Bond length and bond angle
Bond length is the average distance between the nuclei of two bonded atoms. It is measured by X-ray or electron diffraction and is given in picometres (1 pm = 10⁻¹² m). It is roughly the sum of the covalent radii of the two atoms. Examples: H–H 74 pm, O–H 96 pm, C–C 154 pm, Cl–Cl 199 pm.
Bond angle is the angle between two bonds that meet at the same atom. It tells us about the shape of the molecule. Examples: CO₂ 180°, BF₃ 120°, CH₄ 109.5°, NH₃ 107°, H₂O 104.5°.
Bond enthalpy and bond order
Bond enthalpy (bond dissociation enthalpy) is the energy needed to break 1 mol of a given bond in gaseous molecules. Unit: kJ mol⁻¹. H–H needs 435, O=O needs 498, N≡N needs 946. A bigger bond enthalpy means a stronger bond.
For molecules with several identical bonds (like the four C–H in CH₄), each bond breaks with a slightly different energy, so we use the average bond enthalpy (C–H ≈ 414 kJ mol⁻¹).
Bond order (Lewis picture) = number of bonds between two atoms: H₂ 1, O₂ 2, N₂ 3. Molecules with the same number of electrons (isoelectronic), like N₂, CO and NO⁺, have the same bond order (3).
The link
Bond order ↑ → bond enthalpy ↑ → bond length ↓. C–C: 154 pm, 348 kJ; C=C: 134 pm, 614 kJ; C≡C: 120 pm, 839 kJ mol⁻¹.
Resonance
Sometimes a single Lewis structure does not match experiment. In O₃ the Lewis structure shows one O=O (121 pm) and one O–O (148 pm), but experiments show both bonds equal (128 pm).
We then write two or more resonance (canonical) structures joined by a double-headed arrow (↔). The real molecule is the resonance hybrid, a blend of them all. It does not flip between them; it is always the blend.
- Resonance structures have the same positions of atoms; only electrons move.
- The hybrid is more stable (lower energy) than any single structure. The extra stability is the resonance energy.
- Examples: O₃, CO₃²⁻ (each C–O = 1⅓ bond, 129 pm), NO₃⁻, benzene, CO₂.
Bond order in a hybrid = total bonds shared ÷ number of positions. CO₃²⁻: 4 bonds over 3 positions = 1.33.
Polarity of bonds and dipole moment
In H₂ or Cl₂ both atoms pull the shared pair equally: a non-polar covalent bond. In HCl, Cl is more electronegative, so the pair sits nearer to Cl. Cl gets a small negative charge (δ−) and H a small positive charge (δ+): a polar covalent bond.
Polarity is measured by the dipole moment: μ = q × d (charge × distance). Unit: debye (D); 1 D = 3.336 × 10⁻³⁰ C m. It is a vector; chemists draw it as an arrow from + to −, with a small cross at the + end.
Shape decides the total
- CO₂ (linear): two equal C=O dipoles point opposite ways → μ = 0.
- BF₃ (flat triangle), CCl₄, CH₄: dipoles cancel → μ = 0.
- H₂O (bent): dipoles add → μ = 1.85 D.
- NH₃ (1.47 D) vs NF₃ (0.23 D): in NH₃ the lone-pair dipole adds to the bond dipoles; in NF₃ it works against them.
Covalent character in ionic bonds (Fajans' rules)
A small, highly charged cation can pull the electron cloud of a large anion towards itself (polarisation). This gives an ionic bond some covalent character. It is more when the cation is small and highly charged and the anion is large. So LiI is more covalent than NaCl, and AlCl₃ more covalent than NaCl.
Percentage ionic character = (μ observed ÷ μ for a full ionic bond) × 100.
Try it: the comb and water test
Run your comb through dry hair. Open a tap so a very thin stream flows. Bring the comb close without touching. The stream bends! Water molecules are polar and turn their charged ends towards the comb. Predict first: would a stream of a non-polar liquid bend? Then use the free-play step to compare H₂O and CO₂.
