DP Chemistry HL S2.2 The Covalent Model

DP Chemistry HL S2.2 The Covalent Model - Full notes

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DP Chemistry HL | S2.2 The Covalent Model

DP CHEMISTRY · STRUCTURE

S2.2

The Covalent Model

Full notes for HL · IB-style

Contents

1	Covalent bonds and Lewis structures
2	Multiple bonds, bond length and coordination bonds
3	VSEPR theory and molecular shapes
4	Bond polarity and molecular polarity
5	Giant covalent structures
6	Intermolecular forces
7	Physical properties of covalent substances
8	Formal charge, resonance and expanded octets (HL only)
9	Sigma and pi bonds and hybridization (HL only)
10	Lewis formulas: a systematic method
11	Coordination (dative) bonds
12	Bond length and bond strength
13	VSEPR theory: more practice
14	Bond polarity
15	Molecular polarity
16	Intermolecular forces: London dispersion
17	Dipole–dipole forces and hydrogen bonding
18	Comparing boiling points
19	Solubility and intermolecular forces
20	Giant covalent structures
21	Chromatography and intermolecular forces
22	Resonance structures (introduction)
23	Mixed review: worked problems
24	Additional exam-style practice
25	Allotropes of carbon in technology
26	Common misconceptions and links
27	Exam-style question with markscheme
28	Check your understanding
29	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe covalent bonding and draw Lewis structures
relate bond order to bond length and strength, and describe coordination bonds
use VSEPR theory to predict shapes and bond angles for up to four electron domains
explain bond polarity and molecular polarity
describe the structures and properties of diamond, graphite, graphene, C60 and SiO2
explain London forces, dipole-dipole forces and hydrogen bonding
explain trends in boiling point, solubility and conductivity of covalent substances
use formal charge to choose Lewis structures and explain resonance
predict shapes with five and six electron domains
explain sigma and pi bonds and sp, sp2 and sp3 hybridization

Sections marked HL only (Sections 8 and 9) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How do atoms share electrons to form covalent bonds? How do the shapes of molecules and the forces between them determine properties?

How this topic is examined

S2.2 is one of the largest subtopics. Expect Lewis structures, VSEPR shapes with bond angles, polarity, and explanations of boiling points in terms of intermolecular forces. HL questions add formal charge, resonance, expanded octets and hybridization.

Always identify the intermolecular forces present before explaining a physical property.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Covalent bonds and Lewis structures

Sharing pairs of electrons.

Think about it

Two chlorine atoms each need one more electron to fill their outer shell. How can both get what they need at the same time?

A covalent bond is the electrostatic attraction between a shared pair of electrons and the positively charged nuclei of the two bonded atoms. Covalent bonds typically form between non-metal atoms, which have similar, high electronegativities.

In many molecules, each atom achieves a stable octet (eight outer electrons; two for hydrogen). Lewis structures (electron dot structures) show all the valence electrons: shared pairs as lines or pairs of dots between atoms, and lone pairs (non-bonding pairs) on atoms. To draw one: count total valence electrons; join atoms with single bonds; complete octets with lone pairs; form double or triple bonds if octets are incomplete.

Some molecules have incomplete octets: boron in BF3 has only six electrons; beryllium in BeCl2 has four.

Molecule	Total valence electrons	Lewis structure (described)
H2O	8	O with two single bonds to H and two lone pairs
NH3	8	N with three single bonds to H and one lone pair
CO2	16	O=C=O, two lone pairs on each O
N2	10	N≡N, one lone pair on each N
HCN	10	H–C≡N, one lone pair on N

Figure 1. Structural formulas of simple molecules (lone pairs not shown).

Quick check

1. Define covalent bond.

2. How many lone pairs are on the O atom in water?

Answers: 1. attraction between a shared electron pair and the nuclei 2. two

2 Multiple bonds, bond length and coordination bonds

More shared electrons make shorter, stronger bonds.

