CHEM 2060 Lecture 25: MO Theory L25-1 Molecular Orbital Theory (MO theory) VB theory treats bonds as electron pairs. o There is a real emphasis on this point (over-emphasis actually). VB theory is very good for ground state properties (Geometries, Bond Dissociation Energies) o Easily correlated to Lewis Dot structures and VSEPR theory rules. Difficult to apply VB theory to excited states. MO theory is better for spectroscopy (Exited State Properties; Ionization) Molecular orbital (ψ) = Linear Combination of Atomic Orbitals (LCAO) weighted linear sum of the valence orbitals of ALL atoms in the molecule. ψ = C i φ i i a sum of contributions of atomic orbitals C i = coefficient of the i th atomic orbital φ i = i th atomic orbital
CHEM 2060 Lecture 25: MO Theory L25-2 **Molecular Orbitals are delocalized.** In other words one orbital containing two electrons can be delocalized over several bonds, possibly even over the whole molecule! NO emphasis on electron pairs in bonds! The concept of bond order is harder to apply (i.e., single bond, double bond ) First Example H + 2 The figure is meant to illustrate the interaction between an H atom and a bare proton (H + ). There is one electron in a bonding MO.
CHEM 2060 Lecture 25: MO Theory L25-3 H 2 + one electron held in one bonding MO The ground state wavefunction (bonding molecular orbital) is a linear combination of some amount C 1 of the 1s orbital of H atom #1 and some amount C 2 of the 1s orbital of H atom #2 Ground State: ψ = C 1 φ 1 + C 2 φ 2 1s on H 1 1s on H 2 There are 2 possible ways of creating this linear combination: add or subtract ψ = 1s a + 1s b Bonding ψ* = 1s a 1s b Antibonding MIX 2 atomic orbitals (1s a and 1s b ) GET 2 molecular orbitals (ψ and ψ*)
CHEM 2060 Lecture 25: MO Theory L25-4 ψ and ψ* ψ = 1s a +1s b concentrates electron density between nucleii Reduces nuclear-nuclear repulsion BONDING. ψ* = 1s a 1s b e- density concentrated outside inter-nuclear region. Has a NODE between the nuclei ANTIBONDING. ψ 2 (ψ*) 2
CHEM 2060 Lecture 25: MO Theory L25-5 The energy of the bonding MO has a minimum at a specific internuclear distance (bond distance). The energy of the antibonding MO is minimized by pulling the nuclei as far apart as possible. Energy, E Electron in antibonding molecular orbital Separation, R Electron in bonding molecular orbital This is a molecular orbital energy level diagram. Notice that the bonding MO is lower in energy than the 1s AO!
CHEM 2060 Lecture 25: MO Theory L25-6 Just like an atomic orbital, a molecular orbital is a wavefunction, the square of which defines where the electrons can be in space (probability density). In our bonding MO for H 2 +, we can think of the concentration of electrons as an overlap of the two 1s atomic orbitals. MO s can be classified according to symmetry! σ = sigma bonding molecular orbital in a homonuclear diatomic molecule, this has inversion symmetry (g = gerade) σ* = sigma antibonding molecular orbital in a homonuclear diatomic molecule, the wavefunction changes sign upon inversion (u = ungerade) HOMEWORK: Prove to yourself that in a homonuclear diatomic molecule, a π MO should be u and a π* MO should be g.
CHEM 2060 Lecture 25: MO Theory L25-7 Normalized Wavefunctions Bonding MO ψ = C 1 1s a + C 2 1s b ψ 2 dτ =1 ( C 1 1s a + C 2 1s b ) 2 dτ =1 C 1 2 ( ) 2 1s a dτ + C 2 2 1s b dτ + 2C 1C 2 =1 ( ) 2 =1 1s a 1s b dτ S=? overlap integral If the 1s atomic orbitals are already normalized, the integral of their squares is simply equal to 1. In order to proceed any further, we need to know the overlap integral S. In our model H 2 + molecule, S = 0.585. This is actually quite large compared to most homonuclear diatomics. =1
CHEM 2060 Lecture 25: MO Theory L25-8 So C 1 2 + C 2 2 + 2C 1 C 2 (0.585) =1 by symmetry C 1 = C 2 so C 1 2 + C 1 2 + 2C 1 2 (0.585) =1 solving for C 1, we get 3.17C 1 2 =1 C 1 = C 2 = 1 3.17 = 0.562 Therefore ψ = 0.562( 1s a +1s b ) normalized bonding MO This is the actual answer for the H 2 + model. However, most of the time the overlap integral S is much smaller. If you don t know the value of S, a common approximation is S = 0.
CHEM 2060 Lecture 25: MO Theory L25-9 Let s solve for the coefficients (normalize the wavefunction) using the approximation S = 0. **(Note: on an exam, if not given a value of S, this is what you should do.)** C 2 1 + C 2 1 + 2C 2 1 (0) =1 =0 solving for C 1, we get 2C 1 2 =1 ψ = and 1 2 (1s a + 1s b ) ψ* = 1 ( 2 1s a 1s b ) C 1 = C 2 = 1 2 Generic homonuclear diatomic Bonding MO Generic homonuclear diatomic Anti-bonding MO
CHEM 2060 Lecture 25: MO Theory L25-10 SUMMARY: This MO energy level diagram can be used to describe H 2 +, H 2, He 2 +, and He 2. Only the number of electrons changes. RULE: fill from bottom up
CHEM 2060 Lecture 25: MO Theory L25-11 Bond Order (see dek. & G., p. 198) (single, double, triple bonds, etc.) The concept of Bond Order still works well for homonuclear diatomics using MO theory. Electrons in bonding orbitals give stability. Each pair of electrons in a bonding MO contributes 1 to the bond order. Electrons in antibonding orbitals decrease the bond order so cancel the effect of the bonding electrons. Each pair of electrons contributes -1 to the bond order. Bonding Electrons Anti-bonding Electrons Net Bonding Electrons Bond Order Bond Length (Å) He 2 2 2 0 0 - - + H 2 1 0 1 0.5 1.06 256 + He 2 2 1 1 0.5 1.08 231 H 2 2 0 2 1 0.74 432 Single bond = bond order of 1 Double bond = bond order of 2 and so forth Experimental Bond Energy kj/mol