Covalent bonding — bond length, bond energy, bond order: Worked examples

This page is the "throw everything at it" drill for the parent topic on covalent bonding. We build a scenario matrix first — a checklist of every kind of question this topic can generate — then solve one example per cell so you never meet a case you have not already seen.

Definition

$σ$ and $π$ — the two ways orbitals overlap (define before we use them) When two atoms bond, their electron clouds can overlap in two geometrically different ways (built in detail in Sigma and Pi Bonds):

A == $σ$ (sigma) bond== is head-on overlap: the two lobes meet directly along the line joining the nuclei, like two hands pressed palm-to-palm. This gives the densest electron pile-up right between the nuclei, so $σ$ bonds are the strongest. The first bond between any two atoms is always a $σ$ bond.
A == $π$ (pi) bond== is side-on overlap: two parallel $p$ -orbital lobes touch above and below the bond axis, like two hands touching only at the fingertips. Less density sits directly between the nuclei, so a $π$ bond is weaker than a $σ$ bond. Every bond beyond the first (the 2nd in a double, the 2nd and 3rd in a triple) is a $π$ bond.

So: single bond $= 1 σ$ ; double $= 1 σ + 1 π$ ; triple $= 1 σ + 2 π$ . This is why extra bonds add less energy each — you will use it in Ex 1.

The scenario matrix

#	Cell class	Trap / edge it tests	Example
A	Same atoms, different bond order (BO)	ordering length ↔ energy	Ex 1 (N–N series)
B	Different atoms, same BO	size beats bond order	Ex 2 (C–F vs C–C)
C	MO bond order, whole number	counting $N_{b}, N_{a}$	Ex 3 (O $_{2}$ )
D	MO bond order, ion (remove e⁻)	antibonding vs bonding removal	Ex 4 (NO → NO $^{+}$ )
E	Fractional / delocalised BO	resonance, length interpolation	Ex 5 (benzene, O $_{3}$ )
F	Numeric enthalpy	sign convention, arithmetic	Ex 6 (combustion of methane)
G	Real-world word problem	translate words → the formula	Ex 7 (welding gas choice)
H	Degenerate / limiting input	BO = 0, $r \to \infty$ , $r \to 0$	Ex 8 (He $_{2}$ and the curve edges)

Every numeric answer below is machine-checked in the verify block.

Cell A — same atoms, different bond order

Worked example

Ex 1. Rank the three nitrogen–nitrogen bonds N–N, N=N, N≡N by length and by energy. Data: energies $163, 418, 945$ kJ mol $^{- 1}$ ; lengths $146, 124, 110$ pm (order not given — you assign). Forecast: which bond is shortest, and is the strongest one roughly $3 \times$ the weakest? Guess before reading.

Assign bond orders. N–N is BO 1, N=N is BO 2, N≡N is BO 3. Why this step? The number of shared electron pairs is the single control knob for the same pair of atoms.
Order by length (shortest first): N≡N $(110) <$ N=N $(124) <$ N–N $(146)$ pm. Why? More shared pairs pull the nuclei closer, so higher BO = shorter.
Order by energy (strongest first): N≡N $(945) >$ N=N $(418) >$ N–N $(163)$ kJ mol $^{- 1}$ . Why? More bonds dig a deeper potential well; the bond energy is that depth.
Check " $3 \times$ ?" Weakest $\times 3 = 163 \times 3 = 489$ , but the triple bond is $945$ — far bigger. Why? Using the $σ / π$ split just defined: the first ( $σ$ ) bond is the strongest, and the extra $π$ bonds (side-on, weaker) add less each. Here N≡N over-shoots because lone-pair repulsion in the single N–N weakens it. Never assume linearity.

Verify: ratios $945/163 \approx 5.8$ and $418/163 \approx 2.6$ — clearly non-linear, matching the " $σ \neq = π$ " idea. Lengths decrease monotonically as BO rises. ✓

Cell B — different atoms, same bond order (the trap)

Worked example

Ex 2. C–C is $154$ pm, C–F is $135$ pm. Both are single bonds (BO 1). A student says "C–F is shorter, so it must have higher bond order." Correct them. Forecast: if bond order is equal, what else could make C–F shorter?

