4.1
Overlap of Atomic Orbitals to Form Covalent Bonds
Covalent BondA bond formed by the mutual sharing of one or more pairs of electrons between two non-metal atoms. Each atom contributes one electron to each shared pair. The shared electrons are attracted to both nuclei simultaneously, holding the atoms together.
Bond Formation by Orbital Overlap
A covalent bond forms when two atomic orbitals, each containing one electron, overlap. The two electrons (with opposite spins) are shared between both atoms and occupy the overlap region. This lowers the total energy of the system — the bond is more stable than the separated atoms.
The bond order is the number of shared electron pairs: single bond = 1 pair; double bond = 2 pairs; triple bond = 3 pairs.
| Bond | Electron pairs shared | Example | Bond energy trend | Bond length trend |
|---|---|---|---|---|
| Single (C–C) | 1 | H₂, Cl₂, ethane | Weakest | Longest |
| Double (C=C) | 2 | O₂, ethene | Stronger | Shorter |
| Triple (C≡C) | 3 | N₂, ethyne | Strongest | Shortest |
Sigma (σ) and Pi (π) bondsA single bond is one σ bond (end-on overlap). A double bond = one σ + one π (side-on overlap). A triple bond = one σ + two π. σ bonds allow rotation; π bonds restrict rotation around the axis.
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Section 4.1 — Covalent Bonding
10 QuestionsQ1 of 10
A covalent bond is formed by:
Covalent bond: two atoms each contribute one electron → shared pair. Lewis model: electron pair held between nuclei, attracting both → bond.
Q2 of 10
A dative (coordinate) covalent bond differs because:
Dative/coordinate bond: one atom donates BOTH electrons of the shared pair. Example: NH₃→BF₃ (N donates lone pair to B). Once formed, it is identical in strength to a normal covalent bond.
Q3 of 10
Bond length increases as:
Bond order: single < double < triple. More bonding electrons → nuclei pulled closer → shorter bond. C−C = 154 pm > C=C = 134 pm > C≡C = 120 pm.
Q4 of 10
Bond strength (bond enthalpy) increases as:
Higher bond order → more electron density between nuclei → stronger attraction → more energy to break. C≡C (839 kJ/mol) > C=C (614) > C−C (347).
Q5 of 10
The Lewis structure of CO₂ shows:
CO₂: O=C=O. Each C=O is a double bond. O has 2 lone pairs each. C has no lone pairs. Linear molecule (no lone pairs on C).
Q6 of 10
Which molecule has a triple bond?
N₂: N≡N. Triple bond (1σ + 2π). Very strong bond (945 kJ/mol) makes N₂ unreactive. CO₂ has double bonds.
Q7 of 10
How many bonding pairs are in a molecule of H₂O?
H₂O: O forms 2 bonds with 2 H atoms = 2 bonding pairs. Also has 2 lone pairs (not bonding pairs).
Q8 of 10
Electronegativity difference determines:
Large electronegativity difference → polar bond (electrons pulled toward more EN atom). If very large → ionic. Small difference → non-polar covalent.
Q9 of 10
In a polar covalent bond, the δ⁻ charge is on:
More electronegative atom attracts shared electrons more strongly → higher electron density → partial negative charge (δ⁻). Less EN atom has δ⁺.
Q10 of 10
Which molecule is non-polar despite having polar bonds?
CO₂ (O=C=O): two polar C=O bonds are equal and opposite → dipoles cancel → non-polar molecule. Linear geometry = zero net dipole. H₂O, NH₃, HCl all have net dipole moments.
4.2
Theories on the Formation of Covalent Bonds
Lewis Theory
G.N. Lewis (1916) proposed that atoms share electron pairs to complete their octets. Represented by Lewis structures (dot-and-cross diagrams) showing bonding pairs (shared) and lone pairs (non-bonding). Simple and useful but does not explain bond angles or three-dimensional shapes.
Valence Bond Theory (VBT)
Linus Pauling formalised the idea that covalent bonds form by the overlap of half-filled atomic orbitals. The bond consists of a shared pair of electrons in the region of orbital overlap. Predicts bond direction and introduces hybridisation to explain observed geometries.
Molecular Orbital (MO) Theory
In MO theory, atomic orbitals combine to form molecular orbitals spanning the whole molecule. Bonding MOs (lower energy) and antibonding MOs (higher energy) are filled using the same rules as atomic orbitals. MO theory explains properties (like paramagnetism of O₂) that VBT cannot. It is beyond S4 level but good to know exists.
4.3
Coordinate (Dative) Covalent Bonding
Dative (Coordinate) Covalent BondA covalent bond in which both electrons of the shared pair are donated by the same atom (the donor). The receiving atom contributes no electrons. Once formed, a dative bond is indistinguishable from an ordinary covalent bond in terms of length and strength. Represented by an arrow: donor → acceptor.
Examples of Dative Bonds
| Example | Donor (lone pair) | Acceptor | Description |
|---|---|---|---|
| NH₄⁺ (ammonium) | N in NH₃ | H⁺ | NH₃ donates its lone pair to H⁺, which has an empty 1s orbital. |
| H₃O⁺ (oxonium) | O in H₂O | H⁺ | Water donates a lone pair to H⁺. |
| CO → BF₃ | C in CO | B in BF₃ | CO donates to the empty orbital of B (Lewis acid-base adduct). |
| Al₂Cl⁶ | Cl lone pair | Al | Bridging Cl donates a lone pair to the adjacent Al. |
4.4
Valence Bond Theory (VBT)
Core Ideas of VBT
VBT states that a covalent bond forms by the overlap of two half-filled atomic orbitals, one from each atom. The overlapping orbitals must have compatible symmetry. The greater the overlap, the stronger the bond.
