Chemical Bonding and Molecular Structure · Lesson 2 of 12
Drawing and Testing Lewis Structures
“Build Lewis structures systematically, check formal charges, and recognise the limits of the octet rule.”
• Count the available valence electrons in neutral molecules and polyatomic ions. • Construct Lewis structures using a skeleton, lone pairs, and multiple bonds. • Calculate formal charges and distinguish them from actual partial charges. • Recognise incomplete-octet, odd-electron, and expanded-octet exceptions.
A structure must account for every electron
A Lewis structure is an electron-accounting model. It shows which atoms are connected, how many pairs they share, and where the unshared pairs remain. A convincing drawing must use exactly the available number of valence electrons and give the atoms suitable electron counts whenever the octet rule applies.
Begin by adding the valence electrons of all the atoms. Then adjust for the overall charge: add one electron for each negative charge and subtract one for each positive charge. A charge belongs to the entire ion; it is not permission to add an extra electron to every atom.
Problem
How many valence electrons must appear in carbonate and ammonium?
- 1.For , carbon contributes 4 and three oxygen atoms contribute . The two negative charges add 2, giving .
- 2.For , nitrogen contributes 5 and four hydrogens contribute 4. Subtract one electron for the positive charge: .
- 3.These totals count the electrons represented in each complete Lewis structure. Inner-shell electrons are not included.
Build the skeleton before completing octets
The skeleton tells us which atoms are joined. Usually the least electronegative atom occupies the central position, while hydrogen is terminal because it forms only one ordinary covalent bond. Fluorine is also normally terminal. A known chemical structure takes priority over a simple central-atom shortcut.
Join adjacent atoms with single bonds and subtract two electrons for each bond. Complete the octets of terminal atoms other than hydrogen, then place any remaining electrons on the central atom. If a second-period central atom still lacks an octet, convert a neighbouring lone pair into an additional shared pair. This changes the distribution of electrons without changing their total number.
| Stage | What to check |
|---|---|
| Count | Include the contribution of the overall charge. |
| Connect | Choose a chemically sensible skeleton. |
| Distribute | Account for every bonding and nonbonding pair. |
| Complete | Use multiple bonds when required and appropriate. |
| Test | Recount total electrons, local electron counts, and formal charges. |
Problem
Draw and check a Lewis structure for carbon monoxide.
- 1.Carbon contributes 4 valence electrons and oxygen contributes 6, giving 10 in total. The skeleton is C–O.
- 2.A single bond uses 2 electrons. Completing oxygen’s octet uses another 6, leaving 2 as a lone pair on carbon. Carbon then counts only four surrounding electrons.
- 3.Convert two of oxygen’s lone pairs into additional shared pairs. The result is a triple bond, , with one lone pair on each atom.
- 4.The triple bond contains 6 electrons and the two lone pairs contain 4. All 10 electrons are accounted for, and each atom counts an octet. Formal-charge checking, introduced below, gives C −1 and O +1.
Problem
Draw a Lewis structure for the nitrite ion.
- 1.The total is valence electrons. Nitrogen is central, so start with O–N–O.
- 2.The two single bonds use 4 electrons. Complete both oxygen octets with 12 more electrons, then put the final pair on nitrogen.
- 3.Nitrogen now counts only six surrounding electrons. Change a lone pair on one oxygen into a second N–O shared pair.
- 4.The structure has one N=O bond, one N–O bond, and a lone pair on nitrogen. Enclose the complete ion in brackets with a −1 charge. Either oxygen can carry the single bond; the significance of these alternatives is developed through resonance later.
The following reference set extends the same electron-counting method to the species collected in the source’s Lewis table. Bond lines must be supplemented by the stated lone pairs; otherwise the electron picture is incomplete. For second-period central nitrogen, check that a proposed structure does not exceed an octet.
| Species | Complete Lewis accounting |
|---|---|
| H₂ | One H–H shared pair; no lone pairs; each H counts a duplet. |
| O₂ | One O=O double bond and two lone pairs on each oxygen in the simple Lewis picture. Its magnetic limitation is explained later. |
| O₃ | One O–O and one O=O bond; central O has one lone pair; terminal lone-pair counts are three and two. |
| NF₃ | Three N–F single bonds; one central N lone pair; three lone pairs on each F. |
| Carbonate | One C=O and two C–O bonds in one octet contributor; two lone pairs on double-bonded O, three on each single-bonded terminal O; total ion charge −2. |
| HNO₃ | HO–N with two additional O atoms: one terminal N=O and one terminal N–O⁻, with N⁺. OH oxygen has two lone pairs, double-bonded O two, and terminal single-bonded O three. N has no lone pair. |
Problem
Why should nitric acid not be drawn with only three N–O single bonds and no central lone pair?
- 1.Nitric acid has valence electrons. Hydrogen is bonded to oxygen, giving an HO–N skeleton with two other O atoms bonded to N.
- 2.A single-bond-only arrangement with completed terminal oxygen octets leaves nitrogen with six surrounding electrons. Convert a lone pair on a nonprotonated terminal O into an N=O shared pair.
