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 [tex]CH_{4}[/tex], there is no lone pair on C atom. So, no lone pair-bond pair repulsion is present here. Hence [tex]\angle H-C-H[/tex] is [tex]109^{0}5^{'}[/tex]
In [tex]NH_{3}[/tex], there is one lone pair on N atom. So, one lone pair-bond pair repulsion is present here. Hence [tex]\angle H-N-H[/tex] is [tex]107^{0}[/tex]
In [tex]H_{2}O[/tex], there are two lone pairs on O atom. So, two lone pair-bond pair repulsion are present here. Hence [tex]\angle H-O-H[/tex] is [tex]104^{0}5^{'}[/tex]
So, lone pair-bond pair repulsion is best accounted for decrease in bond angle
