Question bank — MO diagrams of H₂, He₂, N₂, O₂, F₂, NO, CO — bond order, magnetism

True — the number of MOs out equals the number of AOs in (LCAO conserves states): one in-phase (bonding) plus one out-of-phase (antibonding). See Molecular Orbital Theory (LCAO basics).

True — that is why the formula subtracts: each bonding electron gives $+\frac{1}{2}$ a bond, each antibonding gives $-\frac{1}{2}$, so a filled ${\sigma }^{\ast }$ wipes out a filled $\sigma$.

False — the ${\sigma }_{1s}$ bonding pair does attract, but the ${\sigma }_{1s}^{\ast }$ antibonding pair cancels it exactly, giving bond order $0$; there is simply no net bond to hold it together.

False — MO theory puts the last two electrons singly into the two degenerate ${\pi }_{2p}^{\ast }$ orbitals (Hund's rule), so O₂ has two unpaired electrons and is paramagnetic. Liquid O₂ sticking to a magnet proves MO right, Lewis wrong. See Paramagnetism and Diamagnetism.

False — odd-electron species like NO (BO $=2.5$), O₂⁺ and O₂⁻ give half-integer bond orders, which are perfectly physical; only Lewis dot-pictures force integers.

False — it depends where they go: electrons entering antibonding MOs lower the bond order. F₂ has more electrons than N₂ yet a far weaker single bond.

Partly — but the essential meaning is a nodal plane between the nuclei; that node removes electron density from where glue is needed, which is why it costs energy. See Bond Order, Bond Length, Bond Energy correlation.

False — for B₂, C₂, N₂ (Z ≤ 7) s–p mixing pushes ${\sigma }_{2p}$ above the ${\pi }_{2p}$ pair; for O₂, F₂ (Z ≥ 8) mixing is weak and ${\sigma }_{2p}$ sits below ${\pi }_{2p}$ (see the on-page summary above). Using the wrong order flips predicted magnetism. See s–p mixing and orbital energy ordering.

Mostly true in electron count and resulting BO/magnetism (both 14 e⁻, BO 3, diamagnetic), but CO's MOs are skewed toward the more electronegative oxygen, so its levels are not perfectly symmetric. See Isoelectronic species (CO, N₂, CN⁻, NO⁺).

True — Hund's rule: parallel spins in separate degenerate orbitals minimise electron repulsion and are lower in energy than pairing up early. See Hund's Rule and Pauli Exclusion.

The error is pairing those last two — the two ${\pi }^{\ast }$ orbitals are degenerate (same energy, see Figure 1), so Hund puts one electron in each with parallel spins, leaving O₂ paramagnetic with 2 unpaired electrons.

Wrong — odd electron count just means an unpaired electron and a half-integer BO: $N_b=10$, $N_a=5$, $\text{BO}=\frac{10-5}{2}=2.5$. Perfectly defined.

Wrong — being high in energy doesn't mean unpaired. ${\sigma }_{2p}^2$ holds two paired electrons and every N₂ MO is filled or empty, so N₂ is diamagnetic.

Wrong reasoning — it does have 8 antibonding electrons ($N_a=8$); they mostly cancel the 10 bonding ones, leaving a net single bond. Low bond order ≠ no antibonding electrons.

Backwards here — the electron removed from NO comes from an antibonding ${\pi }^{\ast }$, so its loss raises BO from 2.5 to 3, making NO⁺ stronger and shorter than NO.

The Aufbau order is right, but the identity of the next orbital changes with the diagram: for B₂–N₂ the ${\pi }_{2p}$ (bonding) fills before ${\sigma }_{2p}$ (also bonding) because of s–p mixing, which even reorders two bonding levels.

In-phase addition piles electron density between the nuclei, screening their mutual repulsion and pulling them together; out-of-phase creates a node there, starving the middle of density so the nuclei repel — anti-glue.

In early elements the 2s and 2p energies are close, so ${\sigma }_{2s}$ and ${\sigma }_{2p}$ mix and repel, shoving ${\sigma }_{2p}$ above ${\pi }_{2p}$. By oxygen the 2s–2p gap is large, mixing is negligible, and the natural order (${\sigma }_{2p}$ below ${\pi }_{2p}$) returns.

Lewis only counts pairs on paper and forces O=O all-paired; MO theory respects orbital degeneracy and Hund's rule, predicting two unpaired ${\pi }^{\ast }$ electrons — which experiment (magnet, spectroscopy) confirms.

More net bonding electron density between nuclei means a deeper, stiffer energy well — stronger bond and shorter distance. Hence N₂(3) > O₂(2) > F₂(1) in strength and reverse in length. See Bond Order, Bond Length, Bond Energy correlation.

Because a single chemical bond is a pair of shared electrons; two net bonding electrons make one bond, so dividing the net electron count by 2 converts electrons into bonds.

Its triple bond leaves a lone pair on carbon (the $\sigma$-donor) plus empty ${\pi }^{\ast }$ orbitals that accept metal electron density ($\pi$-backbonding), a two-way binding that also explains its toxicity toward haemoglobin.

Config ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 1}$, so $\text{BO}=\frac{2-1}{2}=0.5$ — a weak but real half-bond, and paramagnetic from the single unpaired ${\sigma }^{\ast }$ electron. Unlike He₂, it can exist transiently.

Paramagnetic. Under the s–p-mixed order the last two electrons enter the degenerate ${\pi }_{2p}$ pair one each (Hund), giving two unpaired electrons; BO $=\frac{6-4}{2}=1$. This is the trap: without the flipped order you'd wrongly predict a paired ${\sigma }_{2p}^2$ and call it diamagnetic.

Under the mixed order the ${\pi }_{2p}$ pair is now full (${\pi }_{2p}^4$), all paired, so C₂ is diamagnetic with BO $=\frac{8-4}{2}=2$. Note the oddity: C₂'s double bond is made of two $\pi$ bonds and no net $\sigma$ bond in this simple picture.

Removing an antibonding ${\pi }^{\ast }$ electron raises BO, adding one lowers it: O₂⁺ (2.5) > O₂ (2) > O₂⁻ superoxide (1.5). All three are paramagnetic — O₂⁺ and O₂⁻ each carry one unpaired ${\pi }^{\ast }$ electron, O₂ carries two.

With 8 electrons filling ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}$, bonding and antibonding cancel completely: $\text{BO}=\frac{4-4}{2}=0$, so Be₂ is essentially non-bonding, like He₂.

The electron leaves the highest filled orbital, ${\sigma }_{2p}$ (a bonding MO under the mixed order), so BO drops from 3 to 2.5 and N₂⁺ becomes paramagnetic with one unpaired electron.

Not strictly — BO 0 means no net bond, so no stable covalently bonded molecule (He₂, Be₂), but weakly bound van der Waals dimers can flicker into existence; the point is there is no covalent net bonding.

No — an odd total number of electrons guarantees at least one unpaired electron, so any odd-electron species (NO, NO₂, ClO₂) is necessarily paramagnetic.