- σ bond: formed by direct (end-on) overlap along the internuclear axis. Electron density is concentrated between the nuclei. Examples: s-s overlap (H–H), s-p overlap (H–Cl), p-p end-on overlap (Cl–Cl).
- π bond: formed by sideways (side-on) overlap of p orbitals above and below the internuclear axis. Electron density is above and below the axis. Less overlap than σ → weaker than σ. Found in double and triple bonds.
Single vs double vs triple bondsIn ethene (H₂C=CH₂): one σ + one π bond between the carbons. In ethyne (HC≡CH): one σ + two π bonds. The π bonds are responsible for the rigidity and reactivity of double and triple bonds.
4.5
VSEPR Theory — Molecular Shapes
VSEPR — Valence Shell Electron Pair RepulsionThe 3D shape of a molecule is determined by minimising repulsion between all electron pairs (both bonding and lone pairs) around the central atom. Electron pairs arrange themselves as far apart as possible.
Repulsion Strength
Lone pair–lone pair (lp–lp) > lone pair–bonding pair (lp–bp) > bonding pair–bonding pair (bp–bp)
Each lone pair compresses the bond angles by approximately 2° compared to the ideal geometry.
| Electron pairs (total) | Lone pairs | Bonding pairs | Shape | Bond angle | Example |
|---|---|---|---|---|---|
| 2 | 0 | 2 | Linear | 180° | BeCl₂, CO₂ |
| 3 | 0 | 3 | Trigonal planar | 120° | BF₃, SO₃ |
| 3 | 1 | 2 | Bent/V-shaped | <120° | SO₂, O₃ |
| 4 | 0 | 4 | Tetrahedral | 109.5° | CH₄, NH₄⁺ |
| 4 | 1 | 3 | Trigonal pyramidal | <109.5° (~107°) | NH₃, PCl₃ |
| 4 | 2 | 2 | Bent/V-shaped | <107° (~104.5°) | H₂O, H₂S |
| 5 | 0 | 5 | Trigonal bipyramidal | 90° & 120° | PCl₅ |
| 6 | 0 | 6 | Octahedral | 90° | SF₆ |
WORKED EXAMPLE
Predicting Shape of NH₃
1
N has 5 valence electrons. NH₃: 3 bonds with H + 1 lone pair on N = 4 electron pairs around N.
2
4 pairs → tetrahedral electron pair geometry, but one is a lone pair → trigonal pyramidal molecular shape.
3
Bond angle: lp–bp repulsion is greater than bp–bp, so the bond angle is compressed from 109.5° to approximately 107°.
4.6
Hybridisation
HybridisationThe mixing of atomic orbitals of similar energy on the same atom to form new hybrid orbitals with identical energy and shape. Hybridisation explains why carbon forms 4 equivalent bonds in CH₄ (not 2 bonds as predicted by the 2 half-filled p orbitals).
| Hybrid type | Orbitals mixed | No. hybrid orbitals | Geometry | Bond angle | Example |
|---|---|---|---|---|---|
| sp | 1s + 1p | 2 | Linear | 180° | BeCl₂, C₂H₂ (ethyne) |
| sp² | 1s + 2p | 3 | Trigonal planar | 120° | BF₃, C₂H₄ (ethene), graphite |
| sp³ | 1s + 3p | 4 | Tetrahedral | 109.5° | CH₄, NH₃, H₂O, diamond |
| sp³d | 1s + 3p + 1d | 5 | Trigonal bipyramidal | 90°/120° | PCl₅ |
| sp³d² | 1s + 3p + 2d | 6 | Octahedral | 90° | SF₆ |
WORKED EXAMPLE
Hybridisation of Carbon in CH₄, C₂H₄ and C₂H₂
1
CH₄ (methane): C forms 4 single bonds → no π bonds → sp³ hybridised. One 2s and three 2p orbitals mix to give four sp³ hybrids arranged tetrahedrally (109.5°). Each forms a σ bond with H.
2
C₂H₄ (ethene): Each C forms one double bond (= one σ + one π) and two single bonds. C is sp² hybridised: 2s + 2pₙ + 2p™ mix to form three sp² hybrids (trigonal planar, 120°). The remaining unhybridised p𝑧 orbital on each C overlaps sideways to form the π bond. Molecule is planar.
3
C₂H₂ (ethyne): Each C forms one triple bond (= one σ + two π) and one single bond. C is sp hybridised: 2s + 2pₙ mix to form two sp hybrids (linear, 180°). Two remaining unhybridised p orbitals each form a π bond. Linear molecule.
4.7
Polar Covalent Bonds & Molecular Polarity
Polar Covalent BondA covalent bond in which the shared electrons are unequally distributed due to a difference in electronegativity between the two atoms. The more electronegative atom carries a partial negative charge (δ−) and the less electronegative atom carries a partial positive charge (δ⁺). This creates a bond dipole.
Bond Polarity vs Molecular Polarity
A molecule can have polar bonds and still be non-polar overall if the bond dipoles are symmetrically arranged and cancel out. Molecular polarity depends on:
- The polarity of individual bonds (ΔEN)
- The geometry of the molecule (whether dipoles cancel)
| Molecule | Polar bonds? | Shape | Molecular polarity |
|---|---|---|---|
| CO₂ | Yes (C=O) | Linear (symmetric) | Non-polar — dipoles cancel |
| H₂O | Yes (O–H) | Bent (asymmetric) | Polar — net dipole |
| CCl₄ | Yes (C–Cl) | Tetrahedral (symmetric) | Non-polar — dipoles cancel |
| CHCl₃ | Yes (C–Cl, C–H) | Tetrahedral (asymmetric) | Polar — net dipole |
| NH₃ | Yes (N–H) | Trigonal pyramidal | Polar — net dipole |
| BF₃ | Yes (B–F) | Trigonal planar (symmetric) | Non-polar — dipoles cancel |
Quick rule for molecular polarityIf all atoms attached to the central atom are identical AND the molecule has no lone pairs on the central atom, the dipoles cancel → non-polar. Any asymmetry (different atoms or lone pairs on central atom) usually gives a polar molecule.
