Bonding: AQA A-level Chemistry

Understand ionic, covalent and metallic bonding, molecular shapes, bond angles, polarity and intermolecular forces.

Created by Dave Miller, Chemistry teacher, tutor and current AQA examiner with 15+ years of teaching experience. Independent resources for AQA and OCR A.

Ionic bonding & formulae

7 practice questions

Ion formation and ionic bonding

Learning goalDistinguish electron transfer from the attraction between ions.

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There are two ideas to keep separate. Electron transfer makes the ions. The attraction between the resulting positive and negative ions is the ionic bonding.

A metal atom loses electrons and becomes a positive ion. A non-metal atom gains electrons and becomes a negative ion. In a giant ionic lattice, each ion attracts surrounding ions of the opposite charge.

In a dot-and-cross diagram, use dots and crosses consistently to show where the electrons came from. Show outer-shell electrons, put each ion in square brackets and write its charge outside. Check both electron totals and the ratio of ions.

Ion formation: Metals lose electrons to form positive ions; non-metals gain electrons to form negative ions.
Ionic bonding: Strong electrostatic attraction between oppositely charged ions in a lattice.

Group	Electron change	Typical ion charge	Example
1	Loses 1	1+	Na⁺
2	Loses 2	2+	Mg²⁺
3	Loses 3	3+	Al³⁺
5	Gains 3	3−	N^3-
6	Gains 2	2−	O^2-
7	Gains 1	1−	Cl^-

EXAM TRAPElectron transfer makes the ions. It is not the definition of the ionic bond.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 1

Predict the ions formed by potassium, magnesium, nitrogen and sulfur atoms. [4]
Explain why “ionic bonding is the transfer of electrons” is not a correct definition. Give the correct bonding description. [3]

Total: 7 marks

Show answer & marking points

K⁺ [1]; Mg²⁺ [1]; N^3- [1]; S^2- [1].
Electron transfer forms the ions [1]. Ionic bonding is electrostatic attraction [1] between oppositely charged ions in a lattice [1].

✓ DONE WHENYou can predict common ion charges and define ionic bonding precisely.

Ionic formulae and compound ions

Learning goalConstruct neutral ionic formulae using charges and brackets.

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An ionic formula must have no overall charge. Choose the smallest whole-number ratio of ions that makes the positive and negative charges balance.

For example, one 2+ ion needs two 1- ions. Use the charges to find the ratio, then write the subscripts in the formula.

A compound ion stays together as a group. If you need more than one of that ion, put brackets around its formula before adding the subscript. The table below gives common ions to recognise.

Compound ion	Formula
Sulfate	SO₄^2-
Hydroxide	OH^-
Nitrate	NO₃^-
Carbonate	CO₃^2-
Ammonium	NH₄⁺

Worked example

Aluminium sulfate

Construct the formula from Al³⁺ and SO₄^2-.

Worked answer

Two Al³⁺ ions give +6.
Three sulfate ions give −6.
Keep sulfate intact and use brackets: Al₂(SO₄)₃.
Check neutrality and the smallest ratio.

Al₂(SO₄)₃

Ions	Ratio	Formula
Mg²⁺ and Cl^-	1 : 2	MgCl₂
Al³⁺ and O^2-	2 : 3	Al₂O₃
Ca²⁺ and NO₃^-	1 : 2	Ca(NO₃)₂
Ca²⁺ and O^2-	1 : 1	CaO

EXAM TRAP“Drop and swap” can leave an unsimplified formula. Use brackets for repeated compound ions, never alter their internal formulae.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 2

State the formula and charge of sulfate, hydroxide, nitrate, carbonate and ammonium ions. [5]
Write the formulae of barium nitrate, aluminium sulfate, ammonium carbonate, magnesium hydroxide and aluminium oxide. [5]

Total: 10 marks

Show answer & marking points

SO₄^2-; OH^-; NO₃^-; CO₃^2-; NH₄⁺. One mark each.
Ba(NO₃)₂; Al₂(SO₄)₃; (NH₄)₂CO₃; Mg(OH)₂; Al₂O₃. One mark each.

Exam-style question 3

The ionic compounds X₂(SO₄)₃ and Y(OH)₂ contain simple ions X and Y.

Deduce the charges on X and Y. [2]
Write the formula of X nitrate. [1]
Write the formula of Y carbonate. [1]
Write the formula of ammonium sulfate. [1]
A student writes X₃SO₄. Explain why this is incorrect and give the correct sulfate formula. [2]

Total: 7 marks

Show answer & marking points

X³⁺ [1]; Y²⁺ [1].
X(NO₃)₃ [1].
YCO₃ [1].
(NH₄)₂SO₄ [1].
X₃SO₄ is not electrically neutral [1]. X₂(SO₄)₃ is correct [1].

✓ DONE WHENYou can balance charges, use brackets and simplify the ratio.

Giant ionic lattices and conductivity

Learning goalExplain conductivity from the mobility of ions.

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An ionic solid is a giant repeating arrangement of positive and negative ions. Strong electrostatic attractions act throughout the lattice, so a lot of energy is needed to separate the ions on melting.

For conductivity, ask whether charged particles can move. In the solid, the ions are held in fixed positions. When molten, the ions can move and carry charge.

Many ionic compounds dissolve in water, although not all do. If an ionic compound forms an aqueous solution, its mobile ions can carry charge. Do not explain ionic conductivity using moving electrons.

Giant ionic lattice: A regular repeating three-dimensional arrangement of oppositely charged ions.

State	Ions	Conducts?
Solid	Fixed in lattice positions	No: no mobile charge carriers
Molten	Mobile	Yes: ions carry charge
Aqueous, if dissolved	Hydrated and mobile	Yes: ions carry charge

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 4

Describe what is meant by a giant ionic lattice. [2]
Explain why magnesium oxide has a high melting point. [3]
Explain why solid magnesium oxide does not conduct electricity but molten magnesium oxide does. [3]

Total: 8 marks

Show answer & marking points

A regular, repeating three-dimensional arrangement [1] of oppositely charged ions [1].
A giant ionic lattice [1] with strong electrostatic attractions between Mg²⁺ and O^2- ions throughout [1]; much energy is needed to overcome these attractions [1].
Solid ions are fixed in lattice positions [1]. Molten ions are mobile [1] and carry charge [1].

✓ DONE WHENYou can explain solid and molten conductivity by naming the charged particles and their mobility.

Explaining ionic melting points

Learning goalLink ionic structure, attraction and energy to melting point.

🔓 CRACK IT

A high melting point needs an explanation of which particles attract and why separating them needs energy. For an ionic compound, those particles are oppositely charged ions.

Higher ion charges and smaller ionic radii generally strengthen the electrostatic attraction. More energy is then needed to overcome the attractions, so the melting point tends to be higher.

Use the actual ions in a comparison. Saying that one substance has 'stronger bonds' without identifying the attraction leaves out the chemistry.

Structure → particles → attraction → energy → property

Worked example

Why sodium chloride has a high melting point

Explain why sodium chloride has a high melting point. Refer to its structure, the particles present and the attractions between them.

Worked answer

NaCl has a giant ionic lattice.
Strong electrostatic attractions act between oppositely charged Na⁺ and Cl^- ions throughout the lattice.
Much energy is needed to overcome these attractions.

Therefore sodium chloride has a high melting point.

Worked example

Comparing ion charges: MgO and NaF

MgO and NaF have similar ion sizes and comparable lattice arrangements. Why does MgO have the higher melting point?

Worked answer

Both have giant ionic lattices. MgO contains Mg²⁺ and O^2-; NaF contains Na⁺ and F^-.
The ion sizes are similar, so focus on charge: the ions in MgO have larger charges.
The electrostatic attractions between opposite ions are stronger in MgO.
More energy is needed to overcome these attractions on melting.

Higher charges → stronger electrostatic attraction → more energy → higher melting point.

Worked example

Comparing ion sizes: NaCl and KCl

NaCl and KCl have the same ion charges and comparable lattice arrangements. Na⁺ is smaller than K⁺. Why does NaCl have the higher melting point?

Worked answer

Both contain a 1+ metal ion and the same Cl^- ion, so charge does not explain the difference.
The smaller Na⁺ ion allows the centres of opposite ions to be closer together.
The electrostatic attraction between the ions is stronger in NaCl.
More energy is needed to overcome these attractions on melting.

Smaller ions, with charge unchanged → stronger electrostatic attraction → higher melting point.

EXAM TRAP“Lots of strong bonds” does not name the ions or the electrostatic attraction.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Quick check 5

Which has the higher melting point, CaO or KCl? Explain your answer. [3]

Total: 3 marks

Show answer & marking points

CaO has the higher melting point. Ca²⁺ and O^2- have larger charges than K⁺ and Cl^- [1], giving stronger electrostatic attractions between oppositely charged ions [1]. More energy is required to overcome these attractions when CaO melts [1].

