Question

Explain the shape of H2O molecule on the basis of VSEPR theory ?

Answer 1

Answer:

According to the VSEPR theory water has an bond angle of 104.5°

Explanation:

The geometry of water is angular or bent geometry

It has two lone pairs and two band pairs

Therefore lonepair lonepair repulsion and lone pair bond pair repulsion works to reduce the bond angle H-O-H to 104.5°

Please check attachment for structure of water

Answer 2

The shape of H$_2$O molecule on the basis of VSEPR theory

In H$_2$O molecule,

The total number of electrons surrounding the central O-atom

= (6 valence electrons of O-atom + 2 electrons of singly boned H-atom)

= 8 electrons 0r 4 electron pairs

= 2 σ - bond pairs + 2 lone pairs.

• In order to minimise the extent of mutual repulsion, these four electron pairs will be oriented towards the four corners of a tetrahedron.
• The tetrahedron will be somewhat distorted because of the repulsion between two lone pairs.

For this reason, the H-O-H bond angle will be reduced to 104°5' from 109°28'. Hence, the shape of the H$_2$O molecule will be angular or V-shaped.