Number of lone pairs of electrons present is $N{H}_3,H_{2}O, HF$ respectively are 

NH₃​: 3 bond pairs and one lone pair because, out of 5 valence electrons of nitrogen, only 2 electrons are left to form a lone pair.

H₂​O: 2 bond pairs and two lone pairs because, out of 6 valence electrons of oxygen, 4 electrons are left to form 2 lone pairs.

HF: 3 lone pairs and 1 bond pair.