Solutions 4

Q1

1. T
2. T
3. F, the overlap integral is zero. $\beta$ pertains to energy, not overlap.
4. T
5. F, $\beta$ is experimentally found to be negative.
6. F, the resonance integral pertains to energy, so it is $\beta$ .
7. F, non-adjacent resonance integrals are zero.
8. T

Q2

In the ethylene system, there are two p orbitals to consider, $|1\rangle$ and $|p\rangle$, each with energy E.

The Hamiltonian, $\hat{H}$, is given by:

$\begin{bmatrix} \alpha& \beta\\ \beta& \alpha\end{bmatrix}$

We know that $\hat{H} |\psi\rangle = E|\psi\rangle$ . Factoring, we have $\hat{H} -EI|\psi\rangle = | 0 \rangle$, where I is the identity matrix.

This implies that $|\psi\rangle = (\hat{H} -EI)^{-1} |0\rangle$. For the left side to be non-zero, the inverse matrix that acts on the zero vector must be infinite. From linear algebra, the inverse of a matrix is inversely proportionate to its determinant:

$(\hat{H} -EI)^{-1} = \frac{C}{det(\hat{H} -EI)}$. For the inverse to be infinite, we require $det(\hat{H} -EI) = 0$.

Q3

a). $\hat{H}$ for a 5 carbon system by the Huckel Theory, is given by:

$$\begin{bmatrix}\alpha&\beta&0&0&0\\\beta&\alpha&\beta&0&0\\0&\beta&\alpha&\beta&0\\0&0&\beta&\alpha&0\\0&0&0&\beta&\alpha\end{bmatrix}$$

b). Given the following wavefunctions, choose two and show they are normalized:

• $| \pi_1 \rangle = +0.45 | 1 \rangle +0.45 | 2 \rangle + +0.45 | 3 \rangle +0.45 | 4 \rangle +0.45 |5 \rangle$
• $| \pi_2 \rangle = -0.51 | 1 \rangle + 0.63 | 2 \rangle -0.51 | 3 \rangle + 0.20 | 4 \rangle + 0.20 | 5 \rangle$
• $| \pi_3 \rangle = -0.37 | 1 \rangle +  0.37 | 3 \rangle + 0.60 | 4 \rangle + 0.60 | 5 \rangle$
• $| \pi_4 \rangle = -0.21 | 1 \rangle - 0.63 | 2 \rangle -0.20 | 3 \rangle + 0.51 | 4 \rangle + 0.51 | 5 \rangle$
• $| \pi_5 \rangle = +60 | 1 \rangle -0.60 | 3 \rangle + 0.37 | 4 \rangle + 0.37 | 5 \rangle$

The dot products to show normalization are:

• $\langle1|1\rangle$ is: 1.012500e+00
• $\langle2|2\rangle$ is: 9.971000e-01
• $\langle3|3\rangle$ is: 9.938000e-01
• $\langle4|4\rangle$ is: 1.001200e+00
• $\langle5|5\rangle$ is: 9.938000e-01

c). Show that two are orthogonal to each other:

Actually calculating the dot products we see they are not perfectly orthogonal, some with great intersection.

• $\langle1|2\rangle$ is: 4.635000e-01
• $\langle1|3\rangle$ is: 5.400000e-01
• $\langle1|4\rangle$ is: 1.710000e-01
• $\langle1|5\rangle$ is: 3.330000e-01
• $\langle2|3\rangle$ is: 6.174000e-01
• $\langle2|4\rangle$ is: 1.620000e-02
• $\langle2|5\rangle$ is: -4.640000e-01
• $\langle3|4\rangle$is: 7.637000e-01
• $\langle3|5\rangle$ is: 0
• $\langle4|5\rangle$ is: 1.314000e-01

d). Use the Hamiltonian matrix to calculate the distinct energies using these eigenfunctions.

We can find the energies of this system by setting $\hat{H} - EI = 0$.

$$\begin{vmatrix}\alpha-E&\beta&0&0&0\\\beta&\alpha-E&\beta&0&0\\0&\beta&\alpha-E&\beta&0\\0&0&\beta&\alpha-E&0\\0&0&0&\beta&\alpha-E\end{vmatrix} = 0$$

Dividing the matrix by $\beta$ and using the variable $x = \frac{\alpha - E}{\beta}$, we can solve a determinant of the form:

$$\begin{vmatrix}x&1&0&0&0\\1&x&1&0&0\\0&1&x&1&0\\0&0&1&x&0\\0&0&0&1&x\end{vmatrix} = 0$$

Continuing to solve for x by breaking down the determinant into smaller determinants, (also known as its "minors" or Laplace's formula) you will arrive after careful factoring (see https://people.richland.edu/james/le...terminant.html):

$$x(x^2-3)(x^2-1) = 0$$

The energies relate to the other variables via: $E = \alpha - x\beta$

The roots are $x = 0, \pm 1, \pm \sqrt{3}$, so the energies are $\alpha, \alpha \pm \sqrt{3}\beta, \alpha \pm \beta$

Q4

The allyl cation can be described by the Hamtilonian:

$$\hat{H} = \begin{bmatrix}\alpha&\beta&0\\\beta&\alpha&\beta\\0&\beta&\alpha\end{bmatrix}$$

Similar to the previous problem, we must again solve for $det(\hat{H}-EI) = 0$, which results in:

$$\begin{vmatrix}x&1&0\\1&x&1\\0&1&x\end{vmatrix}= x(x^2-2)= 0$$

The roots are $x = 0,\pm\sqrt{2}$, which correspond to energies: $\alpha + \sqrt{2}\beta, \alpha, \alpha - \sqrt{2}\beta$ which correspond to bonding, non-bonding and anti-bonding orbitals. The deolocalization of the electrons lowers the energy of the system to 0.82 $\beta$ when compared to ethylene.