Electric dipole moment

In a nearly perfect ionic bond, such as KF, where electron transfer is almost complete, representing the molecule as

$${\rm K}^+{\rm F}^-$$

is a very good approximation, since the charge on the potassium will be approximately 1$e$ and the charge on the fluorine will be approximately -1$e$.

For a polar covalent bond, such as HF, in which only partial charge transfer occurs, a more accurate representation would be

$${\rm H}^{+\delta}{\rm F}^{-\delta}$$

where $\delta$, expressed in units of $e$, is known as a partial charge. It suggests that a fraction of an electron is transferred, although the reality is that there is simply a little more electron density on the more electronegative atom and a little less on the electropositive atom.

How much charge is actually transferred can be quantified by studying the electric dipole moment of the bond, which is a quantity that can be measured experimentally. The electric dipole moment of an assembly of charges $Q_1,Q_2,...,Q_N$ having positions ${\bf r}_1,{\bf r}_2,...,{\bf r}_N$ is defined to be

$${\bm \mu} = Q_1 {\bf r}_1 + Q_2 {\bf r}_2 + \cdots + Q_N {\bf r}_N =\sum_{i=1}^N Q_i {\bf r}_i$$

Thus, for a diatomic with charges $Q_1 = Q = \delta e$ and $Q_2 = -Q = -\delta e$ on atoms 1 and 2, respectively, the dipole moment, according to the definition, would be

$${\bm \mu} = Q_1 {\bf r}_1 + Q_2 {\bf r}_2$$

$$= Q {\bf r}_1 - Q {\bf r}_2$$

$$= Q ({\bf r}_1 - {\bf r}_2)$$

Hence, the magnitude of the dipole moment is

$$\mu = \vert{\bm \mu}\vert = Q \vert{\bf r}_1 - {\bf r}_2\vert = QR$$

where $R$ is the bond length. The units of the dipole moment are Coulomb$\cdot$ meters. However, this is a very large unit and awkward to work with for molecules. A more convenient unit is the Debye (D), defined to be

$$1 {\rm D} = 3.336\times 10^{-30} {\rm Coulomb}\cdot{\rm meters}$$

1 D is actually the dipole moment of two charges $+e$ and $-e$ separated by a distance of 0.208 Å. Thus, for a diatomic with partial charges $+\delta$ and $-\delta$, the dipole moment in D is given by

$$\mu({\rm D}) = {\delta \times R({\rm\AA}) \over 0.2082 {\rm\AA}{\rm D}^{-1}}$$

and the percent ionic character is defined in terms of the partial charge $\delta$ by

$${\rm percent\ ionic\ character} = 100\% \times \delta$$

As an example, consider HF, for which $\delta$ = 0.41. The bond length is $R$=0.926 Å. Thus, its dipole moment will be

$$\mu({\rm D}) = {0.41\times 0.926 {\rm\AA} \over 0.2082 {\rm\AA}{\rm D}^{-1}} = 1.82 {\rm D}$$

and its percent ionic character is 41%.