Ionic bondingThe simplest type of interaction between atoms that leads to a stable bond is called the ionic bond: two atoms approach, and one or more electrons move from one atom to the other, creating a positive/negative ion pair. The two ions then remain bound to each other through the attractive Coulomb force that always exists between two objects of opposite charge. No ionic bond follows this idealized description completely. The electron that leaves one atom is never completely localized around the other, but many ionic bonds come close to this so-called "100% ionic" picture. Here, we look at the electron configuration requirements for an ionic bond, along with the energy requirements for ionicity that the electron configurations justify. We will take the interaction between Na and F atoms as our model, but as you read this discussion and study the diagrams, compare them to the similar discussion annd diagrams on pages 62 through 65 in the text, where the K + F interaction is described. In every ionic bond, one atom is characterized by a small ionization energy while the other has a large electron affinity. Here, Na has the low IE and F has the high EA. These atoms start our story as neutral, ground-state atoms with the electron configurations and Lewis electron dot diagrams shown below.
When the atoms reach a separation that makes the electron transfer energetically favorable (a distance we will define below very carefully), the ion pair is formed, the ions adopt closed-shell electron configurations, the Na atom becomes a smaller Na+ cation, and the F atom becomes a larger F– anion:
From the point of electron transfer to smaller separations, Coulomb's law governs the attraction and thus the bonding. But at larger separations, we have electrically neutral atoms that do not interact strongly with each other: the potential energy between them is essentially zero at all distances larger than the electron-jump distance. Our first question, therefore, is why does the electron not jump except at a specific distance? The answer is all explained by the fact that the IE of Na is greater than the EA of F. We can understand the energetics of this process if we follow the sequence (1) ionize Na, producing Na+ + e–, followed by (2) attaching the electron to F. Step (1) requires the IE of Na, which is 8.23 x 10–19 J, while (2) releases the EA of F, which is only 5.45 x 10–19 J. This means that electron transfer is not energetically favorable until the energy difference IE – EA = 2.78 x 10–19 J can be overcome by the attractive Coulombic potential energy, which is more and more favorable as the atomic separation decreases. The diagram below illustrates these various energies, showing the non-interacting atoms as a red line at the zero of energy and the ion-pair Coulombic interaction as a blue curve. (As r goes to infinity, it approaches the energy of the green line labeled Na+ + F–, representing the ion pair at infinite separation.) The electron jump distance is at the r value where these two curves cross. You should be able to use the (IE – EA) value quoted above along with the equation for Coulomb's law to calculate that this crossing point is at about 8.30 Å.
The QuickTime movie below animates this process. The black dot follows the potential energy as the two atoms approach. Note the change in the atomic sizes at the curve-crossing point where the electron jumps. What is not shown here is the repulsive force between the ions that only happens at very short distances. (The NaF bond length is about 1.92 Å, much smaller than the electron jump distance.) At distances comparable to and smaller than the bond length, the electrons of both ions are so close that Coulombic attraction is overcome and the ions start repelling each other.  |
