Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
Learn
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
R3.3.1 Radical
Radical
A molecular entity that has an unpaired electron, for example the methyl radical, •CH₃, and the chlorine radical (chlorine atom), Cl•. A radical can be a single atom or a group of atoms, and the radicals met in this topic carry no charge. Radicals such as •CH₃ and Cl• are highly reactive: the unpaired electron readily pairs with an electron from another species, forming a new covalent bond.
Representing a radical
A radical is shown by writing a single dot, •, next to the atom that carries the unpaired electron: Cl• for a chlorine atom and •CH₃ for the methyl radical, where the dot sits on the carbon atom. The dot stands for one electron, not for a charge. A species can be identified as a radical by counting its valence electrons: a neutral species with an odd total, such as CH₃ (4 + 3 = 7) or C₂H₅ (8 + 5 = 13), must have an unpaired electron.
Students often think A radical is an ion: it carries a charge, and the dot in Cl• shows that charge. In fact No. Cl• and •CH₃ are neutral. The dot shows one unpaired electron, not a charge; Cl• has the same seven outer electrons as any chlorine atom.
Students often think A radical is a group of atoms that stays together as a unit in reactions, such as sulfate or ammonium. In fact No. In current chemistry (IUPAC) a radical is a molecular entity with an unpaired electron. A single atom such as Cl• is a radical, and a group such as SO₄²⁻ is not.
R3.3.2 Homolytic fission
Homolytic fission
The breaking of a covalent bond in which each of the two bonded atoms takes one electron of the shared pair, so two radicals form and no ions are produced. Example: Cl₂ → 2Cl•. It contrasts with heterolytic fission, in which one atom takes both electrons and a pair of ions forms.
Initiation step
The first step of a radical chain reaction, in which radicals are produced. In the reaction of an alkane with a halogen, ultraviolet (UV) light or heat supplies the energy to break the halogen–halogen bond homolytically: Cl₂ → 2Cl• or Br₂ → 2Br•. The halogen–halogen bond breaks rather than a C–H bond because it is much weaker (Cl–Cl 242 kJ mol⁻¹; C–H 414 kJ mol⁻¹).
Single-barbed (fish-hook) arrow
A curved arrow with a single barb (half-head), used in a mechanism to show the movement of ONE electron. In the homolytic fission of Cl₂, two fish-hook arrows are drawn, each starting at the Cl–Cl bond and ending at a different chlorine atom, showing one electron of the bonding pair going to each atom. A curly arrow with a double barb (full head) shows the movement of a PAIR of electrons.
Students often think When a bond breaks, it breaks into a positive ion and a negative ion, so homolytic fission of Cl₂ gives Cl⁺ and Cl⁻. In fact No. In homolytic fission each atom takes one electron of the bonding pair, so Cl₂ → 2Cl•: two neutral radicals, no ions.
Students often think Fish-hook and curly arrows mean the same thing: each shows a pair of electrons moving; the shape of the head does not matter. In fact No. A single-barbed (fish-hook) arrow shows the movement of ONE electron; a double-barbed curly arrow shows a PAIR.
R3.3.3 Propagation steps
Propagation steps
Steps in which a radical reacts with a molecule to form a new molecule and a new radical, so each step consumes one radical and produces one. For methane and chlorine: CH₄ + Cl• → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•. The Cl• produced in the second step can start the first step again, so a few initiation events lead to very many product molecules (a chain reaction). Most of the chloroalkane is formed in propagation.
Termination steps
Steps in which two radicals combine to form a molecule, removing radicals from the reaction and ending chains. Any two radicals present can combine, for example Cl• + Cl• → Cl₂, •CH₃ + Cl• → CH₃Cl and •CH₃ + •CH₃ → C₂H₆. Because radicals are present only at low concentration, termination is much less frequent than propagation.
Radical substitution of an alkane
The reaction of an alkane with a halogen in UV light or on heating, in which halogen atoms replace hydrogen atoms by a radical chain mechanism (initiation, propagation, termination). Example overall equation: CH₄ + Cl₂ → CH₃Cl + HCl.
Mixture of products
Radical substitution does not give a single product. Cl• can remove a hydrogen atom from a chloroalkane as well as from the alkane (CH₃Cl + Cl• → •CH₂Cl + HCl; •CH₂Cl + Cl₂ → CH₂Cl₂ + Cl•), so further substitution gives CH₂Cl₂, CHCl₃ and CCl₄; termination steps add products such as C₂H₆. The reaction of methane with chlorine therefore gives a mixture of chloroalkanes, HCl and small amounts of other products.
