Theoretical UV/VIS Spectroscopy

Theoretical UV/VIS Spectroscopy Why is a Ruby Red When Chromium Oxide is Green? How Does a Ruby Laser Work? Goals of this Exercise: - Calculation of the energy of electronically excited states - Understanding problems of calculations of open shell systems - Completing a complex multi-step calculation project - Application of CASSCF to a realistic problem - Use of a truncated model system to explain properties of a bulk system Electronic Ground State of Cr(H2O)6 3+ The coloring components in both chromium(iii)oxide and ruby are Cr 3+ ions. The crystal structure of Chromium(III)oxide is a hexagonal close packing of O 2- ions with Cr 3+ ions placed in 2/3rds of the octahedral gaps. Thus the Cr 3+ ions are embedded in a (quasi- )octahedral environment of O 2- anions. According to ligand field theory, the proximity of the oxygen atoms will cancel the degeneracy of the d-orbitals of chromium ions and split them into three t2g and two eg2 orbitals. The energy difference between those two orbitals corresponds to an energy in the range of visible light, thus the ions can adsorb part of the spectrum and appear colored. We will approximate ligand field environment by looking at a [Cr(H2O)6] 3+ complex. This appears to be a valid choice, since the color of chromium(iii) ions in water is of a similar green than that of chromium(iii)oxide. Following Hund's rule, the three t2g-orbitals should be singly occupied, which means that we would have a quartet state and a so-called open shell system. Open shell calculations can be quite tricky and should always be done with extra care. This electronic ground state (how can we prove that the quartet state is the ground state?) can be calculated using a restricted open shell Hartree-Fock calculation (ROHF). This method is well suited in this case, since we expect the unpaired electrons to be localized on the chromium site. WARNING: For the following steps it is very important to monitor the results at each step very carefully, make sure that important symmetries are preserved and that the unpaired electrons are actually where they are supposed to be. Also it is advisable to make backup copies of checkpoint files at every step or it may be required to re-do the whole procedure in case a calculation goes wrong. Checkpoint files are continuously updated during a calculation and that can be easily "ruined" by an erroneous job. Step 1: Initial Geometry An octahedral geometry is much easier to construct in cartesian coordinates rather than through the z-matrix builder in Molden (you are free to try, it should make no difference to

the calculations how the coordinates are initially created for as long as they are correct). Building the input manually requires using a text editor and writing a file like this: %Chk=cr-h2o-6-quartet.chk %Mem=1GB #P ROHF/6-31G(d,p) 6D 10F Symm=Loose SCF=Tight GFINPUT IOp(6/7=3) Pop=Full [Cr(H2O)6]3+ quartet state. Cr 0.00 0.00 0.00 O 2.00 0.00 0.00 H 2.50 0.75 0.00 H 2.50-0.75 0.00 O -2.00 0.00 0.00... and so on Don't forget to add one empty line at the very end of the file. This input will request a single point calculation, the usual line " 0 1" representing an uncharged molecule in a singlet state, is now " " indicating a total charge of +3 (and reducing the total number of electrons accordingly) and 3 unpaired electrons (a quartet state). Due to the added hydrogen atoms, we don't have perfect octahedral symmetry. Depending on how the hydrogen atoms are placed, the point group should be either D2h or (better) Th. After the single point calculation is finished, inspect the singly occupied orbitals either visually with Molden or in the population analysis of the log file. Orbitals 40, 41, 42 are the singly occupied ones and they should have a very high d-character (one or more coefficients with absolute value larger than 0.3 in the d-symmetry basis functions on the chromium atom). For a perfect input you should get the three highest occupied states with d-character and the exact same (=degenerate) eigenvalue (the three tg2 states) and the two lowest unoccupied (the two eg states) with equal eigenvalue. If your input has D2h symmetry, they will not be identical, but very close and the eg states may lie higher. This is not not a problem per se, but a manifestation of how much the results of this kind of calculation depend in subtle ways on the input. What does that mean about the accuracy of your results, if you investigate a system where you have little a priori knowledge? Step 2: Swap occupied orbitals (only if needed) WARNING: This step is only needed in case not all of your singly occupied orbitals have sufficient d-character. If your initial coordinates are sufficiently close to the ideal octahedral environment, it should not be needed. See below for how to do this, if you need it anyway. Step 3: Partial Geometry Optimization with ROHF So far, we have placed the chromium and oxygen atoms at the experimental distances and then placed the water hydrogen atoms somewhere nearby at some (sloppily) guessed positions. We need to optimize these positions, but this has to be done without destroying the oxygen octahedron. One option to do this is to use ModRedundant option for the Opt keyword in Gaussian. We start from the previously generated checkpoint file and use an input like this:

%Chk=cr-h2o-6-quartet.chk %Mem=2GB #P ROHF/6-31G(d,p) 6D 10F Symm=Loose SCF=VeryTight GFINPUT IOp(6/7=3) Pop=Full Opt=(ModRedundant,Tight) Geom=Check Guess=Read [Cr(H2O)6]3+ quartet state. X 1 F X 2 F X 5 F X 8 F X 11 F X 14 F X 17 F The additional section tells the geometry optimization algorithm, to "fix" (not change) the x-, y-, or z-coordinates of atoms 1, 2, 5, 8, 11, 14, and 17, which are the chromium and the oxygens, respectively. The only movable atoms are the hydrogen atoms. Note that this not only fixes the octahedral shape, but also the Cr-O bond lengths. If you had previously swapped orbitals, you must not do this in this step, as that would undo the previous exchange and the geometry optimization will be not be based on the ground state. Step 4: Swap virtual orbitals (most likely needed) For the subsequent CASSCF calculation also the first two virtual orbitals have to be states with mostly d-character that are located on the chromium. They are most likely not the LUMO and LUMO+1 (orbitals 43 and 44) as required for a CASSCF calculation of this system. To check load the log file from the ROHF partial geometry optimization into Molden, click on the "Movie" button to advance to the final geometry, and then switch to density mode to inspect the lowest unoccupied orbitals that fit the requirements. In case orbitals need to be swapped, prepare another input for an abbreviated calculation will read the final wavefunction from the checkpoint file, modify it, and save it back. %Chk=cr-h2o-6-quartet.chk %Mem=1GB #P ROHF/6-31G(d,p) 6D 10F Symm=Loose GFINPUT IOp(6/7=3) Pop=Full Geom=Check Guess=(Read,Only,Alter,Save) [Cr(H2O)6]3+ quartet state. 47 44 48

