Exercises — Covalent bonding — bond length, bond energy, bond order

To use $\text{BO}=\frac{N_b-N_a}{2}$ you must know which orbitals count as $N_b$ and which as $N_a$, and in what order they fill. For second-row diatomics up to and including ${\text{N}}_2$, the ${\sigma }_{2p}$ orbital sits above the two ${\pi }_{2p}$ orbitals — the famous s–p mixing inversion. From ${\text{O}}_2$ onward the inversion switches off and ${\sigma }_{2p}$ drops below the ${\pi }_{2p}$. Full details live in Molecular Orbital Theory; here is all you need for this page:

• Bonding ($N_b$): ${\sigma }_{2s}$, ${\sigma }_{2p}$, and the two ${\pi }_{2p}$ — together they hold up to 10 electrons for ${\text{O}}_2$/$\text{NO}$ (giving $N_b=10$ when full).
• Antibonding ($N_a$): ${\sigma }_{2s}^{\ast }$ and the two ${\pi }_{2p}^{\ast }$ — the ${\pi }_{2p}^{\ast }$ pair is the highest occupied level in ${\text{O}}_2$ and $\text{NO}$, so electrons are added to or removed from there when we make ions. That single fact — "the topmost electrons of ${\text{O}}_2$ and $\text{NO}$ live in antibonding ${\pi }^{\ast }$" — is what drives every ion result below.

What decides length here? All three are carbon–carbon bonds — the same two elements — so the only thing changing is the bond order (number of shared electron pairs). More shared pairs pull the nuclei closer, so length drops as bond order rises.

Bond orders: $\text{C}-\text{C}$ is 1, $\text{C}=\text{C}$ is 2, $\text{C}\equiv \text{C}$ is 3. Higher bond order ⇒ shorter. $\text{C}\equiv \text{C}(120)<\text{C}=\text{C}(134)<\text{C}-\text{C}(154)\text{pm}$ So increasing length: $\boxed{\text{C}\equiv \text{C}<\text{C}=\text{C}<\text{C}-\text{C}}$.

Definition first. Bond energy is defined as the energy needed to break one mole of bonds in the gas phase. Breaking always costs energy, so the tabulated value is positive and represents energy absorbed.

Forming the same bond does the reverse: it releases $436$ kJ mol${}^{-1}$, i.e. $\Delta H_{\text{form}}=-436$ kJ mol${}^{-1}$. Same magnitude, opposite sign.

$\text{BO}=\frac{N_b-N_a}{2}=\frac{8-2}{2}=3$ Bond order 3 (a triple bond). Any positive bond order means the molecule is bound (net bonding electrons hold it together); only $\text{BO}=0$ means no bond.

Neutral ${\text{O}}_2$ has 16 electrons; the standard filling gives $N_b=10$, $N_a=6$: $\text{BO}({\text{O}}_2)=\frac{10-6}{2}=2.$ ${\text{O}}_2^{+}$ — remove one electron. The highest occupied orbitals of ${\text{O}}_2$ are antibonding ${\pi }^{\ast }$ (see the refresher at the top), so we remove an antibonding electron: $N_a=5$. $\text{BO}({\text{O}}_2^{+})=\frac{10-5}{2}=\mathrm{2.5.}$ ${\text{O}}_2^{-}$ — add one electron into the antibonding ${\pi }^{\ast }$: $N_a=7$. $\text{BO}({\text{O}}_2^{-})=\frac{10-7}{2}=\mathrm{1.5.}$ Length ranking (higher BO ⇒ shorter): ${\text{O}}_2^{+}(\text{BO }2.5)<{\text{O}}_2(\text{BO }2)<{\text{O}}_2^{-}(\text{BO }1.5).$

Bonds broken (cost, +): one H–H ($436$) + one Cl–Cl ($242$) $=678$. Bonds formed (payback, −): two H–Cl $=2\times 431=862$. $\Delta H=\sum BE(\text{broken})-\sum BE(\text{formed})=678-862=-184{\text{kJ mol}}^{-1}.$ Negative ⇒ exothermic: the two new HCl bonds are collectively stronger than the two bonds destroyed.

Count every bond. Broken: 4 × C–H ($4\times 413=1652$) + 2 × O=O ($2\times 498=996$) → total $2648$. Formed: 2 × C=O in ${\text{CO}}_2$ ($2\times 799=1598$) + 4 × O–H in the two waters ($4\times 463=1852$) → total $3450$. $\Delta H=2648-3450=-802{\text{kJ mol}}^{-1}.$ Strongly exothermic, as combustion should be.

The rule "shorter bond ⇒ higher bond order" is only valid for the same pair of atoms. Here the atoms differ: fluorine is a very small atom, so a C–F single bond is naturally short. Bond length is set by both bond order and atomic size (see Electronegativity and Bond Polarity).

• C=C is short because of high bond order (2).
• C–F is short because of small atomic radius, not bond order (still 1). You may only read bond order off length when comparing like with like — e.g. C–C vs C=C vs C≡C.

