# Why is "CHCl"_3 polar even though the electronegativity difference is smaller than 1.0?

Jun 16, 2017

Yeah, when the electronegativity differences get this small, it's no longer practical to depend on the simple cutoffs for what counts as nonpolar, polar, and ionic.

You should start thinking about polarity as a spectrum ($1$ is low polarity, $5$ is high polarity):

1. nonpolar
2. slightly polar
3. polar
4. very polar
5. ionic

In this case, $\text{CH"_3"Cl}$ is around the "polar" side of the spectrum, since you expect it numerically to be nonpolar, but it is asymmetrical with a $\text{Cl}$ atom, whose electronegativity is significantly larger than that of $\text{H}$ (by about $0.9$).

Blue is positive and red is negative in the above diagram.

Even though the typical cutoff assigns it as a nonpolar bond, the fact that molecule is asymmetrical and that a $\text{Cl}$ displaces a $\text{H}$ compared to methane, gives $\text{CH"_3"Cl}$ some significant polarity.

Jun 16, 2017

There is a molecular dipole for any ${H}_{3} C - X$.
The $\stackrel{\delta +}{C} - \stackrel{\delta -}{X}$ forms some form of dipole, i.e. some degree of charge separation operates. When these dipoles are summed geometrically, i.e. $\text{vector addition}$, the resultant dipole enforces molecular polarity.
And thus $C {F}_{4}$ or $C B {r}_{4}$ are NON-POLAR solvents, because even tho the $C - X$ is polar; their vector sum over a tetrahedron gives NO resultant dipole, and thus NO molecular polarity. When we replace the hydrogens of methane with successive chloride ligands, we can assess the effect of polarity, with increasing molecular mass, by examining the normal boiling points of ${\text{CH}}_{4}$, $\text{H"_3"C-Cl}$, ${\text{Cl"_2"CH}}_{2}$, $\text{Cl"_3"CH}$, and ${\text{CCl}}_{4}$ to give $- 164$ ""^@C, $- 24.2$ ""^@C, $+ 39.6$ ""^@C, and $+ 76.7$ ""^@C.
As we add electrons to the molecule (by halogen substitution) the boiling point will naturally grow up because of the increase in intermolecular dispersion forces. And yet an extra force of intermolecular interaction by formation of a dipole. Compare non-polar methane, with slightly polar $\text{chloromethane}$, and $\text{methylene chloride}$. The boiling point increase is dramatic.