃ is sp² hybridized.