Unit 1: Chemical fundamentals (structure, properties and reactions)

QLDChemistrySyllabus dot point

Topic 2: Properties and structure of materials

Describe covalent bonding as the sharing of electron pairs between non-metal atoms, draw Lewis structures for simple molecules and polyatomic ions, predict molecular shape using VSEPR theory, and determine bond polarity and overall molecular polarity from electronegativity differences and geometry

A focused answer to the QCE Chemistry Unit 1 dot point on covalent bonding. Walks through drawing Lewis structures for molecules and polyatomic ions, predicts shapes (linear, bent, trigonal planar, trigonal pyramidal, tetrahedral) using VSEPR theory, and determines bond and overall molecular polarity from electronegativity differences and the symmetry of the structure.

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What this dot point is asking

QCAA wants you to describe covalent bonding (electron-pair sharing between non-metals), draw Lewis structures for simple molecules and common polyatomic ions, predict 3D shape using VSEPR theory, and determine bond polarity (from electronegativity differences) and overall molecular polarity (from the vector sum of bond dipoles, which depends on geometry). All of this feeds into the intermolecular forces dot point.

The answer

A covalent bond is the electrostatic attraction between two nuclei and a shared pair (or pairs) of electrons. Atoms share electrons so that each achieves a closed-shell electron configuration. Lewis structures track the bonding and non-bonding electron pairs; VSEPR translates those electron pair counts into 3D molecular geometry; bond polarity follows from the electronegativity difference; molecular polarity follows from how those bond dipoles add as vectors.

Drawing Lewis structures

Procedure for a simple covalent species:

  1. Count total valence electrons (add electrons for anions; subtract for cations).
  2. Choose the central atom (usually the least electronegative, except H which is never central).
  3. Connect peripheral atoms to the centre with single bonds (2 electrons each).
  4. Place remaining electrons as lone pairs on peripheral atoms first (to satisfy octets), then on the central atom.
  5. If the central atom is short of an octet, convert lone pairs into double or triple bonds.
  6. Check that each atom has a complete octet (duet for H).
  7. For ions, enclose in brackets with the overall charge.

Worked examples:

Species Valence e- Lewis sketch Notes
H_2O 8 H-O-H with 2 lone pairs on O Standard
CO_2 16 O=C=O Double bonds give C an octet
N_2 10 IMATH_0 with 1 lone pair each Triple bond
CH_4 8 H-C-H around C with all 4 H No lone pairs on C
NH_3 8 H-N-H around N with 1 lone pair on N Lone pair on N
NH_4^+ 8 H-N-H around N, no lone pair, +1 charge N has formed 4 bonds
OH- 8 H-O- with 3 lone pairs on O -1 charge
CO_3^2- 24 C central, 3 O; resonance among C=O and two C-O Trigonal planar
NO_3^- 24 N central, 3 O; resonance among N=O and two N-O Trigonal planar

Resonance structures arise when more than one valid Lewis arrangement exists; the actual structure is a weighted average. CO_3^2- and NO_3^- are the standard QCE Chemistry examples.

VSEPR theory

VSEPR (Valence Shell Electron Pair Repulsion) predicts 3D shape from the count of electron regions around the central atom. An "electron region" is a single bond, a double bond, a triple bond, or a lone pair (each counts as one region). Electron regions arrange themselves to be as far apart as possible.

Electron regions Geometry of electron regions Lone pairs Molecular shape Bond angle Example
2 Linear 0 Linear 180 degrees CO_2, BeCl_2
3 Trigonal planar 0 Trigonal planar 120 degrees BF_3, CO_3^2-
3 Trigonal planar 1 Bent ~117 degrees SO_2
4 Tetrahedral 0 Tetrahedral 109.5 degrees CH_4, NH_4^+, CCl_4
4 Tetrahedral 1 Trigonal pyramidal ~107 degrees NH_3, PCl_3
4 Tetrahedral 2 Bent ~104.5 degrees H_2O, OF_2

Bond angles are reduced from the ideal when lone pairs are present, because lone pairs occupy more angular space than bonding pairs (the "lone-pair-bonding-pair" repulsion is stronger than "bonding-pair-bonding-pair").

QCE Chemistry Unit 1 does not require expanded-octet shapes (trigonal bipyramidal, octahedral); these are Unit 4 or beyond.

Bond polarity and electronegativity

A bond is polar if the two atoms have different electronegativities. The more electronegative atom carries a partial negative charge (delta-); the less electronegative atom carries a partial positive charge (delta+).

Electronegativity difference Bond classification
0.0 to 0.4 Non-polar covalent
0.4 to 1.7 Polar covalent
> 1.7 Predominantly ionic

The cut-offs are rules of thumb; the QCAA expectation is that you use the difference plus chemistry context rather than memorising thresholds.

Examples:

  • Cl_2: difference 0. Non-polar.
  • HCl: difference 0.9. Polar covalent.
  • HF: difference 1.9. Polar covalent (some texts call this borderline ionic, but HF is molecular).
  • NaCl: difference 2.1. Ionic.

A bond dipole is a vector: it has magnitude (proportional to the electronegativity difference) and direction (from delta+ to delta-, conventionally drawn with an arrow).

Overall molecular polarity

A molecule is overall polar if its bond dipoles do not cancel as vectors, and overall non-polar if they do cancel. Two questions to ask:

  1. Are the bonds themselves polar? If all bonds are non-polar (e.g. H_2, O_2, Cl_2, CH_4 because C-H is very weakly polar), the molecule is non-polar.
  2. Is the molecule symmetric enough that the polar bonds cancel?