State both bond orders. Both are BO 1 — no difference there. Why this step? Kills the student's premise before we explain the real cause.
Compare atomic sizes. Fluorine's covalent radius $\approx 64$ pm; carbon's $\approx 77$ pm. Why? Bond length is roughly the sum of the two covalent radii. Smaller F ⇒ shorter bond.
Predict the length by radius sum. $77 + 64 = 141$ pm — close to the observed $135$ pm. Why? Electronegativity pull (see Electronegativity and Bond Polarity) contracts it a touch further, but the leading effect is size.
State the rule. BO trends on length/energy are valid only across the same two elements (Cell A). Across different atoms, size and electronegativity dominate.

Verify: predicted $141$ pm vs observed $135$ pm — within $\sim 4%$ , and both are shorter than C–C ( $154$ ) even though BO is identical. The student's reasoning is refuted. ✓

Cell C — bond order from MO electron counts (whole number)

Before touching numbers, look at what $N_{b}$ and $N_{a}$ actually mean. The figure below is a molecular-orbital energy ladder: atomic orbitals on the left and right combine into a lower bonding level (electrons here glue the atoms together) and a higher antibonding level (electrons here push the atoms apart, marked with a star $^{*}$ ). Bond order counts the net glue.

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

Worked example

Ex 3. Find the bond order of O $_{2}$ from Molecular Orbital Theory, given $N_{b} = 10$ bonding electrons and $N_{a} = 6$ antibonding electrons. Forecast: single, double, or triple? And is O $_{2}$ magnetic?

Read the diagram. Ten electrons sit in the lower (bonding) levels, six in the upper (antibonding, starred) levels — the filled circles in the figure. Why this step? Bond order is a net count; you must see which level each electron occupies before subtracting.
Recall the MO formula (see Molecular Orbital Theory): $BO = \frac{N _{b} - N _{a}}{2}$ Why? Every antibonding electron cancels one bonding electron; dividing by $2$ converts electrons to pairs.
Substitute. $BO = \frac{10 - 6}{2} = 2$ . Why? $10 - 6 = 4$ net bonding electrons $= 2$ net pairs.
Interpret. BO 2 means a double bond — consistent with the Lewis O=O picture. Why? The whole-number result matches the simple dot structure, so both models agree here.

Verify: $(10 - 6) /2 = 2$ . ✓ (The two unpaired antibonding electrons — visible as the two single arrows in the top level of the figure — make O $_{2}$ paramagnetic, a famous MO win.)

Cell D — remove an electron (ion), bonding vs antibonding

The same MO ladder tells you which electron leaves when you ionise. Electrons are removed from the highest occupied level first — and in NO that top level is antibonding. The figure above marks the highest occupied ( $π^{*}$ ) level with an arrow: knock that electron out and you delete a bond-weakening electron, so bond order goes up.

Worked example

Ex 4. NO has $N_{b} = 10$ , $N_{a} = 5$ . Compute BO. Then ionise to NO $^{+}$ by removing one electron — from an antibonding orbital. Compute the new BO and predict the change in bond length. Forecast: does the bond get stronger or weaker when we pull an electron out?

BO of NO. $BO = \frac{10 - 5}{2} = 2.5$ . Why this step? The odd electron sits alone in an antibonding $π^{*}$ level (top of the ladder), giving the half.
Remove the antibonding electron. $N_{a}$ falls from $5$ to $4$ ; $N_{b}$ unchanged at $10$ . Why? The highest occupied orbital in NO is antibonding, so that's the electron that leaves first — exactly the level the figure highlights.
New BO. $BO = \frac{10 - 4}{2} = 3$ . Why? Removing a bond-weakening electron raises the net pair count.
Predict length. BO went $2.5 \to 3$ , so NO $^{+}$ is shorter and stronger than NO. Why? Same two atoms, higher BO ⇒ Cell-A trend applies directly.