- 3.The resulting structure gives nitrogen an octet with formal charge +1 and the remaining singly bonded terminal oxygen formal charge −1. All 24 electrons are accounted for.
- 4.A second contributor swaps the double bond between the two nonprotonated terminal oxygens. Moving the H nucleus would be a structural change, not an ordinary resonance-electron shift.
Formal charge: a consistent bookkeeping test
An atom’s formal charge compares its valence-electron count when isolated with the number assigned to it in a Lewis drawing. For this calculation, assign all lone-pair electrons to their atom and half of every shared pair to each bonded atom. The equal split is an accounting convention, even when a real bond is polar.
Add the formal charges of all atoms. Their sum must equal the overall charge of the species. Small formal charges and limited charge separation generally help identify a useful low-energy Lewis description, but this test must be used with electron counts and chemical context. Formal charge alone is not a direct measurement of real charge density.
Problem
Calculate formal charges for an ozone drawing with one double bond and one single bond.
- 1.Each free oxygen atom has 6 valence electrons. The central oxygen has 2 nonbonding electrons and 6 bonding electrons, so .
- 2.The double-bonded terminal oxygen has 4 nonbonding and 4 bonding electrons: .
- 3.The single-bonded terminal oxygen has 6 nonbonding and 2 bonding electrons: .
- 4.The sum is , matching neutral ozone. These are formal charges in one drawing; they do not mean that ozone contains three isolated ions.
Formal charge divides bonding electrons equally. A partial charge describes unequal electron density in a real polar bond. Oxidation-state bookkeeping uses another convention. Similar-looking plus and minus symbols do not make these quantities interchangeable.
The examples specify electron counts and connectivity. Lewis structures alone do not establish the actual molecular angles. A flat drawing of nitric acid or nitrite must not be read as a complete geometrical prediction.
Where the octet rule reaches its limits
The octet rule works particularly well for many second-period compounds, but it cannot describe all species. Some central atoms have fewer than eight surrounding electrons, some molecules have an odd electron total, and some valid Lewis drawings for heavier elements place more than eight electrons around the central atom. These exceptions reveal the limits of the model.
| Exception | Examples | Reason an ordinary octet drawing fails |
|---|---|---|
| Incomplete octet | , , | The central atom can be surrounded by fewer than eight electrons. |
| Odd electron total | and | An odd total cannot be divided into paired electrons giving every atom an octet. |
| Expanded octet | , , in conventional Lewis descriptions | More than four electron pairs are drawn around the heavier central atom. |
Boron in BF₃ forms three B–F bonds and counts six electrons around itself. Nitric oxide has valence electrons, so at least one electron remains unpaired in a simple Lewis description. In PCl₅, five shared pairs place ten electrons around phosphorus. None of these observations authorises an expanded octet on a second-period atom such as carbon, nitrogen, or oxygen.
The rule also leaves major questions unanswered. It does not calculate molecular shape, bond energy, or relative stability, and noble-gas inertness is not absolute: compounds such as XeF₂, KrF₂, and XeOF₂ exist. The source also includes LiCl and monomeric AlCl₃ among simple electron-deficient Lewis examples. Heavier atoms can obey an octet in some compounds and exceed it in other Lewis descriptions. Sulfur in SCl₂, for example, has two bonds and two lone pairs and counts an octet; sulfur is not automatically expanded-octet in every compound.
Expanded-octet Lewis drawings are useful electron-accounting representations. They do not, by themselves, establish the detailed orbital mechanism of bonding. Later geometrical and orbital models answer different questions.
Check your understanding
Work from the total electron count before choosing a drawing. Then use formal charge and the type of central atom to decide whether an apparent octet problem is a drawing error or a genuine limitation of the model.
Quiz
How many valence electrons are available in the nitrite ion?
What is the formal charge on oxygen with three lone pairs and one single bond?
Which change preserves the total number of electrons in a Lewis structure?
Why can NO not give every atom a conventional paired-electron octet?
Which statement about the octet rule is correct?
Practice Problems
- Draw complete Lewis structures for , , , and . Check every electron.
- Calculate the formal charge on each atom in the triple-bonded Lewis structure of CO.
- Draw nitrite, label each atom’s formal charge, and show that their sum matches the ion’s overall charge.
- Explain why forming a double bond by moving a lone pair does not create extra electrons.
- Classify , , and by their different exceptions to the octet rule.
- Correct a proposed Lewis structure of acetic acid in which the carbon attached to oxygen has only three bonds and both oxygen atoms are drawn without their full lone-pair counts.
- State two questions about a molecule that a correct Lewis structure alone cannot answer.
Key Takeaways
• A Lewis structure must use exactly the available valence electrons, including the effect of ionic charge. • Multiple bonds can complete octets without changing the electron total. • Formal charge assigns lone-pair electrons fully and bonding electrons equally. • Formal charges sum to the overall charge but are not actual partial charges. • Incomplete octets, odd-electron species, and expanded octets show why the octet rule has limits.