Atoms can share one, two or three pairs of electrons, forming single, double or triple bonds. As the number of shared pairs (the bond order) increases, the bond becomes shorter and stronger, because more electrons attract the two nuclei together.

A coordination bond (dative covalent bond) is a covalent bond in which both shared electrons come from the same atom. Once formed, it is identical to any other covalent bond. Examples: in the ammonium ion, NH4+, the lone pair on N forms a coordination bond with H+; in the hydronium ion, H3O+; and in carbon monoxide.

Bond	Bond order	Bond length / pm	Bond enthalpy / kJ mol−1
C–C	1	154	346
C=C	2	134	614
C≡C	3	120	839

Quick check

1. How does bond length change from single to triple?

2. Define coordination bond.

Answers: 1. decreases 2. covalent bond where both electrons come from one atom

3 VSEPR theory and molecular shapes

Electron domains repel each other as far apart as possible.

Valence shell electron pair repulsion (VSEPR) theory: electron domains around a central atom (bonding pairs, lone pairs, and multiple bonds each counting as one domain) repel each other and arrange themselves as far apart as possible. The arrangement of domains determines the electron domain geometry; the positions of the atoms determine the molecular geometry (shape).

Lone pairs repel more strongly than bonding pairs, because they are held closer to the central atom. Each lone pair reduces the bond angle by about 2.5°: CH4 is 109.5°, NH3 about 107°, H2O about 104.5°.

Domains	Lone pairs	Shape	Bond angle	Example
2	0	linear	180°	CO2, BeCl2
3	0	trigonal planar	120°	BF3, CO32−
3	1	bent (V-shaped)	about 117°	SO2
4	0	tetrahedral	109.5°	CH4, NH4+
4	1	trigonal pyramidal	about 107°	NH3
4	2	bent (V-shaped)	about 104.5°	H2O

WORKED EXAMPLE 1 Predicting a shape

Predict the shape and bond angle of PCl3.

Solution

P has 5 valence electrons; three bond to Cl, one lone pair remains: 4 domainsM1

Electron domain geometry tetrahedral; molecular shape trigonal pyramidalA1

Bond angle about 107° (lone pair repulsion)A1

Common mistake

Counting a double bond as two domains. A multiple bond counts as one electron domain.

Quick check

1. What is the shape of BF3?

2. Why is the bond angle in water less than 109.5°?

Answers: 1. trigonal planar, 120° 2. two lone pairs repel more than bonding pairs

4 Bond polarity and molecular polarity

Unequal sharing creates dipoles.

Electronegativity is the ability of an atom to attract the shared pair of electrons in a covalent bond. It increases across a period and up a group; fluorine is the most electronegative element (4.0 on the Pauling scale).

When two atoms with different electronegativities form a bond, the electrons are shared unequally: the bond is polar, with partial charges δ− on the more electronegative atom and δ+ on the other. The bond has a dipole moment. The greater the difference in electronegativity, the more polar the bond. Bonding is a continuum: from pure covalent (equal sharing) through polar covalent to ionic.

A molecule is polar if it has polar bonds and its shape is asymmetric, so the bond dipoles do not cancel. CO2 has polar C=O bonds but is linear, so the dipoles cancel and it is non-polar. Water has polar O–H bonds and is bent, so it is polar. CH4 and CCl4 are symmetrical and non-polar; CHCl3 is polar.

Figure 2. Symmetry decides whether bond dipoles cancel.

Quick check

1. Define electronegativity.

2. Why is CO2 non-polar despite having polar bonds?

Answers: 1. ability of an atom to attract a bonding pair 2. linear: bond dipoles cancel

5 Giant covalent structures

Networks of covalent bonds extending throughout the solid.

Some elements and compounds form giant covalent (network) structures, in which atoms are joined by covalent bonds in a continuous three-dimensional (or two-dimensional) network. Allotropes are different structural forms of the same element in the same physical state.

