Question

According to the VSEPR model, the progressive decrease in the bond angles in the series of molecules CH₄, NH₃, and H₂O is best accounted for by the _________.

Answer 1

Answer:

lone pair-bond pair repulsion is best accounted for decrease in bond angle

Explanation:

Bond angle in each compounds depends on presence of lone pair on central atom.

In $CH_{4}$, there is no lone pair on C atom. So, no lone pair-bond pair repulsion is present here. Hence $\angle H-C-H$ is $109^{0}5^{'}$

In $NH_{3}$, there is one lone pair on N atom. So, one lone pair-bond pair repulsion is present here. Hence $\angle H-N-H$ is $107^{0}$

In $H_{2}O$, there are two lone pairs on O atom. So, two lone pair-bond pair repulsion are present here. Hence $\angle H-O-H$ is $104^{0}5^{'}$

So, lone pair-bond pair repulsion is best accounted for decrease in bond angle