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Answer 1 · 2017-03-07T01:28:56

The order is $"NaCl" > "NH"_3 > "NF"_3 > "He"$ .

The most common intermolecular forces in pure substances are, in decreasing order of strength:

$"Ion-ion > dipole-dipole > London dispersion"$

We get an idea of the strengths the forces by looking at the bond electronegativity differences $ΔEN$ and the molecular geometries.

$bb"NaCl"$

For $"NaCl"$ , $ΔEN = |3.16 - 0.93| = 2.23$ .

$ΔEN >1.6$ , so the bond is ionic.

The strongest intermolecular forces are ion-ion attractions, so $"NaCl"$ is first in the list.

$"He"$

Helium is a noble gas. It does not form ionic or covalent bonds.

The only intermolecular forces in helium are London dispersion forces.

Helium is last in the list.

$bb"NH"_3$ and $bb"NF"_3$

Both $"NH_3" and$ "NF"_3# have a trigonal pyramidal geometry (the yellow ball represents a lone pair".

upload.wikimedia.org

For $"NH"_3$ , $ΔEN = |3.04 - 2.20| = 0.84$ .

For $"NF"_3$ , $ΔEN = |3.04 - 3.98| = 0.94$ .

Thus, we would predict the compounds to have about the same polarities, with $"NF"_3$ perhaps being slightly more polar.

However, $"NH"_3$ is much more polar than $"NF"_3$ .

The reason is that the lone pair itself contributes a dipole to the molecule.

img1.mnimgs.com

In $"NH"_3$ the bond dipoles reinforce that of the lone pair, while in $"NF"_3$ the bond dipoles oppose it.

Thus, $"NF"_3$ is less polar than $"NH"_3$ .

The order of polarity for all the compounds is

$"NaCl" > "NH"_3 > "NF"_3 > "He"$ .