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Walsh Diagrams

This is a molecular orbital interpretation for the geometries of the ground and excited states of molecules. We would like to be able to predict whether the molecule XH2 is linear or bent. The idea is to observe that the linear molecule HXH has a higher symmetry (D $_{\infty h}$) than the bent XH2 molecule (C2v). Molecular orbitals that do not interact in D $_{\infty h}$ because they are of different symmetry, may interact in C2v because they then have the same symmetry. We use the following axes
 
 
 
 
 The symmetries of the symmetry orbitals in the two situations are
 
  $_{\infty h}$ C2v
2sX $\sigma_g$ a1
2pzX $\pi_u$ a1
2pxX $\pi_u$ b1
2pyX $\sigma_u$ b2
1sA-1sB $\sigma_u$ b2
1sA+1sB $\sigma_g$ a1
 
The molecular orbitals of the linear molecule are in ascending order of energy $1\sigma_g,1\sigma_u,\pi_u,2\sigma_g^*,2\sigma_u^*$. On bending, one of the $\pi_u$ orbitals, 2pzX, has the same symmetry as the $2\sigma_g^*$, c(1sA+1sB)-2sX, orbital and interacts with it forming a strongly bonding orbital viz
 
 
 Thus the orbital energy ordering of the bent molecule is 1a1,1b2,2a1,1b1,3a1*,2b2*. If the molecule has more than 4 valence electrons its ground state will be bent, the degree of which will depend upon the occupancy of the 2a1 orbital. Thus we have the following results for XH2 molecules.
BeH2....... $1\sigma_g^21\sigma_u^2$....... $^1\Sigma_g^+$ linear
BH2........1a121b222a11.........2A1.....bent, 131o
..............1a121b221b11..........2$_1\equiv\ ^2\Pi_u$ linear
These two states form a Renner-Teller pair, two separate states which become degenerate at linearity.
 
 
  CH2........ 1a121b222a111b11....3B1......bent, 134o, this state is lowest in energy because of the exchange integral (Hund's rule)
..............1a121b222a12.......1A1.......bent, 102o, 9kcal/mol higher in energy.
NH2........1a121b222a121b11.....2B1.....bent, 103o
...............1a121b222a111b12....2A1.....bent, 144o
These are again two components of a Renner-Teller pair; they are degenerate $^2\Pi_u$ at linearity.
 
 
 
 OH2.........1a121b222a121b12......1A1...bent, 105o
The more the 2a1 orbital is occupied, the more strongly the molecule is bent and the higher the bending vibrational frequency. Thus for H2O we have the following configurations and bending vibrational frequency.
H2O........1A1.......1a121b222a121b12...1595cm-1
H2O+......2B1......1a121b222a121b11....1431cm-1
H2O+......2A1......1a121b222a111b12....873cm-1
If an excited state has a very different geometry from the ground state, then the electronic transition will be broad as many vibrational energies of the excited state are accessible. Thus the in photoelectron spectrum a band corresponding to the removal of an electron from the nonbonding 1b1 orbital is narrow, while one corresponding to the removal of an electron from the bonding 2a1 orbital is broad, with many bending states of the ion being formed. ( a vibrational progression). This is demonstrated in the photoelectron spectrum of H2O.
 
 
 The first band on the right corresponds to the removal of an electron from the least stable orbital 1b1. It is fairly sharp, as the geometry of the molecule does not change much on ionisation. The middle band shows many bending vibrations, corresponding to the removal of an electron from the 2a1 orbital which most effects the bending. The lefthand band corresponds to the removal of an electron from the 1b2 bonding orbital, and shows a progression in the stretching vibration which is not very well resolved.


next up previous
Next: Walsh Figure Up: Contents - Previous: Orbital Energy Level Diagrams

Nicholas Handy

1998-09-23