Two ways to answer — and why they disagree.

(a) The theory-first value. Benzene has three $\pi$ bonds spread ("delocalised") over six equivalent C–C links. Sharing 3 double-bond's worth of $\pi$ character across 6 bonds gives each bond one extra half-pair on top of its $\sigma$ single bond: $\text{BO}=1+\frac{3\pi \text{ bonds}}{6\text{links}}=\mathrm{1.5.}$ This is the value chemists actually quote, and it comes straight from resonance / delocalisation — see Resonance and Delocalisation. Picture the six $\pi$ electrons smeared into a ring, not three fixed double bonds.

(b) The naïve "read it off the length" value. If you assume length falls in a straight line with bond order and interpolate between BO 1 (154 pm) and BO 2 (134 pm): ${\text{BO}}_{\text{linear}}\approx 1+\frac{154-139}{154-134}=1+\frac{15}{20}=\mathrm{1.75.}$

Why the two disagree — the key lesson. The length↔order relationship is not linear. Look back at the C–C / C=C / C≡C numbers: length falls $154\to 134\to 120$, i.e. steps of $20$ then $14$ pm — the curve flattens as order rises. Because real length "sags" below a straight line, a true BO of 1.5 produces a length shorter than the straight-line midpoint (144 pm) would predict — closer to 139 pm. So the linear estimate over-reads the bond order (1.75) precisely because it ignores that curvature. Trust the delocalisation value 1.5; treat the linear interpolation only as a rough sanity check that lands "between single and double," never as an exact order.

The figure below plots bond order (horizontal) against bond length (vertical). The blue point marks the C–C single bond at (1, 154 pm) and the orange point the C=C double bond at (2, 134 pm); a grey dashed line joins them as the naïve linear reference. The solid blue curve through C–C, C=C and C≡C (1,154)/(2,134)/(3,120) shows the real, gently curving relationship — it dips below the straight line. The green point marks benzene's true value at (1.5, 139 pm): it sits on the curved line, below the straight-line midpoint, which is exactly why the honest bond order is 1.5, not the 1.75 the straight line would suggest.

Tabulated bond energies are averages taken over many different molecules. The real C–H bond in ${\text{CH}}_4$ is not identical to the "average C–H"; likewise the water and ${\text{CO}}_2$ bonds differ slightly from their averages. Summing many approximate numbers accumulates error, and the bond-energy method also ignores phase changes (the water product being liquid vs gas). So the method is excellent for a quick estimate and the correct sign, but not for precision.

Broken: one N≡N ($945$) + three H–H ($3\times 436=1308$) → $2253$. Formed: two ${\text{NH}}_3$, each with three N–H bonds → $6\times 391=2346$. $\Delta H=2253-2346=-93{\text{kJ mol}}^{-1}.$ Slightly exothermic. Even though N≡N is enormously strong (a big cost to break), we form six N–H bonds whose combined energy ($2346$) just edges past the total broken ($2253$). The lesson: sign is decided by totals on each side, not by any single dramatic bond.

Higher bond order digs a deeper potential-energy well and pulls the nuclei to a shorter length. ${\text{N}}_2$'s triple bond ($945$ kJ mol${}^{-1}$, 110 pm) is one of the deepest wells known, so breaking it to start any reaction demands a huge energy input — hence ${\text{N}}_2$ is inert enough to be the bulk of our atmosphere. ${\text{O}}_2$'s double bond is far shallower ($498$), so it is much easier to activate. Bond order → well depth → activation cost → reactivity is the causal chain.

(a) $\text{BO}(\text{NO})=\frac{10-5}{2}=\mathrm{2.5.}$ The half arises because one lone electron occupies an antibonding ${\pi }^{\ast }$ orbital (the highest occupied level — see the refresher). (b) Ionising removes that antibonding electron: $N_a=4$. $\text{BO}({\text{NO}}^{+})=\frac{10-4}{2}=3.$ Raising BO from 2.5 to 3 makes the bond shorter and stronger: ${\text{NO}}^{+}$ (≈106 pm) has a tighter, deeper-welled bond than $\text{NO}$ (≈115 pm). Losing an antibonding electron helps the bond.

Track only the bonds that change. The four C–H bonds present in ${\text{C}}_2{\text{H}}_4$ survive into ${\text{C}}_2{\text{H}}_6$, so they cancel. What actually changes:

• Broken: one C=C ($614$) and one H–H ($436$) → $1050$.
• Formed: one C–C ($348$) and two new C–H bonds ($2x$). $\Delta H=1050-(348+2x)=-137.$ Solve: $1050-348-2x=-137\Rightarrow 702-2x=-137\Rightarrow 2x=839\Rightarrow x=\mathrm{419.5.}$ So the implied C–H energy is $\approx 420$ kJ mol${}^{-1}$, satisfyingly close to the tabulated $413$ — the small gap is exactly the "average value" scatter from L3.3. ✔

If energy were proportional to bond order we'd expect $614\approx 2\times 348=696$ and $839\approx 3\times 348=1044$. The actual increments:

• $\sigma$ bond (BO 1): $348$.
• Adding the first $\pi$ (BO 1→2): $614-348=266$.
• Adding the second $\pi$ (BO 2→3): $839-614=225$. Each extra bond contributes less than the first, so energy grows sub-linearly. The first bond is a head-on $\sigma$ bond with strong overlap; the extra bonds are side-on $\pi$ bonds with weaker overlap (see Sigma and Pi Bonds). Length, by contrast, falls in near-even 20 pm and 14 pm steps — geometry compresses more smoothly than energy accumulates.

