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

LCAO = Linear Combination of Atomic Orbitals. It builds each molecular orbital by adding (${\psi }_A+{\psi }_B$, in phase → bonding) and subtracting (${\psi }_A-{\psi }_B$, out of phase → antibonding) the atomic-orbital wavefunctions of the joining atoms. Two atomic orbitals in → two molecular orbitals out. This is the machinery behind σ and π bonds in MO language.

(a) σ — two round $s$ clouds overlap head-on, density is symmetric about the axis. (b) π — sideways overlap; the orbital has a nodal plane containing the axis. (c) σ — head-on overlap along the axis. (d) No bond. One orbital is symmetric and the other antisymmetric about the axis, so the positive-lobe overlap exactly cancels the negative-lobe overlap. The overlap integral $S=\int {\psi }_A{\psi }_{B}d\tau$ (defined at the top of the page) works out to $0$: the same-sign overlap on one side is exactly undone by opposite-sign overlap on the other. This is the "same symmetry" condition for atomic orbitals to combine.

Fill the lowest MO first: ${\sigma }_{1s}^2$. $N_b=2$ (both in bonding), $N_a=0$. $\text{B.O.}=\frac{2-0}{2}=1.$ One single bond. All electrons paired → diamagnetic. Matches reality. ✅

${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}$. $N_b=2$, $N_a=2$. $\text{B.O.}=\frac{2-2}{2}=0.$ The antibonding pair cancels the bonding pair → no net bond → He₂ does not exist as a stable molecule. This is MOT's classic win over VBT.

${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 1}$. $N_b=2$, $N_a=1$. $\text{B.O.}=\frac{2-1}{2}=\mathrm{0.5.}$ A half bond — small but positive, so He₂⁺ does exist (it is observed spectroscopically). We removed an antibonding electron, so less "push away" survives, and a weak bond appears.

N₂ has 7 electrons per atom → 14 total. Use the light ladder (N is "nitrogen and below"): ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^2$ Count bonding: ${\sigma }_{1s}(2)+{\sigma }_{2s}(2)+{\pi }_{2p_x}(2)+{\pi }_{2p_y}(2)+{\sigma }_{2p_z}(2)=10$. Count antibonding: ${\sigma }_{1s}^{\ast }(2)+{\sigma }_{2s}^{\ast }(2)=4$. $\text{B.O.}=\frac{10-4}{2}=3.$ A triple bond — that is why N₂ is famously inert and hard to break. Every orbital is doubly filled → no unpaired electrons → diamagnetic.

O has 8 electrons per atom → 16 total. Use the heavy ladder (${\sigma }_{2p_z}$ below ${\pi }_{2p}$): ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\pi }_{2p_x}^{\ast 1}{\pi }_{2p_y}^{\ast 1}$ The last two electrons must go one each into the two equal-energy ${\pi }^{\ast }$ orbitals — that is Hund's rule (spread out before pairing up). Bonding electrons: $2+2+2+2+2=10$. Antibonding electrons: $2+2+1+1=6$. $\text{B.O.}=\frac{10-6}{2}=2.$ Two orbitals hold a lonely electron → 2 unpaired electrons → paramagnetic. VBT's O=O picture wrongly says diamagnetic; MOT nails it. ✅

N₂: B.O. = 3 (from L3.1). O₂: B.O. = 2 (from L3.2). F₂: add two more electrons to O₂'s config, filling both ${\pi }^{\ast }$ to ${\pi }_{2p_x}^{\ast 2}{\pi }_{2p_y}^{\ast 2}$. Bonding = 10, antibonding = $2+2+2+2=8$. ${\text{B.O.}}_{F_2}=\frac{10-8}{2}=1.$ So N₂ (3) > O₂ (2) > F₂ (1). Higher bond order → shorter, stronger bond, so N₂ has the shortest bond, F₂ the longest of the three.

B has 5 electrons per atom → 10 total. B is "nitrogen and below" → light ladder (${\pi }_{2p}$ below ${\sigma }_{2p_z}$): ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^1{\pi }_{2p_y}^1$ After the 8 core/2s electrons, 2 electrons remain. On the light ladder the next level is the degenerate pair ${\pi }_{2p_x},{\pi }_{2p_y}$, so Hund's rule puts one electron in each → 2 unpaired → paramagnetic. ✅ Matches experiment. Bonding = $2+2+1+1=6$; antibonding = $2+2=4$; $\text{B.O.}=(6-4)/2=1$.

Wrong (heavy) ladder: the next level would be the single ${\sigma }_{2p_z}$, so both electrons pair up there → ${\sigma }_{2p_z}^2$, 0 unpaired → "diamagnetic." That contradicts reality — which is precisely how we know B₂ needs the flipped ladder.

Start from O₂'s heavy-ladder valence filling and edit the ${\pi }^{\ast }$ level.

• O₂ (16 e⁻): ${\pi }_{2p_x}^{\ast 1}{\pi }_{2p_y}^{\ast 1}$ → antibonding valence = 2 in ${\pi }^{\ast }$. B.O. $=2$ (from L3.2).
• O₂⁺ (15 e⁻): remove one ${\pi }^{\ast }$ electron → ${\pi }_{2p_x}^{\ast 1}{\pi }_{2p_y}^{\ast 0}$. Now $N_b=10$, $N_a=5$. $\text{B.O.}=(10-5)/2=2.5$. Removing an antibonding electron strengthens the bond.
• O₂²⁻ (18 e⁻): add two → ${\pi }_{2p_x}^{\ast 2}{\pi }_{2p_y}^{\ast 2}$. Now $N_b=10$, $N_a=8$. $\text{B.O.}=(10-8)/2=1$. Bond order: O₂⁺ (2.5) > O₂ (2) > O₂²⁻ (1). Bond length is the reverse: O₂⁺ < O₂ < O₂²⁻ (higher bond order → shorter bond).

Carbon has 6 electrons per atom → C₂ has 12 electrons. C is "nitrogen and below" → light ladder: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2$ After 8 electrons (${\sigma }_{1s},{\sigma }_{1s}^{\ast },{\sigma }_{2s},{\sigma }_{2s}^{\ast }$), 4 electrons remain, and on the light ladder the next level is the degenerate ${\pi }_{2p_x},{\pi }_{2p_y}$ pair — which holds exactly 4. So they fill both π orbitals completely, and ${\sigma }_{2p_z}$ stays empty. Bonding = $2+2+2+2=8$; antibonding = $2+2=4$. $\text{B.O.}=\frac{8-4}{2}=2.$ Correction: the bond order is 2, so the student is right about "double bond" — but both bonds are π bonds, not σ! C₂'s double bond has no net σ contribution from $2p$ (the ${\sigma }_{2p_z}$ is empty; the $2s$ σ and σ* cancel). This is a rare, genuinely surprising MOT prediction. All electrons are paired → diamagnetic.

N₂ is diamagnetic with the top filled orbital ${\sigma }_{2p_z}^2$ (light ladder, L3.1). Removing one electron from N₂⁺ takes it from the highest occupied orbital, which is the bonding ${\sigma }_{2p_z}$: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^1$ Now one orbital (${\sigma }_{2p_z}$) holds a lonely electron → 1 unpaired → paramagnetic. Bonding = $2+2+2+2+1=9$; antibonding = $2+2=4$. $\text{B.O.}=\frac{9-4}{2}=\mathrm{2.5.}$ Why lower than N₂'s 3? Because the electron we pulled out came from a bonding orbital (${\sigma }_{2p_z}$), so we removed glue → bond order drops from 3 to 2.5. (Contrast L4.2, where removing an antibonding electron raised the bond order.)