Quick check 6

NaF and NaCl have comparable lattice arrangements. Explain why NaF has a higher melting point than NaCl. [3]

Total: 3 marks

Show answer & marking points

The smaller F^- ion allows the centres of opposite ions to be closer together [1]. With the same ion charges, electrostatic attractions are stronger in NaF [1], so more energy is needed to overcome them on melting [1].

Exam-style question 7

A solid oxide M₂O₃ melts at 2050 °C, does not conduct as a solid and conducts when molten.

Deduce the charge on M ions. [1]
Deduce the structure type. [1]
Explain all three physical-property observations. [5]
Write the formulae of M nitrate and M hydroxide. [2]

Total: 9 marks

Show answer & marking points

M³⁺ [1].
Giant ionic [1].
A giant lattice of M³⁺ and O^2- ions [1]; strong electrostatic attractions [1] require much energy to overcome [1]. Solid ions are fixed, so no conduction [1]; molten ions are mobile and carry charge [1].
M(NO₃)₃ [1]; M(OH)₃ [1].

✓ DONE WHENYou can explain high melting point using ions, electrostatic attraction and energy.

Covalent & dative bonds

5 practice questions

Covalent bonds and dot-and-cross diagrams

Learning goalRepresent shared pairs using dots, crosses and bond lines.

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In a covalent bond, two atoms share a pair of electrons. The shared electrons are attracted to both positive nuclei, holding the atoms together.

A single bond contains one shared pair, a double bond two and a triple bond three. In a dot-and-cross diagram, put every shared pair in the overlap between the atoms.

A lone pair is not shared in a bond. Show these electrons outside the overlap, and use dots and crosses consistently to show their atom of origin. Count all the outer-shell electrons before finishing.

Single covalent bond: A shared pair of electrons attracted to both positive nuclei.

Bond	Shared pairs	Line example
Single	1	H–H
Double	2	O=O
Triple	3	N≡N

EXAM TRAPCovalent bonds within a molecule remain intact when a molecular substance melts or boils. Forces between molecules are overcome.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 8

Define a single covalent bond. [2]
State the number of shared pairs in Cl₂, O₂ and N₂ and give the line representation of each molecule. [3]
State two pieces of information shown by a dot-and-cross diagram. [2]

Total: 7 marks

Show answer & marking points

A shared pair of electrons [1], attracted to both positive nuclei [1].
Cl₂: one shared pair, Cl–Cl [1]; O₂: two shared pairs, O=O [1]; N₂: three shared pairs, N≡N [1].
The outer-shell electrons [1] and the atom from which each electron originated [1].

Exam-style question 9

Draw a dot-and-cross diagram for Cl₂, showing outer-shell electrons only. [3]
Draw a dot-and-cross diagram for O₂, showing outer-shell electrons only. [3]

Total: 6 marks

Show answer & marking points

One shared pair containing one dot and one cross [1]; three lone pairs on each chlorine [1]; consistent electron-origin symbols [1].
Two shared pairs, each containing one dot and one cross [1]; two lone pairs on each oxygen [1]; consistent electron-origin symbols [1].

✓ DONE WHENYou can draw complete outer-shell diagrams and convert shared pairs into bond lines.

Dative bonds: donor to acceptor

Learning goalIdentify the donated lone pair and draw the arrow correctly.

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A covalent bond contains a shared pair of electrons. Usually, each atom supplies one electron. In a dative covalent bond, one atom supplies both.

The atom supplying the pair is the donor. The species receiving it is the acceptor. Draw the arrow from the donor's lone pair to the acceptor.

For NH₃ reacting with H⁺, nitrogen supplies its lone pair to the hydrogen ion. Once NH₄⁺ has formed, all four N-H bonds are equivalent. The origin of the electrons does not make one of those bonds different afterwards.

Dative covalent bond: A shared pair of electrons with both electrons supplied by one atom.

Role	NH₃ + H⁺ → NH₄⁺
Donor	Nitrogen in NH₃ has an available lone pair
Acceptor	H⁺ accepts the pair

EXAM TRAPThe arrow follows the electron pair: donor → acceptor.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 10

Define a dative covalent bond. [2]
For NH₃ + H⁺ → NH₄⁺, identify the donor and acceptor, state the arrow direction and state how the four N–H bonds compare after formation. [4]
In F^- + BF₃ → BF₄^-, identify the electron-pair donor and state the dative-arrow direction. [2]

Total: 8 marks

Show answer & marking points

A shared pair of electrons [1] with both electrons supplied by one atom [1].
NH₃, through nitrogen, is the donor [1]; H⁺ is the acceptor [1]; the arrow starts at the nitrogen lone pair and points to H⁺ [1]; all four N–H bonds become equivalent [1].
F^- is the donor [1]; the arrow points from a fluoride lone pair towards boron [1].

Exam-style question 11

Ammonia forms a dative covalent bond with AlCl₃. The electron-deficient aluminium atom accepts a lone pair.

Identify the electron-pair donor and acceptor. [2]
Draw the dative arrow on H₃N and AlCl₃. [2]
Explain why ammonia can act as the donor. [1]
Explain why an arrow from aluminium towards nitrogen would be incorrect. [2]

Total: 7 marks

Show answer & marking points

NH₃ through N is the donor [1]; AlCl₃ through Al is the acceptor [1].
The arrow starts at the N lone pair [1] and points towards Al: H₃N → AlCl₃ [1].
N has an available lone pair [1].
N supplies both electrons, not Al [1]; an Al → N arrow reverses donor-to-acceptor direction [1].

Exam-style question 12

Fluoride ions form BF₄^- by donating a lone pair to BF₃. A salt contains Mg²⁺ and BF₄^- ions.

Identify the dative donor and acceptor and describe the arrow fully. [4]
Write the salt’s formula. [1]
State the bonding within BF₄^- and between Mg²⁺ and BF₄^- ions in the solid. [2]
Explain why the solid does not conduct but the molten salt does. [3]
Explain why a high melting point is expected. [3]

Total: 13 marks

Show answer & marking points

F^- donor [1]; B in BF₃ acceptor [1]; arrow starts at fluoride lone pair [1] and points to B [1].
Mg(BF₄)₂ [1].
Covalent within BF₄^- [1]; ionic between Mg²⁺ and BF₄^- [1].
Solid ions are fixed [1]; molten ions are mobile [1] and carry charge [1].
Giant ionic lattice [1]; strong electrostatic attractions between opposite ions [1] require much energy to overcome [1].

✓ DONE WHENYou can identify donor and acceptor and draw an arrow from the lone pair to the acceptor.

Metallic bonding

2 practice questions

Metallic bonding and mobile electrons

Learning goalExplain metallic conductivity and melting using delocalised electrons.

🔓 CRACK IT

A metal contains a giant lattice of positive ions surrounded by delocalised electrons. 'Delocalised' means the electrons are not restricted to one atom or one bond.

The attraction between the positive ions and these electrons holds the metal together. Strong attractions often give metals high melting and boiling points.

The delocalised electrons can move and carry charge in both solid and liquid metals. The positive ions do not travel through the solid to conduct electricity.

Metals do not normally dissolve as intact metallic lattices in molecular solvents. A metal reacting with a substance is a different process.

Metallic bonding: Strong electrostatic attraction between positive metal ions and delocalised electrons.
Delocalised electron: An electron not confined to one atom or one bond, able to move through the structure.

EXAM TRAPName the charge carrier as delocalised electrons, rather than merely “electrons” or “free electrons”.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 13

Define metallic bonding. [3]
Explain why solid magnesium conducts electricity. [2]
Explain why magnesium has a relatively high melting point. [2]

Total: 7 marks

Show answer & marking points

Strong electrostatic attraction [1] between positive metal ions [1] and delocalised electrons [1].
Delocalised electrons are mobile [1] and carry charge through the solid lattice [1].
Strong attractions between positive ions and delocalised electrons [1] require substantial energy to overcome [1].

✓ DONE WHENYou can define metallic bonding and explain conductivity and melting point.

Sodium chloride and magnesium

Learning goalCompare ionic and metallic charge carriers.

🔓 CRACK IT

Sodium chloride and magnesium both have giant structures with strong electrostatic attractions. The difference is what is attracting what.

In sodium chloride, positive sodium ions attract negative chloride ions. In magnesium, positive metal ions attract delocalised electrons.

This explains their different conductivity patterns. Magnesium already has mobile electrons in the solid. Solid sodium chloride has ions, but they cannot move through the lattice until it melts or dissolves.