The figure below puts bond order on the horizontal axis. The blue line with round markers is bond length on the left axis — it slopes steadily downward, $154\to 134\to 120$ pm. The orange line with square markers is bond energy on the right axis, rising $348\to 614\to 839$ but visibly flattening at the top. The grey dashed line is what energy would be if it were simply proportional to bond order ($348\times \text{BO}$, i.e. $348/696/1044$); the real orange curve starts on it at BO 1 but sinks below it at BO 2 and 3. That widening gap — flagged in red — is the visual proof that each extra $\pi$ bond adds less energy than the first $\sigma$ bond, even while length keeps dropping smoothly.

Conclusion: energy is emphatically not proportional to bond order (a proportional model overshoots C≡C by over 200 kJ mol${}^{-1}$), because bonds two and three are weaker $\pi$ bonds. Bond length, however, does fall nearly linearly — the two quantities track bond order but with different mathematical shapes.

Step 1 — get every bond order. Start from neutral ${\text{O}}_2$ ($N_b=10$, $N_a=6$, BO 2) and adjust $N_a$ as electrons enter or leave the topmost orbitals, which for oxygen are the two antibonding ${\pi }^{\ast }$ orbitals (refresher at the top of the page).

• ${\text{O}}_2^{+}$: remove one antibonding electron, $N_a=5$ → $\text{BO}=\frac{10-5}{2}=\mathrm{2.5.}$
• ${\text{O}}_2$: $N_a=6$ → $\text{BO}=\frac{10-6}{2}=2.$
• ${\text{O}}_2^{-}$: add one antibonding electron, $N_a=7$ → $\text{BO}=\frac{10-7}{2}=\mathrm{1.5.}$
• ${\text{O}}_2^{2-}$ (peroxide): add two, $N_a=8$ → $\text{BO}=\frac{10-8}{2}=1.$

Step 2 — rank by length. Why? Same two atoms throughout, so length is governed purely by bond order: higher BO ⇒ shorter. Longest goes with lowest BO: ${\text{O}}_2^{2-}(\text{BO }1)>{\text{O}}_2^{-}(\text{BO }1.5)>{\text{O}}_2(\text{BO }2)>{\text{O}}_2^{+}(\text{BO }2.5).$

Step 3 — find the diamagnetic species by counting unpaired electrons. Why this step? "Diamagnetic" means every electron is paired; the unpaired electrons all live in the two degenerate (equal-energy) ${\pi }^{\ast }$ orbitals, which by Hund's rule fill singly before pairing. So track how many electrons sit in that ${\pi }^{\ast }$ pair:

• ${\text{O}}_2^{+}$: ${\pi }^{\ast }$ holds 1 electron → 1 unpaired → paramagnetic.
• ${\text{O}}_2$: ${\pi }^{\ast }$ holds 2 electrons, one in each orbital (Hund) → 2 unpaired → paramagnetic.
• ${\text{O}}_2^{-}$: ${\pi }^{\ast }$ holds 3 electrons (one orbital full, one half) → 1 unpaired → paramagnetic.
• ${\text{O}}_2^{2-}$: ${\pi }^{\ast }$ holds 4 electrons, filling both orbitals completely → 0 unpaired → diamagnetic.

Answer: length order ${\text{O}}_2^{2-}>{\text{O}}_2^{-}>{\text{O}}_2>{\text{O}}_2^{+}$; the peroxide ion ${\text{O}}_2^{2-}$ is the only diamagnetic member because its filled ${\pi }^{\ast }$ orbitals leave no unpaired electron.

Can you, from memory: read bond order off a length only for the same atom pair? ::: Yes — never across different elements (atomic size interferes). Predict whether ionising raises or lowers bond order? ::: Yes — depends on whether the electron leaves a bonding or antibonding orbital. Write $\Delta H=\sum BE(\text{broken})-\sum BE(\text{formed})$ and cancel spectator bonds? ::: Yes. Explain why energy grows sub-linearly with bond order? ::: Yes — first bond is $\sigma$ (strong overlap), extra bonds are $\pi$ (weaker). Say which ${\text{O}}_2$ species is diamagnetic and why? ::: ${\text{O}}_2^{2-}$ — its two ${\pi }^{\ast }$ orbitals are completely filled, so no electron is unpaired.