5. Answer ALL parts:

(a)    Using Hückel theory, we can describe the 几 electrons in delocalised 几 systems of organic

molecules such as the two structural isomers trimethylenemethane and                             methylenecyclopropene shown in Figure 1. Both molecules are assumed to be planar and neutral.

Figure 1

(i)  Give the secular determinants in terms of Coulomb integrals a and Resonance integrals F for both molecules using the numbering scheme in Figure 1 and the substitution X = (a − E)/F.

(ii) The respective roots of the polynomial equations are:

Trimethylenemethane: X1 = 0, X2 = 0, X3 = 1.73, X4 = − 1.73

Methylenecyclopropene: X1 = 1, X2 = 1.48, X3 = −0.31, X4 = − 2.17

Find the energies of the molecular orbitals in terms of Coulomb integrals a and the resonance integrals F .

(iii)    Draw a molecular orbital diagram showing the energies of the molecular orbitals and how the 4 π electrons occupy these orbitals. (Remember that the resonance integral F has a negative value.)

(iv)    Determine which of the two molecules is alternant or non-alternant. List the

properties of alternant and non-alternant molecules.

(v)    Calculate the total energy and the delocalization energy for both molecules. The       delocalization energy of butadiene is found to be 0.472F . Compare the two compounds against butadiene. Which compound is more stable and why?

[50%]

6. Answer ALL parts:

(a)    Element X forms a simple two-dimensional square lattice with lattice constant a (Figure

3). Element X is a second row element with electrons occupying s and p orbitals. These atomic orbitals linearly combine across unit cells to form delocalised bands in the solid. 

Figure 3

(i)   The reciprocal unit cell corresponding to this structure has the lattice constant b = 2几/a. Sketch the Wigner-Seitz cell in reciprocal space. Indicate the following four high-symmetry points: Γ = (0,0), X = ( , 0) , X, = (0, ) , M = ( , ).

(ii)  For the 9 unit cells and atoms shown in Figure 3, sketch all six different Molecular

Orbitals (MOs) listed below as linear combinations of atomic orbitals by drawing  either s or p orbitals at the positions of the atoms. Indicate the sign of the atomic orbital lobes by using an empty shape for positive and a filled shape for negative:

1.   The MO for the s band at the Γ k-point

2.   The MO for the px band at the Γ k-point

3.   The MO for the s band at the X k-point

4.   The MO for the  px band at the X k-point

5.   The MO for the s band at the M k-point

6.   The MO for the px band at the M k-point

(iii) Rank these six MOs by their expected relative energy. In your ranking refer to the MOs by their number. Justify your ranking based on the nodal structure of the MO visualisations in (ii).

(Hint: all px energies are higher than all s energies.)

(iv) Draw a qualitative band structure diagram with two bands, an s band and a px band that resemble the energy ranking in (iii). The y axis should plot the energy, the x

axis should go from Γ via X to M .                                                                [50%]

Model Answers:

Question 5

(a)

[similar to problems seen in lecture and practised in workshop]

(ii) The MO energies for Trimetylene methane are

E1 = a + 1.73F

E2 = a

E3 = a

E4 = a − 1.73F

The MO energies for Methylene cyclopropene are

E1 = a + 2.17F

E2 = a + 0.31F

E3 = a − F

E4 = a − 1.48F

(iii)

(iv) Trimethylene methane is alternant, methylene cyclopropene is non-alternant. Alternant molecules have higher point symmetries than non-alternant molecules and MO energies, which are              symmetrically paired around alpha. The respective MO coefficients are identical, except for sign   changes. In any neutral alternant molecule, the charge distribution is uniform.

(v) The total energy is

MO

EtOt = x fiEi

i

EdelOc = EtOt − (4a + 4F)

The total energy of trimethylene methane is E = 4a + 3.46F. The total energy of methylene cyclopropene is E = 4a + 4.96F, which means it is more stable.

The respective delocalisation energies are − 0.54F and 0.46F for trimethylene methane and methylene cyclopropene. The order of stability is therefore:

Methylene cyclopropene, butadiene, and trimethylene methane from most stable to least stable.  Methylene cyclopropene, despite being non-alternant shows higher bond orders and lower orbital

energies. Trimethylene methane has a degenerate pair of HOMO MOs, which means it cannot be stable in a planar form.

Question 6

(a) [seen in lecture and handout]

(a) (ii). 1