Feature	Sodium chloride	Magnesium
Particles	Na⁺ and Cl^- ions	Mg²⁺ ions and delocalised electrons
Solid	Does not conduct: ions fixed	Conducts: delocalised electrons mobile
Molten	Conducts: ions mobile	Conducts: delocalised electrons mobile

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 14

State the particles present in barium fluoride, BaF₂, and in potassium metal. [2]
Explain why solid potassium conducts electricity but solid barium fluoride does not. [4]
Both substances conduct when molten. State the mobile charge carrier in each liquid. [2]

Total: 8 marks

Show answer & marking points

BaF₂: Ba²⁺ and F^- ions [1]. Potassium: K⁺ ions and delocalised electrons [1].
In solid BaF₂ the ions are fixed [1], so there are no mobile charge carriers [1]. Potassium has mobile delocalised electrons [1] that carry charge [1].
Ba²⁺ and F^- ions in molten BaF₂ [1]; delocalised electrons in molten potassium [1].

✓ DONE WHENYou can distinguish mobile ions from mobile delocalised electrons.

Structures & properties

12 practice questions

Four structure types: particles and attractions

Learning goalIdentify what is present and what holds it together.

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Start a structure question by identifying the particles: ions, atoms or molecules. Then name the attraction between them. This stops you using the right explanation for the wrong structure.

Choose the structure type from the evidence: giant ionic, metallic, giant covalent or simple molecular.
For melting or boiling, name the attractions that must be overcome and connect their strength to the energy needed.
For conductivity, identify the charged particles and say whether they can move.
Use all the evidence together. A solid conductor could be a metal or graphite, so one property alone may not identify it.

Structure	Particles	Bond or attraction
Giant ionic	Positive and negative ions	Electrostatic attractions between oppositely charged ions
Metallic	Positive metal ions and delocalised electrons	Electrostatic attraction between positive ions and delocalised electrons
Giant covalent	Atoms	Many strong covalent bonds throughout a network
Simple molecular	Molecules	Forces between molecules

EXAM TRAPDo not jump straight from the structure name to the property. The explanation lies in the particles and attractions.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 15

For each structure below, state the particles present and the bond or attraction that holds the structure together. Give one correct entry in each empty cell. [8]

Structure	Particles present	Bond or attraction
Giant ionic	…	…
Metallic	…	…
Giant covalent	…	…
Simple molecular	…	…

Total: 8 marks

Show answer & marking points

Structure	Particles present	Bond or attraction
Giant ionic	Positive and negative ions	Electrostatic attractions between oppositely charged ions
Metallic	Positive metal ions and delocalised electrons	Electrostatic attraction between positive ions and delocalised electrons
Giant covalent	Atoms	Many strong covalent bonds throughout a network
Simple molecular	Molecules	Forces between molecules

Award one mark for each completed cell: 8 marks.

Exam-style question 16

Repair each statement using precise particle language.

“Electron transfer is the ionic bond.” [3]
“Molten iodine contains iodine atoms because its covalent bonds have broken.” [2]
“Solid magnesium conducts because Mg²⁺ ions move through the lattice.” [3]

Total: 8 marks

Show answer & marking points

Transfer forms positive and negative ions [1]. Ionic bonding is electrostatic attraction [1] between oppositely charged ions in a lattice [1].
Molten iodine still contains I₂ molecules [1]; only forces between molecules are overcome, not I–I bonds [1].
Mg²⁺ ions are fixed in the solid [1]; delocalised electrons are mobile [1] and carry charge [1].

✓ DONE WHENYou can identify particles and attractions in all four crystal types.

Identifying structures from physical properties

Learning goalUse melting and conductivity evidence to identify unfamiliar solids.

🔓 CRACK IT

Treat the properties as clues that must fit the same structure. A high melting point suggests strong attractions, but it does not by itself tell you which particles are present.

Conductivity is especially useful. A solid metal conducts through mobile electrons. An ionic compound normally conducts only when its ions can move, such as when molten or dissolved.

Check the full set of observations against the table. Graphite is an important exception to the idea that giant covalent substances never conduct.

Structure	Melting point	Conductivity
Giant ionic	High	Solid: no; molten/aqueous: yes, via ions
Metallic	Often high	Solid and liquid: yes, via delocalised electrons
Giant covalent	Very high	Usually no; graphite is an exception
Simple molecular	Usually low relative to giant structures	No mobile charged particles in the pure substance

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 17

Four solids give the following data.

Solid	Melting point / °C	Conducts as solid?	Conducts when molten?
W	801	No	Yes
X	650	Yes	Yes
Y	3550	No	No
Z	114	No	No

Identify the most likely structure type of W, X, Y and Z. [4]
For each solid, quote the decisive evidence supporting your choice. [4]

Total: 8 marks

Show answer & marking points

W: giant ionic; X: metallic; Y: giant covalent; Z: simple molecular. One mark each.
W conducts only when molten, consistent with mobile ions [1]. X conducts in both states, consistent with mobile delocalised electrons [1]. Y has a very high melting point and no mobile charge carriers [1]. Z has a low melting point and does not conduct [1].

Exam-style question 18

Three samples are sodium chloride, graphite and iodine in an unknown order.

Sample	Thermal observation	Conducts as solid?	Conducts when molten?
P	Melts at 114 °C	No	No
Q	Melts at 801 °C	No	Yes
R	Remains solid above 3000 °C	Yes	Not tested

Identify P, Q and R. [3]
State the structure type of each. [3]
For each, explain one decisive observation using particles, attractions or mobile charge carriers. [3]

Total: 9 marks

Show answer & marking points

P: iodine; Q: sodium chloride; R: graphite. One mark each.
P: simple molecular; Q: giant ionic; R: giant covalent. One mark each.
P melts at relatively low temperature because forces between I₂ molecules are overcome [1]. Q’s molten ions carry charge [1]. R has mobile delocalised electrons [1].

Exam-style question 19

A contains atoms joined in a continuous three-dimensional covalent network. B contains discrete X₈ molecules. Neither conducts electricity.

Classify their structures. [2]
Explain why A has a much higher melting point than B. [5]
Explain why neither conducts. [2]

Total: 9 marks

Show answer & marking points

A: giant covalent [1]; B: simple molecular [1].
A has many strong covalent bonds throughout a network [1]; many must be broken [1], requiring much energy [1]. B has discrete X₈ molecules [1]; melting overcomes forces between them, not X–X bonds [1].
A has no mobile charged particles [1]; neither does B [1].

Exam-style question 20

Samples A–D are diamond, graphite, magnesium and iodine in an unknown order.

Sample	Physical evidence
A	Conducts as solid and liquid; melts at 650 °C
B	Conducts as solid; very high melting point; soft, layered
C	Does not conduct; very high melting point; extremely hard
D	Does not conduct; melts at 114 °C

Identify A–D. [4]
Compare B and C: explain their high melting points and why only B conducts. [5]
Explain why D melts far below C, without referring to weak covalent bonds. [3]

Total: 12 marks

Show answer & marking points

A: magnesium; B: graphite; C: diamond; D: iodine. One mark each.
Both giant covalent with many strong bonds [1], requiring much energy to break [1]. Graphite has delocalised electrons [1] that are mobile and carry charge [1]. Diamond’s outer electrons are used in bonds, so it has no mobile charge carriers [1].
D has discrete I₂ molecules [1]. Melting overcomes forces between molecules [1], requiring much less energy than breaking diamond’s covalent bonds [1].

✓ DONE WHENYou can justify the most likely structure from property evidence.

Diamond: a rigid covalent network

Learning goalRelate diamond’s structure to hardness, melting and conductivity.

🔓 CRACK IT

Diamond is one giant covalent network, rather than a collection of small carbon molecules. Each carbon atom forms four strong covalent bonds in a rigid three-dimensional arrangement.

Melting requires many strong covalent bonds to be broken, so a lot of energy is needed. The rigid network also makes diamond very hard.

All four outer electrons of each carbon are used in bonds. There are no mobile charge carriers, so diamond does not conduct electricity.

EXAM TRAPDiamond is not made of discrete molecules. Do not explain its properties using intermolecular forces.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 21

State the number of covalent bonds formed by each carbon in diamond. [1]
Explain diamond’s high melting point and lack of electrical conductivity. [4]

Total: 5 marks

Show answer & marking points

Four [1].
A giant covalent network [1] has many strong covalent bonds requiring much energy to break [1]. All four outer electrons are used in bonds [1], leaving no mobile charge carriers [1].

✓ DONE WHENYou can explain diamond’s hardness, very high melting point and lack of conductivity.

Graphite: layers and delocalised electrons

Learning goalCompare bonds within layers with forces between them.

🔓 CRACK IT

Graphite contains flat layers of carbon atoms. Each carbon makes three strong covalent bonds, leaving one electron per carbon delocalised.

The strong bonds within the layers explain the high melting point. The delocalised electrons can move along the layers and carry charge, so graphite conducts electricity.

Weak van der Waals attractions between the layers allow them to slide over one another. This explains softness; it is a different explanation from the one for melting point.