4.8
Simple and Giant Covalent Structures
Simple Molecular (Covalent) StructuresConsist of discrete molecules held together by weak intermolecular forces (van der Waals, dipole-dipole, or hydrogen bonds). The covalent bonds within each molecule are strong, but the forces between molecules are weak. Examples: H₂, Cl₂, CO₂, H₂O, CH₄, iodine.
Giant Covalent StructuresAtoms are covalently bonded in a continuous, infinite 3D (or 2D) lattice. All bonds in the structure are strong covalent bonds. Very high melting points. Examples: diamond, graphite, silicon dioxide (SiO₂), silicon.
| Property | Simple molecular | Giant covalent |
|---|---|---|
| Melting point | Low (weak IMF between molecules) | Very high (strong covalent bonds throughout) |
| Electrical conductivity | Non-conductor | Non-conductor (except graphite) |
| Solubility in water | Varies; polar dissolves in polar | Generally insoluble |
| Examples | I₂, CO₂, NH₃ | Diamond, SiO₂, Si |
Diamond vs Graphite
| Property | Diamond | Graphite |
|---|---|---|
| Structure | 3D lattice; each C bonded to 4 others | 2D layers; each C bonded to 3 others in hexagonal rings |
| Hybridisation | sp³ | sp² (one unhybridised p per C) |
| Bond angle | 109.5° | 120° |
| Conductivity | Non-conductor | Conductor (delocalised π electrons between layers) |
| Hardness | Hardest natural substance (all bonds strong, 3D) | Soft/slippery (layers slide — weak forces between layers) |
| Uses | Cutting tools, jewellery | Lubricant, electrodes, pencil |
Silicon dioxide (SiO₂)Giant covalent structure; each Si bonded to 4 O atoms and each O to 2 Si atoms. Very high m.p. (1713°C). Insoluble in water. Used in glass and semiconductor manufacture. Similar to diamond in having a 3D covalent network.
4.9
Intermolecular Forces
Overview
Intermolecular forces (IMF) are the forces of attraction between molecules (not within them). They are much weaker than covalent or ionic bonds. IMF determine physical properties of molecular compounds: melting/boiling points, viscosity, surface tension, and solubility.
1. London Dispersion Forces (Induced Dipole-Induced Dipole)
Present in all molecules — both polar and non-polar. Arise from temporary/instantaneous dipoles caused by random fluctuations in electron distribution. One molecule develops a brief dipole which induces a dipole in a neighbouring molecule → weak, transient attraction.
Factors increasing LDF strength:
- More electrons (higher molar mass) → more polarisable electron cloud → stronger instantaneous dipoles
- Larger surface area (longer chains) → more contact between molecules → greater total LDF
Trend: boiling points of noble gases increase down the group (He < Ne < Ar < Kr < Xe) because molar mass increases → stronger LDF.
2. Permanent Dipole–Dipole Forces
Present in polar molecules. The δ⁺ end of one molecule is attracted to the δ− end of a neighbouring molecule. Stronger than LDF for molecules of similar size. Present alongside LDF.
Example: HCl, SO₂, CHCl₃. Their higher b.p. compared to non-polar molecules of similar mass is due to these additional attractions.
3. Hydrogen Bonding
Hydrogen BondA special, relatively strong type of dipole-dipole interaction between a hydrogen atom bonded to a highly electronegative atom (N, O, or F) and a lone pair on another electronegative atom (N, O, or F) of a neighbouring molecule.
Condition: H must be directly bonded to N, O, or F — these are the only atoms electronegative and small enough to make the bond dipole strong enough and the H atom accessible enough for hydrogen bonding.
Condition: H must be directly bonded to N, O, or F — these are the only atoms electronegative and small enough to make the bond dipole strong enough and the H atom accessible enough for hydrogen bonding.
Effects of Hydrogen Bonding
- Anomalously high boiling point of water (100°C) vs H₂S (−60°C): water molecules form up to 4 H-bonds each → much more energy needed to separate → high b.p.
- Ice floats on water: in ice, H-bonds hold molecules in an open hexagonal lattice (less dense than liquid water where H-bonds are more dynamic). Ice is less dense → floats.
- High surface tension & viscosity of water: strong H-bonds hold surface molecules together.
- DNA double helix: base pairing held together by H-bonds between base pairs (A–T: 2 H-bonds; G–C: 3 H-bonds).
- Protein secondary structure: α-helices and β-sheets stabilised by H-bonds within polypeptide chains.
| Force | Present in | Strength | Condition |
|---|---|---|---|
| London dispersion (LDF) | All molecules | Weakest (increases with molar mass) | Always present; increases with electron count |
| Permanent dipole–dipole | Polar molecules | Moderate | Molecule must have a net dipole moment |
| Hydrogen bond | Molecules with N–H, O–H, or F–H bonds | Strongest IMF | H must be bonded to N, O, or F |
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Section 4.2 — VSEPR Theory
10 QuestionsQ1 of 10
VSEPR theory predicts molecular shape based on:
VSEPR: Valence Shell Electron Pair Repulsion. Electron pairs (bonding and lone) repel → arrange to maximise separation → determines molecular geometry.
Q2 of 10
The shape of BF₃ (3 bond pairs, 0 lone pairs) is:
3 bond pairs, 0 lone pairs → trigonal planar, 120° bond angles. B is electron deficient (6 electrons, not 8 — cannot have lone pairs). Examples: BF₃, AlCl₃.
Q3 of 10
Water (H₂O) has 2 bond pairs and 2 lone pairs. Its shape is:
4 electron pairs (tetrahedral arrangement) but 2 are lone pairs → bent/V-shaped. Actual H-O-H angle = 104.5° (lone pair repulsion reduces from ideal 109.5°).