This will read the wavefunction from the checkpoint file, swap orbitals 43 with 47 and 44 with 48 and then save the wavefunction back to the checkpoint file. Which orbitals will need to be swapped, if at all, depends strongly on the input geometry. Check the result with Molden to make sure that this is not having the orbitals in the desired order. CASSCF Calculations on the Chromium(III)oxide model system Step 1: Quartet state With the previous preparations we are now ready to preform the CASSCF calculation that will allow us to study the d-d-transitions of the Cr3+ ion. Consider which vertical excitations (i.e. those that remain a quartet) between the t2g and eg orbitals are possible. These excited states can no longer be represented by a single Slater determinant but rather a linear combination of multiple determinants. Keep in mind, that CASSCF is effectively a Hartree- Fock calculation and thus has all its limitations (e.g. no dynamical correlation). Thus to get results that compare more favorably with spectroscopic results, more sophisticated calculations are needed. These, however, are outside of the scope of this exercise. Make a backup copy of the checkpoint file and prepare an input for a state averaged 3 electron, 5 orbital CASSCF calculation: %Chk=cr-h2o6-quartet.chk %Mem=2GB #P CASSCF(3,5,StateAverage,Nroot=10)/6-31G(d,p) 6D 10F Symm=Loose GFINPUT IOp(6/7=3) Pop=Reg SCF=VeryTight Geom=Check Guess=(Read) [Cr(H2O)6]3+ cluster, quartet state. 0.100000 0.100000 0.100000 0.100000 0.100000 0.100000 0.100000 0.100000 0.100000 0.100000 The parameter Nroot determines the number of states to average over and the additional section in the input file assigns the individual weights. Unlike the rest of Gaussian this input is in a very strict format, each number and the surrounding blanks have to occupy exactly ten characters and there is a maximum of 8 numbers per line (For experts, this is a socalled Fortran 8F10.6 fixed format). Step 2: Doublet State We will later have a look at doublet states of this configuration. Since we have already a suitable wavefunction in our checkpoint file, we can start the CASSCF calculation of the doublet state right away. Why can we do this? What changes between the quartet and the doublet state of [Cr(H2O)6] 3+? We - again - have to perform a state averaged calculation, but for the doublet, we consider only 8 states to average over. Don't forget to adjust the weights.

Step 3: The Color of [Cr(H2O)6]3+ Color λ/nm ν/10 14 Hz ν/10 4 cm 1 E/eV E/kJ mol 1 Infrared >1000 <3.00 <1.00 <1.24 <120 Red 700 4.28 1.43 1.77 171 Orange 620 4.84 1.61 2.00 193 Yellow 580 5.17 1.72 2.14 206 Green 530 5.66 1.89 2.34 226 Blue 470 6.38 2.13 2.64 254 Violet 420 7.14 2.38 2.95 285 Near ultraviolet 300 10.0 3.3.15 400 Far ultraviolet <200 >15.0 >5.00 >6.20 >598 The energies of the individual states are given in the log file in the section EIGENVALUES AND EIGENVECTORS OF CI-MATRIX as eigenvalues. Compute the d-d excitation energies from that, convert them to more suitable units and check in the list from above what color of light chromium(iii)oxide should approximately adsorb. Keep in mind that since we compute the adsorption energy, so the color that we would see is the part that does not get adsorbed, i.e. the complementary color. The Color of Ruby Ruby is α-aluminium(iii)oxide (Corundum) where about 8% of the Al 3+ ions are replaced by Cr 3+ ions. The crystal structure is effectively the same as in chromium(iii)oxide only that in ruby the distance between the chromium cation and the oxygen anions is determined by the embedding Al2O3 lattice, i.e. shorter. To compute the color of ruby repeat the procedure from above for a Cr-O distance of 1.85 angstrom. Also note that CASSCF is not a terribly accurate method, and we likely benefit from some fortuitous cancellation of errors in these calculations. What type of errors are present in these calculations? The Ruby Laser Now compare the previously computed energy levels for the quartet and the doublet states. What happens to the lowest lying doublet states when the Cr-O distance shrinks from 2.0 to 1.85 angstrom? In the ruby laser, there is first a vertical excitation of the quartet ground state which then can relax into the lowest doublet state. Why does is this property required? Draw a diagram of the computed energy levels and explain briefly the 3-level solid state laser.

Questions 1) What happens to the total energy and the relative energy levels of the d-d transitions when the Cr-O distance shrinks from 2.0 to 1.85? Please also see the article L.E. Orgel, Nature, 179, p1348 (1957) on the subject. Why is it that Cr2O3 is green but ruby is red? 2) What happens to the lowest lying doublet states when the Cr-O distance shrinks from 2.0 to 1.85 angstrom? In the ruby laser, there is first a vertical excitation of the quartet ground state which then can relax into the lowest doublet state. Why does is this property required? Draw a diagram of the computed energy levels and explain briefly the 3-level solid state laser. 3) What sources of error are present in CASSCF calculations?