Verify: $(10 - 5) /2 = 2.5$ and $(10 - 4) /2 = 3$ . BO increased by $0.5$ on removing an antibonding electron. ✓ (Contrast: removing a bonding electron would have lowered BO — always check which orbital.)

Cell E — fractional / delocalised bond order

Worked example

Ex 5. (a) Benzene C–C is measured at $139$ pm. Given C–C(single) $= 154$ pm and C=C(double) $= 134$ pm, what effective bond order does that imply? (b) In ozone O $_{3}$ , two equivalent resonance forms each show one O–O single and one O=O double bond. What is each O–O bond order? Forecast: will the benzene BO be closer to $1$ or to $2$ ?

Justify the linear assumption first. From the parent note's C–C data, length falls $154 \to 134 \to 120$ pm as BO goes $1 \to 2 \to 3$ — steps of $20$ then $14$ pm, roughly equal. So over the narrow $1 \to 2$ window, treating length as a straight-line function of BO is a fair local approximation (its limitation: the real relation curves slightly, and it only holds across the same two atoms, so this is an estimate, not a law). Why this step? You must never interpolate without first checking the data actually looks linear over the range you use.
Set up the interpolation for benzene. $BO \approx 1 + \frac{154 - 139}{154 - 134} = 1 + \frac{15}{20} = 1.75 (rough)$ Why? Length between single and double implies a BO between $1$ and $2$ ; the fraction says how far along.
Locate 139 pm relative to the midpoint. The midpoint of $154$ and $134$ is $\frac{154 + 134}{2} = 144$ pm. Since $139 < 144$ , the measured length lies on the double-bond side of the midpoint. Why this step? It sanity-checks the reading: shorter than the midpoint means BO above $1.5$ , consistent with the $1.75$ estimate — the bond has more than half double-bond character.
Compare with the resonance count. Six electrons delocalised over six bonds (see Resonance and Delocalisation) give the textbook value $BO = 1.5$ . Why? Interpolation is only a linear guess; the structural answer from delocalisation is $1.5$ . The two estimates bracket reality — benzene's true C–C character sits between $1.5$ and $\sim 1.75$ , which is why $139$ pm falls just below the midpoint.
Ozone bond order. Averaging the two resonance forms, each O–O is $\frac{1}{2} (1 + 2)$ : $BO_{O-O} = \frac{1 + 2}{2} = 1.5$ Why? Each bond is a single in one structure and a double in the other; the real bond is the average.

Verify: ozone BO $= (1 + 2) /2 = 1.5$ ; benzene interpolation $= 1 + 15/20 = 1.75$ ; midpoint length $= 144$ pm and $139 < 144$ confirms the double-bond side. All fractional, all between single and double. ✓

Cell F — numeric enthalpy from bond energies

The figure below is an energy-level (enthalpy) diagram for the reaction. The vertical axis is energy in kJ mol $^{- 1}$ (higher = more energy stored); the horizontal axis is just reaction progress (left = start, right = end, no scale). We first pay to break every reactant bond, climbing from the blue reactant level up to the yellow "free atoms" level; then energy is returned as product bonds form, dropping to the pink product level. The net drop from blue to pink is $Δ H$ : because pink sits below blue, energy is released and $Δ H < 0$ .

Worked example

Ex 6. Estimate $Δ H$ for the combustion of methane: $CH_{4} + 2 O_{2} \to CO_{2} + 2 H_{2} O$ Bond energies (kJ mol $^{- 1}$ ): C–H $= 414$ , O=O $= 498$ , C=O $= 799$ (in CO $_{2}$ ), O–H $= 463$ . Forecast: exothermic or endothermic? (You burn it for heat — so guess the sign.)

