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Note that the chemical bond in Fig. 1b results from the electron’s position between the nuclei. On first thought, this appears to answer our question of what we mean by “sharing an electron pair” to form a chemical bond. An electron positioned between two nuclei is “shared” to the extent that its potential energy is lowered due to attraction to both nuclei simultaneously.

On second thought, though, this description must be inaccurate. We have learned our study of Energy Levels in Atoms that an electron must obey the uncertainty principle and that, as a consequence, the electron does not have a definite position, between the nuclei or otherwise. We can only hope to specify a probability for observing an electron in a particular location. This probability is, from quantum mechanics, provided by the wave function. What does this probability distribution look like for the H 2 + size 12{H rSub { size 8{2} } rSup { size 8{+{}} } } {} molecular ion?

To answer this question, we begin by experimenting with a distribution that we know: the 1s electron orbital in a hydrogen atom. This we recall has the symmetry of a sphere, with equal probability in all directions away from the nucleus. To create an H 2 + size 12{H rSub { size 8{2} } rSup { size 8{+{}} } } {} molecular ion from a hydrogen atom, we must add a bare second hydrogen nucleus (an H + size 12{H rSup { size 8{+{}} } } {} ion). Imagine bringing this nucleus closer to the hydrogen atom from a very great distance (see Fig. 2a). As the H + size 12{H rSup { size 8{+{}} } } {} ion approaches the neutral atom, both the hydrogen atom’s nucleus and electron respond to the electric potential generated by the positive charge. The electron is attracted and the hydrogen atom nucleus is repelled. As a result, the distribution of probability for the electron about the nucleus must become distorted, so that the electron has a greater probability of being near the H + size 12{H rSup { size 8{+{}} } } {} ion and the nucleus has a greater probability of being farther from the ion. This distortion, illustrated in Fig. 2b, is called “polarization”: the hydrogen atom has become like a “dipole”, with greater negative charge to one side and greater positive charge to the other.

This polarization must increase as the H + size 12{H rSup { size 8{+{}} } } {} ion approaches the hydrogen atom until, eventually, the electron orbital must be sufficiently distorted that there is equal probability for observing the electron in proximity to either hydrogen nucleus (see Fig. 2c). The electron probability distribution in Fig. 2c now describes the motion of the electron, not in a hydrogen atom, but in an H 2 + size 12{H rSub { size 8{2} } rSup { size 8{+{}} } } {} molecular ion. As such, we refer to this distribution as a “molecular orbital.”

We note that the molecular orbital in Fig. 2c is more delocalized than the atomic orbital in Fig. 2a, and this is also important in producing the chemical bond. We recall from the discussion of Atomic Energy Levels that the energy of an electron in an orbital is determined, in part, by the compactness of the orbital. The more the orbital confines the motion of the electron, the higher is the kinetic energy of the electron, an effect we referred to as the “confinement energy.” Applying this concept to the orbitals in Fig. 2, we can conclude that the confinement energy is lowered when the electron is delocalized over two nuclei in a molecular orbital. This effect contributes significantly to the lowering of the energy of an electron resulting from sharing by two nuclei.

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Source:  OpenStax, General chemistry i. OpenStax CNX. Jul 18, 2007 Download for free at http://cnx.org/content/col10263/1.3
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