Key formulas and definitions
- Bond order (Lewis) = number of shared pairs between two atoms
- Bond order of a resonance hybrid = total bonds ÷ number of bonded positions
- Dipole moment μ = q × d; 1 D = 3.336 × 10⁻³⁰ C m
- % ionic character = (μ observed ÷ μ ionic) × 100, μ ionic = e × d
- Bond order ↑ ⇒ bond length ↓ and bond enthalpy ↑
Worked examples
1. Find the bond order of N₂, O₂ and F₂ from their Lewis structures.
Step 1: N≡N has 3 shared pairs → bond order 3. Step 2: O=O has 2 shared pairs → 2. Step 3: F–F has 1 shared pair → 1. Answer: N₂ 3, O₂ 2, F₂ 1. So N₂ is the strongest and shortest bond.
2. Find the C–O bond order in the carbonate ion, CO₃²⁻.
Step 1: One resonance structure has one C=O and two C–O: 4 bonds in total. Step 2: These are spread over 3 equal positions. Step 3: Bond order = 4 ÷ 3. Answer: 1.33.
3. Find the N–O bond order in NO₃⁻ and in NO₂⁻.
Step 1: NO₃⁻: one N=O + two N–O = 4 bonds over 3 positions → 4/3 = 1.33. Step 2: NO₂⁻: one N=O + one N–O = 3 bonds over 2 positions → 3/2 = 1.5. Answer: 1.33 and 1.5. So the N–O bond in NO₂⁻ is shorter.
4. Arrange C–C bonds in ethane, ethene and ethyne by bond length and by bond enthalpy.
Step 1: Bond orders: ethane 1, ethene 2, ethyne 3. Step 2: Higher order → shorter: ethyne (120 pm) < ethene (134) < ethane (154). Step 3: Higher order → stronger: ethane (348) < ethene (614) < ethyne (839 kJ mol⁻¹). Answer: length ethyne < ethene < ethane; enthalpy ethane < ethene < ethyne.
5. HCl has a bond length of 127 pm and a dipole moment of 1.03 D. Find its percentage ionic character. (e = 1.602 × 10⁻¹⁹ C)
Step 1: If fully ionic, μ = e × d = 1.602 × 10⁻¹⁹ × 127 × 10⁻¹² = 2.03 × 10⁻²⁹ C m. Step 2: In debye: 2.03 × 10⁻²⁹ ÷ 3.336 × 10⁻³⁰ = 6.10 D. Step 3: % ionic = 1.03 ÷ 6.10 × 100 ≈ 16.9 %. Answer: about 17 % ionic.
6. Which has a dipole moment: CO₂, H₂O, BF₃, NH₃? Explain.
Step 1: CO₂ is linear: two equal dipoles in opposite directions cancel → μ = 0. Step 2: BF₃ is a flat triangle: three equal dipoles at 120° cancel → μ = 0. Step 3: H₂O is bent and NH₃ is pyramidal: the dipoles add up. Answer: H₂O (1.85 D) and NH₃ (1.47 D) are polar; CO₂ and BF₃ are not.
7. Energy needed to break all bonds in 1 mol of CH₄ is 1656 kJ. Find the average C–H bond enthalpy.
Step 1: CH₄ has 4 C–H bonds. Step 2: Average = 1656 ÷ 4. Answer: 414 kJ mol⁻¹.
Common mistakes
- Thinking a molecule flips back and forth between resonance structures. It is always the single blended hybrid.
- Saying a molecule with polar bonds must be polar. CO₂ and CCl₄ have polar bonds but μ = 0 because of their symmetric shape.
- Mixing up bond length trends: a higher bond order gives a SHORTER bond, not a longer one.
- Forgetting to convert pm to m (× 10⁻¹²) when calculating dipole moment.