<p>Show that this action leaves the s-state $|s \rangle$ unaltered, while a p-state $|p \rangle = \sum_{i \in \{x, y, z \}} a_i |p_i \rangle$ is transformed to $|p' \rangle = \sum_{i, j \in \{x, y, z\}} (R_{ij}{a_j}) |p_i \rangle = \sum_{i \in \{x, y, z \}} a_i' |p_i \rangle$. In other words taking the coefficients $a_i$ as a vector, $R$ acts as a matrix multiplication from the left side on $\vec{a}$ giving the new vector $\vec{a}'$.</p>
<p>$\textit{Hint:}$ The $s$-state $|s(\vec{r})\rangle = s(r)$ doesn't have an angular dependence, while the $p$-states can be written as $|p_i(\vec{r}) \rangle = r_ip(r)$ (with $i = x, y, z$), where $p(r)$ doesn't show any angular dependence.</p>
<p>b) We first look at the $sp$ hybridization, where we construct two hybrid orbitals out of a $|s \rangle$ and a $|p_z \rangle$ orbital</p>
<p>This hybridization is used, if a central atom forms bonds with two equal partners on both its sides of a given axis (here z-axis). The symmetry constraint is hence that $|sp_2 \rangle$ can be obtained from $|sp_1 \rangle$ by a $180^°$ rotation. Using this together with the orthonormality condition, as well as the requirement that all coefficients are real, derive coefficients $s_1, s_2, p_{z1}, p_{z2}$.</p>
<p>c) Simiarly the $sp^2$ hybridization constructs the 3 hyrbid orbitals out of the $|s \rangle$, $|p_x \rangle$, $p_y \rangle$ orbitals</p>
<p>This hybridization is used, if the central atom forms 3 equivalent bonds with partners in the same plane. Proceed as in exercise b) and find the coefficients $s_i, p_{xi}, p_{yi}$.</p>
<p>$\textit{Remark:}$ Note, there is an additional degree of freedom to choose one of the hybridized orbitals to point along a specific direction in the plane. We will put $p_{y1} = 0$, which means that $|sp^2_1 \rangle$ will point along the $x$-direction.</p>
<p>d) Finally we look at the $sp^3$ hybridization, which constructs the 4 hyrbid orbitals out of the $|s \rangle$, $|p_x \rangle$, $|p_y \rangle$, $|p_z \rangle$ orbitals.</p>
<p>This hybridization is used, if the central atom forms 4 equivalent bonds with partners yielding a tetrahedral symmetry. Proceed as in exercise b) and find the coefficients.</p>
<p>$\textit{Remark:}$ Keep in mind that there are three rotational degrees of freedom. You may choose them such that the calculation will be as simple as possible. We propose to start with $p_{x1} = p_{y1} = p_{z1}$, which fixes two of the three degrees of freedom.</p>
<p>e) Plot the orbitals of the $sp$, $sp^2$ and $sp^3$ hybridization.</p>
<div class="highlight hl-ipython3"><pre><span></span><span class="c1">#%matplotlib widget use matplotlib widget if you want to interactively rotate the orbitals</span>
<div class="highlight hl-ipython3"><pre><span></span><span class="c1">#%matplotlib widget use matplotlib widget if you want to interactively rotate the orbitals</span>
<div class="highlight hl-ipython3"><pre><span></span><span class="c1">#%matplotlib widget use matplotlib widget if you want to interactively rotate the orbitals</span>
<p>$$\textit{In this exercise we determine the structure of different molecules using the knowledge on the different forms of hybridization from exercise 1.}$$</p>
<p>We are interested in the structure of the following molecules $\text{C}_2\text{H}_6 (\text{H}_3 \ce{C-C} \text{H}_3)$, $\text{N}_2\text{H}_4 (\text{H}_2 \ce{N-N} \text{H}_2)$, $\text{H}_2\text{O}_2 (\text{H} \ce{O-O}\text{H})$, $\text{H}\text{C}\text{N} (\text{H}\ce{C#N})$ and $\text{H}_2\text{C}\text{C}\text{C}\text{H}_2 (\text{H}_2 \ce{C=C=C}\text{H}_2)$.</p>
<p>a) For every atom ($\text{C, N, O}$) determine the hybridization, such that the bond orders given in the brackets can be satisfied.</p>
<p>b) Sketch the 3D structure of the molecules.</p>
<p>$\textit{Remark:}$ You can use this interactive website <a href="https://app.molview.com/">https://app.molview.com/</a> to visualize your molecules.</p>
