In the Valence Bond (VB) theory an atom may rearrange its atomic orbitals prior to the bond formation. Instead of using the atomic orbitals directly, mixtures of them (hybrids) are formed. For carbon (and other elements of the second row) the hybridization is limited to mixing one 2s and three 2p orbitals, as appropriate.
We recognize three basic types of hybridization: sp3, sp2 and sp. These terms specifically refer to the hybridization of the atom and indicate the number of p orbitals used to form hybrids.
The hybrids with larger s character have bigger front lobes (and smaller "tails") than the hybrids with smaller s character, as illustrated above for an sp (left) and sp3 (right) hybrids. The s/p ratio is, thus, responsible for the bonding properties of the hybrid. Increased s contribution brings electrons closer to the nuclei, increasing stabilizing Coulomb interactions. The more s character the hybrid orbital has the lower its energy, the better its overlap (with bonding partners), and the stronger (and shorter) bonds it can form.
It is important not to confuse the hybridization type that applies to the atom (see above) with the individual hybrid character that is described by the hybridization index m. The first (indirectly) indicates the number of p orbitals set aside (for participation in π systems), the second precisely describes the specific mixture of s and p used to construct the given hybrid.
The mixing of s and p orbitals in different ratios also results in changes of the angle between the resulting hybrids. Since the atomic p orbitals are 90° from each other, their various degree of participation in the mixing will yield hybrids separated by different angles, depending on their p character. For two identical hybrids, in general, the more p character in the hybrids the smaller the angle between them. Thus, two pure p orbitals (100% p) are 90° apart, two sp3 hybrids are 109.5° apart, two sp2 hybrids are 120° apart, and two sp hybrids are 180° apart. More generally, the angle (α) between any two hybrids (spm and spn) is given by cosα =–1/(m·n)0.5.
An atom will adjust its hybridization in such a way as to form the strongest possible bonds and keep all its bonding and lone-pair electrons in as low-energy hybrids as possible, and as far from each other as possible (to minimize electron-electron repulsions). This adjustment is accomplished by varying s and p characters of individual mixtures, but "moving" s character between hybrids (to lower energy of some) also changes the angles between them (potentially increasing electron-electron repulsion). Thus, it is all a compromise game.
Let us look at some examples. Something simple to start: methane. The carbon atom forms four identical bonds using four identical hybrid orbitals. These orbitals are the result of sp3 hybridization (here we talk about the hybridization type), i.e. one s and three p orbitals are mixed to form four sp3 hybrids (here we talk about the composition, or character, of each hybrid). Each of these hybrids is composed of ¼ of s and ¾ of p (the p/s ratio is 3, i.e. m = 3). The angle between any two such hybrid orbitals is (yes... that's the cosine formula above) 109.5o. The table below gives more examples of different hybrid orbitals involved in making C-H bonds. Note that for the same hybridzation type (sp3 in our example) one can have quite different hybrids involved in making C-H bonds.
Table 1. Hybrids Involved in Making C-H Bonds in Simple Hydrocarbons
When we say that bonds made out of hybrids containing more s character are stronger, we have to be very precise in our meaning. Bond strength are commonly measured by bond dissociation energies (BDE's) that reflect enthalpies of homolysis (i.e. the energy required to break a bond, forming two radicals). What counts in such considerations is the actual strength of the bond broken (that is related to the s character of the hybrid) and the stability of the radicals formed which is influenced strongly by other effects (such as hyperconjugation or resonance) that may not even be present in the molecule before homolysis. In general, the stability of the radicals is a more important contribution to BDE's than the hybridization effect (i.e. s character).
Similarly the s character of the orbital containing the lone electron pair will influence the stability of the anion, and therefore, the pKa value of the hydrocarbon precursors of this anion. If the geometry of the hydrocarbon is similar to that of the anion, the more s character in the hybrid involved in making the "acidic" C-H bond, the lower the pKa value of the hydrocarbon (and the more stable the anion). But, again caution is required in making comparisons: the stability of the anion also strongly depends on other effects (such as inductive and resonance stabilization, ion pairing and solvation, i.e. interactions with solvent molecules). In particular, if the resonance stabilization is involved, the hybridization of the hydrocarbon and the anion derived from it are going to be quite different.
In general, very rarely all hybrids formed by an atom are exactly equal. We may have an infinite number of combinations of mixtures. For example, let us assume that carbon forms two sp2 hybrids (with an angle of 120°) to two identical substituents and additional two (equal to each other) spx hybrid orbitals that are going to be used to form two additional bonds, as shown in A below. How can we find x? It is very simple, just a little fraction arithmetic! Each of the two sp2 hybrids contains 1/3 of s (total of 2/3 s). The remaining 1/3 s must be divided between the two identical spx hybrids; i.e. 1/6 s per orbital. To make the "full" hybrid the "missing" 5/6 of the hybrid must be composed of p. So, the ratio of p/s = 5 (or x = 5) and we have our answer (sp5). If you do not believe it, you may check the math doing the balance for the p orbitals. Here, how it goes: the two sp2 hybrids contain 2/3 of p each (total 4/3 p). Since all three p orbitals participate in hybridization, we have 3 – 4/3 = 5/3 of p left to be used in the two spx orbitals. That leaves (5/3)/2, or 5/6 of p per hybrid; exactly the same answer we got doing the balance for the s orbital.
Now we can look at the hybridization of nitrogen in ammonia (B) or oxygen in water (C) with more precision. Since each of these central atoms has four electron pairs around, and no π bonds (that information is available from a simple Lewis structure) we may say that oxygen and nitrogen are sp3 hybridized. We mean that each atom uses three p's and one s atomic orbitals to make four hybrids. But none of these hybrids is an sp3 hybrid! The angle between hydrogens in :NH3 is 107° Since all the hydrogens are identical, we can calculate that the hybrids used to make N-H bonds are sp3.42 (cos(107°) = –1/m or m = 3.42); i.e they have 1/(1+3.42) = 22.6% s character each). That leaves 32.2% s character for the hybrid containing the lone pair, or the lone pair is an sp2.10 hybrid (the p character in the lone-pair hybrid must be 100 –-32.2 = 67.8%, or m/(m+1) = 0.678 from where m = 2.10). For a moment, forget the arithmetic! The point is that the lone pair hybrid has increased its s character in comparison to what it would be in the pure sp3 hybrid (32.2% vs 25% s) stabilizing the lone pair (more s character more stabilization). And the lone pair needs more stabilization: these are, by definition, unshared electrons! The price to be paid is the decrease in the H-N-H angle from the ideal tetrahedral 109.5o to 107o and the increased repulsion between the bonding pairs. The observed situation is the compromise between these two trends. You may recognize this concept as being equivalent to saying that the lone pair needs "more space" than the bonding pairs. Similar situation is found in H2O. Here, the H-O-H bond is 104.5°. Now, you do the math! The O-H bonds use sp4 hybrids and the lone pairs (two here) are sp2.3, or have about 30% s character each. Still, that is better than the 25% s of the pure sp3 hybrids.