Stability of alkanes
Alkanes react with few reagents because their C–C and C–H bonds are strong (bond enthalpies about 346 and 414 kJ mol⁻¹) and essentially non-polar: carbon and hydrogen have similar electronegativities (2.6 and 2.2), so the bonds offer no significant partial charges for polar reagents to attack. Their reactions, such as combustion and radical substitution, need a high-energy input or a highly reactive species such as a radical.
Students often think Radicals are used up in propagation, so every product molecule needs a new radical from the initiation step. In fact No. Each propagation step consumes one radical and produces one, so the number of radicals does not change. The Cl• used in CH₄ + Cl• → •CH₃ + HCl is regenerated in •CH₃ + Cl₂ → CH₃Cl + Cl•.
Students often think The chloroalkane is made mainly in a termination step, •CH₃ + Cl• → CH₃Cl, because that is where the final product is formed. In fact No. Most CH₃Cl is formed in the propagation step •CH₃ + Cl₂ → CH₃Cl + Cl•. The termination step •CH₃ + Cl• → CH₃Cl does occur but is rare, because radicals are present only at very low concentration.
Diagnostic a bearings check, not a test
6 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement defines a radical?
Answer and reasoning
A molecular entity that has an unpaired electron — This is the definition: a radical has an unpaired electron, as in •CH₃ and Cl•. The unpaired electron is what makes radicals highly reactive.
A molecular entity that carries an electric charge — You chose the definition of an ion. Radicals such as Cl• and •CH₃ are neutral; the dot shows an unpaired electron, not a charge. Cl⁻ is charged but has all its electrons paired, so it is not a radical.
A group of atoms that stays together in reactions — This is an older meaning of 'radical', as in the 'sulfate radical' of older textbooks. In IB chemistry a radical is a species with an unpaired electron: Cl• is a single atom yet is a radical, and SO₄²⁻ is not a radical.
A molecular entity with an incomplete octet — An incomplete octet is not the test. BF₃ and CH₃⁺ have only six electrons around the central atom but all are paired, so they are not radicals. What matters is an unpaired electron.
2 In the homolytic fission of Cl₂, two single-barbed (fish-hook) arrows are drawn, each starting at the Cl–Cl bond and ending at a different chlorine atom. What does each arrow represent?
Answer and reasoning
The movement of an electron pair — A pair is shown by a double-barbed curly arrow. If each of the two arrows moved a pair, four electrons would leave a bond that holds only two.
The movement of a chlorine atom — Mechanism arrows show electrons, not atoms. The atoms separate because the electrons move; each arrow tracks one electron.
The release of stored bond energy — Arrows do not show energy, and no energy stored in the bond is released: breaking the Cl–Cl bond absorbs energy (242 kJ mol⁻¹, supplied by UV light or heat).
The movement of one electron — A single barb means one electron. Each arrow moves one electron of the Cl–Cl bonding pair to a different chlorine atom, so each atom becomes Cl•.
3 Alkanes react with very few reagents. Which statement explains the low reactivity of alkanes?
Answer and reasoning
Their C–C and C–H bonds are strong and are essentially non-polar. — Strong bonds need a large energy input to break, and because carbon and hydrogen have similar electronegativities the bonds have no significant partial charges for polar reagents to attack.
They are saturated, and saturated compounds are unreactive. — Many saturated compounds react readily: halogenoalkanes and alcohols are saturated. Saturation rules out addition, but alkanes' low reactivity is due to their strong, essentially non-polar bonds.
Each atom has a full outer shell, so the molecule is unreactive. — Full outer shells are found in nearly every stable molecule, including reactive ones such as CH₃Cl and HCl. Reactivity depends on bond strength and polarity, not on full shells.
Strong intermolecular forces hold the molecules together. — Reactions break covalent bonds within molecules, not forces between them. In any case the forces between alkane molecules are weak London forces; methane boils at −162 °C.
4 Which species is a radical, and for what reason? Any charge on a species is shown, but unpaired electrons are not marked.
Answer and reasoning
BF₃, because boron has only 6, not 8, outer-shell electrons — Boron does have only six outer electrons in BF₃, an incomplete octet, but that is not what makes a radical. BF₃ has 3 + 3 × 7 = 24 valence electrons, all paired, so it has no unpaired electron.
C₂H₅, because it has an odd number (13) of valence electrons — Count the valence electrons: 2 × 4 + 5 × 1 = 13. An odd number of electrons cannot all be paired, so one is unpaired: this is the ethyl radical, •C₂H₅ (usually written •CH₂CH₃).
