Lecture 13 Diatomic orbitals o Hydrogen molecule ion

Lecture 13: Diatomic orbitals o Hydrogen molecule ion (H 2+) o Overlap and exchange integrals o Bonding/Anti-bonding orbitals o Molecular orbitals PY 3 P 05

Schrödinger equation for hydrogen molecule ion - o Simplest example of a chemical bond is the hydrogen molecule ion (H 2+). o Consists of two protons and a single electron. o If nuclei are distant, electron is localised on one nucleus. Wavefunctions are then those of atomic hydrogen. o a is the hydrogen atom wavefunction of electron belonging to nucleus a. Must therefore ra rb a + +b rab satisfy (1) Ha and correspondingly for other wavefunction, b. The energies are therefore Ea 0 = Eb 0 = E 0 PY 3 P 05

Schrödinger equation for hydrogen molecule ion o If atoms are brought into close proximity, electron localised on b will now experience an attractive Coulomb force of nucleus a. o Must therefore modify Schrödinger equation to include Coulomb potentials of both nuclei: (2) where o To find the coefficients ca and cb, substitute into Eqn. 2: Ha Hb PY 3 P 05

Solving the Schrödinger equation o Can simplify last equation using Eqn. 1 and the corresponding equation for Ha and Hb. o By writing Ea 0 a in place of Ha a gives o Rearranging, o As Ea 0 = Eb 0 by symmetry, we can set Ea 0 - E = Eb 0 - E = - E, (3) PY 3 P 05

Overlap and exchange integrals o a and b depend on position, while ca and cb do not. Now we know that for orthogonal wavefunctions Normalisation integral but a and b are not orthoganal, so o Overlap integral If we now multiply Eqn. 3 by a and integrate the results, we obtain (4) (5) o As -e| a|2 is the charge density of the electron => Eqn. 4 is the Coulomb interaction energy between electron charge density and nuclear charge e of nucleus b. o The -e a b in 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. PY 3 P 05

Overlap and exchange integrals o Eqn. 4 can be visualised via figure at right. o Represents the Coulomb interaction energy of an electron density cloud in the Coulomb field on the nucleus. o Eqn. 5 can be visualised via figure at right. o Non-vanishing contributions are only possible when the wavefunctions overlap. o See Chapter 24 of Haken & Wolf for further details. PY 3 P 05

Orbital energies o If we mutiply Eqn 3 by a and integrate we get o Collecting terms gives, o Similarly, (6) (7) o Eqns. 6 and 7 can be solved for c 1 and c 2 via the matrix equation: o Non-trivial solutions exist when determinant vanishes: PY 3 P 05

Orbital energies o Therefore (- E + C)2 - (- E·S + D)2 = 0 => C - E = ± (D - E ·S) o This gives two values for E: o As C and D are negative, Eb < Ea o Eb correspond to bonding molecular orbital energies. Ea to anti-bonding MOs. o H 2+ atomic and molecular orbital energies are shown at right. and Energy (e. V) -13. 6 PY 3 P 05

Bonding and anti-bonding orbitals o Substituting for Ea into Eqn. 6 => cb = -ca = c. The total wavefunction is thus Anti-bonding orbital o Similarly for Eb => ca = cb = c, which gives Bonding orbital o For symmetric case (top right), occupation probability for is positive. Not the case for a asymmetric wavefunctions (bottom right). o Energy splits depending on whether dealing with a bonding or an anti-bonding wavefunction. PY 3 P 05

Bonding and anti-bonding orbitals o The energy E of the hydrogen molecular ion can finally be written o ‘+’ correspond to anti-bonding orbital energies, ‘-’ to bonding. o Energy curves below are plotted to show their dependence on rab PY 3 P 05

Molecular orbitals PY 3 P 05

Molecular orbitals PY 3 P 05

Molecular orbitals o Energies of bonding and anti-bonding molecular orbitals for first row diatomic molecules. o Two electrons in H 2 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. PY 3 P 05

Molecular orbitals PY 3 P 05