Q4 of 10
Lone pairs repel more than bonding pairs because:
Lone pairs: held by one nucleus only → spread out more → greater repulsion than bonding pairs (shared between two nuclei → more compressed). LP-LP > LP-BP > BP-BP repulsion.
Q5 of 10
The bond angle in CH₄ is:
CH₄: 4 bond pairs, 0 lone pairs → tetrahedral → 109.5° bond angles. All 4 bonds equivalent.
Q6 of 10
Which molecule has a linear shape?
BeCl₂: 2 bond pairs, 0 lone pairs → linear, 180°. Be has only 4 electrons in bonds (electron deficient). CO₂ and C₂H₂ also linear.
Q7 of 10
NH₃ has bond angle of 107° rather than 109.5° because:
NH₃: 4 electron pairs (tetrahedral base), 1 lone pair. Lone pair repels 3 bonding pairs → H-N-H compresses from 109.5° to 107°.
Q8 of 10
PCl₅ has what molecular geometry?
PCl₅: 5 bond pairs, 0 lone pairs → trigonal bipyramidal. P uses d orbitals (expanded octet). 3 equatorial bonds (120°) and 2 axial bonds (90° to equatorial).
Q9 of 10
SF₆ has 6 bond pairs and 0 lone pairs. Shape is:
SF₆: 6 bond pairs → octahedral. All bond angles 90°. S uses d orbitals. Very stable molecule.
Q10 of 10
ClF₃ has 3 bond pairs and 2 lone pairs. Shape is:
5 electron pairs (trigonal bipyramidal arrangement), 2 lone pairs in equatorial positions → T-shaped. 90° and 180° bond angles (approximately). ClF₃ shape is T.
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Exercises
- Use VSEPR theory to predict the shape and bond angles of: (a) PCl₃, (b) SF₂, (c) BCl₃, (d) XeF₄ (hint: Xe has 2 lone pairs + 4 bonds).
(a) PCl₃: P has 3 bonds + 1 lone pair = 4 pairs → trigonal pyramidal, ~107°.
(b) SF₂: S has 2 bonds + 2 lone pairs = 4 pairs → bent/V-shaped, ~103° (2 lone pairs compress angle further).
(c) BCl₃: B has 3 bonds + 0 lone pairs = 3 pairs → trigonal planar, 120°.
(d) XeF₄: 4 bonds + 2 lone pairs = 6 pairs → octahedral electron geometry; 2 lone pairs are opposite each other → square planar molecular shape, 90°. - Explain why H₂O has a bond angle of 104.5° rather than 109.5°, and why this differs from the angle in NH₃ (107°).
Both have 4 electron pairs around the central atom (tetrahedral electron geometry, ideal 109.5°). In NH₃: 3 bonds + 1 lone pair. The one lone pair compresses bond angles by ~2.5° → 107°. In H₂O: 2 bonds + 2 lone pairs. Two lone pairs cause greater repulsion (lp–lp > lp–bp > bp–bp), compressing angles by ~5° total → 104.5°. H₂O has two lone pairs vs one in NH₃ → greater compression → smaller bond angle.
- Explain the hybridisation of carbon in (a) diamond and (b) graphite, and relate this to their different physical properties.
(a) Diamond: each C is sp³ hybridised → forms 4 σ bonds in a 3D tetrahedral network. All outer electrons used in bonding → no free electrons → non-conductor. All bonds are strong → extremely high m.p. and hardness. (b) Graphite: each C is sp² hybridised → forms 3 σ bonds in planar hexagonal layers. The remaining unhybridised p𝑧 electron on each C forms a delocalised π system across each layer → conducts electricity. Layers are held together only by weak London dispersion forces → layers slide easily → soft/lubricant.
- State which type of intermolecular force is present in each: (a) I₂, (b) HCl, (c) NH₃, (d) H₂O. Explain why the boiling point of H₂O (100°C) is much higher than H₂S (−60°C) despite S being in a higher period.
(a) I₂: London dispersion only. (b) HCl: LDF + permanent dipole-dipole. (c) NH₃: LDF + dipole-dipole + hydrogen bonds (N–H). (d) H₂O: LDF + dipole-dipole + hydrogen bonds (O–H).
H₂O vs H₂S: S is less electronegative than O and larger, so S–H bonds are not polar enough and H is not accessible enough for H-bonding. H₂S has only LDF and weak dipole-dipole → low b.p. H₂O forms strong O–H···O hydrogen bonds (up to 4 per molecule) that require much more energy to overcome → much higher b.p. - CO₂ has polar C=O bonds yet is a non-polar molecule. Explain, with a diagram, why this is the case.
CO₂ is linear (sp hybridised C, no lone pairs on C). The two C=O bond dipoles point in opposite directions (each O pulls electrons towards itself). Because the molecule is symmetric and linear, the two dipole vectors are equal in magnitude and exactly opposite in direction → they cancel to give zero net dipole moment. Diagram: O(δ−)=C=O(δ−) with arrows pointing towards each O, cancelling. Overall the molecule is non-polar despite having polar bonds.
- Distinguish between a covalent bond and a dative (coordinate) bond. Give one example of a dative bond and explain how it forms.
Covalent bond: formed when each atom contributes one electron to the shared pair. Dative (coordinate) bond: formed when one atom (donor) contributes both electrons of the shared pair to an atom with an empty orbital (acceptor). Once formed, both bonds are identical in strength and length — the distinction is only in how they formed. Example: NH₃ + H⁺ → NH₄⁺. The nitrogen in NH₃ has a lone pair; H⁺ has an empty 1s orbital. N donates both electrons of its lone pair to H⁺ → dative bond forms → all four N–H bonds in NH₄⁺ are equivalent.
