Examination of the electronic configuration of beryllium [1s22s22px02py02pz0] with its paired s-electrons might suggest that beryllium is incapable of bonding to other atoms. Although beryllium has three vacant 2p orbitals, it does not bond with ions, i.e., fluoride, to form BeF-1, BeF2-2 or BeF3-3. Likewise, boron [1s22s22px12py02pz0] has one unpaired 2px electron but does not form linear BF with a fluorine atom. Carbon [1s22s22px12py12pz0] does not produce CF2 with a 900 bond angle as a stable compound. [Note: Dumbbell-shaped 2p orbitals are orthogonal to one another.] In the real world beryllium difluoride in the gas phase is linear, boron trifluoride is planar and carbon tetrafluoride is tetrahedral. How can these observations be reconciled with the orbital energy and occupation shown on the right?

In 1940 Sidgwick and Powell suggested that the geometry of non-transition metal compounds can be determined by considering the repulsion of electron pairs, whether bonding or non-bonding. This concept became known as valence shell electron pair repulsion (VSEPR). Imagine you have two sticks (bonds) which are flexibly attached at one end (central atom) to one another with repulsive charges (pairs of electrons) at the end of each stick. What is the best spatial arrangement of the sticks? The repulsive charges want to be as far apart from one another as possible resulting in a linear arrangement of the sticks. Applying this logic to three sticks, they will arrange themselves in a trigonal, planar configuration while four sticks will result in a tetrahedral arrangment. Thus, BeF₂ (Fig. 1), BF₃ (Fig. 2) and CF₄ (Fig. 3) are arranged respectively in a linear (sp); trigonal , planar (sp²); and tetrahedral (sp³) arrangement in the gas phase and they bear no net charge. The remaining two unoccupied p-orbitals of beryllium in BeF₂ lie along the orthogonal X- and Y-axes. Similarly, the unoccupied p-orbital of boron in BF₃ lies along the Z-axis, orthogonal to the XY-plane of the atoms. [Continued below the chart.]

Both BeF₂ and BF₃ qualify as Lewis acids because they bear unfilled p-orbitals. Imagine adding a fluoride ion to one of the p-orbitals of BeF₂ to form BeF₃^-. Will the trivalent species be planar having a "T" shape with the remaining p-orbital orthogonal to the plane of the atoms of the anion? Au contraire! The anion rehybridizes to an sp² geometry (Fig. 4) which, on the surface looks no different from BF₃ (Fig. 2) except it bears a negative charge with a vacant p-orbital along the Z-axis orthogonal to the XY-plane of the atoms. If the monoanion BeF₃^- is in turn treated with fluoride ion, the dianion BeF₄^-- forms and, as you might have guessed by now, a tetrahedral dianion results (Fig. 6), which is akin to the hybridization of carbon tetrafluoride.