Graphite is a giant covalent structure. The diagram's forces between layers do not mean that each layer is a collection of small molecules.

EXAM TRAPSliding layers do not require the strong covalent bonds within each layer to break.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 22

State the structure type shared by diamond and graphite. [1]
State how many covalent bonds each carbon atom forms in diamond and graphite. [2]
Explain why both have very high melting points. [3]
Explain why graphite conducts electricity but diamond does not. [3]
Explain why graphite’s layers can slide over one another. [1]

Total: 10 marks

Show answer & marking points

Giant covalent [1].
Diamond: four [1]; graphite: three [1].
Many strong covalent bonds [1] extend throughout giant structures [1], so much energy is required to break them [1].
Graphite has one mobile delocalised electron per carbon [1] that carries charge [1]. Diamond’s four outer electrons are used in bonds, so it has no mobile charge carriers [1].
Weak forces between the layers allow them to slide [1].

Exam-style question 23

C consists of flat covalent layers. Each atom forms three covalent bonds and contributes one delocalised electron. Weak forces act between the layers.

Classify C’s structure. [1]
Predict and explain its melting point. [3]
Predict and explain its electrical conductivity. [2]
Predict and explain whether its layers can slide. [2]
Explain why C should not be described as simple molecular. [2]

Total: 10 marks

Show answer & marking points

Giant covalent [1].
Very high [1]; many strong covalent bonds within continuous layers [1] require much energy to break [1].
Conducts [1] because mobile delocalised electrons carry charge [1].
Layers can slide [1] because forces between them are weak [1].
Atoms form continuous covalent layers [1], not separate molecules [1].

✓ DONE WHENYou can compare diamond and graphite using bonding, electron mobility and layers.

Molecular crystals: iodine and ice

Learning goalExplain changes of state without breaking bonds inside molecules.

🔓 CRACK IT

Iodine and ice are made of separate molecules arranged in a solid. There are covalent bonds inside each molecule and attractions between neighbouring molecules.

Melting partly overcomes the attractions between molecules. Boiling separates the molecules further. The molecules themselves remain intact.

For iodine, name van der Waals forces between I₂ molecules. For ice, name hydrogen bonds between water molecules. Do not say that melting breaks the I-I or O-H covalent bonds.

Crystal	Particles	Within molecules	Between molecules
Iodine	I₂ molecules	Strong I–I covalent bonds	van der Waals forces
Ice	H₂O molecules	Strong O–H covalent bonds	Hydrogen bonds

Change	Energy
Melting or boiling	Energy absorbed to overcome attractions
Freezing or condensing	Energy released as attractions form

EXAM TRAPDo not call iodine’s covalent bonds weak or claim that melting ice breaks O–H bonds.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 24

A student writes: “Iodine melts at a low temperature because the weak covalent bonds in I₂ molecules break.”

Identify two errors. [2]
Explain correctly what happens when solid iodine melts. [3]
Explain why diamond melts at a much higher temperature than iodine. [2]

Total: 7 marks

Show answer & marking points

The I–I covalent bonds are not broken on melting [1]; iodine’s melting point is not low because those bonds are weak [1].
Iodine is a molecular crystal of I₂ molecules [1]. Melting overcomes forces between molecules [1]; the molecules remain intact [1].
Diamond is a giant covalent network. Melting iodine overcomes only weak intermolecular (van der Waals) forces between I₂ molecules [1]; breaking the strong C–C covalent bonds in diamond requires much more energy [1].

Exam-style question 25

State the particles present in ice. [1]
State which attractions are overcome when ice melts and which bonds are not broken. [2]
State whether melting absorbs or releases energy. [1]
Describe the attraction and energy changes when water freezes. [2]
Explain why each water molecule’s chemical identity is unchanged during melting. [1]

Total: 7 marks

Show answer & marking points

H₂O molecules [1].
Forces between water molecules are partly overcome [1]; O–H covalent bonds remain intact [1].
Energy is absorbed [1].
An ordered network of attractions forms between water molecules [1]; energy is released [1].
The O–H bonds remain intact, so the molecules remain H₂O [1].

✓ DONE WHENYou can identify molecules and the attractions involved in changes of state.

Drawing exact numbers of particles

Learning goalDraw the requested entity and count it accurately.

🔓 CRACK IT

Before drawing, underline what you have been asked to count: atoms, ions or molecules. The number of circles you need depends on that choice.

For I₂, draw separate touching pairs. One pair is one molecule, so four molecules need eight atoms.
For NaCl, show an alternating arrangement of touching positive and negative ions, with the charges labelled.
Count the requested entities again at the end. Check that the overall composition and charge match the substance.

EXAM TRAPFour I₂ molecules contain eight iodine atoms. A correct pattern with the wrong number of particles can lose marks.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 26

Draw exactly five I₂ molecules. [2]
Draw a touching, alternating lattice containing exactly four Na⁺ ions and four Cl^- ions. Show every charge. [3]
State the particle represented in each diagram. [1]

Total: 6 marks

Show answer & marking points

Five distinct I–I pairs [1], with ten atoms arranged as five molecules [1].
Exactly four Na⁺ and four Cl^- ions [1]; opposite charges alternate in a touching lattice [1]; all charges correct [1].
Molecules for iodine; ions for sodium chloride [1].

✓ DONE WHENYour drawing has the requested entity, exact count and correct charges.

Shapes & bond angles

8 practice questions

VSEPR: count charge clouds, then minimise repulsion

Learning goalCount bonding regions and lone pairs before naming a shape.

🔓 CRACK IT

The shape comes from electron regions repelling one another around the central atom. They arrange as far apart as possible. Count bonding regions and lone pairs before naming the shape.

Find the central atom's outer-shell electron number: for example, 5 for N or 6 for O.
For a suitable species with single bonds to H or halogens, add one electron for each attached atom.
Add electrons for a negative charge and subtract them for a positive charge.
Divide by two to get the total pairs. Subtract the bonding pairs to find the lone pairs.
Choose the parent electron arrangement. Then name the molecular shape using the positions of the atoms.

The shortcut has limits. With multiple bonds or a supplied electron diagram, count regions directly: a single, double or triple bond is one bonding region. Lone pairs count as regions too.

VSEPR: Valence Shell Electron Pair Repulsion: outer-shell electron pairs repel and arrange as far apart as possible.
Charge cloud: One region of electron density: each bond region and each lone pair counts as one.

Total pairs = (central valence electrons + singly bonded atoms + negative charge − positive charge) ÷ 2

Worked example

Ammonia versus ammonium

For NH₃ and NH₄⁺, calculate the total number of electron pairs around nitrogen, state how many are bonding pairs and lone pairs, and predict the shape.

Worked answer

NH₃: (5 + 3)/2 = 4 pairs; 3 bonding and 1 lone.
NH₄⁺: (5 + 4 − 1)/2 = 4 pairs; all 4 bonding.

NH₃: trigonal pyramidal. NH₄⁺: tetrahedral.

EXAM TRAPThis shortcut is for suitable singly bonded species. For multiple bonds or supplied electron diagrams, count regions directly: a single, double or triple bond is one bonding region.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 27

Explain why electron pairs around a central atom adopt particular positions in space. [2]
Place bond pair–bond pair, lone pair–bond pair and lone pair–lone pair repulsions in decreasing strength. [2]
Explain why a lone pair repels more strongly than a bonding pair. [2]

Total: 6 marks

Show answer & marking points

Charge clouds repel [1] and arrange as far apart as possible to minimise repulsion [1].
Lone pair–lone pair > lone pair–bond pair [1] > bond pair–bond pair [1].
A lone pair is attracted to only one nucleus [1], so its electron density occupies more space near the central atom and repels more strongly [1].

✓ DONE WHENYou can count regions, identify lone pairs and explain minimisation of repulsion.

Two and three charge clouds

Learning goalRecognise linear and trigonal-planar arrangements.

🔓 CRACK IT

With two electron regions and no lone pairs, the greatest separation is opposite sides of the central atom. The shape is linear, 180°.

With three electron regions and no lone pairs, the regions spread into one plane. The shape is trigonal planar, 120°.

Count regions, not individual shared pairs. Each double bond in CO₂ counts as one region, so CO₂ has two regions around carbon and is linear.

Species	Bonding regions	Shape	Angle
CO₂	2	Linear	180°
BeCl₂	2	Linear	180°
BF₃	3	Trigonal planar	120°
SO₃	3	Trigonal planar	120°

Worked example

Sulfur trioxide

Deduce SO₃’s shape from three bonding regions and no lone pairs.

State the bond angle and explain the arrangement of the bonding regions.

Worked answer

The three charge clouds repel equally into one plane.

