Lewis Diagrams: Study Guide (Chemistry 2.5)

What a Lewis diagram shows

In a covalent molecule, atoms are held together by shared pairs of valence electrons. A Lewis diagram (also called a Lewis structure or electron-dot diagram) is a drawing of those valence electrons. It shows two things:

• Which atoms are bonded to which. This is the skeleton, or connectivity.
• Where every valence electron is. Electrons are either shared between two atoms in a bond or held by one atom as a lone pair.

The conventions are simple:

• A bonding pair (two shared electrons) is drawn as a line. A double bond is two lines (four shared electrons); a triple bond is three lines (six shared electrons).
• A lone pair (two electrons that belong to one atom and are not shared) is drawn as a pair of dots.
• Core electrons are never drawn. The element symbol stands for the nucleus plus its core electrons.

Counting electrons “around” an atom

When you check whether an atom has an octet, count every electron in its bonds plus its own lone pairs. In water, oxygen has 2 bonds (4 electrons) and 2 lone pairs (4 electrons), so it is surrounded by 8. Each shared electron is counted for both atoms in the bond. That is why the octet check for each atom can add up to more than the total in the molecule.

Most second-period atoms (C, N, O, F) end up with 8 electrons around them: the octet rule. Hydrogen is full with 2 electrons (one bond).

A step-by-step method

Lewis diagrams follow a fixed set of principles. This method works for nearly every molecule and ion in the course.

1. Count the total valence electrons. Add the valence electrons of every atom (group number for groups 1 and 2; group number minus 10 for groups 13 to 18). For a negative ion, add one electron per unit of negative charge. For a positive ion, subtract one per unit of positive charge.
2. Choose the skeleton. Put the least electronegative atom in the centre (usually the atom there is only one of, such as C in CCl₄ or S in SO₃). Hydrogen is always on the outside because it forms only one bond. Fluorine is also always on the outside: it needs only one more electron, so it forms one bond, and it is the most electronegative element. If the formula is written to show connectivity, such as CH₃OH, follow it.
3. Join the atoms with single bonds. Each bond uses 2 electrons. Subtract them from the total.
4. Complete the octets of the outer atoms with lone pairs (hydrogen needs none). Subtract these electrons.
5. Put any electrons that are left on the central atom as lone pairs.
6. Check the central atom. If it has fewer than 8 electrons, move a lone pair from a neighbouring outer atom into the bond to make a double bond. Repeat (a triple bond if needed) until the central atom has 8. Use lone pairs from C, N, O or S for this; halogens on the outside almost always keep one single bond.
7. Recount. The diagram must contain exactly the number of electrons from step 1. For an ion, draw square brackets around the whole diagram and write the charge outside the top right.

A quick check for multiple bonds

When every atom obeys the octet rule (8 for each atom, 2 for each hydrogen), you can predict how many bonds there are before you draw:

• electrons needed = 8 × (number of non-H atoms) + 2 × (number of H atoms)
• electrons available = total valence electrons from step 1
• shared electrons = needed − available, so the number of bonds = (needed − available) ÷ 2

If that number of bonds is more than the number of lines in your skeleton, the extra bonds must be double or triple bonds. This shortcut does not work for molecules with an incomplete or expanded octet.

Typical bonding patterns

When you read or check a Lewis diagram, these patterns appear again and again for neutral atoms with formal charge zero (Topic 2.6 explains formal charge).

Atom	Bonds	Lone pairs	Example
H	1	0	every H in CH₄
C	4	0	CH₄, CO₂
N	3	1	NH₃, N₂ (one triple bond)
O	2	2	H₂O, the O in C=O
F, Cl, Br, I (outer atom)	1	3	each Cl in CCl₄

A double bond counts as 2 bonds and a triple bond as 3. If an atom in your diagram does not fit its usual pattern, check your electron count again.

Worked example 1: a molecule that needs a double bond

Question. Draw the Lewis diagram of phosgene, COCl₂, in which the carbon atom is bonded to the oxygen atom and both chlorine atoms.

1. Count. C has 4, O has 6, each Cl has 7: 4 + 6 + 2(7) = 24 valence electrons.
2. Skeleton. Carbon is the least electronegative atom and the only one that can make four bonds, so it goes in the centre with O and two Cl around it.
3. Single bonds. Three bonds use 6 electrons. 24 − 6 = 18 left.
4. Outer octets. O needs 3 lone pairs (6 electrons) and each Cl needs 3 lone pairs (6 each): 6 + 6 + 6 = 18. 18 − 18 = 0 left.
5. Central atom. Nothing is left for carbon.
6. Check carbon. C has only three bonds, so 6 electrons. Move one lone pair from oxygen into the C–O bond to make a C=O double bond. Carbon now has 8, and oxygen keeps 2 lone pairs plus the double bond (8).
7. Recount. 4 bonds (C=O counts as two) × 2 = 8 electrons, plus 2 lone pairs on O (4) and 3 lone pairs on each Cl (12): 8 + 4 + 12 = 24. ✓

Quick check. Needed = 8 × 4 atoms = 32. Available = 24. Shared = 32 − 24 = 8 electrons = 4 bonds. The skeleton has only 3 lines, so one of them must be a double bond. ✓

Why the double bond is to oxygen, not chlorine. Oxygen fits its usual pattern (2 bonds, 2 lone pairs) and each chlorine fits its pattern (1 bond, 3 lone pairs). A C=Cl double bond would put chlorine out of its usual pattern. In Topic 2.6 you will use formal charge to justify this choice in numbers.

The finished diagram is part (a) of Figure 2.

Worked example 2: a polyatomic ion with a triple bond

Question. Draw the Lewis diagram of the cyanide ion, CN⁻.

