I know when a carbonyl group is attacked by a nucleophile, the electrons of the nucleophile enter the antibonding π orbital of the $\ce{C=O}$ bond, which breaks the π bond. My textbook says that it enters the antibonding π orbital instead of the antibonding σ orbital, because the antibonding π orbital is lower in energy. This is my point of confusion: why is it lower in energy?


1 Answer 1


When two atoms form a bond, if two atomic orbitals mix, then two molecular orbitals are created, one of lower and one of higher energy than that of the atomic orbitals.

Now, just how much the energy changes is dictated by how well the atomic orbitals overlap. Overlapping depends on three things:

  • Orbital energy
  • Distance between atoms
  • Orbital symmetry (they must be "aligned")

Now, assuming that the atomic orbitals are similar in energy and the symmetries are correct, which is the case for C and O.

The sigma bond is a result of two orbitals very close together, directly facing each other frontally. The overlap is great. The pi bond emerges from two orbitals overlapping side-by-side, which isn't as efficient because it happens at a greater distance.

Here is a pic to exemplify:

Up: sigma bond. Down: pi bond.

Thus, not only is the bonding combination of a sigma bond lower in energy, but also the antibonding combination is higher in energy.

  • $\begingroup$ Correction: s orbitals could bond with f orbitals, if they are aligned right. // 'assuming that the atoms are similar in energy' I assume you mean the orbitals of the atoms? // This is a very simplified viewpoint. Sigma orbitals are lower in energy, because the maximum of the electron density is aligned on the bonding axis, which is not the case for pi, delta, phi, ... orbitals (where it is exactly 0). $\endgroup$ Commented Feb 5, 2019 at 9:49
  • $\begingroup$ @Martin Thanks for the corrections. Indeed it is a very simple explanation. I felt the need to write the answer in a very basic, introductory way. Once OP understands the core concept he can perfect it by reading more about the topic. $\endgroup$ Commented Feb 5, 2019 at 11:28

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