List bonds broken (reactants). $4 \times$ C–H $+ 2 \times$ O=O. Why this step? The formula $Δ H = \sum B E_{broken} - \sum B E_{formed}$ needs every broken bond (see Hess's Law and Enthalpy). This is the upward climb to the yellow "free atoms" level in the figure.
Sum broken. $4 (414) + 2 (498) = 1656 + 996 = 2652$ kJ. Why? Breaking always costs energy (positive), so this is the "spend" side — the length of the upward arrow.
List bonds formed (products). CO $_{2}$ has $2 \times$ C=O; each H $_{2}$ O has $2 \times$ O–H, and there are $2$ waters ⇒ $4 \times$ O–H. Why? Count every new bond in the products; watch the coefficients.
Sum formed. $2 (799) + 4 (463) = 1598 + 1852 = 3450$ kJ. Why? Forming pays energy back (negative side) — the length of the downward arrow to the pink level.
Combine. $Δ H = 2652 - 3450 = - 798$ kJ mol $^{- 1}$ . Why? The downward arrow is longer than the upward one, so pink sits below blue ⇒ net release ⇒ negative.

Verify: $2652 - 3450 = - 798$ ; the negative sign confirms combustion is exothermic, and the magnitude is the right order for a real methane value ( $\approx - 802$ kJ mol $^{- 1}$ tabulated). ✓

Cell G — real-world word problem

Worked example

Ex 7. A workshop must choose a fuel gas for cutting steel and wants the flame that releases the most heat. Compare ethyne (acetylene, HC≡CH) with ethane (CH $_{3}$ –CH $_{3}$ ) burning to CO $_{2}$ and H $_{2}$ O. Use the bond-energy method $Δ H = \sum B E_{broken} - \sum B E_{formed}$ . Data (kJ mol $^{- 1}$ ): C–H $= 414$ , C–C $= 348$ , C≡C $= 839$ , O=O $= 498$ , C=O $= 799$ , O–H $= 463$ . Forecast: which has the higher bond order in its carbon backbone, and does higher fuel bond order help or hurt heat output?

Balance ethyne combustion. $C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 CO_{2} + H_{2} O$ . Why this step? You cannot count bonds until the equation is balanced.
Ethyne — bonds broken. $1 \times$ C≡C $+ 2 \times$ C–H $+ \frac{5}{2} \times$ O=O $= 839 + 2 (414) + 2.5 (498) = 839 + 828 + 1245 = 2912$ kJ. Why? Every reactant bond must be broken; the fuel's C≡C (BO 3) is a large single term.
Ethyne — bonds formed. $2$ CO $_{2}$ have $4 \times$ C=O; $1$ H $_{2}$ O has $2 \times$ O–H $= 4 (799) + 2 (463) = 3196 + 926 = 4122$ kJ. Why? Count all product bonds with coefficients.
Ethyne $Δ H$ . $2912 - 4122 = - 1210$ kJ mol $^{- 1}$ . Why? Broken − formed; strongly negative ⇒ very exothermic.
Balance ethane combustion. $C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 CO_{2} + 3 H_{2} O$ . Why? Same procedure for the fair comparison.
Ethane — bonds broken. $1 \times$ C–C $+ 6 \times$ C–H $+ \frac{7}{2} \times$ O=O $= 348 + 6 (414) + 3.5 (498) = 348 + 2484 + 1743 = 4575$ kJ. Why? Ethane's backbone is C–C (BO 1) but it carries six C–H bonds.
Ethane — bonds formed. $2$ CO $_{2}$ ( $4 \times$ C=O) $+ 3$ H $_{2}$ O ( $6 \times$ O–H) $= 4 (799) + 6 (463) = 3196 + 2778 = 5974$ kJ. Why? Product bond count with coefficients.
Ethane $Δ H$ . $4575 - 5974 = - 1399$ kJ mol $^{- 1}$ . Why? Broken − formed; even more exothermic per mole than ethyne.
Resolve the twist. Ethane releases more total heat per mole ( $- 1399$ vs $- 1210$ ), yet the oxy-acetylene flame is hotter in practice ( $\sim 3300^{\circ}$ C). Why? Ethyne is a high-energy, endothermic-to-form molecule that needs less oxygen per fuel carbon, so its heat is dumped into a smaller mass of product gas — that concentrates the energy into a higher temperature even though the molar total is smaller. "Most total heat" (ethane) and "hottest flame" (ethyne) are different questions; the workshop wants the latter.