Diamond: each carbon atom is covalently bonded to four others in a tetrahedral arrangement. Very hard, very high melting point, does not conduct electricity (no mobile electrons), good thermal conductor.

Graphite: each carbon is bonded to three others in flat hexagonal layers; the fourth electron of each atom is delocalized across the layer. Layers are held together by weak London forces, so they slide over each other: graphite is soft and used as a lubricant and in pencils. It conducts electricity along the layers. High melting point.

Graphene: a single layer of graphite, one atom thick. Extremely strong, very light, excellent conductor of electricity and heat.

Fullerene C60: a molecule of 60 carbon atoms arranged in hexagons and pentagons forming a sphere. Because it is a simple molecular substance, it has a lower melting point and is a poor conductor.

Silicon dioxide (quartz), SiO2: each silicon is bonded to four oxygen atoms tetrahedrally, and each oxygen to two silicon atoms. Very hard, high melting point, non-conductor.

Structure	Bonding	Electrical conductivity	Hardness
Diamond	each C bonded to 4 C, tetrahedral	none	very hard
Graphite	each C bonded to 3 C in layers; delocalized electrons	good (along layers)	soft; layers slide
Graphene	single layer of graphite	excellent	very strong
C60	molecules; each C bonded to 3	poor	soft
SiO2	each Si bonded to 4 O	none	very hard

Quick check

1. Why does graphite conduct electricity?

2. What is an allotrope?

Answers: 1. delocalized electrons move along layers 2. different structural form of the same element

6 Intermolecular forces

Forces between molecules determine physical properties.

Intermolecular forces are attractions between molecules. They are much weaker than covalent bonds within molecules. There are three main types.

London (dispersion) forces exist between all molecules. Electrons move randomly, creating an instantaneous dipole in one molecule, which induces a dipole in a neighbouring molecule; the two attract. London forces increase with the number of electrons (and so with molar mass) and with surface area: straight-chain alkanes have stronger London forces than their branched isomers.

Dipole-dipole forces act between polar molecules, which have permanent dipoles: the δ+ end of one molecule attracts the δ− end of another. They are stronger than London forces in molecules of similar size.

Hydrogen bonding is a particularly strong dipole-dipole attraction between a hydrogen atom bonded to a very electronegative atom (N, O or F) and a lone pair on an N, O or F atom of another molecule. It explains the high boiling points of water, ammonia and HF compared with other hydrides in their groups, and is vital in biology (DNA base pairs, protein structure).

Relative strengths: London < dipole-dipole < hydrogen bonding < covalent bonds, for molecules of similar size.

Figure 3. Water's boiling point is anomalously high because of hydrogen bonding.

Common mistake

Saying that boiling breaks covalent bonds or that hydrogen bonds are bonds within a molecule. Hydrogen bonds act between molecules.

Quick check

1. Name the three types of intermolecular force.

2. Which elements must be bonded to H for hydrogen bonding?

3. Why does pentane boil higher than 2,2-dimethylpropane?

Answers: 1. London, dipole-dipole, hydrogen bonding 2. N, O or F 3. larger surface area: stronger London forces

7 Physical properties of covalent substances

Simple molecules versus giant structures.

Simple molecular substances (such as water, methane, iodine) have low melting and boiling points, because only weak intermolecular forces need to be overcome, not covalent bonds. They do not conduct electricity (no mobile charged particles). Solubility follows 'like dissolves like': polar molecules dissolve in polar solvents such as water (especially if they can hydrogen bond, such as ethanol), and non-polar molecules dissolve in non-polar solvents such as hexane.

Giant covalent substances have very high melting points, because many strong covalent bonds must be broken. They are insoluble in water, and do not conduct (except graphite and graphene).

Chromatography separates substances by their different attractions to a stationary and a mobile phase, which depend on polarity and intermolecular forces (S1.1).