Trigonal planar, 120°.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Quick check 28

Deduce CS₂’s shape and bond angle around carbon. [2]
Deduce BCl₃’s shape and angle around boron. [2]

Total: 4 marks

Show answer & marking points

Carbon forms two C=S double bonds. Each double bond is one bonding region; carbon has no lone pairs. The two regions repel as far apart as possible, giving a linear shape [1] and 180° bond angle [1].
Three bonding regions and no lone pairs: trigonal planar [1], 120° [1].

✓ DONE WHENYou can recognise two- and three-region shapes without counting double bonds twice.

Four bonding regions: tetrahedral shapes

Learning goalDraw four bonding regions in three dimensions.

🔓 CRACK IT

Four bonding regions spread out in three dimensions, giving a tetrahedral shape with angles of 109.5°. Four flat bonds at right angles would put the regions too close together.

CH₄, SiCl₄, NH₄⁺ and BF₄⁻ all have four bonding regions and no lone pairs around the central atom.

An ordinary line shows a bond in the plane of the paper.
A solid wedge shows a bond coming towards you.
A hashed wedge shows a bond going away from you.
For an ion, put brackets and the overall charge around the whole species.

EXAM TRAPDo not draw tetrahedral as four flat bonds at right angles.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 29

Use solid and hashed wedges where needed to show three-dimensional arrangement.

Draw CH₄. [2]
Draw NH₃, including its lone pair. [3]
Draw H₂O, including both lone pairs, and label the H–O–H angle. [3]

Total: 8 marks

Show answer & marking points

Four bonds in a tetrahedral arrangement [1], using valid wedge-and-dash representation [1].
Three bonds in a trigonal-pyramidal arrangement [1], one N lone pair [1], valid 3D representation [1].
Bent arrangement [1], two O lone pairs [1], 104.5° [1].

✓ DONE WHENYou can draw tetrahedral molecules and ions using valid wedge-and-dash bonds.

Lone pairs and compressed bond angles

Learning goalExplain how lone pairs affect shape and angle.

🔓 CRACK IT

A lone pair takes up more space near the central atom than a bonding pair. It is attracted to one nucleus rather than shared between two nuclei. It repels nearby bonding regions more strongly and pushes them closer together.

CH₄, NH₃ and H₂O each have four electron regions around the central atom. Replacing bonds with lone pairs changes the positions of the atoms and therefore the molecular shape.

CH₄: four bonds, no lone pairs; tetrahedral, 109.5°.
NH₃: three bonds, one lone pair; pyramidal, about 107°.
H₂O: two bonds, two lone pairs; bent/non-linear, about 104.5°.

Subtracting about 2.5° for each lone pair is a memory aid for this familiar comparison. It is not a universal formula for every molecule or parent arrangement.

Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair

Species	Bonding regions	Lone pairs	Shape	Angle
SO₂	2	1	Bent	Less than 120°
NH₃	3	1	Trigonal pyramidal	107°
H₂O	2	2	Bent	104.5°
H₃O⁺	3	1	Trigonal pyramidal	Use supplied data; do not automatically assume 107°

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 30

CH₄, NH₃ and H₂O each have four electron pairs around the central atom.

State the shape and bond angle of each molecule. [6]
Explain the angle decrease from CH₄ to NH₃ to H₂O. [2]

Total: 8 marks

Show answer & marking points

CH₄: tetrahedral, 109.5° [2]; NH₃: trigonal pyramidal, 107° [2]; H₂O: bent, 104.5° [2].
Lone pair–bond pair repulsion is stronger than bond pair–bond pair repulsion [1]. Water has two lone pairs, compared with ammonia’s one, compressing the bonding pairs further [1].

✓ DONE WHENYou can explain the familiar angle sequence and avoid treating 2.5° as a universal rule.

Five and six charge clouds

Learning goalRecognise trigonal-bipyramidal and octahedral arrangements.

🔓 CRACK IT

Five regions arrange in a trigonal bipyramid. Three equatorial positions lie in one plane; two axial positions point above and below it. Six regions arrange octahedrally.

Lone pairs still occupy positions in those arrangements, even though they are left out when naming the molecular shape. That gives the seesaw, T-shaped, linear, square-pyramidal and square-planar shapes in the table.

With five regions, lone pairs take equatorial positions to reduce the number of 90° repulsions. With six regions and two lone pairs, as in XeF₄, the lone pairs sit opposite one another.

Use the table's angles and any data supplied. Do not apply the methane/ammonia/water subtraction shortcut to these shapes.

Total regions	Bonding + lone	Example	Molecular shape	Angles
5	5 + 0	PCl₅	Trigonal bipyramidal	90°, 120°, 180°
5	4 + 1	SF₄	Seesaw	Distorted below ideal 90°, 120°, 180°
5	3 + 2	ClF₃	T-shaped	Approximately 90°, 180°, slightly compressed
5	2 + 3	XeF₂	Linear	180°
6	6 + 0	SF₆	Octahedral	90°, 180°
6	5 + 1	BrF₅	Square pyramidal	Approximately 90°, 180°
6	4 + 2	XeF₄	Square planar	90°, 180°

EXAM TRAPFive bonds are not a flat pentagon, and six bonds are not a flat hexagon.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 31

Complete all seven rows with the molecular shape and characteristic angle(s). Treat each multiple bond as one bonding region. [14]

Species	Bonding regions	Lone pairs	Shape	Angle(s)
CO₂	2	0	…	…
BF₃	3	0	…	…
CH₄	4	0	…	…
NH₃	3	1	…	…
H₂O	2	2	…	…
PCl₅	5	0	…	…
SF₆	6	0	…	…

Total: 14 marks

Show answer & marking points

Species	Bonding regions	Lone pairs	Shape	Angle(s)
CO₂	2	0	Linear	180°
BF₃	3	0	Trigonal planar	120°
CH₄	4	0	Tetrahedral	109.5°
NH₃	3	1	Trigonal pyramidal	107°
H₂O	2	2	Bent / non-linear	104.5°
PCl₅	5	0	Trigonal bipyramidal	90°, 120° (also 180° between axial bonds)
SF₆	6	0	Octahedral	90° (also 180° between opposite bonds)

One mark per correct shape and one per correct angle entry: 14 marks.

Exam-style question 32

In AX_mE_n, A is the central atom, X a bonded atom and E a lone pair on A.

Species	AXₘEₙ
A SF₄	AX₄E
B ClF₃	AX₃E₂
C XeF₂	AX₂E₃
D BrF₅	AX₅E

Determine the molecular shape of A–D. [4]
Explain why the lone pairs in A–C occupy equatorial positions. [2]
Give characteristic angles, using “less than” where lone-pair repulsion compresses ideal angles. [4]
Predict whether C and D have permanent dipoles and explain. [3]

Total: 13 marks

Show answer & marking points

A seesaw; B T-shaped; C linear; D square pyramidal. One mark each.
Equatorial positions [1] have fewer 90° interactions and minimise strong lone-pair repulsion [1].
SF₄: <90°, <120° and <180° [1]; ClF₃: <90° and <180° [1]; XeF₂: 180° [1]; BrF₅: approximately 90° and 180°, with distortions from lone-pair repulsion [1].
XeF₂ is linear [1], so its equal opposite dipoles cancel [1]. BrF₅ is asymmetric square pyramidal, so its dipoles do not cancel and it is polar [1].

✓ DONE WHENYou can name and justify the five- and six-region shapes, including lone-pair variants.

Deducing unfamiliar molecular shapes

Learning goalApply the region-counting method to unfamiliar species.

🔓 CRACK IT

An unfamiliar formula does not need a new method. Work out its electron regions, place any lone pairs, then look at where the atoms sit.

Count the bonding regions and lone pairs around the central atom.
Choose the parent electron arrangement from the total number of regions.
Place lone pairs where repulsion is minimised.
Ignore the lone-pair positions when naming the molecular shape, but keep their effect on the arrangement.

For XeF₄, six regions include four bonds and two opposite lone pairs. The atoms left in the middle plane form a square, so the shape is square planar.

Central atom → bonding regions → lone pairs → parent geometry → molecular shape

Species	Bonding regions	Lone pairs	Reasoning	Shape
NO₂⁺	2	0	Opposite regions	Linear
SO₂	2	1	Three-region parent; omit lone-pair position	Bent
XeF₄	4	2	Six-region parent; opposite lone pairs	Square planar

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 33

Use the supplied counts; each multiple bond counts as one bonding region.

Species	Bonding regions	Lone pairs
NH₄⁺	4	0
SO₂	2	1
I₃^-	2	3
XeF₄	4	2

Determine each molecular shape. [4]
State one characteristic angle for each. [4]
Draw XeF₄, showing the two Xe lone pairs relative to the square plane. [2]

Total: 10 marks

Show answer & marking points

NH₄⁺: tetrahedral; SO₂: bent; I₃^-: linear; XeF₄: square planar. One mark each.
109.5°; slightly less than 120°; 180°; 90°, respectively. One mark each.
Four F atoms in a square plane [1]; opposite lone pairs perpendicular to it [1].