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Multiple Choice Quiz — 25 Questions
Unit 4 Quiz
Select one answer per questionQ1
A covalent bond forms by:
Covalent bond: mutual sharing of electron pair(s). Both atoms contribute one electron each (or both from donor in dative bond). Attraction to both nuclei holds atoms together.
Q2
A dative bond differs from a normal covalent bond because:
Dative bond: one atom (donor) donates both electrons. Once formed, it is identical in strength/length to a regular covalent bond.
Q3
The shape of H₂O is:
O has 2 bonds + 2 lone pairs = 4 electron pairs → tetrahedral electron geometry → bent molecular shape, angle 104.5°.
Q4
The bond angle in CH₄ is:
CH₄: C has 4 bonds, 0 lone pairs → tetrahedral → bond angle 109.5°. No lone pairs to compress angle.
Q5
Which molecule has a trigonal pyramidal shape?
NH₃: N has 3 bonds + 1 lone pair → trigonal pyramidal shape (~107°). BF₃ = trigonal planar; CO₂ = linear; H₂O = bent.
Q6
sp³ hybridised carbon forms:
sp³: 4 hybrid orbitals → tetrahedral geometry, bond angles 109.5°. Examples: CH₄, diamond.
Q7
A double bond consists of:
Double bond = one σ (end-on overlap, stronger) + one π (side-on overlap, weaker). Triple bond = one σ + two π.
Q8
Which has the strongest intermolecular forces?
H₂O has hydrogen bonds (O–H) — the strongest type of intermolecular force. CH₄ and Ar have only LDF; HCl has LDF + dipole-dipole.
Q9
Why does CO₂ have polar bonds but is a non-polar molecule?
CO₂ is linear (sp C, no lone pairs on C). The two C=O dipoles point in exactly opposite directions → vector sum = zero → no net dipole → non-polar molecule.
Q10
Hydrogen bonds can form between molecules containing:
H-bond requirement: H directly bonded to N, O, or F (the only atoms electronegative and small enough). Cl is electronegative but too large; H–Cl forms only dipole-dipole, not true H-bonds.
Q11
The hybridisation of carbon in ethene (C₂H₄) is:
In ethene, each C makes 3 σ bonds (1 to other C, 2 to H) + 1 π bond. → sp²: 3 hybrid orbitals (trigonal planar, 120°) + unhybridised p forms π bond.
Q12
Which property does graphite have that diamond does not?
Graphite conducts electricity due to delocalised π electrons between layers. Diamond: all 4 valence electrons in σ bonds → no free electrons → non-conductor.
Q13
VSEPR theory predicts bond angles by:
VSEPR: electron pairs (bonding + lone) repel each other and adopt positions of minimum repulsion → determines the geometry and bond angles.
Q14
Why does the boiling point of noble gases increase from He to Xe?
Noble gases are non-polar → only LDF. Larger atoms (more electrons) develop stronger instantaneous dipoles → stronger LDF → higher b.p. going down the group.
Q15
Simple molecular substances have low melting points because:
Melting a molecular substance only requires overcoming the weak IMF between molecules, not the strong covalent bonds within each molecule.
Q16
The bond angle in NH₃ (≈107°) is less than the tetrahedral angle (109.5°) because:
A lone pair occupies more space than a bonding pair (concentrated closer to N, less directed). lp–bp repulsion > bp–bp → bond angles compressed below 109.5°.
Q17
Which molecule is polar?
CHCl₃ (chloroform): tetrahedral but asymmetric (one H, three Cl) → dipoles do not cancel → net dipole → polar. CCl₄, CO₂, BF₃ are all symmetric → dipoles cancel → non-polar.
Q18
Which species contains a dative covalent bond?
NH₄⁺ forms when NH₃ donates its lone pair to H⁺ via a dative bond. H₂ and Cl₂ are normal covalent; NaCl is ionic.
Q19
Diamond has a very high melting point because:
Diamond: giant covalent 3D lattice; every C forms 4 strong σ bonds (sp³). All bonds must be broken to melt → extremely high m.p. (~3550°C).
Q20
The hybridisation of carbon in ethyne (HC≡CH) is:
Ethyne: each C makes 2 σ bonds (triple bond has one σ; other single bond to H) → 2 hybrid orbitals → sp. Linear, 180°. Two unhybridised p orbitals form 2 π bonds.
Q21
Which correctly orders the repulsion strength?
Lone pairs are more concentrated (not shared between two nuclei) → exert greater repulsion. Order: lp–lp > lp–bp > bp–bp.
Q22
Ice is less dense than liquid water because:
In ice, every H₂O forms 4 H-bonds in a rigid, open hexagonal lattice. In liquid water, H-bonds are dynamic and molecules pack more closely → liquid water is denser → ice floats.
Q23
Which substance has the highest boiling point?
H₂O: each molecule can form up to 4 hydrogen bonds (2 donor O–H, 2 acceptor lone pairs). More H-bonds per molecule than HF (1 donor) or NH₃ (1 acceptor + limited donors) → highest b.p. (100°C).
Q24
Silicon dioxide (SiO₂) is best described as:
SiO₂ forms a giant 3D covalent lattice (similar to diamond) with each Si bonded to 4 O atoms. Very high m.p., hard, insoluble. Giant covalent structure.
Q25
A π bond is weaker than a σ bond because:
π bond: side-on overlap of p orbitals above and below the axis → less overlap than σ (direct end-on). Less overlap → weaker bond. π bonds are also more reactive (accessible to electrophiles).
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Unit Test — 50 Marks
Section A — Short Answer
20 marksQ1 [4 marks]
Use VSEPR theory to predict the shape and bond angle of each molecule: (a) BCl₃ [1]; (b) PCl₅ [1]; (c) SF₆ [1]; (d) SO₂ [1].