Examples:

  • CO_2. Two polar C=O bonds pointing in opposite directions. Linear shape. Bonds cancel. Non-polar overall.
  • H_2O. Two polar O-H bonds at approximately 104.5 degrees. Bonds do not cancel. Polar overall.
  • CH_4. Four C-H bonds (very weak polarity). Tetrahedral, perfectly symmetric. Bond dipoles cancel. Non-polar overall.
  • CCl_4. Four polar C-Cl bonds, tetrahedral, perfectly symmetric. Bond dipoles cancel. Non-polar overall. (Worth noting: a molecule with polar bonds can be non-polar overall.)
  • CHCl_3. Three C-Cl bonds and one C-H bond. Tetrahedral but no longer symmetric. Bond dipoles do not cancel. Polar overall.
  • NH_3. Three polar N-H bonds plus a lone pair on N. Trigonal pyramidal (not planar). Bond dipoles plus lone pair direction add. Polar overall.

A workflow QCAA rewards

For each molecule in a question, work in this order:

  1. Draw the Lewis structure (electron count, lone pairs).
  2. Count electron regions on the central atom and apply VSEPR.
  3. Identify polar bonds using electronegativity differences.
  4. Decide whether the geometry allows the bond dipoles to cancel.

Stating each step explicitly is how the EA short response awards full marks.

Common traps

Forgetting lone pairs on the central atom. Lone pairs on the central atom change the shape (H_2O is bent, not linear).

Predicting bond angles without adjusting for lone pairs. A tetrahedral electron-region geometry with two lone pairs (H_2O) is not 109.5 degrees; it is approximately 104.5 degrees.

Calling CCl_4 polar because the bonds are polar. Symmetric geometry cancels the bond dipoles. Always check both bond polarity and geometry.

Mis-counting electron regions in resonance structures. A double bond counts as one region, not two. C in CO_2 has two regions, not four.

Confusing electronegativity with ionisation energy or electron affinity. They correlate but are distinct concepts. Electronegativity refers specifically to attraction in a bond.

Treating "ionic versus covalent" as a sharp threshold. The two are extremes of a continuum. Polar covalent bonds (like H-Cl) share the spectrum.

In one sentence

A covalent bond is a shared pair of electrons between non-metal atoms; Lewis structures track bonding and non-bonding pairs, VSEPR predicts molecular shape (linear, bent, trigonal planar, trigonal pyramidal, tetrahedral) from the number of electron regions on the central atom, and overall molecular polarity follows from whether the polar bonds (set by electronegativity difference) cancel as vectors under the molecule's geometry.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2024 QCAA-style5 marksConsider the molecules CO_2, H_2O and NH_3. For each: (a) draw the Lewis structure, (b) predict the shape using VSEPR theory, (c) state whether the molecule is overall polar or non-polar and justify.
Show worked answer →

A 5-mark answer needs all three molecules treated.

CO_2. Lewis: O=C=O with two double bonds, no lone pairs on C. Electron pair count on C: 2 regions. VSEPR shape: linear, bond angle 180 degrees. Each C=O bond is polar (electronegativity O 3.5, C 2.5, difference 1.0). The two polar bonds point directly opposite and cancel as vectors. Overall non-polar.

H_2O. Lewis: H-O-H with 2 lone pairs on O. Electron pair count on O: 4 regions (2 bonding pairs, 2 lone pairs). VSEPR shape: bent (V-shaped), bond angle approximately 104.5 degrees. Each O-H bond is polar (electronegativity O 3.5, H 2.1, difference 1.4). The two polar bonds do not cancel because the molecule is bent. Overall polar, with the negative pole on O.

NH_3. Lewis: H-N-H bonds with one lone pair on N. Electron pair count on N: 4 regions (3 bonding, 1 lone pair). VSEPR shape: trigonal pyramidal, bond angle approximately 107 degrees. Each N-H bond is polar (electronegativity N 3.0, H 2.1, difference 0.9). The three polar bonds do not cancel because the lone pair lifts symmetry. Overall polar.

Markers reward correct Lewis structures including lone pairs, the right shape name, the bond angle in the right range, and a clear vector-cancellation argument for polarity.

2023 QCAA-style3 marksDraw the Lewis structure of the carbonate ion (CO_3^2-) and predict its shape. Explain whether the ion is overall polar or non-polar.
Show worked answer →

A 3-mark answer needs the Lewis structure with resonance, the shape, and the polarity reasoning.

Lewis structure. Total valence electrons: 4 (C) + 6 x 3 (O) + 2 (negative charge) = 24 electrons (12 pairs). C in the centre; three O atoms around it. One C=O double bond and two C-O single bonds (each O- with negative charge on the singly bonded O). The double bond can be drawn to any of the three O atoms: three resonance structures contribute equally.

Shape. Three regions of electron density around C, no lone pairs on C. VSEPR shape: trigonal planar, bond angles 120 degrees. The three C-O bonds are equivalent on average.

Polarity. Each C-O bond is polar. The three bonds are equivalent and arranged symmetrically at 120 degrees, so the bond dipoles cancel as vectors. The carbonate ion is therefore overall non-polar in shape, although it carries a net 2- charge.

Markers reward the resonance, the trigonal planar shape with 120 degree angles, and the symmetric-cancellation argument for polarity.

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