Question

Compare the Lewis structures of ${\mathrm{CH}}_4$ and ${\mathrm{H}}_2\mathrm{O}$. Why do these molecules have similar bond angles but different molecular shapes?

Short answer

${\mathrm{CH}}_4$and ${\mathrm{H}}_2\mathrm{O}$ molecules have similar bond angles but different molecular shapes due to BP-LP and LP-LP repulsion.

Step-by-step solution

Given information

We are given that ${\mathrm{CH}}_4$and ${\mathrm{H}}_2\mathrm{O}$ have similar bond angles but different molecular shapes.

Simplify

Bond pairs and lone pairs have different repulsive forces, according to VSEPR theory.
However, repulsive force works as follows:
B.P - B.P < B.P - L.P < L.P - L.P
As a result of the changing repulsive force, the bond pairs and lone pairs deviated from their normal shape.
There are four bond pairs and no lone pair of electrons in ${\mathrm{CH}}_4$
As a result, it maintains its regular geometry, namely, tetrahedral with $109.5$degrees.${\mathrm{H}}_2\mathrm{O}$ has two bond pairs and two lone pairings.
According to VSEPR, there is a strong attraction between LP-LP.
Again, LP-BP repulsion is greater than that of two bond pairs.
As a result, two bonds become further constricted, resulting in a V-shaped molecule with a bond angle of $104.5°$.