MO theory correctly predicts that O₂ is paramagnetic, while valence bond theory (Lewis structures) predicts it is diamagnetic. Which feature of the MO energy-level diagram explains paramagnetism?
AO₂ has a triple bond, giving it more electrons than expected.
BTwo electrons occupy degenerate π* antibonding orbitals singly, with parallel spins, following Hund's rule.
CThe σ bonding MO is completely filled before any electrons enter antibonding orbitals.
DThe overlap integral S for the 2p orbitals is negative, destabilizing the bonding MO.
After filling σ and π bonding MOs, O₂ has two electrons remaining to place in two degenerate π* orbitals. By Hund's rule, one electron goes into each π* orbital with parallel spins rather than pairing in one. Parallel (unpaired) electrons create a net magnetic moment, making O₂ paramagnetic. A Lewis structure cannot represent degenerate orbitals or Hund's rule, so it draws a double bond with all electrons paired and incorrectly predicts diamagnetism.
Question 2 True / False
In LCAO-MO theory, the coefficients c_A and c_B in the expansion φ = c_A χ_A + c_B χ_B represent the probability of finding an electron on atom A or B, respectively.
TTrue
FFalse
Answer: False
The coefficients c_A and c_B are amplitude coefficients, not probabilities, and can be negative (as in antibonding MOs where the combination is φ* = c_A χ_A − c_B χ_B). The probability density contribution of atom A is related to c_A² (and a cross term involving the overlap S), not c_A itself. This distinction matters: a negative coefficient means the wavefunction has a node between the atoms — the defining feature of an antibonding orbital.
Question 3 Short Answer
Using MO theory, calculate the bond order of He₂ and state what this predicts about the molecule's stability. Identify which electrons are responsible for this result.
Think about your answer, then reveal below.
Model answer: Bond order = (bonding electrons − antibonding electrons) / 2 = (2 − 2) / 2 = 0. A bond order of zero predicts He₂ does not form a stable bond and should not exist as a molecule. The two electrons in the σ(1s) bonding MO provide stabilization, but this is exactly canceled by the two electrons in the σ*(1s) antibonding MO, which destabilize the molecule by roughly the same amount. He₂ is indeed not observed under normal conditions.
This is one of MO theory's most elegant predictions. Both bonding and antibonding MOs form from the 1s atomic orbitals of the two helium atoms. The bonding energy gained from filling σ is nearly perfectly canceled by the antibonding energy cost of filling σ*. The net stabilization is essentially zero, and the molecule does not form. The analogous argument explains why He₂⁺ (bond order = 1/2) is weakly stable — removing one electron from σ* tips the balance toward bonding.