Account for every outer electron when drawing bonds, including multiple bonds, electron-deficient centres and coordinate donation.
A shared pair attracts both nuclei
A covalent bond is the strong electrostatic attraction between two nuclei and a shared pair of electrons between them. The shared electrons are negative and attracted to both positive nuclei. A single bond contains one shared pair; double and triple bonds contain two and three pairs respectively.
A dot-and-cross diagram records which atom supplied each electron before bonding. Dots and crosses are not different kinds of electrons. Show outer electrons, including non-bonding lone pairs, and check the total against the group numbers and overall charge. Hydrogen usually has two electrons around it in a bond; the usual eight-electron pattern for second-period atoms is not a universal rule for every element.
Worked electron inventories: H₂, O₂ and N₂
H₂ contains two outer electrons, so one shared dot–cross pair gives each hydrogen a filled first shell. O₂ contains 12 outer electrons: two shared pairs use four electrons, leaving eight as two lone pairs on each oxygen. N₂ contains ten outer electrons: three shared pairs use six, leaving one lone pair on each nitrogen.
For methane, carbon supplies four electrons and the four hydrogens supply one each: four shared pairs, no lone pair on carbon. For carbon dioxide, carbon supplies four and each oxygen six: 16 outer electrons. Draw O=C=O with two shared pairs in each C=O bond and two lone pairs on each oxygen; carbon has no lone pair.
Swipe horizontally to view the whole diagram.
A coordinate bond uses a pair from one donor
A dative covalent or coordinate bond forms when both electrons in the shared pair come from one atom. The donor has a lone pair; the acceptor has an available orbital. A coordinate arrow points from the lone-pair donor towards the acceptor, not from a positive ion towards the donor.
Worked example: ammonia has three N–H shared pairs and one nitrogen lone pair. H⁺ has no electron to contribute. Nitrogen donates its lone pair to H⁺ to form NH₄⁺. Draw the new pair with two of nitrogen’s electron symbols, enclose the whole ion in brackets and show charge +. Nitrogen now has four bonding pairs and no lone pair.
After formation, all four N–H bonds in ammonium are equivalent; the coordinate description tracks how the pair was supplied. For shape, count it as one ordinary bonding region. NH₄⁺ is tetrahedral, while NH₃ is pyramidal because it still has a lone pair.
Al₂Cl₆ contains two bridging chlorine atoms
An AlCl₃ molecule has only six electrons around aluminium in three bonds. Two molecules can combine so a lone pair on a chlorine of each molecule is donated to the electron-deficient aluminium of the other. The dimer Al₂Cl₆ has two bridging chlorines and four terminal chlorines, and each aluminium is surrounded by four bonding regions.
To build its dot-and-cross diagram, draw two aluminium centres and two chlorine bridges between them. Each bridging chlorine has one ordinary Al–Cl pair and donates one lone pair to the other Al. The two donation arrows therefore come from different bridging chlorines and point to different aluminium atoms. Each terminal Cl has three lone pairs; each bridging Cl has two lone pairs after donation.
Check 2(3) + 6(7) = 48 outer electrons. Eight bonding pairs use 16; four terminal chlorines contribute 24 non-bonding electrons and two bridges contribute eight, giving 48 altogether. A drawing with a donor arrow from a terminal chlorine fails to represent the two bridges. The analogous Al₂Br₆ drawing was tested in 8CH0/01 June 2023 Q3(a)(iii).
Diagram placeholder
Dot-and-cross construction of Al₂Cl₆
Labels to include:
- Two Al atoms; four terminal Cl atoms; two bridging Cl atoms
- Six ordinary Al–Cl pairs: one dot plus one cross each
- Two donated pairs, both electrons from their bridging chlorine
- Donation arrows Cl → Al, one to each aluminium
- Three lone pairs per terminal Cl; two per bridging Cl
- 48 outer electrons in total; neutral dimer
The two bridges join both aluminium centres. Count all pairs after donation, and keep each arrow’s origin at the donating chlorine. This labelled placeholder does not claim that a finished three-dimensional drawing is present.
Shorter comparable bonds are usually stronger
For bonds between the same pair of elements, greater bond order usually produces a shorter, stronger bond: C≡C is shorter and stronger than C=C, which is shorter and stronger than C–C. Greater shared electron density and orbital overlap strengthen attraction between the nuclei and bonding electrons. A stronger bond requires more energy to break.
Across comparable bonds to larger atoms, increased bond length and more diffuse orbitals often reduce overlap and attraction. Bond strength and bond enthalpy are linked, but not interchangeable words in an explanation: identify why attraction or overlap changes, then say more or less energy is required. Do not claim every short bond of any element pair must be stronger than every longer bond.
The bond enthalpy comparison in 8CH0/01 June 2023 Q3(b)(i) required a physical explanation connecting length and attraction/overlap. That supports naming the relevant electron–nucleus attraction rather than only repeating the observed data trend.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.
Q1. How many bonding and lone pairs are present in one O₂ molecule?Show answer
There are two shared bonding pairs in O=O and four lone pairs in total, two on each O. The inventory is 2(2) + 4(2) = 12 outer electrons.
Q2. Explain the coordinate bond in NH₄⁺, including donor, acceptor and overall charge.Show answer
Nitrogen in NH₃ donates its lone pair to H⁺. The arrow is N → H. The product is bracketed with overall charge +; nitrogen has four bonding pairs and no lone pair. Both donated electrons originated on nitrogen.
Q3. A proposed Al₂Cl₆ diagram has six chlorines each with three lone pairs and eight bonding pairs. What is wrong?Show answer
It contains 36 + 16 = 52 outer electrons, but Al₂Cl₆ has only 48. The two bridging chlorines each use a lone pair to make a coordinate bond and therefore retain two lone pairs, not three.
Q4. Predict the bond-length order C–C, C=C and C≡C and connect it to strength.Show answer
C≡C is shortest, then C=C, then C–C. For this same element pair, higher bond order generally gives greater electron density/overlap between the nuclei, stronger attraction and more energy needed for bond breaking.
Q5. BCl₃ has only six electrons around boron. Why should its diagram not be given a lone pair on B just to reach eight?Show answer
B contributes three outer electrons and three chlorines contribute 21, for 24 total. Three bonds and three lone pairs on each Cl already use all 24. Adding a boron lone pair invents electrons; boron is electron-deficient in BCl₃.
Sources
Sources and examiner guidance (reviewed 9 October 2026)
- Pearson Edexcel 9CH0 specification, Issue 3 — Topic 2, printed pp. 9–11, checked against 8CH0 printed pp. 7–9. Reviewed 9 October 2026.
- Chemrevise — Edexcel Bonding — All 14 pages reviewed as a secondary coverage check. Explanations, worked applications and exercises are original.
- Pearson 8CH0/01 June 2023 mark scheme — Q3(a)(iii), Q3(b)(i), Q4(a), Q5(c–d), Q7(a–b); PDF pp. 11–12, 14, 18–19, 22–23. Question-specific evidence.
- Pearson 8CH0/01 June 2023 examiner report — Q3–5 and Q7, PDF pp. 4–6. Reviewed with the mark scheme and question context on 9 October 2026.
- Pearson 8CH0/01 June 2023 question paper — Context for the cited bonding questions; no official question reproduced here.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.