OH⁻, because it has one more electron than a neutral OH group — A charge does not make a radical. OH⁻ has 6 + 1 + 1 = 8 valence electrons, all paired, so it is an ion but not a radical. It is the neutral OH species, with 7 valence electrons, that is a radical.
Cl₂, because it reacts vigorously with many substances — Chlorine is highly reactive, but reactivity is not the test for a radical. Cl₂ has 14 valence electrons, all paired (one bonding pair and six lone pairs). It forms radicals only when UV light or heat breaks the Cl–Cl bond.
5 Methane reacts with bromine in ultraviolet (UV) light. Which equation represents the initiation step?
Answer and reasoning
Br₂ → Br⁺ + Br⁻ — This is heterolytic fission, with one atom taking both electrons. The initiation step is homolytic: each Br atom takes one electron, giving two neutral radicals and no ions.
CH₄ → •CH₃ + H• — The C–H bond in methane is much stronger than the Br–Br bond, so UV light breaks Br–Br, not C–H. The C–H bond is broken later, in propagation, by a Br• radical, and no H• is formed.
Br₂ → Br• + Br• — UV light supplies the energy to break the Br–Br bond homolytically: one electron of the bonding pair goes to each atom, giving two bromine radicals. The Br–Br bond is much weaker than a C–H bond, so it is the bond that breaks.
CH₄ → CH₃⁻ + H⁺ — Methane does not ionize: carbon and hydrogen have similar electronegativities, so the C–H bond is essentially non-polar. The initiation step is the homolytic fission of the much weaker Br–Br bond.
6 Ethane reacts with chlorine in UV light. Which equation represents a propagation step of this reaction?
Answer and reasoning
CH₃CH₃ + Cl• → CH₃CH₂Cl + H• — Cl• does not swap directly for a hydrogen atom. It removes a hydrogen atom to form HCl, leaving an ethyl radical: C₂H₆ + Cl• → •C₂H₅ + HCl. No H• radical forms.
•C₂H₅ + Cl₂ → C₂H₅Cl + Cl• — A propagation step has one radical on each side: •C₂H₅ reacts with a Cl₂ molecule to form chloroethane and a new Cl• radical, which carries the chain on. The other propagation step is C₂H₆ + Cl• → •C₂H₅ + HCl.
•C₂H₅ + Cl• → CH₃CH₂Cl — Two radicals combining is a termination step, not propagation: radicals are removed and none is formed. It does make some chloroethane, but most forms in propagation, •C₂H₅ + Cl₂ → C₂H₅Cl + Cl•.
C₂H₆ + Cl⁺ → C₂H₅Cl + H⁺ — This step uses ions, but radical substitution involves no ions. Initiation is homolytic fission, which gives Cl• radicals rather than Cl⁺, and each propagation step has one radical on each side.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
6 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Methane and chlorine react in UV light or on heating. Bond enthalpies: Cl–Cl 242 kJ mol⁻¹; C–H 414 kJ mol⁻¹. Which bond breaks in the initiation step, and why?
Answer and reasoning
C–H breaks, because breaking the stronger bond releases the greater amount of energy — Breaking a bond never releases energy; it always requires energy. Breaking C–H would need 414 kJ mol⁻¹, far more than the 242 kJ mol⁻¹ needed for Cl–Cl, so Cl–Cl is the bond that breaks.
Cl–Cl breaks, because the bond splits readily into Cl⁺ and Cl⁻ ions — The right bond, but the wrong products. Initiation is homolytic fission: each chlorine atom takes one electron, giving two Cl• radicals, not ions.
Cl–Cl breaks, because less energy is needed to break it than to break C–H — Breaking a bond requires energy equal to its bond enthalpy. The Cl–Cl bond (242 kJ mol⁻¹) needs far less than the C–H bond (414 kJ mol⁻¹), so UV light or heat breaks Cl–Cl homolytically: Cl₂ → 2Cl•.
C–H breaks, because the alkane must lose an H atom before chlorine can react — Methane does not need to be broken first. The weaker Cl–Cl bond breaks in initiation; the Cl• formed then removes an H atom from CH₄ in propagation, CH₄ + Cl• → •CH₃ + HCl.
2 A student claims: 'Chlorine radicals are used up in the propagation steps, so every molecule of CH₃Cl formed from methane and chlorine needs a new Cl• from the initiation step.' Which evaluation of this claim is correct?
Answer and reasoning
Incorrect: the H• formed in the first step reacts with Cl₂ to re-form Cl• for the chain — The claim is wrong, but not for this reason. No H• forms: the first propagation step is CH₄ + Cl• → •CH₃ + HCl. The Cl• is regenerated when •CH₃ reacts with Cl₂.