Property	Simple molecular	Giant covalent
Melting point	low: weak intermolecular forces	very high: covalent bonds broken
Conductivity	none	none (except graphite, graphene)
Solubility in water	polar ones dissolve; non-polar do not	insoluble
Examples	H2O, CO2, I2	diamond, SiO2

8 Formal charge, resonance and expanded octets (HL only)

Choosing the best Lewis structure.

HL only

This section is Additional Higher Level content. SL students do not need it.

Formal charge (FC) helps decide between possible Lewis structures: FC = valence electrons − (non-bonding electrons + ½ bonding electrons). The preferred structure has formal charges as close to zero as possible, with any negative formal charge on the most electronegative atom.

Resonance occurs when more than one valid Lewis structure can be drawn by moving electrons but not atoms. The real structure is a resonance hybrid: the electrons are delocalized over several atoms. In the carbonate ion, the three C–O bonds are all the same length, intermediate between single and double (bond order 1⅓). In benzene, C6H6, the six C–C bonds are identical (bond order 1.5) because six electrons are delocalized around the ring; this gives benzene extra stability.

Elements in period 3 and beyond can have expanded octets, using d orbitals. Molecules with five electron domains: trigonal bipyramidal (PCl5, 90° and 120°); see-saw (SF4); T-shaped (ClF3); linear (XeF2). Six domains: octahedral (SF6, 90°); square pyramidal (BrF5); square planar (XeF4).

Domains	Lone pairs	Shape	Example
5	0	trigonal bipyramidal	PCl5
5	1	see-saw	SF4
5	2	T-shaped	ClF3
5	3	linear	XeF2
6	0	octahedral	SF6
6	1	square pyramidal	BrF5
6	2	square planar	XeF4

WORKED EXAMPLE 2 Formal charge

Use formal charges to show why the structure O=C=O is preferred over O≡C–O for CO2.

Solution

O=C=O: each O has 4 non-bonding and 4 bonding electrons: FC = 6 − (4 + 2) = 0; C: 4 − (0 + 4) = 0M1

O≡C–O: triple-bonded O: 6 − (2 + 3) = +1; single-bonded O: 6 − (6 + 1) = −1M1

O=C=O has all formal charges zero, so it is preferredR1

9 Sigma and pi bonds and hybridization (HL only)

How orbitals overlap to form bonds.

HL only

This section is Additional Higher Level content. SL students do not need it.

A sigma (σ) bond forms by head-on (axial) overlap of orbitals; electron density lies along the axis between the nuclei. Every single bond is a σ bond. A pi (π) bond forms by sideways overlap of parallel p orbitals; electron density lies above and below the axis. A double bond consists of one σ and one π bond; a triple bond of one σ and two π bonds. π bonds prevent rotation about the bond.

Hybridization is the mixing of atomic orbitals to form new, equivalent hybrid orbitals. sp3: one s and three p orbitals mix to give four orbitals pointing to the corners of a tetrahedron (109.5°), as in methane. sp2: one s and two p give three orbitals in a plane (120°), with one unhybridized p orbital forming a π bond, as in ethene. sp: one s and one p give two orbitals at 180°, with two p orbitals forming two π bonds, as in ethyne.

Hybridization	Domains	Geometry	Example
sp3	4	tetrahedral, 109.5°	CH4, C in ethane
sp2	3	trigonal planar, 120°	C in ethene, benzene
sp	2	linear, 180°	C in ethyne, CO2

Quick check

1. How many σ and π bonds are in ethyne, C2H2?

2. What is the hybridization of carbon in ethene?

Answers: 1. 3 σ and 2 π 2. sp2

10 Lewis formulas: a systematic method

Count, connect, complete.

1. Count the total valence electrons (add one for each negative charge, subtract one for each positive charge). 2. Connect atoms with single bonds (the least electronegative atom is usually central). 3. Complete octets on outer atoms with lone pairs. 4. Place remaining electrons on the central atom. 5. If the central atom lacks an octet, form multiple bonds.

