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Chemistry LibreTexts

Solutions 4

  • Page ID
    52462
  • 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.