Draw a Jablonski diagram for formaldehyde? Label all the radiative and radiationless processes. Assume the extinction coefficients and quantum yields for all the absorptions and emissions (respectively) are equal. Sketch the UV-Vis spectrum.

Well, let me give this a shot...

DISCLAIMER: LONG ANSWER (obviously)!

#a)#

First of all, a Jablonski Diagram shows the vibrational states (the denser energy levels) amongst the relevant electronic states (#S_0, S_1, S_2, T_1#, i.e. each entire group of vibrational energy levels):

• an electronic ground-state singlet #S_0#.
• an electronic excited-state triplet #T_1# higher in energy.
• an electronic excited-state singlet #S_1# higher in energy than that.
• another electronic excited-state singlet #S_2# of higher energy than that.

In this case you are asked for only the first four vibrational levels, rather than the five or six shown here. The key features of the diagram are outlined below.

Make sure you know the terminology for the types of transitions.

• A bold line indicates the ground vibrational level for that particular molecular state (#S_1#, #T_1#, etc).
• An upwards solid arrow indicates an excitation (absorption, radiative).
• A downwards solid arrow indicates a relaxation (fluorescence or radiative emission).
• Any wavy arrow indicates a radiationless process (internal conversion, vibrational relaxation, and intersystem crossing).
• A dashed arrow indicates a very slow radiative process (phosphorescence).

In using formaldehyde as an example, I would use NIST to look at the vibrational states.

This reference seems to go over the electronic and vibrational states in more detail.

The first four vibrational modes, and hence energy levels are:

• #nu_4 = "1167 cm"^(-1)# (#"CH"_2# wagging mode under #B_2# symmetry)
• #nu_6 = "1249 cm"^(-1)# (#"CH"_2# rocking mode under #B_1# symmetry)
• #nu_3 = "1500 cm"^(-1)# (#"CH"_2# scissoring mode under #A_1# symmetry)
• #nu_2 = "1746 cm"^(-1)# (#"C"="O"# #A_1# stretching mode under #A_1# symmetry)

i.e. #E_"vib"(nu_2) > E_"vib"(nu_3) > E_"vib"(nu_6) > E_"vib"(nu_4)#.

I can't find exact electronic energy levels for it, although I can give you the names of those states:

• #""^1 A_1# ground singlet state
• #""^3 A_2# excited triplet state
• #""^1A_2# excited singlet state

i.e. #E_"elec"(""^1A_2) > E_"elec"(""^3 A_2) > E_"elec"(""^1 A_1)#.

I think that's about fine... just be sure that the #""^1 A_2# manifold of vibrational states is higher in energy than the #""^3 A_2# manifold of vibrational states, which is higher in energy than the #""^1 A_1# manifold of vibrational states.

In other words, let #S_0 = ""^1A_1#, #T_1 = ""^3 A_2#, and #S_1 = ""^1A_2#, omitting #S_2#.

#b)#

I'll let you do this one yourself. I think the hard part is outlining the blank diagram.

Follow the Jablonski diagram example, and note that

• The radiative processes to include are absorption, fluorescence, and phosphorescence.
• The radiationless processes to include are vibrational relaxation and intersystem crossing. There is not a third singlet state so you cannot have internal conversion.

Again, let #S_0 = ""^1A_1#, #T_1 = ""^3 A_2#, and #S_1 = ""^1A_2# in the example Jablonski Diagram, omitting #S_2#.

#c)#

In assuming the extinction coefficients and quantum yields for all the absorptions and emissions (respectively) are equal, you are only asked for the approximate shape and relative wavelengths.

I would expect vibronic excitations---absorbances---of (in order of increasing energy in the order as listed below):

#DeltaE_((nu_4, ""^1A_1) -> (nu_4, ""^1A_2)) < DeltaE_((nu_4, ""^1A_1) -> (nu_6, ""^1A_2))#

#< DeltaE_((nu_4, ""^1A_1) -> (nu_3, ""^1A_2)) < DeltaE_((nu_4, ""^1A_1) -> (nu_2, ""^1A_2))#

with the y-axis being % absorbance.

Thus,

#lambda_((nu_4, ""^1A_1) -> (nu_4, ""^1A_2)) > lambda_((nu_4, ""^1A_1) -> (nu_6, ""^1A_2))#

#> lambda_((nu_4, ""^1A_1) -> (nu_3, ""^1A_2)) > lambda_((nu_4, ""^1A_1) -> (nu_2, ""^1A_2))#

For the vibronic relaxations---fluorescences---simply flip your absorbance graph upside-down (keep the axes where they are), then change the y-axis to % transmission.

Keeping in mind that there are

• symmetry-forbidden transitions (#DeltaS ne 0# is not allowed)
• spin-forbidden transitions (#g -> g#, #u -> u# are not allowed)

and that formaldehyde has no inversion symmetry, you can therefore assume that the absorption and fluorescence transitions are spin-allowed, but some are symmetry-forbidden.

• The symmetry-forbidden ones should not be drawn (#""^1 A_2 -> ""^3 A_2# is forbidden, since #DeltaS = +2#).
• The symmetry-allowed ones are based on #""^1 A_2 -> ""^1 A_1#, which are precisely the four shown above.

#d)#

In this case we are asked for mechanistic sequences (one or two steps). Let #Psi^("*"(1))# be the #""^1 A_2# state, #Psi^("*"(3))# be the #""^3 A_2# state, and #Psi^0# be the #""^1 A_1# ground state.

Recall which ones are radiative and which are radiationless:

RADIATIVE

• absorption
• fluorescence
• phosphorescence

Therefore, an example is (DRAW THIS!):

#Psi^("*"(1))(nu_2, ""^(1) A_2) -> Psi^0(nu_4, ""^(1) A_1)#
(fluorescence, from fourth vibrational level in second excited state to first vibrational level in ground state) [radiative]

RADIATIONLESS

• internal conversion
• intersystem crossing
• vibrational relaxation

Here is an example (DRAW THIS!):

1. #Psi^("*"(1))(nu_4, ""^(1) A_2) -> Psi^("*"(3))(nu_4, ""^(3) A_2)#
   (intersystem crossing between vibrational ground states from the second excited state to the first excited state) [radiationless]

2. #Psi^("*"(3))(nu_4, ""^(3) A_2) -> Psi^0(nu_4, ""^(1) A_1)#
   (phosphorescence between vibrational ground states from the first excited state to the ground state) [radiative]