Covalent bonding and Lewis structures

IB Chemistry HL· 2.2 Covalent bonding· 25 min read

1. Nature of Covalent Bonding★★☆☆☆⏱ 5 min

📘 Definition

Covalent bond

A chemical bond formed by the electrostatic attraction between a shared pair of electrons and the positively charged nuclei of two bonded atoms.

Example:

In a hydrogen molecule ( $H_{2}$ ), two hydrogen atoms share one pair of electrons to form a single covalent bond.

Covalent bonds typically form between nonmetal atoms, where the difference in electronegativity between the bonded atoms is too small for electron transfer (the process that forms ionic bonds). Electrons are shared rather than transferred between atoms.

Single bond: 1 shared pair of electrons (bond order = 1)
Double bond: 2 shared pairs of electrons (bond order = 2)
Triple bond: 3 shared pairs of electrons (bond order = 3)

📐 Worked Example

Identify the number of bonding pairs and lone pairs in an oxygen molecule ( $O_{2}$ ).

1
Each oxygen atom has 6 valence electrons, so total valence electrons = $6 + 6 = 12$ .
2
Two oxygen atoms share 2 pairs (4 electrons) to form a double bond, satisfying the octet rule for both atoms.
3
Each oxygen has 4 remaining unbonded electrons, equal to 2 lone pairs per atom. Total bonding pairs = 1 double bond = 1 bonding pair, total lone pairs = 4 across the molecule.

2. Step-by-Step Lewis Structure Drawing★★★☆☆⏱ 10 min

Lewis structures are 2D representations of molecules that show all valence electrons as bonding pairs (lines between atoms) or lone pairs (dots on individual atoms). Follow this consistent method to avoid mistakes:

Count the total number of valence electrons for all atoms. Add 1 electron for each negative charge on an ion, subtract 1 for each positive charge.
Arrange atoms with the least electronegative atom as the central atom (hydrogen is never central).
Connect each pair of bonded atoms with a single bond (2 electrons per bond).
Distribute remaining electrons as lone pairs first to terminal atoms to satisfy their octet.
If the central atom does not have a full octet, convert lone pairs from terminal atoms into double or triple bonds.

📐 Worked Example

Draw the Lewis structure of the hydroxide ion ( $O H^{-}$ ).

1
Count total valence electrons: O has 6, H has 1, add 1 for the -1 charge. Total = $6 + 1 + 1 = 8$ .
2
Arrange atoms: O is central, H is terminal: $H - O$ .
3
The single bond uses 2 electrons. Remaining electrons = $8 - 2 = 6$ .
4
Distribute 6 electrons as 3 lone pairs on O. H has 2 electrons (full outer shell, only needs 2), O has 8 electrons (full octet).
5
Enclose the ion in square brackets and add the negative charge at the top right to get the final structure.

Exam tip:

Always remember to adjust your total valence electron count for the charge of polyatomic ions — this is one of the most commonly missed marks in IB exams.

3. Formal Charge and Resonance★★★★☆⏱ 6 min

📘 Definition

Formal charge

$F C = V - L - \frac{B}{2}$

The charge assigned to an atom in a Lewis structure, calculated where $V$ = valence electrons of the neutral atom, $L$ = number of lone pair electrons, $B$ = number of bonding electrons.

Example:

Used to select the most stable Lewis structure when multiple valid structures exist.

The most stable Lewis structure follows two rules for formal charge: 1) formal charges are as close to zero as possible, 2) any negative formal charge is located on the most electronegative atom.

📘 Definition

Resonance

Occurs when multiple valid Lewis structures exist for a molecule, differing only in the position of double bonds and lone pairs (not atomic position). The actual molecule is an average of all resonance forms.

📐 Worked Example

Calculate formal charges for the cyanate ion $N C O^{-}$ and identify the more stable structure.

1
Total valence electrons: $5 (N) + 4 (C) + 6 (O) + 1 (c ha r g e) = 16$ .
2
Structure 1 (N-C, C≡O): $F C (N) = 5 - 6 - 2/2 = - 1$ , $F C (C) = 4 - 0 - 8/2 = 0$ , $F C (O) = 6 - 2 - 6/2 = + 1$
3
Structure 2 (N≡C, C-O): $F C (N) = 5 - 2 - 6/2 = 0$ , $F C (C) = 4 - 0 - 8/2 = 0$ , $F C (O) = 6 - 6 - 2/2 = - 1$
4
Structure 2 is more stable: all formal charges are closest to zero, and the negative charge is on O, which is more electronegative than N.