(a) BCl₃: B has 3 bonds, 0 lone pairs → 3 electron pairs → trigonal planar, 120°. [1]
(b) PCl₅: P has 5 bonds, 0 lone pairs → 5 electron pairs → trigonal bipyramidal, 90° and 120°. [1]
(c) SF₆: S has 6 bonds, 0 lone pairs → 6 electron pairs → octahedral, 90°. [1]
(d) SO₂: S has 2 bonds + 1 lone pair = 3 electron pairs → bent/V-shaped, <120° (~119°). [1]
(b) PCl₅: P has 5 bonds, 0 lone pairs → 5 electron pairs → trigonal bipyramidal, 90° and 120°. [1]
(c) SF₆: S has 6 bonds, 0 lone pairs → 6 electron pairs → octahedral, 90°. [1]
(d) SO₂: S has 2 bonds + 1 lone pair = 3 electron pairs → bent/V-shaped, <120° (~119°). [1]
Q2 [4 marks]
Explain the hybridisation of carbon in diamond and graphite. Relate each to the electrical conductivity of the two allotropes. [4]
Diamond: each C is sp³ hybridised; four sp³ hybrid orbitals form four σ bonds to neighbouring C atoms in a giant 3D tetrahedral lattice. All four outer electrons are used in σ bonds → no delocalised electrons → non-conductor. [2]
Graphite: each C is sp² hybridised; three sp² hybrid orbitals form σ bonds in planar hexagonal layers. Each C has one remaining unhybridised p𝑧 orbital that overlaps with adjacent p𝑧 orbitals → delocalised π electron system throughout each layer. These mobile π electrons carry charge → conducts electricity. [2]
Graphite: each C is sp² hybridised; three sp² hybrid orbitals form σ bonds in planar hexagonal layers. Each C has one remaining unhybridised p𝑧 orbital that overlaps with adjacent p𝑧 orbitals → delocalised π electron system throughout each layer. These mobile π electrons carry charge → conducts electricity. [2]
Q3 [4 marks]
For each pair, state which has the higher boiling point and explain why in terms of intermolecular forces: (a) CH₄ vs CCl₄ [2]; (b) HF vs HCl [2].
(a) CCl₄ (b.p. 77°C) > CH₄ (b.p. −162°C). Both are non-polar → only London dispersion forces. CCl₄ (M = 154 g/mol) has far more electrons than CH₄ (M = 16 g/mol) → more polarisable → stronger LDF → more energy to overcome → higher b.p. [2]
(b) HF (b.p. 19.5°C) > HCl (b.p. −85°C). Despite HCl having more electrons (higher M), HF forms strong hydrogen bonds (F is highly electronegative, small, H is accessible). HCl forms only LDF + dipole-dipole (Cl too large for H-bonds). H-bonds are much stronger than LDF/dipole-dipole → HF has higher b.p. [2]
(b) HF (b.p. 19.5°C) > HCl (b.p. −85°C). Despite HCl having more electrons (higher M), HF forms strong hydrogen bonds (F is highly electronegative, small, H is accessible). HCl forms only LDF + dipole-dipole (Cl too large for H-bonds). H-bonds are much stronger than LDF/dipole-dipole → HF has higher b.p. [2]
Q4 [4 marks]
Explain, using the concept of dipole moments and molecular geometry, why: (a) CCl₄ is non-polar [2]; (b) CHCl₃ is polar [2].
(a) CCl₄: four C–Cl bonds, each polar (δ⁺C–Clδ−). The molecule is perfectly tetrahedral (all 4 substituents identical). The four bond dipole vectors are equal in magnitude and point symmetrically outward → they cancel completely → zero net dipole moment → non-polar. [2]
(b) CHCl₃: three C–Cl and one C–H bond. C–Cl dipoles point towards Cl; C–H dipole is smaller and points in a different direction. The geometry is tetrahedral but asymmetric (mixed substituents) → dipoles do not cancel → net dipole towards Cl end → polar molecule. [2]
(b) CHCl₃: three C–Cl and one C–H bond. C–Cl dipoles point towards Cl; C–H dipole is smaller and points in a different direction. The geometry is tetrahedral but asymmetric (mixed substituents) → dipoles do not cancel → net dipole towards Cl end → polar molecule. [2]
Q5 [4 marks]
State and explain two anomalous properties of water that arise directly from its ability to form hydrogen bonds. [4]
(1) Anomalously high boiling point [2]: Expected b.p. of H₂O from the H₂S/H₂Se/H₂Te trend would be approximately −80°C. Actual b.p. is 100°C. Each H₂O can form up to 4 H-bonds. These bonds require much more energy to break than LDF alone → much higher b.p. than predicted. [2]
(2) Ice less dense than liquid water (ice floats) [2]: In ice, H-bonds lock molecules into a rigid, open hexagonal lattice where each molecule is tetrahedrally bonded to 4 neighbours → large spaces in the lattice → lower density than liquid water. In liquid water, H-bonds are more dynamic and molecules pack more closely → higher density → ice floats. This is ecologically vital (aquatic life survives under ice). [2]
(2) Ice less dense than liquid water (ice floats) [2]: In ice, H-bonds lock molecules into a rigid, open hexagonal lattice where each molecule is tetrahedrally bonded to 4 neighbours → large spaces in the lattice → lower density than liquid water. In liquid water, H-bonds are more dynamic and molecules pack more closely → higher density → ice floats. This is ecologically vital (aquatic life survives under ice). [2]
Section B — Extended Answer
30 marksQ6 [8 marks]
Compare simple molecular structures and giant covalent structures in terms of: bonding, melting point, electrical conductivity, and solubility. Illustrate with specific examples. Use the structures of ice (H₂O) and silicon dioxide (SiO₂) to explain the difference in their melting points. [8]
