How can the value of charge on an atom be less than that of a proton? The dipole moment of an $\text{HCl}$ molecule is about $3.6×10^{-30} \ \mathrm{Cm}$ and the separation between the atoms of it is about $1.27×10^{-10}\ \mathrm m$. We know that dipole moment, $p=qd$. So substituting the values of $p$ and $d$, we can calculate the value of charge on the atoms of the $\text{HCl}$ molecule and it will be approximately $q=p/d=2.8×10^{-20}\ \mathrm C$. This value is definitely smaller than the charge of a proton. How is it possible?
 A: Whether you believe that this is some sort of time-averaging of point-like electrons moving very quickly, or that the electrons are actually smeared out by the wave function, this is because the electrons in orbit around the atoms are smeared out in orbitals. So, even though we describe the Cl atom as "stealing" the electron from the H, it doesn't, completely. The reality is more akin to a water molecule - where outer-most filled electron orbital is more concentrated around the more electro-negative atom than the lesser one. It just so happens that for water this imbalance is relatively small, for HCl it's big, and for NaCl it's a massive $8.971$ Debye ($2.992 \times 10^{-29}\,\mathrm{C\, m}$). Dividing that by the spacing, from the same web page, of $2.361\,$Å gives $1.267\times 10^{-19}\,\mathrm{C}$ (about 79% of the charge of an electron).
A: If the molecule consisted of an unperturbed Cl$^-$ ion and a proton the dipole moment would indeed be $p=ed$. However Cl$^-$, especially its valence shell, is polarized by the proton, which results in the much smaller effective dipole moment.
