Lecture 13: Diatomic orbitals

1. Lecture 13: Diatomic orbitals • Hydrogen molecule ion (H2+) • Overlap and exchange integrals • Bonding/Anti-bonding orbitals • Molecular orbitals PY3P05

2. Schrödinger equation for hydrogen molecule ion - • Simplest example of a chemical bond is the hydrogen molecule ion (H2+). • Consists of two protons and a single electron. • If nuclei are distant, electron is localised on one nucleus. Wavefunctions are then those of atomic hydrogen. ra rb + a + b rab • ais the hydrogen atom wavefunction of electron belonging to nucleus a. Must therefore satisfy and correspondingly for other wavefunction, b. The energies are therefore Ea0= Eb0= E0 (1) Ha PY3P05

3. Schrödinger equation for hydrogen molecule ion • If atoms are brought into close proximity, electron localised on b will now experience an attractive Coulomb force of nucleus a. • Must therefore modify Schrödinger equation to include Coulomb potentials of both nuclei: where • To find the coefficients caand cb, substitute into Eqn. 2: (2) Ha Hb PY3P05

4. Solving the Schrödinger equation • Can simplify last equation using Eqn. 1 and the corresponding equation for Ha and Hb. • By writing Ea0ain place of Haa gives • Rearranging, • As Ea0 = Eb0by symmetry, we can set Ea0 - E = Eb0 - E = -E, (3) PY3P05

5. Overlap and exchange integrals • a and bdepend on position, while caand cbdo not. Now we know that for orthogonal wavefunctions but a andbare not orthoganal, so • If we now multiply Eqn. 3 by a and integrate the results, we obtain • As -e|a|2is the charge density of the electron => Eqn. 4 is the Coulomb interaction energy between electron charge density and nuclear charge e of nucleus b. • The -ea bin Eqn. 5 means that electron is partly in state a and partly in state b => an exchange between states occurs. Eqn. 5 therefore called an exchange integral. Normalisation integral Overlap integral (4) (5) PY3P05

6. Overlap and exchange integrals • Eqn. 4 can be visualised via figure at right. • Represents the Coulomb interaction energy of an electron density cloud in the Coulomb field on the nucleus. • Eqn. 5 can be visualised via figure at right. • Non-vanishing contributions are only possible when the wavefunctions overlap. • See Chapter 24 of Haken & Wolf for further details. PY3P05

7. Orbital energies • If we mutiply Eqn 3 bya and integrate we get • Collecting terms gives, • Similarly, • Eqns. 6 and 7 can be solved for c1 and c2via the matrix equation: • Non-trivial solutions exist when determinant vanishes: (6) (7) PY3P05

8. Orbital energies • Therefore (-E + C)2 - (-E·S + D)2 = 0 => C - E = ± (D - E ·S) • This gives two values for E: and • As C and D are negative, Eb < Ea • Ebcorrespond to bonding molecular orbital energies. Eato anti-bonding MOs. • H2+ atomic and molecular orbital energies are shown at right. Energy (eV) -13.6 PY3P05

9. Bonding and anti-bonding orbitals • Substituting for Ea into Eqn. 6 => cb = -ca = -c.The total wavefunction is thus • Similarly for Eb => ca = cb = c, which gives • For symmetric case (top right), occupation probability for  is positive. Not the case for a asymmetric wavefunctions (bottom right). • Energy splits depending on whether dealing with a bonding or an anti-bonding wavefunction. Anti-bonding orbital Bonding orbital PY3P05

10. Bonding and anti-bonding orbitals • The energy E of the hydrogen molecular ion can finally be written • ‘+’ correspond to anti-bonding orbital energies, ‘-’ to bonding. • Energy curves below are plotted to show their dependence on rab PY3P05

11. Molecular orbitals PY3P05

12. Molecular orbitals PY3P05

13. Molecular orbitals • Energies of bonding and anti-bonding molecular orbitals for first row diatomic molecules. • Two electrons in H2 occupy bonding molecular orbital, with anti-parallel spins. If irradiated by UV light, molecule may absorb energy and promote one electron into its anti-bonding orbital. PY3P05

14. Molecular orbitals PY3P05