Simple molecular [3]: Discrete molecules held together by weak intermolecular forces (LDF, dipole-dipole, H-bonds). Covalent bonds within molecules are strong. Melting: only IMF must be overcome → low m.p. (e.g. I₂ 114°C, CH₄ −182°C). Conductivity: no charged particles free to move → non-conductor (except some dissolved in water, e.g. HCl). Solubility: “like dissolves like” — polar dissolves in polar (NH₃ in H₂O), non-polar in non-polar (I₂ in CCl₄). [3]
Giant covalent [3]: All atoms covalently bonded in an infinite 3D (or 2D) network. Every bond is a strong covalent bond. Melting: must break many strong covalent bonds → very high m.p. (diamond ~3550°C, SiO₂ ~1710°C). Conductivity: no free electrons or ions → non-conductor (except graphite which has delocalised π electrons). Solubility: generally insoluble in water. [3]
Ice vs SiO₂ [2]: Ice (H₂O) is a simple molecular solid. In ice, molecules are held by H-bonds (relatively weak, ~20–40 kJ/mol) → low m.p. (0°C). SiO₂ is a giant covalent network; each Si is covalently bonded to 4 O atoms and each O to 2 Si atoms with strong Si–O covalent bonds (~450 kJ/mol) throughout the lattice → very high m.p. (~1710°C). The difference arises because ice must overcome only weak H-bonds, while SiO₂ requires breaking a vast number of strong covalent bonds. [2]
Giant covalent [3]: All atoms covalently bonded in an infinite 3D (or 2D) network. Every bond is a strong covalent bond. Melting: must break many strong covalent bonds → very high m.p. (diamond ~3550°C, SiO₂ ~1710°C). Conductivity: no free electrons or ions → non-conductor (except graphite which has delocalised π electrons). Solubility: generally insoluble in water. [3]
Ice vs SiO₂ [2]: Ice (H₂O) is a simple molecular solid. In ice, molecules are held by H-bonds (relatively weak, ~20–40 kJ/mol) → low m.p. (0°C). SiO₂ is a giant covalent network; each Si is covalently bonded to 4 O atoms and each O to 2 Si atoms with strong Si–O covalent bonds (~450 kJ/mol) throughout the lattice → very high m.p. (~1710°C). The difference arises because ice must overcome only weak H-bonds, while SiO₂ requires breaking a vast number of strong covalent bonds. [2]
Q7 [8 marks]
Describe and explain, using intermolecular force theory, the following trends: (a) boiling points of the halogens: F₂ (−188°C) < Cl₂ (−34°C) < Br₂ (59°C) < I₂ (184°C) [3]; (b) boiling points across Group 16 hydrides: H₂O (100°C) >> H₂Te (−2°C) > H₂Se (−41°C) > H₂S (−60°C) [3]; (c) why ethanol (C₂H₅OH, b.p. 78°C) has a much higher b.p. than dimethyl ether (CH₃OCH₃, b.p. −23°C) despite having the same molecular formula. [2]
(a) Halogens [3]: All are non-polar diatomic molecules → only London dispersion forces act between molecules. Going down the group: molar mass increases (F₂ 38, Cl₂ 71, Br₂ 160, I₂ 254 g/mol) → more electrons → larger, more polarisable electron clouds → stronger instantaneous dipoles → stronger LDF → more energy needed to vaporise → boiling points increase progressively. [3]
(b) Group 16 hydrides [3]: H₂Te, H₂Se, H₂S follow expected trend: increasing molar mass going down → stronger LDF → rising b.p. (H₂S < H₂Se < H₂Te). H₂O is the anomaly: O is highly electronegative and small, so O–H bonds are very polar and H is accessible → each H₂O forms up to 4 strong hydrogen bonds. H-bonds require much more energy to overcome than the LDF/dipole-dipole in H₂S/H₂Se/H₂Te → H₂O has anomalously high b.p. (100°C vs expected ~−80°C). S, Se, Te are too large for H-bonds despite being electronegative. [3]
(c) Ethanol vs DME [2]: Same molecular formula (C₂H⁶O) but different connectivity. Ethanol has an O–H group → forms hydrogen bonds (O–H···O). DME has no O–H group (O bonded to two C atoms) → cannot form H-bonds; has only LDF + dipole-dipole forces. H-bonds in ethanol are much stronger than dipole-dipole in DME → ethanol has much higher b.p. (+101°C difference). [2]
(b) Group 16 hydrides [3]: H₂Te, H₂Se, H₂S follow expected trend: increasing molar mass going down → stronger LDF → rising b.p. (H₂S < H₂Se < H₂Te). H₂O is the anomaly: O is highly electronegative and small, so O–H bonds are very polar and H is accessible → each H₂O forms up to 4 strong hydrogen bonds. H-bonds require much more energy to overcome than the LDF/dipole-dipole in H₂S/H₂Se/H₂Te → H₂O has anomalously high b.p. (100°C vs expected ~−80°C). S, Se, Te are too large for H-bonds despite being electronegative. [3]
(c) Ethanol vs DME [2]: Same molecular formula (C₂H⁶O) but different connectivity. Ethanol has an O–H group → forms hydrogen bonds (O–H···O). DME has no O–H group (O bonded to two C atoms) → cannot form H-bonds; has only LDF + dipole-dipole forces. H-bonds in ethanol are much stronger than dipole-dipole in DME → ethanol has much higher b.p. (+101°C difference). [2]
Q8 [6 marks]
Using VSEPR and hybridisation theory, fully describe the bonding and geometry of: (a) ethene (C₂H₄) [3]; (b) water (H₂O) [3]. In each case state: hybridisation, bond angles, shape, number of σ and π bonds, and whether the molecule is polar. [6]
(a) Ethene [3]: Each C has 3 groups around it (1 double bond counts as one electron group) + 0 lone pairs → 3 electron pairs → sp² hybridised. Each C forms three sp² σ bonds: one C–C and two C–H. Each C has one unhybridised p𝑧 orbital perpendicular to the molecular plane; these overlap side-on to form one C–C π bond. Bond count: 5 σ bonds (1 C–C, 4 C–H) + 1 π bond (C=C). Bond angles: ~120° (H–C–H and H–C–C). Shape: planar (all atoms in one plane, rigid due to π bond). Polarity: C–H bonds slightly polar, but molecule is symmetric → non-polar overall. [3]
