Question

How does Buta-1,3 diene have 9 σ MOs when all 4 carbons have sp3?

Looking again at the Lewis structure of buta-1,3-diene, notice that there are 22 total valence electrons. The first 18 of those electrons occupy the nine σ MOs. The remaining four electrons occupy π1 and π2, the two lowest-energy π MOs. The remaining two MOs of π symmetry, π3* and π4*, are empty. Therefore, π2 is the HOMO, and π3 is the LUMO.

Answer 1