WORKED EXAMPLE 3 Carbon dioxide (standard)

Draw the Lewis formula of CO2.

Solution

16 valence electronsA1

O=C=O with two lone pairs on each OA1

WORKED EXAMPLE 4 An ion (standard)

Draw the Lewis formula of NH4+.

Solution

5 + 4 − 1 = 8 electronsA1

N with four single N–H bonds, in square brackets with + chargeA1

Quick check

1. Draw the Lewis formula of HCN.

2. Draw the Lewis formula of the hydroxide ion.

3. How many lone pairs are on the oxygen in water?

Answers: 1. H–C≡N with one lone pair on N 2. [O–H]− with three lone pairs on O 3. 2

11 Coordination (dative) bonds

Both electrons from one atom.

In a coordination bond, both shared electrons come from the same atom, which must have a lone pair; the acceptor must have an empty orbital. Once formed, it is identical to other covalent bonds. Examples: NH4+, H3O+, CO, and metal complexes.

WORKED EXAMPLE 5 Hydronium ion (standard)

Explain the formation of H3O+ from H2O and H+.

Solution

A lone pair on the oxygen of waterR1

forms a coordination bond to H+, which has an empty orbitalR1

Quick check

1. Identify the coordination bond in carbon monoxide.

2. Which species can act as an acceptor: BF3 or NH3?

Answers: 1. one of the three C≡O bonds; both electrons from O 2. BF3 (B has an empty orbital)

12 Bond length and bond strength

Multiple bonds are shorter and stronger.

More shared electron pairs pull the nuclei closer together, so triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds.

Bond	Length / pm	Enthalpy / kJ mol−1
C–C	154	346
C=C	134	614
C≡C	120	839
C–O	143	358
C=O	120	804

WORKED EXAMPLE 6 Comparing (standard)

Explain why the N≡N bond is much stronger than the N–N bond.

Solution

Three shared pairs versus oneR1

Greater electrostatic attraction between the nuclei and the shared electronsR1

Quick check

1. Predict which is shorter: the C–O bond in methanol or the C=O bond in methanal.

Answers: 1. C=O

13 VSEPR theory: more practice

Electron domains determine shape.

Count electron domains (bonding pairs, lone pairs and multiple bonds each count as one domain) around the central atom. Domains repel to maximize separation. Lone pairs repel more than bonding pairs, reducing bond angles by about 2.5° per lone pair.

Domains	Lone pairs	Shape	Angle	Example
2	0	linear	180°	CO2
3	0	trigonal planar	120°	BF3
3	1	bent	≈117°	SO2
4	0	tetrahedral	109.5°	CH4
4	1	trigonal pyramidal	107°	NH3
4	2	bent	104.5°	H2O

WORKED EXAMPLE 7 Predicting shape (standard)

Predict the shape and bond angle of PCl3.

Solution

4 domains: 3 bonding pairs, 1 lone pairM1

Trigonal pyramidal, about 107°A1

Quick check

1. Predict the shape of H2S.

2. Predict the shape of the nitrate ion.

3. Predict the shape of HCN.

Answers: 1. bent, about 104.5° 2. trigonal planar, 120° 3. linear, 180°

14 Bond polarity

Unequal sharing.

A covalent bond is polar when the atoms differ in electronegativity: the more electronegative atom has a partial negative charge (δ−). The larger the difference, the more polar the bond.

WORKED EXAMPLE 8 Partial charges (basic)

Show the partial charges in H–Cl and C–F.

Solution

Hδ+–Clδ−A1

Cδ+–Fδ−A1

Quick check

1. Which bond is more polar: O–H or N–H?

Answers: 1. O–H (larger electronegativity difference)

15 Molecular polarity

Shape decides.

A molecule is polar if it has polar bonds and an asymmetric shape, so the bond dipoles do not cancel. Symmetrical molecules such as CO2, BF3 and CCl4 are non-polar despite polar bonds.