1. Count. C has 4 and N has 5. The 1− charge means one extra electron: 4 + 5 + 1 = 10 valence electrons.
2. Skeleton. There are only two atoms, so they are simply joined: C–N.
3. Single bond. 2 electrons used; 8 left.
4. Outer octet. Give N three lone pairs (6 electrons); 2 left.
5. Central atom. Put the last 2 electrons on C as a lone pair. C now has only 4 electrons (one bond and one lone pair).
6. Check carbon. Move one N lone pair into the bond (C=N): C has 6. Move another (C≡N): C has 8, and N has a triple bond plus one lone pair (8).
7. Recount. Triple bond 6 + lone pair on C 2 + lone pair on N 2 = 10. ✓ Draw brackets and write the 1− charge outside.

Quick check. Needed = 8 × 2 = 16. Available = 10. Shared = 6 electrons = 3 bonds: a triple bond. ✓

Interpretation. Cyanide has the same number of valence electrons (10) as N₂ and CO, and all three have the same Lewis pattern: a triple bond with one lone pair on each atom. Species with equal numbers of electrons are called isoelectronic.

When the central atom does not end with eight

The octet rule is a strong guide for C, N, O and F, but a valid Lewis diagram does not always give the central atom eight electrons.

Fewer than eight: boron and beryllium

Boron has only 3 valence electrons. In boron trifluoride, BF₃ (3 + 3 × 7 = 24 electrons), three B–F single bonds and three lone pairs on each fluorine use all 24 electrons. Boron is left with 6. You could move a fluorine lone pair to make a B=F double bond, but that puts fluorine out of its usual pattern, and the 6-electron diagram is the accepted best model (Topic 2.6 explains why with formal charge). Beryllium compounds such as BeCl₂ in the gas phase similarly leave Be with 4.

More than eight: period 3 and beyond

Atoms from period 3 onwards (P, S, Cl, Se, Br, I, Xe and others) are larger. In some molecules they are surrounded by 10 or 12 electrons: an expanded octet. This happens when step 5 leaves electrons over after every outer atom has its octet, or when the central atom must bond to more than four atoms (PCl₅ has five P–Cl bonds; SF₆ has six S–F bonds).

Period 2 atoms (C, N, O, F) never have more than 8 electrons around them. Their valence shell (n = 2) has room for only 8.

Worked example 3: an expanded octet

Question. Draw the Lewis diagram of sulfur tetrafluoride, SF₄.

1. Count. S has 6 and each F has 7: 6 + 4(7) = 34 valence electrons.
2. Skeleton. S in the centre with four F around it (F is always outside).
3. Single bonds. Four bonds use 8 electrons; 34 − 8 = 26 left.
4. Outer octets. Each F needs 3 lone pairs: 4 × 6 = 24. 26 − 24 = 2 left.
5. Central atom. Put the last 2 electrons on S as one lone pair.
6. Check sulfur. S has 4 bonds (8 electrons) plus 1 lone pair (2): 10 electrons. This is allowed because sulfur is in period 3. No double bonds are needed.
7. Recount. 4 bonds × 2 = 8, plus 12 lone pairs on F (24), plus 1 lone pair on S (2): 34. ✓

Common error. Stopping after step 4 and forgetting the 2 leftover electrons gives a diagram with only 32 electrons. The lone pair on sulfur matters a great deal in Topic 2.7, where it changes the shape of the molecule.

One formula, more than one skeleton

A molecular formula does not always fix the skeleton. C₂H₆O has 2(4) + 6(1) + 6 = 20 valence electrons, and two different valid Lewis diagrams:

• Ethanol, CH₃CH₂OH: the two carbons are bonded to each other, and the oxygen sits between a carbon and a hydrogen.
• Dimethyl ether, CH₃OCH₃: the oxygen sits between the two carbons.

Both diagrams have every C with 4 bonds, the O with 2 bonds and 2 lone pairs, and every H with 1 bond. They are different compounds with different properties. That is why questions about larger molecules either give the skeleton or write the formula in a way that shows it (CH₃OH, not CH₄O).

Common misconceptions

• Forgetting the charge on an ion. NH₄⁺ has 5 + 4 − 1 = 8 valence electrons, not 9. CN⁻ has 10, not 9. Negative charge adds electrons; positive charge removes them.
• Putting hydrogen in the middle. Hydrogen can form only one bond, so it is always an outer atom.
• Using the most electronegative atom as the centre. The central atom is normally the least electronegative (other than H).
• Leaving electrons over. Every valence electron must appear in the diagram. Leftover electrons go on the central atom, even if this gives it more than 8, provided it is in period 3 or beyond.
• Giving C, N, O or F more than 8 electrons. Period 2 atoms cannot have an expanded octet.
• Making double bonds to halogens on the outside. Outer F, Cl, Br and I nearly always keep one single bond and three lone pairs.
• Thinking the drawing shows the shape. A Lewis diagram of water drawn in a straight line does not mean water is linear. Shape comes from VSEPR (Topic 2.7).
• Counting a shared pair once when checking octets. For the octet check, a bonding pair counts for both atoms. For the total electron count, it counts once.

Where this leads

Lewis diagrams build on the valence electron counts from Topic 1.8 and the bond types of Unit 2, most recently Topic 2.4, Structure of Metals and Alloys. Next, Topic 2.6, Resonance and Formal Charge, shows what to do when more than one valid diagram can be drawn, and how to choose between them. Topic 2.7 then uses Lewis diagrams to predict shapes, bond angles and polarity. Try the practice questions now, then use the revision notes and the checklist to consolidate.