# 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?

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.

• My guess would be: Given that electrons and protons have the same charge and there are only integer numbers of electrons and protons, one might assume that the total charge difference also need to be an integer. But somehow the atoms overall charge is not an integer charge, how can this be? – Natanael Feb 28 at 8:22

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.

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 $$\delta^+= 3.4 \cdot 10^{-20} C$$