Exercises — Why O₂ is paramagnetic (MOT prediction)

WHAT: count protons/electrons. Each oxygen atom is element 8, so it has 8 electrons. WHY: MOT fills orbitals with the total electron count, exactly like Aufbau does for atoms. $8+8=16\text{ electrons.}$ Filling the ladder, the last 2 electrons land in the degenerate ${\pi }^{\ast }$ pair, one in each (Hund). Answer: 16 total electrons, 2 unpaired.

Attracted → paramagnetic → it has at least one unpaired electron. A diamagnetic substance (all electrons paired) is instead weakly pushed out of the field. See Paramagnetism and Diamagnetism.

The degenerate antibonding pair ${\pi }_{2p_x}^{\ast }$ and ${\pi }_{2p_y}^{\ast }$ — one electron in each. That single occupancy is the whole reason O₂ is paramagnetic.

$\text{KK}({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p_z}{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\pi }_{2p_x}^{\ast }{)}^1({\pi }_{2p_y}^{\ast }{)}^1$ Here the superscript after the bracket is the occupancy (number of electrons), so $({\pi }_{2p_x}^{\ast }{)}^1$ means "one electron in the ${\pi }_{2p_x}^{\ast }$ orbital." The two ${\pi }^{\ast }$ electrons are written as one-each because the orbitals are degenerate and Hund's rule maximises spin. So $n=2$ → paramagnetic.

Bonding electrons $N_b$: ${\sigma }_{2s}(2)+{\sigma }_{2p_z}(2)+\pi (4)=8$ in the valence shell, plus the ${\sigma }_{1s}(2)$ core = $10$. Antibonding electrons $N_a$: ${\sigma }_{1s}^{\ast }(2)+{\sigma }_{2s}^{\ast }(2)+{\pi }^{\ast }(2)=6$. $\text{B.O.}=\frac{N_b-N_a}{2}=\frac{10-6}{2}=2.$ A double bond, matching Lewis — see Bond Order.

$n=2$, so $\mu =\sqrt{n(n+2)}=\sqrt{2(2+2)}=\sqrt{8}\approx 2.83\text{BM}.$ A non-zero $\mu$ confirms paramagnetism.

Total electrons $=16+2=18$. The two extra electrons fill the previously half-filled ${\pi }^{\ast }$: $\dots ({\pi }_{2p_x}^{\ast }{)}^2({\pi }_{2p_y}^{\ast }{)}^2$ Now every orbital is doubly filled → $n=0$ → diamagnetic. $N_b=10$ (unchanged — the extra electrons went into antibonding ${\pi }^{\ast }$, so no bonding orbital gained electrons), $N_a={\sigma }_{1s}^{\ast }(2)+{\sigma }_{2s}^{\ast }(2)+{\pi }^{\ast }(4)=8$: $\text{B.O.}=\frac{10-8}{2}=1.$ See Superoxide and Peroxide ions.

Total $=17$. The extra electron pairs up in one of the ${\pi }^{\ast }$ orbitals, leaving the other singly occupied: $\dots ({\pi }_{2p_x}^{\ast }{)}^2({\pi }_{2p_y}^{\ast }{)}^1\Rightarrow n=1$ Paramagnetic, with $\mu =\sqrt{1(1+2)}=\sqrt{3}\approx 1.73\text{BM}$. Now count the antibonding electrons. Why $N_b$ is still 10: the added electron went into an antibonding ${\pi }^{\ast }$ orbital, so no bonding orbital changed — $N_b$ is identical to neutral O₂. The antibonding total is $N_a=\underset{2}{\underbrace{{\sigma }_{1s}^{\ast }(2)}}+\underset{2}{\underbrace{{\sigma }_{2s}^{\ast }(2)}}+\underset{3}{\underbrace{{\pi }^{\ast }(2+1)}}=7,$ where the "$3$" is the ${\pi }^{\ast }$ count: 2 electrons in one ${\pi }^{\ast }$ orbital plus 1 in the other (${\pi }^{\ast 2}+{\pi }^{\ast 1}$). So $\text{B.O.}=\frac{10-7}{2}=\mathrm{1.5.}$

Bond orders: O₂ = 2, ${\text{O}}_2^{-}$ = 1.5, ${\text{O}}_2^{2-}$ = 1. Higher bond order → shorter, stronger bond, because more net bonding electrons pull the nuclei closer. ${\text{O}}_2(121\text{ pm})<{\text{O}}_2^{-}(128\text{ pm})<{\text{O}}_2^{2-}(149\text{ pm}).$ Adding electrons goes into antibonding ${\pi }^{\ast }$ orbitals, which weakens the bond each time.

N₂ order: $\dots {\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^2$ (the π's fill first, then $\sigma$). All 14 electrons are paired → $n=0$ → diamagnetic. $N_b={\sigma }_{1s}(2)+{\sigma }_{2s}(2)+\pi (4)+{\sigma }_{2p_z}(2)=10$, $N_a=2+2=4$: $\text{B.O.}=\frac{10-4}{2}=3.$ Contrast: N₂ has a triple bond and no unpaired electrons; O₂ has a double bond but two unpaired electrons in ${\pi }^{\ast }$. See N2 vs O2 MO diagram. Notice the reordering does not change O₂'s ${\pi }^{\ast }$ occupancy — O₂ stays paramagnetic either way.

Total $=15$. Remove one electron from the highest occupied MO, a ${\pi }^{\ast }$: $\dots ({\pi }_{2p_x}^{\ast }{)}^1({\pi }_{2p_y}^{\ast }{)}^0\Rightarrow n=1$ Paramagnetic, $\mu =\sqrt{1(3)}=\sqrt{3}\approx 1.73\text{BM}$. Why $N_b$ is still 10: we removed the electron from an antibonding ${\pi }^{\ast }$ orbital, so every bonding orbital is untouched — $N_b$ is the same as neutral O₂. The antibonding total drops by one: $N_a={\sigma }_{1s}^{\ast }(2)+{\sigma }_{2s}^{\ast }(2)+{\pi }^{\ast }(1)=5.$ $\text{B.O.}=\frac{10-5}{2}=\mathrm{2.5.}$ So ${\text{O}}_2^{+}$ has a stronger, shorter bond than neutral O₂ (2.5 vs 2) — removing an antibonding electron strengthens the bond.

Count electrons and fill the ladder each time; only the ${\pi }^{\ast }$ occupancy changes.

The bond-length trend that follows from these bond orders is drawn below.

The amber curve is bond order (falling as we add antibonding electrons) and the cyan curve is the measured O–O bond length (rising in step) — read them together to see that more antibonding electrons means a weaker, longer bond.

(b) Bond strength (∝ bond order): ${\text{O}}_2^{+}>{\text{O}}_2>{\text{O}}_2^{-}>{\text{O}}_2^{2-}.$ Paramagnetic: ${\text{O}}_2^{+},{\text{O}}_2,{\text{O}}_2^{-}$. Diamagnetic: ${\text{O}}_2^{2-}$. Every extra electron sits in an antibonding ${\pi }^{\ast }$ orbital, so bond order falls by $\frac{1}{2}$ each step from the cation down.

The single fact that unlocks every problem above ::: The number of unpaired electrons $n$ (found by Hund-filling the degenerate π* level) decides magnetism; $N_b-N_a$ over 2 decides bond order.