Calculating the Energies of Molecular Orbitals Friday we solved for the energies of the HF molecule using the basis set ψ = c 1φH1s + c 2 φF2p and obs...
Calculating the Energies of Molecular Orbitals Friday we solved for the energies of the HF molecule using the basis set ψ = c 1φH1s + c 2 φF2p and observed how the atomic orbital coefficients can be determined. The choice of basis set is completely up to the researcher. In this exercise you will extend our examination of the bonding in HF by increasing the size of the basis set to ψ = c 1φH1s + c 2 φF2p + c 3 φF2s. Start this exercise by writing on your Mathcad page the energies of the HF orbitals we obtained using the ψ = c 1 φH1s + c 2φF2p basis set. Look in our class notes from Friday. Enter the numerical values of the coulomb and resonance integrals in electron volts (eV).
EH1s := −13.6
E2p := −18.6
E2s := −40.2
β := −1.0
SAB := 0
SAA = 1
Create and solve the 3x3 secular determinant created by H1s,F2p , and F2s. The resonance integrals between F2s and F2p is set to zero for reasons of symmetry. Entering a determinant in Mathcad can be done from the x button on the Matrix tool bar. Inside the determinant enter a matrix by typing cntrl+m. To determine the energy E use Symbolic, Variable, S olve with the blue cursor on an E in the determinant
Solve the 3 x 3 determinant for the energies of the sigma orbitals created from Ψ σ = φΗ1s + φF2p + φF2s
Η 1s Η 1s F2p F2s
F2p
F2s
−1 −1 −13.6 − E −1 =0 −18.6 − E 0 0 −40.2 − E −1
At what decimal place is there a change in the orbital energies compared to the energies of the MO's created from the ψΗ1s + ψF2p wavefunction?
We have just solved for the energies of the three sigma orbitals in HF. Now we want to know how much of each atomic coefficient is required to make each of the molecular orbitals. The coefficients for the highest energy MO wavefunction is solved below. Repeat these calculations for determining the coefficients of this molecular orbital, but then go on to calculate the coefficients for the other two MO's in our HF molecule.
C H := .5
C 2p := .5
C 2s := .5
Initial guess
E := −13.37146 Given 2
2
2
Normalization condition
C H + C2p + C2s = 1 Secular equations −1 −1 −13.6 − E CH −1 ⋅ C2p = 0 −18.6 − E 0 0 −40.2 − E C2s −1
You now need to determine the coefficients for the 2σ and 1σ molecular orbitals. How do these compare to the coefficients of the two term basis set? Would there be a significant improvement in the energies if the fluorine 1s orbital or the hydrogen 2s orbital were included in the basis-set? 3σ C H = 0.982 C 2p = −0.188 C 2s = −0.037
2σ
1σ
For the work we have just done it is pretty nice to have used a computer to solve these simultaneous equations. However, MathCad has built-in-functions that make solving these equations really quick. Below is how the functions: Eigenvals and Eigenvecs, solve for the energies and coefficients of molecular orbitals. Create a matrix, M, that is the secular determinant. This is done without including the energy terms along the diagonal. The secular equations for HF can be given in a matrix in the following way. −1 −13.6 −1 M := −1 −18.6 0 0 −40.2 −1
Notice that the diagonal terms do not have the molecular orbital energies, E.
The function eigenvals(M) returns the molecular orbital energies.
−40.238 eigenvals( M ) = −18.791 −13.371 Hey that was quick. The coefficients are much more easily obtained using the eigenvecs(M) function. Enter the eigenvecs function as is done below.
0.188 −0.982 0.038 −3 eigenvecs( M ) = 1.737 × 10 0.982 0.188 −3 −8.759 × 10 0.037 0.999 Reading this output requires some explanation. Each column is a specific molecular orbital and each row is for a specific atomic orbital.
You can, of course, use these two new matrix functions for the rest of our secular equation calculations.
Huckle calculations of butadiene and cyclobutadiene Butadiene has four π electrons. Create the 4 x 4 secular determinants for π orbitals of butadiene and cyclobutadiene. Use the Huckel approximations for the terms of the secular determinant.
Solve the secular determinant for the four π-molecular orbital energies in terms of α and β for both compounds. You can compare your results 1,3 butadiene with Model 2 in ChemActivity 15. Recall that both α and β are negative numbers and positive β terms will be of lower energy. For the carbon 2p orbital the values of α and β in electron volts (eV) are: α := −11.26
β := −2.5
Secular equations matrix for butadiene: C1 C2 C3 C4
α β M := 0 0
β 0 0 C1
α β 0 C2 β α β C 3 0 β α C 4
butadiene molecular orbitals
Molecular orbital energies for butadiene are calculated with the eigenvals function
Solve the same secular equations for cyclobutadiene. Create the molecular orbital diagram (as is done above) for both butadiene and cyclobutadiene. Use your molecular orbital diagram to answer the following questions: Which molecular orbital is lowest in energy?
Place the four π-electrons into the orbitals for butadiene and cyclobutadiene. What is the spin multiplicity of butadiene and cyclobutadiene? Add up the energies of each electron. What are the total energies for butadiene and cylcobutadiene? Which compound is more stable? Which cation would you predict to be more stable, butadiene or cyclobutadiene? Explain your reasoning. ChemActivity 15 provides the LCAO coefficients for butadiene on page 137. Calculate the LCAO coefficients for cyclobutadiene. Make a sketch (by hand) of the four molecular orbitals of cyclobutadiene similar to the second MO in butadiene is sketched in ChemActivity 15 CTQ-2. Two of the orbitals in cyclobutadiene are considered non-bonding. Explain how these orbitals are non-bonding from the AO coefficients of these two orbitals.
