# Question d4c10

Mar 7, 2017

The order is $\text{NaCl" > "NH"_3 > "NF"_3 > "He}$.

#### Explanation:

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

$\text{Ion-ion > dipole-dipole > London dispersion}$

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

$\boldsymbol{\text{NaCl}}$

For $\text{NaCl}$, ΔEN = |3.16 - 0.93| = 2.23.

ΔEN >1.6, so the bond is ionic.

The strongest intermolecular forces are ion-ion attractions, so $\text{NaCl}$ is first in the list.

$\text{He}$

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.

${\boldsymbol{\text{NH}}}_{3}$ and ${\boldsymbol{\text{NF}}}_{3}$

Both $\text{NH_3} \mathmr{and}$"NF"_3 have a trigonal pyramidal geometry (the yellow ball represents a lone pair". For ${\text{NH}}_{3}$, ΔEN = |3.04 - 2.20| = 0.84.

For ${\text{NF}}_{3}$, ΔEN = |3.04 - 3.98| = 0.94.

Thus, we would predict the compounds to have about the same polarities, with ${\text{NF}}_{3}$ perhaps being slightly more polar.

However, ${\text{NH}}_{3}$ is much more polar than ${\text{NF}}_{3}$.

The reason is that the lone pair itself contributes a dipole to the molecule. In ${\text{NH}}_{3}$ the bond dipoles reinforce that of the lone pair, while in ${\text{NF}}_{3}$ the bond dipoles oppose it.

Thus, ${\text{NF}}_{3}$ is less polar than ${\text{NH}}_{3}$.

The order of polarity for all the compounds is

$\text{NaCl" > "NH"_3 > "NF"_3 > "He}$.