The arithmetic behind hybrids is rather simple, like that of mixing drinks. In all cases we start with one s orbital (think of it  as 100 mL of rum) and three p orbitals (three times 100-mL of pepsi).  Now what remains is to make some mixtures (hybrids or drinks).  At the end, there always has to be four of them (since we have started with four orbitals), although some p may remain unmixed (depending on hybridization type).

Let's start with some basics. The angle between two bonds formed by two hybrids with hybridization indices of mi and mj is quantitatively defined by these indices through the cosine formula on the left.  The formula is based on simple geometrical considerations (which we are going to skip here).  The important point is that the bond angles can be measured experimentally and provide information about specific hybrids used to make bonds or to hold lone pairs. If the bonds are equivalent, the situation is even simpler. It leads, for example, to the familiar angles of 109.5o between two sp3 hybrids, 120o between two sp2 hybrids, and 180o between two sp hybrids.

For any individual hybrid orbital the fraction of s (fs) and the fraction of p (fp) must add up to 1 (i.e. whole orbital), and–by definitionthe ratio of p to s is equal to the hybridization index, m.  A simple rearrangement of these two conditions tells us how to calculate fs and fp when m is known. On a given atom, all fractions of s in all hybrid orbitals must add up to one, and all fractions of p must add up to the number of p orbitals used in a given hybridization type (3 for sp3, 2 for sp2 and 1 for sp).  In other words, we have to deal with "full" orbitals, and no orbital, or even its piece, can be "lost" or "gained".

Let's consider a simple example of water molecule. The oxygen atom is sp3 hybridized and forms two equivalent bonds to hydrogen atoms using spm1hybrids (m1 to be solved for). The two equivalent lone pairs reside in spm2 hybrids (m2 to be solved for).  From the experimental observation the HOH bond angle is 104.5o. Using the cosine formula for two equivalent hybrids we can find that cos(104.5o) = 1/m1= 0.25, or m1 = 4.0.

The table below summarizes all the arithmetic manipulation needed to calculate all the relevant fractions. The calculation can be performed for fs's or fp's (or both if internal check is desired).

Hybrid hybridization index



OH bond m1 = 4.0 1/(m1+ 1) = 1/5 = 0.20 m1/(1+ m1) = 4/5 = 0.80
OH bond m1 = 4.0 1/(m1+ 1) = 1/5 = 0.20 m1/(1+ m1) = 4/5 = 0.80
Total used for bonds:   2/(m1+ 1) = 2/5 = 0.40 2m1/(1+ m1) = 8/5 = 1.60
Remaining per lone-pair hybrid:   (1 - 0.40)/2 = 0.30 (3 - 1.60)/2 = 0.70
lone-pair hybrid m2 = 0.70/0.30 = 2.3 0.30 0.70
lone-pair hybrid m2 = 0.70/0.30 = 2.3 0.30 0.70
Total used for all hybrids:   2 x 0.2 + 2 x 0.3 = 1.0 2 x 0.8 + 2 x 0.7 = 3.0

In conclusion, the s-character is withdrawn from the OH bonds and increased in the lone-pair hybrids. This transfer helps to stabilize the lone pairs by bringing them closer to the oxygen's nucleus (on average). The bonding electrons get stabilization from both nuclei (O and H) and of course are in a bond built out in part of H's s orbital.

Molecular Gallery  Last updated 07/29/10  Copyright 1997-2013
Advanced Quantum Lighter side Reactions Connections Tutorials Comments To