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

$\text{BO}=\frac{N_b-N_a}{2}$ We divide by two because one chemical bond is one shared pair of electrons. Counting raw electrons $(N_b-N_a)$ counts individuals; dividing by 2 converts individuals into pairs, i.e. into bonds.

Only O₂ and F₂ ($Z\ge 8$, so total valence electrons put them at $\ge 15$ total electrons). For B₂, C₂, N₂ the s–p mixing pushes ${\sigma }_{2p}$ above ${\pi }_{2p}$.

The asterisk marks an antibonding orbital: it has a node (a plane of zero electron density) between the two nuclei, it sits higher in energy, and filling it weakens the bond (it subtracts from $N_b$).

Two electrons, both drop into the single lowest bonding MO, paired: ${\sigma }_{1s}^2$ $N_b=2,N_a=0$, so $\text{BO}=\frac{2-0}{2}=1$. All paired → diamagnetic. A normal H–H single bond. ✔

${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}$ The third and fourth electrons are forced into the antibonding ${\sigma }_{1s}^{\ast }$. $N_b=2,N_a=2$, so $\text{BO}=\frac{2-2}{2}=0$. No net bond → He₂ does not exist. Diamagnetic (all paired). ✔

Using ${\sigma }_{2p}$ below ${\pi }_{2p}$: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p}^2{\pi }_{2p}^4{\pi }_{2p}^{\ast 4}$ Bonding electrons: ${\sigma }_{1s}(2)+{\sigma }_{2s}(2)+{\sigma }_{2p}(2)+{\pi }_{2p}(4)=10$. Antibonding: ${\sigma }_{1s}^{\ast }(2)+{\sigma }_{2s}^{\ast }(2)+{\pi }_{2p}^{\ast }(4)=8$. $\text{BO}=\frac{10-8}{2}=1$. Every orbital is fully filled → all paired → diamagnetic. ✔

$\dots {\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p}^2{\pi }_{2p}^4{\pi }_{2p}^{\ast 2}$ The final two electrons enter the two degenerate ${\pi }_{2p}^{\ast }$ orbitals. By Hund's rule they go one each with parallel spin → 2 unpaired electrons → paramagnetic. $N_b=10,N_a=6\Rightarrow \text{BO}=\frac{10-6}{2}=2$ (a double bond). A Lewis dot structure has no way to represent two singly-occupied degenerate orbitals, so it wrongly predicts diamagnetic. Liquid O₂ clinging to a magnet is the experimental proof MO theory wins here. See Valence Bond Theory vs MO Theory. ✔

• ${\text{N}}_2$ (14 e⁻, mixed order): ${\pi }_{2p}^4{\sigma }_{2p}^2$ full, $N_b=10,N_a=4\Rightarrow \text{BO}=3$.
• ${\text{O}}_2$: $\text{BO}=2$ (from L3.1).
• ${\text{F}}_2$: $\text{BO}=1$ (from L2.3). Bond order: $3>2>1$. Higher bond order pulls nuclei closer, so bond length: ${\text{N}}_2<{\text{O}}_2<{\text{F}}_2$ (shortest to longest), and bond energy runs the opposite way (strongest N₂). This is the Bond Order, Bond Length, Bond Energy correlation. ✔

In ${\text{N}}_2$ the highest occupied MOs are ${\pi }_{2p}^4$ and ${\sigma }_{2p}^2$ — completely filled, so all electrons are paired → diamagnetic. ${\text{O}}_2$'s two extra electrons cannot fit in those filled orbitals; they must go into the next level, the two degenerate ${\pi }_{2p}^{\ast }$ orbitals — singly, parallel spins (Hund) → 2 unpaired → paramagnetic. The tiny difference of two electrons landing in degenerate antibonding orbitals flips the magnetism entirely. ✔

${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p}^4{\sigma }_{2p}^2{\pi }_{2p}^{\ast 1}$ The 15th electron is a lone electron in a ${\pi }_{2p}^{\ast }$ orbital. $N_b=10,N_a=5\Rightarrow \text{BO}=\frac{10-5}{2}=2.5$. One unpaired electron → paramagnetic. ✔

