# Polar Covalent Bonds

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 12:46

Figure 1 Comparison of bonding electron densities in (a) LiH band; (b) CH bond; and (c) HF bond. Note the considerable shift of electron density away from hydrogen as the electronegativity of its bonding partner increases.
We could consider the three bonds shown in Fig. 1 by assuming that each had been distorted from a purely covalent, electron-pair sharing situation. In that case the H―C bond is closest to pure covalency―in it the H has roughly the same electron density it would have in an H2 moleculeA set of atoms joined by covalent bonds and having no net charge.. In the H—Li bond, however, the H has almost complete control over both electrons and hence has a negative charge. This situation is often indicated as follows:

The Greek letter δ (delta) is used here to indicate that electron transfer is not complete and that some sharing takes place. The dipole momentThe magnitude of the separation of electrical charge in a molecule that makes the molecule polar; the partial positive charge times the partial negative charge divided by the distance by which the charges are separated. of LiH shows that in effect only 77 percent of a full electronic charge has been transferred to H, and so δ = 0.77. If the transfer had been complete, δ would have been 1.0. Because the Li—H bond is only partially negative at the one end and partially positive at the other, we often say that the bond is polar or polar covalent, rather than 100 percent ionic. The H―F bond is also a polar covalent bond, but in this case F rather than H has the partial negative charge, and we can write

Again the value of δ can be found from a dipole-moment measurement, as the following example illustrates.

EXAMPLE 1 The dipole moment of the HF molecule is found to be 6.37 × 10–30 C m, while the H―F distance is 91.68 pm. Find the partial charge on the H and F atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume..

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. Rearranging Eq. (1) from Polarizability, we have

$Q=\frac{\mu }{r}$

Thus the apparent charge on each end of the molecule is given by

$Q=\frac{\text{6}\text{.37 }\times \text{ 10}^{-\text{30}}\text{ C m}}{\text{91}\text{.68 }\times \text{ 10}^{-\text{12}}\text{ C m}}=\text{6}\text{.95 }\times \text{ 10}^{-\text{20}}\text{ C}$

Since the charge on a single electron is 1.6021 × 10-19 C, we have

$\delta =\frac{\text{6}\text{.95 }\times \text{ 10}^{-\text{20}}}{\text{1}\text{.6021 }\times \text{ 10}^{-\text{19}}}=\text{0}\text{.43}$

It is worth noting in the above example that the dipole moment measures the electrical imbalance of the whole molecule and not just that of the H―F bonding pairA pair of electrons between atoms joined by a covalent bond.. In the HF molecule there are four valence electronIn a neutral atom, any of the electrons found in the highest occupied shell as well as any electrons in incompletely filled subshells of lower shells. pairs:

The three lone pairs sticking out on the right of the F atom also contribute to the overall negative charge of F.