MO theory for polyatomics
Orbitals are treated as objects that can be labeled by their irreducible representations
This can be done within either the VBT or LCAOMO approaches
VBT: Find the irreducible reps of the bond pairs; this allows us to label the bonds into different energy groups but gives no indication of the relative energy of each type of irred. rep.
CH_{4}
T_{d} 






a_{1} 







a_{2} 






e 







t_{1} 







t_{2} 







bonds 





These transform as a_{1} + t_{2}
a_{1} is also an s orbital; t_{2} is also p_{x},
p_{y}, p_{z} orbitals which implies an sp^{3} hybrid
SF_{4} find irred reps of bonds; find hybrid
C_{2v} 





a_{1} 






a_{2} 






b_{1} 






b_{2} 






bonds 





lp 




bonds 2a_{1} + b_{1} + b_{2 }sp^{3}
hybrid?
lp a_{1} therefore dsp^{3} hybrid!
LCAO MO: attempt to use symmetry to convert polyatomic problems into something like a diatomic problem.
1. Break the molecule up into two parts: the center atom and all the atoms bonded to it.
2. Find the irreducible representations of the basis orbitals on the center atom, generally by inspection of the character table.
3. Create "Group Orbitals" from the basis orbitals on the peripheral atoms using symmetry operations.
4. Estimate overlaps remembering that only orbitals of the same irreducible rep can overlap with each other.
H_{2}O
C_{2v} 





a_{1} 






a_{2} 






b_{1} 






b_{2} 






Basis set: H, 1s; O 2s, 2p
Irreducible reps of the O basis orbitals are read directly from the character table:
2s – a_{1}
2p_{z} – a_{1}
2p_{x} – b_{1}
2p_{y} – b_{2}
The irreducible reps of the H orbitals must be done as a group since they are not located at the "center" of the molecule.
Reducible rep of H 1s orbitals:
E  C_{2}  _{v}(xz)  _{v}'(yz)  
1s Group  2  0  0  2 
This transforms as a_{1} + b_{2}
What do the a_{1} and b_{2} group orbitals look like?
Use projection operators:
Note that the sum is over all rotations, not just classes!
Use the rotation operator on the H s orbitals:
E  C_{2}  _{v}(xz)  _{v}'(yz)  
s_{1}  s_{2}  s_{2}  s_{1} 
Since s_{1} and s_{2} are connected by symmetry, we do not need to use the rotation operator on s_{2}.
Now, use the projection operator for each group orbital:
The wavefunctions are the result of the projection operator normalized:
(a_{1}) = 2N(s_{1}+s_{2}), where N is the normalization constant.
Normalized wavefunctions are found by squaring the function and setting it equal to 1:
^{2} = 1 = 4N^{2}(s_{1}+s_{2})^{2}
s_{1}^{2} = s_{2}^{2} = 1; s_{1}s_{2} = 0, so
Likewise