Chemical Bonding and Molecular Structure Part 3

Concept:Dipole moment depends on the magnitude of bond dipoles and molecular geometry.A molecule with bond dipoles that nearly cancel gives a very low net dipole moment.Explanation:Among $H_2S$, $H_2O$, $NF_3$, $NH_3$, and $CHCl_3$, the dipole moments are:$H_2O$ ≈ 1.85 D (bent, two lone pairs), $H_2S$ ≈ 0.97 D (bent but weaker bonds), $NH_3$ ≈ 1.47 D (trigonal pyramidal, one lone pair), $CHCl_3$ ≈ 1.04 D.$NF_3$ has a very low dipole moment (≈ 0.24 D).In $NF_3$, nitrogen has one lone pair, and the vector sum of the three N–F bond dipoles opposes the lone pair moment, resulting in near cancellation.Thus the molecule $X$ with the lowest dipole moment is $NF_3$.The central atom (N) in $NF_3$ has one lone pair of electrons.From the given values, the dipole moment is the smallest for $\mathrm{NF}_3$ (it is $0.23$), so molecule $(X)$ is $\mathrm{NF}_3$.In $\mathrm{NF}_3$, the central atom is nitrogen $(\mathrm{N})$.Nitrogen has $5$ valence electrons. It makes $3$ $\mathrm{N{-}F}$ bonds, so $3$ electrons are used for bonding.The remaining $2$ valence electrons stay on nitrogen as one lone pair.So, the number of lone pairs on the central atom in $(X)=\mathrm{NF}_3$ is $1$.Answer:1 lone pair (Option D).