WORKED EXAMPLE 9 Polarity (exam level)

Explain why CCl4 is non-polar but CHCl3 is polar.

Solution

Both have polar bondsR1

CCl4 is symmetrical tetrahedral: dipoles cancelR1

CHCl3: different bonds, dipoles do not cancelR1

Quick check

1. Is SO2 polar?

2. Is BF3 polar?

Answers: 1. yes (bent) 2. no (trigonal planar, symmetrical)

16 Intermolecular forces: London dispersion

Temporary dipoles.

London (dispersion) forces exist between all molecules. Random electron movement creates temporary dipoles, which induce dipoles in neighbours. They increase with the number of electrons (molar mass) and with surface area (straight chains more than branched).

WORKED EXAMPLE 10 Boiling points (standard)

Explain why pentane (36 °C) boils at a higher temperature than 2,2-dimethylpropane (10 °C).

Solution

Same number of electronsR1

Pentane is straight-chain: larger contact area, stronger London forcesR1

Quick check

1. Explain the increase in boiling points of the halogens down group 17.

Answers: 1. more electrons, stronger London forces

17 Dipole–dipole forces and hydrogen bonding

Permanent dipoles.

Polar molecules attract each other through dipole–dipole forces. Hydrogen bonding is an especially strong dipole–dipole attraction between a hydrogen atom bonded to N, O or F and a lone pair on another N, O or F atom.

WORKED EXAMPLE 11 Hydrogen bonding (exam level)

Explain why water (100 °C) boils much higher than hydrogen sulfide (−60 °C).

Solution

Water molecules form hydrogen bonds (O–H···O)R1

H2S has only weaker dipole–dipole and London forcesR1

More energy needed to separate water moleculesR1

Common mistake

Saying that boiling breaks the covalent O–H bonds. Boiling only overcomes intermolecular forces; covalent bonds inside molecules are not broken.

Quick check

1. Which can form hydrogen bonds: CH3OH, CH3OCH3, CH4?

2. Why is ice less dense than liquid water?

Answers: 1. CH3OH 2. hydrogen bonds hold molecules in an open lattice

18 Comparing boiling points

Identify the strongest force.

To compare boiling points: identify the intermolecular forces in each substance (London, dipole–dipole, hydrogen bonding), compare their strength, and consider molar mass. Hydrogen bonding usually dominates for small molecules.

WORKED EXAMPLE 12 Ranking (exam level)

Rank in order of increasing boiling point: propane, ethanol, methoxymethane (all similar M).

Solution

Propane: London only; methoxymethane: dipole–dipole; ethanol: hydrogen bondingR1

propane < methoxymethane < ethanolA1

Quick check

1. Rank HF, HCl, HBr by boiling point and explain.

Answers: 1. HCl < HBr < HF: HF has hydrogen bonding; HBr has more London forces than HCl

19 Solubility and intermolecular forces

Like dissolves like.

Polar and ionic substances dissolve in polar solvents such as water; non-polar substances dissolve in non-polar solvents such as hexane. Small alcohols and acids are soluble in water because they hydrogen bond with it; solubility falls as the non-polar chain gets longer.

WORKED EXAMPLE 13 Alcohol solubility (standard)

Explain why ethanol is completely miscible with water but hexan-1-ol is only slightly soluble.

Solution

Both can hydrogen bond with water through OHR1

Hexan-1-ol's long non-polar chain disrupts hydrogen bonding with waterR1

Quick check

1. Explain why iodine dissolves in hexane but not well in water.

Answers: 1. I2 is non-polar; London forces with hexane

20 Giant covalent structures

Diamond, graphite, silicon dioxide.

In giant covalent (network) structures, atoms are linked by covalent bonds throughout. Diamond: each C bonded to four others tetrahedrally, very hard, high melting point, non-conductor. Graphite: layers of hexagonal rings, each C bonded to three others, delocalized electrons allow conduction; weak forces between layers make it soft and slippery. SiO2: each Si bonded to four O, each O to two Si.

