1. Theoretical Foundations
The Schrödinger equation for any multi-electron system cannot be solved exactly because of electron-electron repulsion terms. Even for the simplest molecule-like system, the hydrogen molecular ion \(\text{H}_2^+\) (which consists of two nuclei and one electron), approximations are necessary to obtain analytical expressions. The Linear Combination of Atomic Orbitals (LCAO) approximation is a key variational method to construct molecular wavefunctions.
Under the Born-Oppenheimer approximation, we assume the nuclei are clamped in space relative to the much faster-moving electron. The molecular orbital wavefunction \(\psi\) is expressed as a linear combination of normalized atomic orbitals \(\phi_A\) and \(\phi_B\) centered on nuclei A and B: \[\psi = c_A \phi_A + c_B \phi_B\] where \(c_A\) and \(c_B\) are variational coefficients determined by minimizing the total energy.
2. Deriving the Energy States (Secular Equations)
To find the coefficients \(c_A\) and \(c_B\) and the corresponding energies, we use the variational principle:
\[E = \frac{\langle \psi | \hat{H} | \psi \rangle}{\langle \psi | \psi \rangle} = \frac{\int \psi^* \hat{H} \psi d\tau}{\int \psi^* \psi d\tau}\]
Substituting \(\psi = c_A \phi_A + c_B \phi_B\) yields:
\[E = \frac{c_A^2 H_{AA} + c_B^2 H_{BB} + 2c_A c_B H_{AB}}{c_A^2 S_{AA} + c_B^2 S_{BB} + 2c_A c_B S_{AB}}\]
where we define the following integrals:
• Coulomb Integral: \(H_{AA} = \int \phi_A^* \hat{H} \phi_A d\tau\) and \(H_{BB} = \int \phi_B^* \hat{H} \phi_B d\tau\). Since the molecule is homonuclear, \(H_{AA} = H_{BB}\).
• Resonance (Exchange) Integral: \(H_{AB} = H_{BA} = \int \phi_A^* \hat{H} \phi_B d\tau\). This represents the interaction energy when the electron is shared between nuclei.
• Overlap Integral: \(S_{AB} = S_{BA} = S = \int \phi_A^* \phi_B d\tau\) (since atomic wavefunctions are normalized, \(S_{AA} = S_{BB} = 1\)).
Minimizing the energy with respect to the coefficients (\(\frac{\partial E}{\partial c_A} = 0\) and \(\frac{\partial E}{\partial c_B} = 0\)) leads to the set of linear secular equations: \[c_A(H_{AA} - E) + c_B(H_{AB} - ES) = 0\] \[c_A(H_{AB} - ES) + c_B(H_{BB} - E) = 0\] For a non-trivial solution (\(c_A, c_B \neq 0\)), the determinant of the coefficients must vanish: \[\begin{vmatrix} H_{AA} - E & H_{AB} - ES \\ H_{AB} - ES & H_{AA} - E \end{vmatrix} = 0\] Solving this quadratic equation for \(E\) yields two solutions: \[(H_{AA} - E)^2 - (H_{AB} - ES)^2 = 0 \implies H_{AA} - E = \pm(H_{AB} - ES)\] This results in the bonding (\(E_+\)) and antibonding (\(E_-\)) energy levels: \[E_+ = \frac{H_{AA} + H_{AB}}{1 + S} \quad \text{and} \quad E_- = \frac{H_{AA} - H_{AB}}{1 - S}\]
3. Bonding vs. Antibonding Molecular Orbitals
Substituting the energy values back into the secular equations allows us to determine the coefficients:
• For \(E_+\), we obtain \(c_A = c_B\). The normalized bonding wavefunction is:
\[\psi_+ = \frac{1}{\sqrt{2(1+S)}}(\phi_A + \phi_B)\]
This wavefunction exhibits constructive interference, resulting in an accumulation of electron probability density between the nuclei, which screens the nuclear-nuclear repulsion and facilitates bonding.
• For \(E_-\), we obtain \(c_A = -c_B\). The normalized antibonding wavefunction is:
\[\psi_- = \frac{1}{\sqrt{2(1-S)}}(\phi_A - \phi_B)\]
This wavefunction exhibits destructive interference, resulting in a nodal plane (zero probability density, \(\psi_- = 0\)) right in the middle of the bond, leading to high nuclear repulsion and instability.
Because the overlap integral \(S\) is positive (\(0 < S < 1\)), the antibonding orbital is destabilized by a greater amount than the bonding orbital is stabilized relative to the isolated atomic orbitals: \[|E_- - H_{AA}| > |E_+ - H_{AA}|\] This asymmetry explains why molecular helium (\(\text{He}_2\), which would occupy both \(\sigma\) and \(\sigma^*\) orbitals) is unstable and does not form.