(b) Water [3]: O has 2 bonds + 2 lone pairs = 4 electron pairs → sp³ hybridised. O forms 2 σ bonds (O–H) using two sp³ hybrid orbitals; two sp³ orbitals hold lone pairs. Bond count: 2 σ bonds, 0 π bonds. Electron pair geometry: tetrahedral. Molecular shape: bent/V-shaped. Bond angle: 104.5° (reduced from 109.5° due to 2 lone pairs exerting greater repulsion than bonding pairs). Polarity: O is more electronegative than H → O–H bonds are polar; bent shape means dipoles do not cancel → polar molecule (net dipole towards O). [3]
(b) Water [3]: O has 2 bonds + 2 lone pairs = 4 electron pairs → sp³ hybridised. O forms 2 σ bonds (O–H) using two sp³ hybrid orbitals; two sp³ orbitals hold lone pairs. Bond count: 2 σ bonds, 0 π bonds. Electron pair geometry: tetrahedral. Molecular shape: bent/V-shaped. Bond angle: 104.5° (reduced from 109.5° due to 2 lone pairs exerting greater repulsion than bonding pairs). Polarity: O is more electronegative than H → O–H bonds are polar; bent shape means dipoles do not cancel → polar molecule (net dipole towards O). [3]
Q9 [8 marks]
Define and give an example of each type of intermolecular force. Explain why London dispersion forces exist in all molecules. Compare the magnitudes of the three types and explain how each contributes to boiling point. Use specific examples to illustrate. Explain why HF has a higher b.p. than HI despite HI having more electrons. [8]
Three IMF types [3]:
(1) London dispersion forces (LDF): present in all molecules. Arise from instantaneous dipoles caused by random electron movement creating a brief uneven charge distribution; this induces a dipole in an adjacent molecule → transient attraction. Example: Ar, I₂, CH₄. [1]
(2) Permanent dipole-dipole: present in polar molecules. δ⁺ end of one molecule attracted to δ− end of neighbour. Example: HCl, SO₂. Present alongside LDF. [1]
(3) Hydrogen bonds: strongest IMF; between H bonded to N/O/F and a lone pair on N/O/F of adjacent molecule. Example: H₂O, NH₃, HF. [1]
Why LDF in all molecules [1]: All molecules have electrons, which are in constant motion. At any instant, electron distribution is uneven → temporary dipole. This induces an opposite dipole in a neighbouring molecule → brief attraction. These fluctuations are present regardless of whether the molecule has a permanent dipole. [1]
Magnitudes and b.p. contribution [2]: LDF: smallest (0.1–10 kJ/mol per interaction); increases with electron count/molar mass and molecular surface area. Dipole-dipole: 5–20 kJ/mol; adds to LDF in polar molecules. H-bonds: 10–40 kJ/mol per H-bond; much stronger than other IMF → significantly raises b.p. Examples: noble gas b.p. series (LDF only); HCl b.p. (−85°C) vs H₂S (−60°C) shows LDF effect; H₂O (100°C) shows H-bond effect. [2]
HF vs HI [2]: HI (M=128) has far more electrons than HF (M=20) → stronger LDF. Yet HF (b.p. +19.5°C) > HI (b.p. −35.4°C). This is because F is the most electronegative element and is very small → the H–F bond is strongly polar and H is accessible → HF molecules form strong hydrogen bonds (F–H···F). I is large and insufficiently electronegative for H-bond formation → HI has only LDF + weak dipole-dipole. H-bonds in HF outweigh the LDF advantage of HI → HF has the higher b.p. [2]
(1) London dispersion forces (LDF): present in all molecules. Arise from instantaneous dipoles caused by random electron movement creating a brief uneven charge distribution; this induces a dipole in an adjacent molecule → transient attraction. Example: Ar, I₂, CH₄. [1]
(2) Permanent dipole-dipole: present in polar molecules. δ⁺ end of one molecule attracted to δ− end of neighbour. Example: HCl, SO₂. Present alongside LDF. [1]
(3) Hydrogen bonds: strongest IMF; between H bonded to N/O/F and a lone pair on N/O/F of adjacent molecule. Example: H₂O, NH₃, HF. [1]
Why LDF in all molecules [1]: All molecules have electrons, which are in constant motion. At any instant, electron distribution is uneven → temporary dipole. This induces an opposite dipole in a neighbouring molecule → brief attraction. These fluctuations are present regardless of whether the molecule has a permanent dipole. [1]
Magnitudes and b.p. contribution [2]: LDF: smallest (0.1–10 kJ/mol per interaction); increases with electron count/molar mass and molecular surface area. Dipole-dipole: 5–20 kJ/mol; adds to LDF in polar molecules. H-bonds: 10–40 kJ/mol per H-bond; much stronger than other IMF → significantly raises b.p. Examples: noble gas b.p. series (LDF only); HCl b.p. (−85°C) vs H₂S (−60°C) shows LDF effect; H₂O (100°C) shows H-bond effect. [2]
HF vs HI [2]: HI (M=128) has far more electrons than HF (M=20) → stronger LDF. Yet HF (b.p. +19.5°C) > HI (b.p. −35.4°C). This is because F is the most electronegative element and is very small → the H–F bond is strongly polar and H is accessible → HF molecules form strong hydrogen bonds (F–H···F). I is large and insufficiently electronegative for H-bond formation → HI has only LDF + weak dipole-dipole. H-bonds in HF outweigh the LDF advantage of HI → HF has the higher b.p. [2]