This is a question on polarity from the 313 Supplement:
F is more electronegative than H and so the N−F bond dipole should be larger than the N−H bond dipole. However, the molecular dipole moments for :NH3 and :NF3 are 1.47 D and 0.24 D, respectively. Explain. [Hint: the non-bonded electrons are important to the explanation, so draw the correct geometries for the molecules and explicitly determine the direction of the bond dipoles and then the effect of the non-bonded pair.]
While you are working out an answer, here is a graphic of the electrostatic potential maps of the two molecules.
Each of these nitrogens has 3 sigma bonds and 1 non-bonded pair of electrons and so 4 molecular orbitals are required to accommodate the 4 electron pairs. Thus, the nitrogen atoms are sp3-hybridized, approximately tetrahedral geometry and ~109 deg. bond angles. Draw the two molecules in their correct geometry -- one good way is shown below.
The electronegativity order of the atoms is F > N > H, so now draw in the bond dipole arrows as shown in the picture. (Vector magnitudes are not drawn to scale.) We might expect the N-F dipoles to be larger than the N-H dipoles (in opposite directions) because the E.N. difference between N-F is larger than the E.N. difference between N-H (you could look at Pauling E.N. numbers to confirm this.)
Next, determine the net dipole for the bonds. The answer is shown and labeled next to each molecule.
If only the bond dipoles were important in the overall dipole moment, then we might expect NF3 to be more polar because of its more polar N-F bonds. But this is not in accord with the facts.
Now, take into account the non-bonding pair on each N. There is a large amount of electron density in the sp3 non-bonded pair orbitals on each nitrogen.