<p>$$\textit{We use the Hückel model for computing of the energy spectra and charge distributions of $\pi$-conjugated molecules.}$$</p>
<p>In first part of this exercise we will study the butadiene molecule $\text{C}_4\text{H}_6$ (see Figure 1).</p>
<p>ai) According to the Hückel model, set up the Hamiltonian for butadiene describing the $p_z$ orbitals of the carbon chain. Write the Hamiltonian in terms of the onsite energies $\alpha$ and the hoppings between two neighbouring carbon atoms $\beta < 0$.</p>
<p>aii) Calculate the eigenvalues of the matrix.</p>
<p>aiii) Keeping in mind that each molecular orbital can be populated by two electrons, what is the energy of the HOMO (highest occupied molecular orbital) and the LUMO (lowest unoccupied molecular orbital) in the ground state. What is the HOMO-LUMO energy difference (HOMO-LUMO gap)?</p>
<p>aiv) We will now examine the eigenvectors: $(a, b, b, a), (a, -b, b, -a), (b, a, -a, -b), (b, -a, -a, b)$ with $a, b > 0$. Without calculating them explicitely, can you order them in energetically ascending order.</p>
<p>The second part of this exercise will be concerned with the benzene molecule (see Figure 2)</p>
<p>b) Numerically calculate the molecular orbital energies and wavefunctions. To do so, use the provided code and setup the corresponding Hückel matrix. We will take the common approximation $\beta = -2.7$ eV for the hoppings, while $\alpha$ is set to zero. You can use the second python cell to visualize your results.</p>
<p>The last part of this exercise is concerned with naphtalene and azulene(see Figure 3 and 4) that are isomers, This means that both have the same molecular formula $\text{C}_{10} \text{H}_8$, but different structures. Furhtermore all carbon atoms in the two molecules have $sp^2$ hybridization. Nevertheless we will see that their energy spectra and molecular orbitals, and hence their physical properties, are quite different. We will demonstrate this with the example of the electrical dipole moment.</p>
<p>ci) Set up the Hückel Hamiltonian and calculate the spectra and eigenvectors for both molecules. Use $\beta = -2.7$ eV and onsite energy $\alpha = 0$ eV.</p>
<p>cii) Compare the sizes of the HOMO-LUMO energy gaps of the two molecules.</p>
<span class="nb">print</span><span class="p">(</span><span class="s1">'The band gap of azulene is'</span><span class="p">,</span> <span class="p">,</span> <span class="s1">'eV'</span><span class="p">)</span>
<span class="nb">print</span><span class="p">(</span><span class="s1">'The band gap of naphtalene is'</span><span class="p">,</span> <span class="p">,</span> <span class="s1">'eV'</span><span class="p">)</span>
<p>The azulene molecule has a remarkably big dipole moment, which we will try to estimate in the following</p>
<p>ciii) Numerically define the positions of the carbon atoms assuming all (nearest neighbour) $\text{C}-\text{C}$ distances are $l = 1.39$ Ang. You can use the provided code snipplets.</p>
<div class="highlight hl-ipython3"><pre><span></span><span class="c1">## If the hybridization orbitals are displayed (from exercise 1), please rerun the cell </span>
<p>civ) Use the results from ci) to calculate the electron occupation for each carbon atom in the ground state. Deduce from this the dipole moment of azulene.</p>
<p>$\text{Hint:}$ The electron occupation $\rho_i$ of an atom $i$ is given by</p>
<p>$$\rho_i = 2\sum_j |\psi_j(i)|^2$$</p>
<p>where the sum extends over the occupied molecular orbitals $\psi_j$ and the factor 2 is accounting for the spin degeneracy. In the exercise, $\psi_j(i)$ corresponds to the $i$-th entry of the $j$-th eigenvector. Using the electron occupation $\rho_i$ the dipole moment is furthermore given by</p>
<div class="highlight hl-ipython3"><pre><span></span><span class="c1">#cv) ## If the hybridization orbitals are displayed (from exercise 1), please rerun the cell </span>