Exam-style question 34

U is ClF₃ (AX₃E₂, 44 electrons per molecule). V is XeF₄ (AX₄E₂, 90 electrons per molecule). V has the higher melting point. A is the central atom, X a bonded atom and E a lone pair.

Determine U’s shape and the positions of its two lone pairs. [2]
Determine V’s shape and the relative positions of its lone pairs. [2]
Predict each molecule’s polarity and explain. [2]
Suggest why V can have the higher melting point despite being non-polar. [4]
Draw both molecules with all central lone pairs and appropriate 3D conventions. [3]

Total: 13 marks

Show answer & marking points

U is T-shaped [1], with two equatorial lone pairs in a trigonal-bipyramidal arrangement [1].
V is square planar [1], with two opposite lone pairs perpendicular to the plane [1].
U is polar: its T-shaped dipoles do not cancel [1]. V is non-polar: its square-planar dipoles cancel [1].
XeF₄ has more electrons [1], a more polarisable cloud [1] and stronger van der Waals forces [1], so more energy is needed [1]. This explains why a non-polar substance can melt higher; crystal packing can also affect melting points.
ClF₃: AX₃E₂ with two equatorial lone pairs [1]. XeF₄: AX₄E₂ with opposite lone pairs [1]. Correct 3D conventions and atom labels [1].

✓ DONE WHENYou can work from supplied bonding regions and lone pairs without guessing.

Electronegativity & polarity

4 practice questions

Electronegativity and polar bonds

Learning goalAssign partial charges using attraction for bonding electrons.

🔓 CRACK IT

In a covalent bond, both atoms attract the shared electrons. Electronegativity describes how strongly an atom attracts that bonding pair.

If one atom attracts the pair more strongly, the electrons are shared unequally. That atom becomes δ− and the other becomes δ+. These are partial charges, not the full charges of ions.

Use the difference between the supplied electronegativity values. A larger difference generally gives a more polar bond, but there is no single numerical cut-off that works for every bond.

Electronegativity: The power of an atom to attract the pair of electrons in a covalent bond.
Polar covalent bond: Unequal sharing of bonding electrons produces δ+ and δ− ends.

Electronegativity difference = |value for atom 1 − value for atom 2|

EXAM TRAPSubtract the electronegativities; do not average them. δ denotes partial, not full ionic charge.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 35

Define electronegativity. [2]
Define a polar covalent bond. [2]
State the three types of intermolecular force named in the AQA specification. [3]
State which type acts between all atoms and molecules. [1]

Total: 8 marks

Show answer & marking points

The power of an atom to attract the bonding pair of electrons [1] in a covalent bond [1].
A covalent bond with unequal distribution of bonding electrons [1], producing δ+ and δ− ends [1].
Induced dipole–dipole (van der Waals) forces [1]; permanent dipole–dipole forces [1]; hydrogen bonding [1].
Induced dipole–dipole (van der Waals) forces [1].

Exam-style question 36

Use Pauling electronegativity values: H 2.1, C 2.5, Br 2.8, Cl 3.0, O 3.5, F 4.0.

Add δ+ and δ− to H–Cl. [1]
Add δ+ and δ− to C=O. [1]
Add δ+ and δ− to O–H. [1]
Explain how partial charges arise in a polar covalent bond. [2]
Place H–F, H–Cl and H–Br in decreasing bond polarity. Justify the first position. [2]

Total: 7 marks

Show answer & marking points

H^δ+–Cl^δ− [1].
C^δ+=O^δ− [1].
O^δ−–H^δ+ [1].
The more electronegative atom attracts the bonding pair more strongly [1], producing unsymmetrical electron distribution and partial charges [1].
H–F > H–Cl > H–Br [1]; H–F has the largest electronegativity difference [1].

✓ DONE WHENYou can define electronegativity, compare differences and assign partial charges.

From bond polarity to molecular polarity

Learning goalUse shape to decide whether bond dipoles cancel.

🔓 CRACK IT

A molecule can contain polar bonds and still be non-polar overall. The answer depends on whether the bond dipoles cancel in its three-dimensional shape.

Identify which bonds are polar and mark their dipole directions.
Work out the molecular shape, including the effect of lone pairs.
Check whether the bond dipoles balance in all directions.
If they cancel completely, the molecule has no permanent dipole. If they do not, the molecule is polar.

CO₂ is linear, so its equal C=O dipoles cancel. H₂O is bent, so its O-H dipoles do not cancel. Draw the shape before deciding.

Worked example

SO3 and SO2

SO₃ and SO₂ both contain polar S–O bonds. Explain why SO₂ is polar overall but SO₃ is not.

Worked answer

SO₃ is symmetrical trigonal planar, so equivalent dipoles cancel. SO₂ is bent, so its dipoles do not cancel.

SO₃ is non-polar; SO₂ has a permanent dipole.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 37

The figure shows A: CO₂, B: BF₃, C: H₂O and D: CH₂Cl₂. Every bond to O, F or Cl shown is polar.

State whether each molecule A–D has a permanent dipole. [4]
Explain why A and B do not. [2]
Explain why C and D do. [2]

Total: 8 marks

Show answer & marking points

A: no; B: no; C: yes; D: yes. One mark each.
CO₂ is linear with equal opposite dipoles that cancel [1]. BF₃ is symmetrical trigonal planar, so its three dipoles cancel [1].
H₂O is bent, so dipoles do not cancel [1]. CH₂Cl₂ has unlike outer atoms in a tetrahedral arrangement, so its dipoles do not cancel [1].

✓ DONE WHENYou can distinguish a polar bond from a polar molecule.

Symmetry and cancellation of dipoles

Learning goalExplain polarity through the complete three-dimensional arrangement.

🔓 CRACK IT

To decide whether a molecule is polar, look at the whole arrangement of bond dipoles. Their sizes and directions must balance for complete cancellation.

BF₃, CO₂ and CCl₄ have equivalent surrounding atoms arranged so the dipoles cancel. NH₃ and H₂O have shapes in which they do not.

A tetrahedral shape alone does not guarantee cancellation. In CH₂Cl₂ the surrounding atoms are different, so its bond dipoles do not balance and the molecule is polar.

Molecule	Shape	Permanent dipole?
BF₃	Symmetrical trigonal planar	No: equal dipoles cancel
CO₂	Symmetrical linear	No: equal dipoles cancel
CCl₄	Symmetrical tetrahedral	No: equal dipoles cancel
H₂O	Bent	Yes: dipoles do not cancel
NH₃	Trigonal pyramidal	Yes: dipoles do not cancel
CH₂Cl₂	Tetrahedral with unlike surrounding atoms	Yes: dipoles do not cancel

EXAM TRAPTetrahedral does not automatically mean non-polar. Cancellation depends on the sizes and directions of all bond dipoles.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 38

The figure shows A: CF₄, B: NF₃, C: SF₄ and D: SF₆. Each bond to F is polar.

Identify the two molecules without permanent dipoles. [2]
Identify the two with permanent dipoles. [2]
Explain cancellation in the non-polar molecules. [2]
Explain why dipoles do not cancel in NF₃ and SF₄. [2]
Explain why polar bonds alone cannot establish molecular polarity. [1]

Total: 9 marks

Show answer & marking points

A CF₄ [1] and D SF₆ [1].
B NF₃ [1] and C SF₄ [1].
Symmetrical arrangements of identical polar bonds [1] give cancellation [1].
The lone pair produces an asymmetric molecular shape [1], so bond dipoles do not cancel [1].
The shape and vector directions of all bond dipoles must also be considered [1].

✓ DONE WHENYou can justify polarity using shape and whether dipoles cancel.

Intermolecular forces

5 practice questions

Identifying all intermolecular forces

Learning goalDistinguish the three forces between molecules.

🔓 CRACK IT

Keep the location of an attraction clear. Covalent bonds hold atoms together within a molecule. Intermolecular forces attract one molecule to another.

Start with induced dipole-dipole attractions, which act between all atoms and molecules. Polar molecules also have permanent dipole-dipole attractions. Suitable molecules can form hydrogen bonds as well.

List all the forces present before comparing substances. Naming hydrogen bonding does not mean the other applicable forces disappear.

London dispersion is another name for induced dipole-dipole interactions. Use the specific force name when explaining an attraction.

