Both example molecules described earlier, CH4 and NH3, involve sp3 hybridized orbitals on the central atom. In the case of CH4, a 1s orbital on each of the four H atoms overlaps with each of the four sp3 hybrid orbitals to form four σ bonds. All angles between pairs of C–H bonds are 109.5°. In NH3 the situation is different in that there are only three H atoms. Three of the four sp3 hybrid orbitals form three bonds to H atoms, but the fourth sp3 hybrid orbital contains the lone pair. The lone pair is different from the H atoms, and this is important.
Earlier we described each of a set of four sp3 hybridized orbitals as having ¼ s character and ¾ p character. This is what happens in CH4. Each of the four C–H bonds involves a hybrid orbital that is ¼ 2s and ¾ 2p. Summing over the four bonds gives 4 × ¼ = one 2s orbital and 4 × ¾ = three 2p orbitals—exactly the number and type of atomic orbitals from which the hybrid orbitals were formed.
In NH3, however, three of the four sp3 hybrid orbitals form bonds to H atoms and the fourth holds a lone pair. In this and similar situations, the partial s and p characters in all sp3 hybrid orbitals must still sum to one and three, but each hybrid orbital does not have to be the same as the others. That is, a hybrid orbital forming an N–H bond can have more p character (and less s character) compared to the hybrid orbital involving the lone pair. The only requirement is that the total s character and the total p character, summed over all four hybrid orbitals, must be one s and three p.
A different ratio of s character and p character gives a different bond angle. For example, in sp2 hybridized orbitals (with ⅓ s character and ⅔ p character) the angle between bonds is 120°, whereas, for sp3 the angle is 109.5°. More p character results in a smaller bond angle. (This makes sense, because for the maximum p character, that is, for two unhybridized p orbitals, the bond angle would be 90° because the p orbitals are at 90°.)
How can you tell how much s character and how much p character is in a specific hybrid orbital? A molecular orbital theory calculation can provide this information. But for our purposes, a qualitative rule that indicates where there will be more p character is sufficient. Bent’s rule says that a hybrid orbital on a central atom has greater p character the greater the electronegativity of the other atom forming a bond. A lone pair is assigned zero electronegativity because there is no atom attracting electrons in the bond away from the central atom.
Applying Bent’s rule to NH3, the three bonded H atoms have higher electronegativity than the lone pair (no atom) so we expect more p character in the hybrid orbitals that form the bond pairs. Thus, the angle between any two N–H bonds should be less than the tetrahedral angle. The experimentally measured angle is 106.7°, a bit less than the expected 109.5°.
In the H2O molecule, two of the O’s sp2 hybrid orbitals are involved in forming the O-H σ bonds. One of O lone pairs is in the other sp2 hybrid orbital; the other O lone pair is in the unhybridized 2p AO. The overall molecular geometry is bent. If O had perfect sp2 hybridization, the H-O-H angle would be 120°, but because the three hybrid orbitals are not equivalent, the angle deviates from ideal. To obtain an accurate bond angle requires an experimental measurement or a high-level molecular orbital calculation.
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