Question bank — Molecular Orbital Theory (MOT) — LCAO, bonding - antibonding orbitals

False. Orbitals are conserved: N atomic orbitals always give exactly N molecular orbitals. Two in, two out — one bonding (in-phase), one antibonding (out-of-phase). Overlap strength changes the energy gap, never the count.

True for identical atoms. The in-phase combination piles density between the nuclei, so an electron there feels attraction from both nuclei → energy drops below the isolated atomic orbital level. (Edge case: in a heteronuclear molecule the two atomic orbitals start at different energies, and the bonding MO can lie between them, closer to the lower atom — see the edge-cases section.)

False. The antibonding orbital is raised slightly more than the bonding one is lowered. The reason is the overlap $S=\int {\psi }_A{\psi }_{B}d\tau >0$ in the normalisation: it shrinks the bonding stabilisation (denominator $1+S$) but inflates the antibonding destabilisation (denominator $1-S$). This asymmetry is why filling both (He₂) leaves a net destabilisation, not a clean zero.

False. An orbital is just an allowed energy level; it exists whether or not electrons occupy it. Empty antibonding orbitals (${\pi }^{\ast }$ in N₂, say) are perfectly real — they matter for excited states and reactivity.

False. A node is a surface where the probability of finding the electron is zero — the electron is simply never located there, it lives on either side. Nodes cost energy because they force density away from the stabilising between-nuclei region.

True as a general rule for the same orbitals, because head-on overlap concentrates more density directly on the internuclear axis. But this "σ looks stronger" instinct is exactly what makes people wrongly order the MOs — see the s-p mixing trap below.

False. Bond order can be a half-integer, e.g. ${\mathrm{O}}_2^{-}$ or $\mathrm{NO}$ have B.O. = 1.5 or 2.5. It just counts $\frac{N_b-N_a}{2}$; an odd electron count naturally gives a fractional answer.

True. Paramagnetism requires unpaired electrons. Fully paired configurations (like ${\mathrm{N}}_2$, ${\mathrm{H}}_2$) are diamagnetic — repelled by a magnetic field.

Error: it ignores the two antibonding electrons in ${\sigma }_{1s}^{\ast 2}$. So $N_b=2$, $N_a=2$, and B.O. = $(2-2)/2=0$ → no net bond, He₂ does not exist. Antibonding electrons subtract — never leave them out.

Error: this is the VBT picture, and it's wrong. MOT places the last two electrons singly into the two degenerate ${\pi }^{\ast }$ orbitals (Hund's rule) → 2 unpaired electrons → O₂ is paramagnetic. This is MOT's headline triumph over VBT.

Error: this only holds for O₂, F₂, Ne₂. For B₂, C₂, N₂ the 2s and 2p energies are close, so s-p mixing pushes ${\sigma }_{2p_z}$ above ${\pi }_{2p}$. Using the wrong order makes B₂ and C₂ look diamagnetic when they are actually paramagnetic.

Error: they have different symmetry about the axis. The positive-lobe overlap exactly cancels the negative-lobe overlap, so the overlap integral $S=\int {\psi }_A{\psi }_{B}d\tau =0$ → no bond, no matter how close they get.

Error: same symmetry is not enough. The energies are wildly mismatched (H 1s is far lower than Na 3s), violating the comparable-energy condition, so the combination is negligible.

Error: antibonding electrons actively destabilise the molecule — they concentrate density away from the middle and let the nuclei repel. That's why they carry a minus sign in bond order.

Error: bond order correlates with bond strength and length (higher B.O. → shorter, stronger bond), but it isn't an energy value itself. It's a count of net bonding pairs.

Because a wavefunction is a wave, and two waves can meet in phase (add → constructive, density between nuclei) or out of phase (subtract → destructive, a node). Nature provides both combinations, which is also why orbital count is conserved.

An electron sitting between two positive nuclei is pulled by both at once, lowering its potential energy. The $+2{\psi }_A{\psi }_B$ cross-term in $|{\psi }_b{|}^2$ (where ${\psi }_A,{\psi }_B$ are the two atoms' orbitals) is precisely this build-up of "glue" density.

Because one full covalent bond is one electron pair (2 electrons). Dividing the net bonding-electron count $N_b-N_a$ by 2 converts electrons into number-of-bonds.

Because the overlap $S$ enters the two normalised energies unequally: bonding gets divided by $1+S$ (softening the drop) while antibonding gets divided by $1-S$ (amplifying the rise). Physically, the antibonding node throws density outside the nuclei, letting them repel harder than the shared middle can glue them.

Mixing happens between orbitals of the same symmetry. The ${\sigma }_{2s}$ and ${\sigma }_{2p_z}$ are both σ-symmetric, so they interact and repel each other in energy, pushing ${\sigma }_{2p_z}$ up. The $\pi$ orbitals have different symmetry and don't mix with the 2s.

MOT fills delocalised molecular orbitals following Hund's rule, naturally landing two electrons unpaired in degenerate ${\pi }^{\ast }$ orbitals. VBT's localised O=O double-bond picture pairs everything up and simply can't produce the observed unpaired electrons.

The strength of the interaction depends on the overlap integral $S$. If the orbitals barely reach each other, $S\approx 0$, the energy split between bonding and antibonding vanishes, and effectively no molecular orbital forms.

All bonding orbitals up to ${\sigma }_{2p}$ are matched by their filled antibonding partners: $N_b=N_a$, so B.O. = 0. Like He₂, Ne₂ does not exist as a bound molecule.

Yes. With one electron in ${\sigma }_{1s}$: $N_b=1$, $N_a=0$, so B.O. = $(1-0)/2=0.5$. A single electron between two nuclei still provides net attraction, so ${\mathrm{H}}_2^{+}$ is a genuine (weak) one-electron bond — proof a "bond" needn't be a full pair.

Not below both atomic orbitals. Since F's orbital is much lower in energy than H's, the bonding MO lies between the two atomic energies (closer to F, the more electronegative atom) while the antibonding MO sits above both. The "below both" rule is a homonuclear-only shortcut.

Removing an electron from an antibonding ${\pi }^{\ast }$ raises B.O. (${\mathrm{O}}_2^{+}$: 2.5), while adding one lowers it (${\mathrm{O}}_2^{-}$: 1.5), vs neutral O₂ at 2. So ${\mathrm{O}}_2^{+}$ has the shortest, strongest bond — because you took away destabilising density.

Yes — O₂ is the classic case with 16 electrons yet 2 unpaired in degenerate ${\pi }^{\ast }$ orbitals. Paramagnetism is about unpaired electrons, not about parity of the total count.

Each such electron increases $N_a$, so $N_b-N_a$ shrinks and bond order falls. If you keep going, bond order can reach zero and the molecule dissociates — the antibonding electrons win the tug-of-war.

Not by itself. You must also count the antibonding occupancy: if an equally filled antibonding orbital cancels it (He₂), the net bond order is zero and there is no stable molecule.