	Diamond	Graphite	Graphene	C60 fullerene
Bonds per C	4	3	3	3
Conducts?	no	yes	yes (excellent)	poorly
Structure	3D tetrahedral network	layers	single layer	molecular sphere

WORKED EXAMPLE 14 Graphite (standard)

Explain why graphite conducts electricity and is used as a lubricant.

Solution

Each C uses three electrons in bonds; the fourth is delocalizedR1

Layers held by weak London forces slide over each otherR1

Quick check

1. Why does silicon dioxide have a very high melting point?

Answers: 1. many strong covalent bonds must be broken

21 Chromatography and intermolecular forces

Separating by polarity.

In chromatography, the stationary phase (paper or polar silica) and the mobile phase (solvent) interact differently with each component through intermolecular forces. Polar compounds are held more strongly by a polar stationary phase and travel less far (lower Rf) with a non-polar solvent.

WORKED EXAMPLE 15 Predicting Rf (standard)

On a silica plate with hexane as solvent, which travels further: hexane-soluble naphthalene or polar benzoic acid?

Solution

Naphthalene: weak interaction with polar silicaR1

It has the higher RfA1

22 Resonance structures (introduction)

Delocalized electrons.

When more than one valid Lewis formula can be drawn, the real structure is a resonance hybrid. In the carbonate ion, the three C–O bonds are identical (bond order 4/3), not one double and two single. Delocalization stabilizes the ion.

WORKED EXAMPLE 16 Carbonate (standard)

Explain why all C–O bonds in CO32− have the same length.

Solution

Three resonance structures with the double bond in different positionsR1

Electrons delocalized: all bonds identical, intermediate between single and doubleR1

Quick check

1. Draw two resonance structures of the nitrite ion, NO2−.

Answers: 1. O=N–O− and −O–N=O

23 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 17 Mixed problem (exam level)

For NF3 and BF3: (a) draw Lewis formulas; (b) predict shapes and bond angles; (c) state which is polar.

Solution

(a) NF3: N with one lone pair; BF3: B with no lone pairsA1

(b) NF3 trigonal pyramidal 107°; BF3 trigonal planar 120°A1 A1

Quick check

1. Explain why propanone (56 °C) boils lower than propan-1-ol (97 °C).

Answers: 1. propan-1-ol forms hydrogen bonds; propanone only dipole–dipole

24 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 18 Paper 1 question (standard)

Which molecule is linear? A. H2O B. SO2 C. CO2 D. O3

Solution

C: two domains, no lone pairs on CA1

WORKED EXAMPLE 19 Paper 2 question (exam level)

(a) Draw the Lewis formula of ethanoic acid. (b) Explain why ethanoic acid has a higher boiling point than ethanal. (c) Explain why ethanoic acid dissolves in water.

Solution

(a) CH3–C(=O)–O–H with lone pairs on OA1

(b) hydrogen bonding between ethanoic acid moleculesR1

Quick check

1. State the bond angle in ethene.

Answers: 1. 120°

25 Allotropes of carbon in technology

Structure gives properties.

Diamond is used in cutting tools; graphite in electrodes, pencils and lubricants; graphene (a single layer of graphite) is extremely strong and an excellent conductor, with uses in electronics and composites; fullerenes and nanotubes are being developed for drug delivery and materials.

WORKED EXAMPLE 20 Explaining a use (standard)

Explain why diamond is used in drill tips.

Solution

Each C covalently bonded to four others in a rigid 3D networkR1

Extremely hardA1

In the real world

Carbon nanotubes are stronger than steel by mass and conduct heat and electricity well; they are used to strengthen sports equipment such as bicycle frames and tennis rackets.

