Question

please explain your answer. This is quantum chemistry.

12. The electronic term energy, Te is the energy of the potential energy minimum of an excited state relative to the potential energy minimum of the ground state. Essentially, it is the electronic energy required to excite a molecule to an electronically excited state. The ground and some of the lowest-lying excited states of CO are Configuration Term Electronic cmm 48687 55825 A 61120 64230 65076 Δ 165928 Il (a). Which (if any) of these excited states can be reached by optical excitations from the ground state? (b). Estimate the wavelength (in nm) at which a sample of CO gas will begin to absorb. Do you expect this gas to be transparent in the region of the ultraviolet spectrum that is transmitted through air (200-400 nm)?

Answer 1

a) Electron transition obey selection rules.

In this case two selection rules will have to be applied

Angular momentum selection rule:

The transitions with   $\Delta \Lambda = 0,\pm 1$ are allowed.

For example the following transition will be allowed

$\Sigma \leftrightarrow \Sigma$ , $\Sigma \leftrightarrow \Pi$ , $\Delta \leftrightarrow \Pi$ etc.

Spin selection rule:

The transitions with $\Delta S = 0$ are allowed. This means singlet to singlet and triplet<->triplet transition are allowed.

Applying these two rules for the given states we will find that only $^{1}\Sigma \rightarrow ^{1}\Pi$ transition is allowed.

b) The energy difference for the $^{1}\Sigma \rightarrow ^{1}\Pi$ transition is 65076 cm^-1

Wavelength for this transition = 1/ 65076 cm = 153.66 nm

Therefore gas will be transparent in the 200-400 nm region.