If one divides the molecular dipole of HCl by the interatomic separation, the result is not a multiple of the electron charge. Why is that?

Question

I'm having some trouble understanding the reasoning behind the final note in the following worked problem:

An HCl molecule has a dipole moment of $3.4\times10^{-30}\:\rm C\:m$. Assuming that equal and opposite charges lie on the two atoms to form a dipole, what is the magnitude of this charge? The separation between the two atoms of HCl is $1.0\times 10^{-10}\:\rm m$.

Solution: if the charges on the two atoms are $q$, $-1$, \begin{align} q(1.0\times 10^{-10}\:\mathrm m) & = 3.4\times10^{-30}\:\mathrm{ C\:m} \\ & = 3.4\times10^{20}\:\mathrm{C} \end{align}

Note that this is less than the charge of a proton. Can you explain, how such a charge can appear on an atom?

I think it might be electronegativity difference.

Answer 1

In simple terms Chlorine is more electronegative than Hydrogen.

In combination this results in the Chlorine distorting the electron orbitals resulting in a slight residual positive charge in the region of the hydrogen nucleus and a slight residual negative charge in the region of the Chlorine nucleus.
The Chlorine has a greater share of an electron than the Hydrogen which is like a covalent bond and unlike an ionic bond where a whole electron would have been transferred.

The magnitude of the residual changes does no have to be a multiple of the charge on the electron.

Answer 2

The bond is not 100% ionic, but rather covalent with a difference in electronegativity of about 0.9 thus you don't have an H that has become $H^+$ (a proton). There are only partial partial charges on both atoms $H^{\delta +}$ and $Cl^{\delta -}$ with $\delta^+= 3.4 \cdot 10^{-20} C$