Force	When present	Attraction
Induced dipole–dipole (van der Waals)	Between all atoms and molecules	Opposite instantaneous and induced partial charges
Permanent dipole–dipole	Also between polar molecules	Opposite permanent partial charges
Hydrogen bonding	H bonded to N/O/F and a neighbouring N/O/F lone pair	δ+ H bonded to N/O/F and a lone pair on N/O/F

EXAM TRAPA molecule with oxygen but no O–H/N–H/F–H may form hydrogen bonds to δ+ H atoms in water while being unable to hydrogen-bond to itself.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 39

State the strongest intermolecular force in HCl. [1]
State the strongest intermolecular force in CH₄. [1]
State the strongest intermolecular force in NH₃. [1]
State the strongest intermolecular force in CH₃OCH₃. [1]
Explain why all four have van der Waals forces. [2]

Total: 6 marks

Show answer & marking points

Permanent dipole–dipole [1].
Induced dipole–dipole (van der Waals) [1].
Hydrogen bonding [1].
Permanent dipole–dipole [1].
Continuously moving electrons produce instantaneous dipoles [1], which induce dipoles in neighbouring molecules and cause attraction [1].

✓ DONE WHENYou can identify every applicable force rather than naming only the strongest.

Induced dipole–dipole (van der Waals) forces

Learning goalExplain electron-number and branching effects.

🔓 CRACK IT

Electrons are always moving. For an instant, they can be distributed unevenly, giving one side of a particle a small negative charge and the other a small positive charge.

This instantaneous dipole distorts the electron cloud of a neighbouring particle.
That distortion creates an induced dipole in the neighbour.
Opposite partial charges attract. New dipoles continually form as the electrons move.

A larger electron cloud is usually more easily distorted, or more polarisable. This tends to give stronger van der Waals attractions and a higher boiling point.

For isomers with the same electron count, more branching usually reduces surface contact between molecules. The overall attractions are then weaker, so less energy is needed to separate them.

Factor	Consequence
More electrons / larger electron cloud	Greater polarisability, stronger temporary dipoles
Greater molecular surface contact	More neighbouring electron clouds can interact

Worked example

Halogen boiling-point trend

Explain Cl₂ < Br₂ < I₂ in boiling point.

Worked answer

Electron number and polarisability increase down the group. van der Waals forces strengthen, requiring more energy to separate molecules.

Boiling point increases.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 40

The table gives Group 0 data.

Element	Number of electrons	Boiling point / °C
Ne	10	−246
Ar	18	−186
Kr	36	−153
Xe	54	−108

Name the force between the atoms. [1]
Explain the trend from Ne to Xe. [4]
Predict how Rn’s boiling point compares with Xe’s and justify your answer. [3]

Total: 8 marks

Show answer & marking points

Induced dipole–dipole (van der Waals) forces [1].
More electrons [1] give a more polarisable cloud [1], stronger temporary dipoles and attractions [1], requiring more energy to separate atoms [1].
Above −108 °C [1]; more electrons / greater polarisability [1] give stronger van der Waals forces [1].

Exam-style question 41

The three C₅H₁₂ isomers have the same electron count but different shapes.

Compound	Boiling point / °C
Pentane	36.1
2-Methylbutane	27.8
2,2-Dimethylpropane	9.5

Place them in increasing volatility. [1]
Explain why boiling points differ despite identical molecular formulae. [4]

Total: 5 marks

Show answer & marking points

Pentane < 2-methylbutane < 2,2-dimethylpropane [1].
Same electron count / similar potential for van der Waals forces [1]. More branching makes molecules compact and reduces surface contact [1]; attractions become weaker [1], so less energy is needed and boiling point falls [1].

✓ DONE WHENYou can explain trends through polarisability or surface contact and energy.

Permanent dipole–dipole attractions

Learning goalIdentify attractions between neighbouring polar molecules.

🔓 CRACK IT

A polar molecule has a permanent uneven distribution of charge. The δ+ end of one molecule attracts the δ− end of a neighbouring molecule.

These permanent dipole-dipole attractions act as well as induced dipole-dipole attractions. For otherwise comparable molecules, the extra attractions can increase the energy needed to separate them.

HCl has an overall dipole because it has one polar bond. SO₂ is bent, so its bond dipoles do not cancel. Use the shape as well as bond polarity when deciding whether this force is present.

Do not rank boiling points from polarity alone. A large non-polar molecule can have stronger overall attractions than a small polar molecule.

Permanent dipole–dipole force: Electrostatic attraction between the δ+ end of one polar molecule and the δ− end of a neighbouring molecule.

Worked example

HCl versus F2

HCl and F₂ have similar relative molecular masses. Explain why HCl has the higher boiling point.

Worked answer

Both have van der Waals forces. HCl is polar and has additional permanent dipole–dipole attractions, increasing the energy needed to separate molecules.

HCl has stronger overall attractions and the higher boiling point.

EXAM TRAPDo not rank boiling points from dipole size alone. A larger non-polar molecule can have stronger overall forces than a smaller polar molecule.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 42

A student claims that the molecule with the largest permanent dipole must have the highest boiling point.

Molecule	Dipole moment / D	Boiling point / °C
CH₃F	1.85	−78
CH₃Cl	1.87	−24
CH₃Br	1.82	4
CH₃I	1.62	42

Explain why all four have permanent dipoles. [2]
Explain the boiling-point trend from CH₃F to CH₃I. [4]
Evaluate the claim using the data. [2]

Total: 8 marks

Show answer & marking points

C–X is polar [1]; unlike outer atoms make the tetrahedral molecule asymmetric, so dipoles do not cancel [1].
More electrons [1], greater polarisability [1], stronger van der Waals forces [1], greater energy needed to separate molecules [1].
False: CH₃Cl has the largest dipole but CH₃I boils highest [1]. van der Waals forces can dominate the total attraction [1].

Exam-style question 43

All C–Cl bonds are polar.

Molecule	Permanent dipole / D	Boiling point / °C
CH₄	0	−162
CH₃Cl	1.87	−24
CH₂Cl₂	1.60	40
CHCl₃	1.04	61
CCl₄	0	77

Identify the two non-polar molecules. [2]
Explain why CCl₄ has no permanent dipole despite polar C–Cl bonds. [2]
Explain why boiling point rises from CH₃Cl to CCl₄ as permanent dipole decreases. [4]
Evaluate “CCl₄ has weaker intermolecular forces than CH₃Cl because CCl₄ has no permanent dipole.” [2]

Total: 10 marks

Show answer & marking points

CH₄ [1] and CCl₄ [1].
Symmetrical tetrahedral arrangement [1] makes four equal C–Cl dipoles cancel [1].
Replacing H by Cl increases electron number [1] and polarisability [1], strengthening van der Waals forces [1] and increasing the energy needed to boil [1].
False: both have van der Waals forces [1]; CCl₄’s more polarisable cloud gives stronger overall attractions and a higher boiling point [1].

✓ DONE WHENYou can explain permanent-dipole effects while also considering van der Waals forces.

Hydrogen bonding & properties

6 practice questions

Hydrogen bonding: conditions and diagrams

Learning goalIdentify and draw a hydrogen bond correctly.

🔓 CRACK IT

Hydrogen bonding needs a particular arrangement. Find a δ+ hydrogen directly bonded to N, O or F, then a lone pair on N, O or F in a neighbouring molecule.

Find a δ+ H atom directly bonded to N, O or F in one molecule.
Find a lone pair on N, O or F in a neighbouring molecule.
Draw a dotted attraction from that lone pair to the δ+ H. The covalent bond from H to its own atom stays in place.
Add the appropriate partial charges and keep the hydrogen-bond arrangement clear.

Water, ammonia and HF can hydrogen-bond between their own molecules. An oxygen lone pair can attract a δ+ H atom in water, even when the oxygen-containing molecule has no O-H bond and cannot hydrogen-bond with itself. H bonded only to carbon does not meet the hydrogen-bonding condition here.

Hydrogen bond: An attraction between a lone pair on N, O or F and a δ+ H atom covalently bonded to N, O or F in a neighbouring molecule.

Substance	Self hydrogen-bonding?	Reason
HF	Yes	H–F bonds and F lone pairs
H₂O	Yes	O–H bonds and O lone pairs
NH₃	Yes	N–H bonds and an N lone pair
CH₃F	No	H is bonded to C, not F
CH₃OCH₃	No	O lone pairs but no H directly bonded to O

Worked example

Propan-1-ol and methoxyethane

Propan-1-ol and methoxyethane have the same molecular formula. Explain why propan-1-ol has the higher boiling point.

Worked answer

Both have van der Waals and permanent dipole–dipole forces. Propan-1-ol also self hydrogen-bonds through O–H groups; methoxyethane has no H directly bonded to N, O or F.

Stronger overall attractions in propan-1-ol require more energy to overcome.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 44

Draw a hydrogen bond between two water molecules, showing relevant partial charges and a lone pair on oxygen. [4]
Explain why HCl molecules do not hydrogen-bond to one another. [1]
State why a hydrogen bond must not be drawn as a solid covalent bond. [1]

Total: 6 marks

Show answer & marking points

O–H covalent bond [1]; Hδ+ and Oδ− on the neighbouring molecule [1]; oxygen lone pair [1]; dotted attraction from Hδ+ to that lone pair [1].
H is not directly bonded to N, O or F [1].
It is an intermolecular electrostatic attraction, not a shared pair of electrons [1].

