Module 2 : Molecular Structure
Lecture 8 : Hetenuclear Diatomics
 

It should be apparent even from one example such as this as to how important and wide ranging the concept of hybridization is in chemistry. We will have more on this in the next chapter. While closing this lecture, you may be wondering how the energy levels such as the ones in fig 8.5 (a) and fig 8.5 (b) are obtained. These are all obtained as solutions of the Schrödinger equation for the molecule.

 
It is only from the coefficients of the AOs in the MO (Eq 8.1) that we can conclude that there is hybridization. The energy of hybrid h1= 0.8 (2s) + 0.6 (2p) is

= (0.8)2 E2s + (0.6)2 E2p = 0.64E2s + 0.36 E2p. The contribution of the AO in the hybrid is given by the square of the coefficient of AOs participating in the hybrid. The above hybrid may actually be represented as 2s 0.64 2p 0.36. The other feature you may have noticed is that starting with two AOs, you can create two and only two hybrids.

 

Orbitals can neither be created nor destroyed! The other thing that you might be wondering is s whether you can understand molecules using MOs alone. There is another approach called the valence bond approach (which uses electron pair [valency!] functions or non-linear combination of atomic orbitals to begin with) but that is perhaps more complicated to introduce than the MO approach. However, both approaches or theories give the same results which agree closely with experimental energy levels of most small molecules.

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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