1. Molecules respond to the many wavelengths of light. The wavelengths that are absorbed and then re-emitted determine an object’s color, while the wavelengths that are NOT re-emitted raise the temperature of the object. Molecular Orbitals -- The VSEPR and valence-bond theories don’t explain the excited states of molecules, which come into play when molecules absorb and emit light. -- This is one thing that the molecular orbital (MO) theory attempts to explain.

2. molecular orbitals: wave functions that describe the locations of electrons in molecules -- these are analogous to atomic orbitals in atoms (e.g., 2s, 2p, 3s, 3d, etc.), but MOs are possible locations of electrons in molecules (not atoms) -- MOs, like atomic orbitals, can hold a maximum of two e – with opposite spins -- but MOs are for entire molecules MO theory is more powerful than valence-bond theory; its main drawback is that it isn’t easy to visualize.

3. Hydrogen (H 2 ) The overlap of two atomic orbitals produces two MOs. 1s H atomic orbitals + H 2 molecular orbitals  1s (bonding MO) (antibonding MO)  * 1s -- The lower-energy bonding molecular orbital concentrates e – density between nuclei. -- For the higher-energy antibonding molecular orbital, the e – density is concentrated outside the nuclei. -- Both of these are  molecular orbitals.

4. Energy-level diagram (molecular orbital diagram) 1s  1s  * 1s H atom H 2 m’cule H atom 1s H atomic orbitals + H 2 molecular orbitals  1s (bonding MO) (antibonding MO)  * 1s

5. Consider the energy-level diagram for the hypothetical He 2 molecule… 1s  1s  * 1s He atom He 2 m’cule He atom 2 bonding e –, 2 antibonding e – No energy benefit to bonding. He 2 molecule won’t form.

6. no bond bond order = ½ (# of bonding e – – # of antibonding e – ) -- the higher the bond order, the greater the bond stability -- a bond order of... 0 = single bond 1 = double bond 2 = triple bond 3 = -- MO theory allows for fractional bond orders as well. What is the bond order of H 2 + ? 1 e – total 1 bonding e –, zero antibonding e – BO = ½ (1–0) = ½ James Bond 7 =007 =

7. Second-Row Diatomic Molecules 1. # of MOs = # of combined atomic orbitals 2. Atomic orbitals combine most effectively with other atomic orbitals of similar energy. 3. As atomic orbital overlap increases, bonding MO is lowered in energy, and the antibonding MO is raised in energy. 4. Both the Pauli exclusion principle and Hund’s rule apply to MOs. Li 3 6.941 Be 4 9.012 B 5 10.811 C 6 12.011 N 7 14.007 O 8 15.999 F 9 18.998 Ne 10 20.180

8. Li 2 Li Li  1s  * 1s 1s 2s  * 2s  2s Use MO theory to predict whether Li 2 and/or Be 2 could possibly form. BO = ½ (4–2) = 1“YEP.”

9. Be 2 Be Be 2s  * 2s  2s BO = ½ (4–4) = 0 “NOPE.” Bonding and antibonding e – cancel each other out in core energy levels, so any bonding is due to e – in bonding orbitals of outermost shell.