The electron removed is the highest-energy one — the single ${\pi }_{2p}^{\ast }$ antibonding electron. Removing an antibonding electron decreases $N_a$ by 1: $N_b=10,N_a=4\Rightarrow \text{BO}=\frac{10-4}{2}=3.$ Bond order rises $2.5\to 3$. Removing anti-glue should strengthen the bond — consistent. ${\text{NO}}^{+}$ is now isoelectronic with N₂ and CO (14 e⁻), all-paired → diamagnetic. See Isoelectronic species (CO, N₂, CN⁻, NO⁺). ✔

14 electrons, same count as ${\text{N}}_2$, so same filling (mixed order): $\dots {\pi }_{2p}^4{\sigma }_{2p}^2$ $N_b=10,N_a=4\Rightarrow \text{BO}=3$. All paired → diamagnetic. The triple bond plus a lone pair on carbon lets CO donate that lone pair into a metal's empty orbitals — this is why CO is such a strong ligand (and why it poisons haemoglobin). ✔

Start from neutral ${\text{O}}_2$: $\dots {\sigma }_{2p}^2{\pi }_{2p}^4{\pi }_{2p}^{\ast 2}$, with $N_b=10$ and the ${\pi }_{2p}^{\ast }$ occupancy varying.

• ${\text{O}}_2^{+}$ (15 e⁻): ${\pi }_{2p}^{\ast 1}$. $N_b=10,N_a=5\Rightarrow \text{BO}=2.5$. 1 unpaired → paramagnetic.
• ${\text{O}}_2$ (16 e⁻): ${\pi }_{2p}^{\ast 2}$ (one each). $N_a=6\Rightarrow \text{BO}=2$. 2 unpaired → paramagnetic.
• ${\text{O}}_2^{-}$ (17 e⁻): ${\pi }_{2p}^{\ast 3}$. $N_a=7\Rightarrow \text{BO}=1.5$. 1 unpaired → paramagnetic.
• ${\text{O}}_2^{2-}$ (18 e⁻): ${\pi }_{2p}^{\ast 4}$ (both full). $N_a=8\Rightarrow \text{BO}=1$. All paired → diamagnetic.

Bond order: ${\text{O}}_2^{+}(2.5)>{\text{O}}_2(2)>{\text{O}}_2^{-}(1.5)>{\text{O}}_2^{2-}(1)$. Each added electron goes into antibonding ${\pi }^{\ast }$, so each step lowers bond order by $0.5$. Only peroxide is diamagnetic. ✔

Mixed order (${\pi }_{2p}$ below ${\sigma }_{2p}$), 12 electrons: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p}^4$ The ${\pi }_{2p}$ pair of orbitals fills completely (4 electrons), and ${\sigma }_{2p}$ stays empty. $N_b=8,N_a=4\Rightarrow \text{BO}=\frac{8-4}{2}=2$. All paired → diamagnetic. What's unusual: the double bond is made of two π bonds and no σ bond between the 2p orbitals — a direct consequence of s–p mixing raising ${\sigma }_{2p}$ above ${\pi }_{2p}$. (Had we wrongly used the unmixed order, we'd predict two unpaired electrons and paramagnetism — which is wrong for C₂.) ✔

Mixed order, 10 electrons: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p}^2$ The last two electrons enter the two degenerate ${\pi }_{2p}$ orbitals — by Hund's rule, one each, parallel spin → 2 unpaired → paramagnetic. $N_b=6,N_a=4\Rightarrow \text{BO}=\frac{6-4}{2}=1$ (a single bond, but made of two half-π bonds). The student's claim is wrong: like O₂, B₂ is paramagnetic, and again this is the direct signature of degenerate orbitals + Hund's rule — invisible to Lewis structures. ✔

Species → (electrons, bond order, magnetism):

• H₂ ::: 2 e⁻, BO 1, diamagnetic
• He₂ ::: 4 e⁻, BO 0, does not exist
• B₂ ::: 10 e⁻, BO 1, paramagnetic
• C₂ ::: 12 e⁻, BO 2, diamagnetic
• N₂ ::: 14 e⁻, BO 3, diamagnetic
• O₂ ::: 16 e⁻, BO 2, paramagnetic
• F₂ ::: 18 e⁻, BO 1, diamagnetic
• NO ::: 15 e⁻, BO 2.5, paramagnetic
• CO ::: 14 e⁻, BO 3, diamagnetic
• O₂⁺ ::: 15 e⁻, BO 2.5, paramagnetic
• O₂⁻ ::: 17 e⁻, BO 1.5, paramagnetic
• O₂²⁻ ::: 18 e⁻, BO 1, diamagnetic