Answer the following questions for the pi molecular orbitals of 1,3,5,7-octatetr

Question

Answer the following questions for the pi molecular orbitals of 1,3,5,7-octatetraene. 1. Which are the bonding MOs, which are the antibonding MOs? 2. Which MOs are symmetric and which are antisymmetric? 3. Which MO is the HOMO and which is the LUMO in the ground state? 4. Which MO is the HOMO and which is the LUMO in the excited state? 5. How many nodes does the highest-energy molecular orbital of 1,3,5,7-octatetraene have between the nuclei? Please briefly explain how you arrive at each answer. Thanks!

Explanation / Answer

A simplified answer (MO theory) is that the orbitals of the atoms come together to form molecular orbitals. Much like classical waves, the atomic orbitals (wave functions) can then interfere constructively or destructively. Considering, say, H2, your two atomic s-orbitals come together constructively to form a s orbital, where you have increased density between the two atoms (s1 + s2 -> s) - thus it's called a bonding orbital. But they can also come together 'destructively' to form a s*, where you have a decreased density between the two atoms (s1 - s2 -> s*), hence anti-bonding. The s orbital is greater than zero everywhere (because the s orbitals are), whereas the s* is zero (= has a node) in the plane that's equidistant from both atoms, where the two s-orbitals cancel out completely. What you can infer from this, is that going from one atom to the other, the anti-bonding orbital has a sharper curvature (passes through a greater range of values) than the bonding orbital, and with quantum-mechanical wave functions (as well with classical waves), a sharper curvature means higher energy. So this tells us that anti-bonding orbitals are usually higher in energy than the corresponding bonding orbitals.

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