4. Exceptions to the Octet Rule★★★★☆⏱ 4 min

The octet rule is a general guideline, not a law. There are three common categories of exceptions:

Incomplete octet: Central atom has fewer than 8 electrons. Common for Be, B, Al (e.g. $B e C l_{2}$ , $B F_{3}$ )
Expanded octet: Central atom has more than 8 electrons. Only possible for Period 3+ central atoms with empty d-orbitals (e.g. $P C l_{5}$ , $S F_{6}$ )
Odd electron species: Total valence electrons is odd, so one atom has an unpaired electron (e.g. $N O$ , $C l O_{2}$ )

📐 Worked Example

Explain why $S F_{6}$ exists but $O F_{6}$ does not.

1
Sulfur is a Period 3 element. It has a vacant 3d subshell that can accommodate extra electrons, allowing an expanded octet with 12 valence electrons around S.
2
Oxygen is a Period 2 element. It only has 2s and 2p subshells, which hold a maximum of 8 electrons. There is no 2d subshell to accommodate extra electrons.
3
Therefore $S F_{6}$ is a stable molecule, but $O F_{6}$ cannot exist.

5. Common Pitfalls

Wrong move:

Forgetting to adjust the total valence electron count for the charge of a polyatomic ion

Why:

This leads to an incorrect number of electrons, so the final Lewis structure will be wrong

Correct move:

Always add 1 electron per negative charge, subtract 1 per positive charge when counting total valence electrons

Wrong move:

Putting the most electronegative atom as the central atom

Why:

This leads to incorrect bonding and unnecessary multiple bonds

Correct move:

Always place the least electronegative atom as the central atom (hydrogen is always terminal)

Wrong move:

Drawing different atomic positions for resonance structures

Why:

Resonance only differs in electron position, not atomic position. Different positions mean different isomers, not resonance

Correct move:

Keep all atoms in the same position, only move electrons between bonds and lone pairs for resonance forms

Wrong move:

Forcing all atoms to follow the octet rule even when exceptions apply

Why:

Forcing an octet leads to incorrect formal charge and wrong structure selection

Correct move:

Always check for incomplete octets, expanded octets, and odd-electron species when drawing structures

Wrong move:

Placing negative formal charge on a less electronegative atom

Why:

This violates the stability rules for formal charge, leading to selection of the wrong structure

Correct move:

If formal charges are non-zero, place negative formal charge on the most electronegative atom in the molecule

6. Quick Reference Cheatsheet

Step	Action
1	Count total valence electrons, adjust for ion charge
2	Place least electronegative atom as central
3	Add single bonds between all connected atoms
4	Add lone pairs to terminal atoms first
5	Form multiple bonds if central atom lacks octet
6	Calculate formal charge to select most stable structure
Formal charge	$F C = V - L - B /2$

7. Frequently Asked

Why do some molecules break the octet rule?

Molecules with central atoms from Period 3 and below have empty d-orbitals that can accommodate more than 8 electrons. Small atoms like beryllium and boron often have incomplete octets due to their low number of valence electrons.

Do I always need to include formal charges in my exam answer?

Only when the question explicitly asks you to use formal charge to determine the most stable structure. Always check the command term in the question.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Lewis structure of sulfur dioxide
2024 · Paper 2
Draw Lewis structure of nitrate ion
2023 · Paper 1
Formal charge stability question

Going deeper

simulationPhET Lewis Structure BuilderInteractive practice drawing structures

What's Next

Understanding covalent bonding and Lewis structures is the foundation for learning about VSEPR theory and molecular geometry, which builds directly on the Lewis drawing skills you developed here. It also underpins understanding of bond order, bond enthalpy, and intermolecular forces, which are frequently tested in both Paper 1 and Paper 2 of IB Chemistry HL. Mastery of Lewis structures is also required for drawing and understanding organic reaction mechanisms later in the course, making this a core skill for all subsequent topics.