Question

Compared to the ideal bond angle, the actual/real bond angle in a molecule of water will be:______.

Answer 1

Answer: d. smaller.

Explanation:

Hybridization :

$N=\frac{1}{2}[V+N+A-C]$

N = number of electrons

where, V = number of valence electrons present in central atom i.e. oxygen = 6

N = number of monovalent atoms bonded to central atom=32

C = charge of cation = 0

A = charge of anion = 0

The number of electrons is 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.  The bond angle for tetrahedral geometry is $109.5^0$

But as there are two atoms around the central oxygen, the third and 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 bent and the bond angle will be smaller than expected and is $104.5^0$