Exam-style question 45

Hydride	H₂O	H₂S	H₂Se	H₂Te
Boiling point / °C	100	−60	−41	−2

Explain the trend from H₂S to H₂Te. [3]
Explain water’s anomalously high boiling point. [3]
Draw one hydrogen bond between water molecules, with partial charges and an oxygen lone pair. [4]

Total: 10 marks

Show answer & marking points

Electron count and polarisability increase [1]; van der Waals forces strengthen [1], so more energy is required [1].
Water forms hydrogen bonds [1]; its overall attractions are stronger than those in the other hydrides [1], so more energy is needed [1].
Hδ+ in an O–H bond [1]; Oδ− on the neighbouring molecule [1]; O lone pair [1]; dotted attraction from H to the lone pair [1].

✓ DONE WHENYou can identify the required H atom and lone pair and draw the correct intermolecular attraction.

Hydrogen bonding, ice and anomalous boiling points

Learning goalLink hydrogen bonding to boiling point and ice density.

🔓 CRACK IT

Hydrogen bonds help explain two unusual properties of water. They make water's boiling point higher than the simple trend for comparable hydrides suggests, and they give ice its open structure.

In ice, hydrogen bonds hold water molecules further apart than they are on average in liquid water. The same mass occupies a larger volume, so ice is less dense and floats.

On melting, some hydrogen bonds are overcome and the open arrangement partly collapses. The O-H covalent bonds inside each water molecule remain intact.

Use the same link for boiling-point comparisons: stronger overall attractions between molecules mean more energy is needed to separate them.

Comparison	Explanation
NH₃ versus PH₃	NH₃ forms hydrogen bonds; its boiling point is unusually high
HF versus HCl	HF forms hydrogen bonds; more energy is required to separate molecules
H₂O versus other Group 6 hydrides	Water’s hydrogen-bonded network raises its boiling point

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 46

The figure shows a simplified open structure of ice.

Identify the dotted attraction and the solid bond. [2]
Explain why ice is less dense than liquid water. [3]
State what is overcome on melting and what remains unbroken within each molecule. [2]

Total: 7 marks

Show answer & marking points

Dotted: hydrogen bonds [1]; solid O–H: covalent bonds [1].
Hydrogen bonds hold an open lattice [1]; molecules are further apart and the same mass occupies a larger volume [1], so density is lower [1].
Some intermolecular hydrogen bonds are overcome as the lattice collapses [1]; O–H covalent bonds remain intact [1].

✓ DONE WHENYou can explain floating ice and anomalous boiling points with a complete force-to-property chain.

Boiling, melting and competing intermolecular effects

Learning goalCompare overall attractions and connect them to energy and volatility.

🔓 CRACK IT

To compare boiling points, explain what has to be overcome when the particles separate. Stronger overall intermolecular attractions need more energy, so boiling point tends to be higher.

Identify the molecules and all the forces between them.
Compare electron count, polarisability, shape and any extra permanent-dipole or hydrogen-bonding attractions.
Decide which substance has stronger overall attractions.
Link that to more energy needed and the higher boiling point.

A more volatile substance evaporates more readily. Stronger attractions generally reduce volatility at a given temperature. Melting points also depend on how well particles pack in the crystal, so avoid using one simple force ranking for every comparison.

Particles → all forces → overall strength → energy required → boiling point

Process	Overcome	Remains intact
Boiling water	Intermolecular hydrogen bonds	O–H covalent bonds
Melting iodine	Intermolecular van der Waals forces	I–I covalent bonds

Worked example

Ethanol and ethane

Explain why ethanol has a higher boiling point than ethane.

Worked answer

Both have van der Waals forces. Ethanol is polar and also forms hydrogen bonds. Its overall attractions are stronger, so more energy is needed to separate molecules.

Ethanol has the higher boiling point.

EXAM TRAP“Hydrogen bonding beats every other molecule” is not a universal rule. Consider electron count, shape and every force present; melting points also depend on crystal packing.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 47

Substance	Formula	Mᵣ	Boiling point / °C
A	CH₃CH₂CH₃	44	−42
B	CH₃CHO	44	21
C	CH₃CH₂OH	46	78
D	HOCH₂CH₂OH	62	197

State the strongest intermolecular force in each. [4]
Explain why B boils above A despite equal Mᵣ. [2]
Explain why C boils above B. [2]
Explain why D boils above C. [3]
Place A–D in increasing volatility. [1]

Total: 12 marks

Show answer & marking points

A van der Waals; B permanent dipole–dipole; C hydrogen bonding; D hydrogen bonding. One mark each.
B has permanent dipole–dipole attractions as well as van der Waals forces [1], giving stronger overall attractions and greater energy needed than A [1].
C forms hydrogen bonds [1], so its overall attractions require more energy to overcome than B’s [1].
D has two O–H groups and a more extensive hydrogen-bonded network [1], plus more electrons and stronger van der Waals forces [1], so greater energy is needed [1].
D < C < B < A [1].

Exam-style question 48

Three liquid jets are tested with the same charged rod.

Jet	Deflection	Mᵣ	Boiling point / °C
P	Strong	18	100
Q	Moderate	58	56
R	Negligible	86	69

Identify P, Q and R from water, propanone and hexane. [3]
Explain why a charged rod deflects a polar liquid jet. [2]
Explain why deflection alone does not prove self hydrogen bonding. [2]
Explain why P boils highest despite having the lowest Mᵣ. [2]
Explain why non-polar R boils above Q. [3]

Total: 12 marks

Show answer & marking points

P water [1]; Q propanone [1]; R hexane [1].
The electric field aligns / attracts permanent molecular dipoles [1], giving a net sideways force [1].
In this comparison, deflection indicates polarity rather than uniquely identifying hydrogen bonding [1]. Propanone deflects but has no H directly bonded to N, O or F, so it cannot hydrogen-bond with itself [1].
Water forms hydrogen bonds [1], requiring more energy to overcome its overall attractions [1].
Hexane has more electrons and greater polarisability [1]; its stronger van der Waals forces [1] outweigh propanone’s permanent dipole contribution [1].

✓ DONE WHENYou can evaluate competing effects without claiming that boiling breaks covalent bonds.

Solubility: attractions disrupted and formed

Learning goalUse solute–solvent interactions to explain mixing.

🔓 CRACK IT

Dissolving means particles of the solute mix with the solvent. Some attractions within the solute and between solvent particles are disrupted, while new attractions form between solute and solvent.

Name those new attractions. Ethanol can hydrogen-bond with water. Propanone can accept hydrogen bonds from water through its oxygen lone pairs, even though propanone does not hydrogen-bond to itself.

Hexane cannot form hydrogen bonds that adequately replace water-water attractions, so its solubility in water is very low. Iodine mixes more readily with non-polar hexane through favourable van der Waals interactions.

'Like dissolves like' is a useful starting point. An exam explanation needs the actual particles and attractions, rather than the slogan alone.

Pair	Interaction reasoning	Prediction
Ethanol + water	O–H groups and O lone pairs allow hydrogen bonds between the two substances	Mix readily
Hexane + water	Hexane cannot form hydrogen bonds that adequately replace water–water attractions	Very low solubility
Iodine + hexane	Both are non-polar; favourable van der Waals interactions form	Iodine dissolves better than in water
Propanone + water	O lone pairs attract δ+ H atoms in water	Mix readily

EXAM TRAPOxygen-containing molecules may form hydrogen bonds to δ+ H atoms in water without hydrogen-bonding to themselves. Polar does not automatically mean water-soluble.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 49

Ethanol, CH₃CH₂OH, boils at 78 °C. Methoxymethane, CH₃OCH₃, boils at −25 °C. Both have formula C₂H₆O.

State the strongest intermolecular force in each. [2]
Explain the boiling-point difference. [3]
Both hydrogen-bond with water. Explain and identify one difference in their hydrogen-bonding behaviour. [2]

Total: 7 marks

Show answer & marking points

Ethanol: hydrogen bonding [1]; methoxymethane: permanent dipole–dipole [1].
Ethanol’s O–H groups allow self hydrogen bonding [1]. Methoxymethane has no H bonded to N, O or F, so it cannot self hydrogen-bond [1]. Stronger overall attractions in ethanol require more energy to overcome [1].
Both have O lone pairs that attract δ+ H atoms in water [1]. Ethanol also has a δ+ H in an O–H bond that can hydrogen-bond to a water oxygen lone pair; methoxymethane has no O–H bond [1].

✓ DONE WHENYou can name attractions disrupted and formed rather than relying on a slogan.

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