Verify: ethyne $Δ H = 2912 - 4122 = - 1210$ ; ethane $Δ H = 4575 - 5974 = - 1399$ ; both exothermic, ethane larger in magnitude — the non-obvious "hottest ≠ most total heat" conclusion holds. ✓

Cell H — degenerate and limiting inputs

The Morse potential is a standard model curve for the energy $E$ of two bonded atoms as a function of their separation $r$ : $E (r) = D_{e} (1 - e^{- a (r - r_{e})})^{2} - D_{e}$ Here $D_{e}$ = the well depth (equal to the bond energy $B E$ ), $r_{e}$ = the equilibrium separation (the bond length, where the curve bottoms out), and $a$ = a stiffness constant setting how steeply the walls rise. This is exactly the curve introduced in the parent note. The figure below marks all three edge behaviours we test in this example: the steep pink wall at small $r$ , the flat blue tail at large $r$ , and the pink dot at the bottom of the well.

Worked example

Ex 8. The edge cases. (a) He $_{2}$ : $N_{b} = 2$ , $N_{a} = 2$ — compute BO and say whether the molecule exists. (b) On the Morse curve above, what is the energy as $r \to \infty$ , and what happens to the curve as $r \to 0$ ? Forecast: can bond order be zero? What does a "bond" of order zero even mean?

He $_{2}$ bond order. $BO = \frac{2 - 2}{2} = 0$ . Why this step? The two bonding electrons are exactly cancelled by two antibonding electrons — net zero pairs.
Interpret BO = 0. No net bond ⇒ no stable He $_{2}$ molecule. This is the degenerate input: the bond well vanishes entirely, so there is no curve like the figure at all. Why? With no net electron glue between the nuclei, repulsion wins at every separation.
Limit $r \to \infty$ . In the Morse form as $r \to \infty$ the exponential $e^{- a (r - r_{e})} \to 0$ , so $E \to D_{e} (1 - 0)^{2} - D_{e} = 0$ — the blue dashed line and the flat right tail in the figure. Why? Two separated atoms at infinity are defined as the zero of energy — the atoms feel nothing, so no bonding occurs.
Limit $r \to 0$ . As $r \to 0$ , $(r - r_{e}) < 0$ so $e^{- a (r - r_{e})}$ grows huge; $E \to + \infty$ — the steep pink left wall in the figure. Why? Nucleus–nucleus repulsion ( $\propto 1/ r$ ) blows up — no two nuclei can sit on top of each other.

Verify: He $_{2}$ BO $= (2 - 2) /2 = 0$ ; Morse limit at infinity $= D_{e} (1)^{2} - D_{e} = 0$ ; and $E (r) \to + \infty$ as the exponent $\to + \infty$ . All three edges behave. ✓

Recall Quick self-test — cover the answers

Which cell is "C–F shorter than C–C despite equal BO"? ::: Cell B — size beats bond order. BO of O $_{2}^{+}$ if O $_{2}$ is BO 2 (with $N_{b} = 10, N_{a} = 6$ ) and you remove an antibonding electron? ::: $(10 - 5) /2 = 2.5$ . What is the energy of the Morse curve at infinite separation? ::: Zero — the defined reference. Sign of $Δ H$ when more/stronger bonds form than break? ::: Negative (exothermic). Bond order of any molecule with $N_{b} = N_{a}$ ? ::: Zero — no stable molecule. Why is the first bond of any pair a $σ$ and the strongest? ::: Head-on overlap piles the most electron density directly between the nuclei.

Related builds: Sigma and Pi Bonds · VSEPR and Molecular Geometry · Molecular Orbital Theory.