4. Solved CSIR-NET & GATE Questions
Question 1 (CSIR-NET Chemical Sciences)
Show that the normalization constant \(N\) for the LCAO bonding wavefunction \(\psi = N(\phi_A + \phi_B)\) is equal to \(\frac{1}{\sqrt{2(1+S)}}\), given that the atomic orbitals \(\phi_A\) and \(\phi_B\) are normalized and their overlap integral is \(S\).
Detailed Solution:
- For \(\psi\) to be normalized, we require the total probability integral to be 1: \[\int \psi^* \psi d\tau = 1\]
- Substitute \(\psi = N(\phi_A + \phi_B)\): \[\int N^*(\phi_A^* + \phi_B^*) N(\phi_A + \phi_B) d\tau = 1\] \[|N|^2 \left[ \int \phi_A^* \phi_A d\tau + \int \phi_B^* \phi_B d\tau + \int \phi_A^* \phi_B d\tau + \int \phi_B^* \phi_A d\tau \right] = 1\]
- Given that the atomic orbitals are normalized: \[\int \phi_A^* \phi_A d\tau = \int \phi_B^* \phi_B d\tau = 1\] And the overlap integral definition: \[\int \phi_A^* \phi_B d\tau = \int \phi_B^* \phi_A d\tau = S\]
- Substitute these properties into the expression: \[|N|^2 [1 + 1 + S + S] = 1 \implies |N|^2 [2 + 2S] = 1\] \[|N|^2 = \frac{1}{2(1+S)} \implies N = \frac{1}{\sqrt{2(1+S)}}\]
Answer: The normalization constant is verified as \(N = \frac{1}{\sqrt{2(1+S)}}\).
Question 2 (GATE Chemistry)
For a homonuclear diatomic molecule, the Coulomb integral is \(H_{AA} = -13.6\text{ eV}\), the resonance integral is \(H_{AB} = -3.2\text{ eV}\), and the overlap integral is \(S = 0.25\). Calculate the energy values (in eV) of the bonding and antibonding molecular orbitals.
Detailed Solution:
- Calculate the bonding energy level \(E_+\): \[E_+ = \frac{H_{AA} + H_{AB}}{1 + S} = \frac{-13.6 + (-3.2)}{1 + 0.25} = \frac{-16.8}{1.25} = -13.44\text{ eV}\]
- Calculate the antibonding energy level \(E_-\): \[E_- = \frac{H_{AA} - H_{AB}}{1 - S} = \frac{-13.6 - (-3.2)}{1 - 0.25} = \frac{-10.4}{0.75} = -13.87\text{ eV}\] Wait, check: \(H_{AA} - H_{AB} = -13.6 - (-3.2) = -13.6 + 3.2 = -10.4\text{ eV}\). And \(1 - S = 1 - 0.25 = 0.75\). So \(E_- = \frac{-10.4}{0.75} = -13.87\text{ eV}\).
- Note: In physical chemistry calculations, the resonance integral \(H_{AB}\) is a negative quantity representing stabilizing interaction energy. When checking destabilization relative to isolated atomic orbitals:
• Bonding energy change: \(\Delta E_+ = E_+ - H_{AA} = -13.44 - (-13.6) = +0.16\text{ eV}\) (less stable than \(H_{AA}\)? No, wait! Let's re-verify the signs: \(H_{AA} = -13.6\) eV is negative. \(H_{AB} = -3.2\) eV is also negative. \(E_+ = \frac{-13.6 - 3.2}{1.25} = \frac{-16.8}{1.25} = -13.44\) eV. Since \(-13.44 > -13.6\), it is actually higher in energy than \(H_{AA}\). Wait! This occurs when the overlap integral \(S\) is very large relative to the resonance integral! Typically, if \(S\) is neglected (\(S = 0\)), \(E_+ = H_{AA} + H_{AB} = -13.6 - 3.2 = -16.8\) eV (stabilized). If \(S = 0.2\), \(E_+ = \frac{-16.8}{1.2} = -14.0\) eV (stabilized by \(0.4\) eV). For \(S = 0.25\), \(E_+ = \frac{-16.8}{1.25} = -13.44\) eV (which is actually higher than \(-13.6\) eV, showing that a very high overlap value without corresponding increase in \(H_{AB}\) can prevent stabilizing bonding).
Answer: Bonding orbital energy \(E_+ = -13.44\text{ eV}\), Antibonding orbital energy \(E_- = -13.87\text{ eV}\).
Vishnudas
Senior Chemistry Faculty | IIT Madras Alumni | JRF AIR-18 | PhD SpecialistVishnudas is a PhD researcher from IIT Madras who secured CSIR JRF AIR 18. He teaches Quantum Chemistry, Molecular Spectroscopy, and Advanced Physical Chemistry modules at Benzil Academy.