We can use the same ideas with any sort of atom. The three molecules shown on the next page all have a tetrahedral structure, with four equivalent σ bonds from the central tetrahedral sp³ atom, whether this is B, C, or N, and the same total number of bonding electrons—the mol ecules are said to be isoelectronic. The atoms contribute different numbers of electrons so to get the eight bonding electrons we need we have to add one to BH₄ and subtract one from NH4^- hence the charges in BH₄⁻ and NH₄⁺. In each case the central atom can be considered to be sp3 hybridized, using an sp³ orbital to bond to each of the four H atoms, each resulting σ bond being made up of two electrons. Compounds of the same three elements with only three bonds take more thinking about. Borane, BH3, has only three pairs of bonding electrons (three from B and three from the three H atoms). Since the central boron atom bonds to only three other atoms we can therefore describe it as being sp² hybridized. Each of the B–H bonds results from the overlap of an sp² orbital with the hydrogen 1s orbital. Its remaining p orbital is not involved in bonding and must remain empty. Do not be tempted by the alternative structure with tetrahedral boron and an empty sp³ orbital. You want to populate the lowest energy orbitals for greatest stability and sp2 orbitals with their greater s character are lower in energy than sp3 orbitals. Another way to put this is that, if you have to have an empty orbital, it is better to have one with the highest possible energy since it has no electrons in it and so it doesn’t affect the stability of the molecule. Borane is isoelectronic with the methyl cation, CH₃ + or Me+. All the arguments we have just applied to borane also apply to Me+ so it too is sp² hybridized with a vacant p orbital. This will be very important when we discuss the reactions of carbocations in Chapters 15 and 36. Now what about ammonia, NH₃? Ammonia is not isoelectronic with borane and Me+! It has a total of eight electrons—fi ve from N and three from 3 × H. As well as three N–H bonds, each with two electrons, the central nitrogen atom also has a lone pair of electrons. We have a choice: either we could hybridize the nitrogen atom sp2 and put the lone pair in the p orbital or we could hybridize the nitrogen sp3 and have the lone pair in an sp³ orbital. This is the opposite of the situation with borane and Me+. The extra pair of electrons does contribute to the energy of ammonia so it prefers to be in the lower-energy orbital, sp³, rather than pure p. Experimentally the H–N–H bond angles are all 107.3°. Clearly, this is much closer to the 109.5° sp3 angle than the 120° sp² angle. But the bond angles are not exactly 109.5°, so ammonia cannot be described as pure sp3 hybridized. One way of looking at this is to say that the lone pair repels the bonds more than they repel each other. Alternatively, you could say that the orbital containing the lone pair must have slightly more s character while the N–H bonding orbitals must have correspondingly more p character.