26 Common misconceptions and links

Students often think…	Actually…
boiling breaks covalent bonds	boiling overcomes intermolecular forces
a molecule with polar bonds is always polar	symmetry can cancel dipoles
lone pairs don't affect shape	they repel more strongly and reduce bond angles
benzene has alternating single and double bonds	electrons are delocalized; all C–C bonds equal

How S2.2 connects to the rest of the course

Topic	Connection
S2.1 Ionic model	electronegativity difference
S1.1 Particulate nature	changes of state
S3.2 Functional groups	organic molecules
S2.4 Materials	bonding triangle
R1.2 Energy cycles	bond enthalpies

27 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

Figure 4. Boiling points of group 15 hydrides (NH3 to SbH3).

(a) Draw the Lewis structure of NH3. [1]

(b) Predict the shape and bond angle of NH3. [2]

(d) Explain why ammonia has a higher boiling point than phosphine (PH3). [3]

(e) Compare the electrical conductivity of diamond and graphite. [2]

(f) State the hybridization of the C atoms in ethene and the number of σ and π bonds. [2]

Markscheme

Part	Answer	Marks
(a)	N with three N–H bonds and one lone pair	A1
(b)	trigonal pyramidal	A1
	about 107°	A1
(c)	polar N–H bonds	R1
	asymmetric shape: dipoles do not cancel	R1
(d)	NH3 forms hydrogen bonds	R1
	PH3 only London / dipole-dipole forces	R1
	hydrogen bonds stronger; more energy needed	R1
(e)	diamond: no delocalized electrons; non-conductor	R1
	graphite: delocalized electrons between layers; conducts	R1
(f)	sp2	A1
	5 σ and 1 π	A1

Examiner's comments. In (d), weaker answers said hydrogen bonds 'break' during boiling without comparing the forces in the two substances.

28 Check your understanding

Section A: multiple choice

1. The shape of CH4 is

A. tetrahedral B. square planar C. trigonal pyramidal D. bent

2. The bond angle in H2O is about

A. 109.5° B. 120° C. 104.5° D. 180°

3. Which is non-polar?

A. H2O B. CCl4 C. NH3 D. CHCl3

4. Hydrogen bonding occurs in

A. HCl B. CH4 C. H2S D. HF

5. Each carbon in diamond is bonded to

A. three carbons B. four carbons C. two carbons D. six carbons

6. A coordination bond is found in

A. NH4+ B. CH4 C. Cl2 D. H2O

7. The shape of SF6 is

A. trigonal bipyramidal B. square planar C. tetrahedral D. octahedral

8. A triple bond contains

A. three σ bonds B. two σ and one π C. one σ and two π bonds D. three π bonds

Section B: structured

9. Explain the shapes of CO2, BF3 and NH3. [3]

10. Explain the origin of London forces. [3]

11. Describe and explain the properties of graphite. [4]

12. Explain resonance in the carbonate ion. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. CO2: 2 domains, linear; BF3: 3 domains, trigonal planar; NH3: 4 domains with one lone pair, pyramidal.

10. Random electron movement creates instantaneous dipole; induces dipole in neighbour; attraction.

11. Layers of hexagons; delocalized electrons conduct; weak forces between layers slide; high mp.

12. Three equivalent Lewis structures; electrons delocalized; all C–O bonds equal, bond order 1⅓.

29 Summary and topic map

Idea	Remember
Covalent bond	shared pair attracted to both nuclei
Lewis structures	bonding and lone pairs; octet
Bond order	higher: shorter and stronger
VSEPR	domains repel; lone pairs repel more
Polarity	electronegativity; symmetry
Giant covalent	diamond, graphite, graphene, SiO2; C60 molecular
Intermolecular forces	London < dipole-dipole < hydrogen bonding
HL	formal charge; resonance; 5 and 6 domains; σ, π; hybridization

Before the exam, make sure you…

•count domains, not atoms
•state shape and angle
•identify forces before explaining boiling points
•distinguish intra- and intermolecular forces

The Covalent Model on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 5. Topic map for The Covalent Model.

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