Covalent bonding, Lewis structures, VSEPR and polarity (QCE Chemistry Unit 1)

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:

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.

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+).

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.

Examples in context

Example 1. Methane combustion at Carnarvon basin gas plants. Queensland's Curtis LNG terminal in Gladstone liquefies methane piped from the Carnarvon and Surat basins. Methane is a tetrahedral covalent molecule: a central $\text{C}$ atom shares four bonding pairs with $\text{H}$ atoms, giving $\text{H-C-H}$ bond angles of $109.5^{\circ}$. The $\text{C-H}$ bond enthalpy near $413 \, \text{kJ mol}^{-1}$ keeps methane stable at room temperature, but ignition supplies the activation energy. Because $\text{C}$ and $\text{H}$ have similar electronegativities, the molecule is non-polar despite four polar bonds; symmetry cancels dipoles, explaining why $\text{CH}_4$ liquefies only below $-162^{\circ}\text{C}$ under modest pressure.

Example 2. Ammonia in Gibson Island fertiliser plant. Incitec Pivot's Brisbane facility manufactures ammonia by Haber-Bosch synthesis for downstream urea production. The $\text{NH}_3$ molecule has a trigonal pyramidal shape: nitrogen carries three bonding pairs to hydrogen and one lone pair, giving $\text{H-N-H}$ angles of $107^{\circ}$ (slightly less than the tetrahedral $109.5^{\circ}$ because the lone pair repels more strongly). The lone pair plus polar $\text{N-H}$ bonds make $\text{NH}_3$ a polar molecule, allowing hydrogen bonding to water and a high solubility of $700 \, \text{L}$ ammonia per litre water at $20^{\circ}\text{C}$, which is critical for transporting nitrogen to crops.

Try this

Q1. Use VSEPR theory to predict the shape and bond angle of $\text{H}_2 \text{O}$. State whether the molecule is polar and justify. [4 marks]

• Cue. Bent / V-shaped, $104.5^{\circ}$, two bonding two lone pairs, lone-pair repulsion. Polar: $\text{O-H}$ polar, shape does not cancel dipoles.

Q2. A molecule has central atom with three bonding pairs and one lone pair. Identify the shape, bond angle and one Queensland-relevant example. [3 marks]

• Cue. Trigonal pyramidal, $\approx 107^{\circ}$, $\text{NH}_3$ from Gibson Island plant or $\text{PH}_3$.

Q3. Compare $\text{CO}_2$ and $\text{SO}_2$. (a) Draw Lewis structures. (b) Predict the shape of each. (c) Explain why one is non-polar and the other polar. [2+2+3 marks]

• Cue. (a) $\text{O=C=O}$ and bent $\text{S}$ with two double bonds and one lone pair. (b) Linear vs bent. (c) $\text{CO}_2$ symmetrical dipoles cancel; $\text{SO}_2$ bent so dipoles add. Application criterion.