Correct: each propagation step uses up a radical, so each CH₃Cl needs a new initiation — Count the radicals on each side of a propagation step: one on the left and one on the right. Propagation does not change the number of radicals; only termination removes them.
Correct: CH₃Cl forms when •CH₃ and Cl• combine, which uses up two radicals each time — That termination step does occur, but it is rare because radicals are present at very low concentration. Most CH₃Cl forms in propagation, •CH₃ + Cl₂ → CH₃Cl + Cl•, which regenerates Cl•.
Incorrect: the Cl• used in the first step is re-formed in the second, so it can react again — In CH₄ + Cl• → •CH₃ + HCl a Cl• is used, and in •CH₃ + Cl₂ → CH₃Cl + Cl• a Cl• is formed. Each cycle regenerates the radical, so one initiation event can lead to many CH₃Cl molecules: a chain reaction.
3 The products of the reaction of methane with chlorine in UV light include CH₂Cl₂, CHCl₃ and CCl₄. How are most of these more highly substituted products formed?
Answer and reasoning
Cl• removes an H atom from a chloromethane; the radical then reacts with Cl₂ — Cl• attacks C–H bonds in chloromethanes as it does in methane, e.g. CH₃Cl + Cl• → •CH₂Cl + HCl, then •CH₂Cl + Cl₂ → CH₂Cl₂ + Cl•. Repeating these propagation steps gives CHCl₃ and CCl₄.
Cl• replaces an H atom in a chloromethane directly, releasing a free H• radical — Cl• does not swap for H. It removes a hydrogen atom as HCl, leaving a radical such as •CH₂Cl, which then takes a Cl atom from Cl₂. No H• radical is formed.
Chlorinated methyl radicals combine with Cl• in termination, the main source of products — Radical–radical combinations are rare because radicals are present at very low concentration. Termination makes a little of these products, but most form in propagation steps with Cl₂.
Each forms in one step, e.g. CH₃Cl + Cl₂ → CH₂Cl₂ + HCl, as its equation shows — An overall equation is the sum of mechanism steps, not a single collision. Each further substitution happens by the same radical propagation steps: Cl• removes an H atom, then the radical reacts with Cl₂.
4 A small amount of ethane, C₂H₆, is found in the product mixture when methane reacts with chlorine in UV light. Which statement explains why ethane is present?
Answer and reasoning
A methyl radical replaces H in methane: •CH₃ + CH₄ → C₂H₆ + H• — Radicals in this reaction do not swap for H atoms, and no H• forms. Ethane forms when two •CH₃ radicals combine in a termination step.
A methyl cation and methyl anion combine: CH₃⁺ + CH₃⁻ → C₂H₆ — No ions form in this reaction: the C–H and Cl–Cl bonds break homolytically, giving radicals. Ethane forms when two •CH₃ radicals combine in termination.
Two methyl radicals combine in a termination step: •CH₃ + •CH₃ → C₂H₆ — In termination any two radicals combine. When two •CH₃ radicals meet, their unpaired electrons pair to form a C–C bond, giving ethane. Such collisions are rare, so only a little ethane forms.
It is not a reaction product; only CH₃Cl and HCl form, so it is an impurity — Radical substitution gives a mixture of products. The •CH₃ radicals formed in propagation can combine with each other in termination, so ethane is a genuine, if minor, product.
5 Bromine reacts with methane in ultraviolet (UV) light. The diagram shows two drawings of the Br–Br bond in Br₂ breaking. Which statement about the drawings is correct?
Answer and reasoning
Br₂ splits into Br⁺ and Br⁻ in the initiation step, as Drawing Q shows — Drawing Q does show a bond splitting into ions: the double-barbed arrow moves BOTH bonding electrons to the right-hand Br, giving Br⁻ and Br⁺ (heterolytic fission). But that is not how Br₂ breaks in UV light. The initiation step is homolytic: the pair is shared out one electron to each atom, and no ions form.
Drawings P and Q show the same process, as each moves two electrons — The number of barbs matters. Drawing P moves one electron to each atom (two radicals); Drawing Q moves the pair to one atom (two ions). Two electrons leave the bond in both, but they end up in different places, so the products differ.
In Drawing P, each arrow shows one bromine atom moving to a new position — A student who reads mechanism arrows as atom movements picks this. Each arrow in Drawing P starts at the bond, where the bonding electrons are, and ends on the atom that receives one of them: the arrows track electrons, not atoms.
Drawing P shows the initiation step, forming two Br• radicals — Each single-barbed arrow moves ONE electron of the Br–Br bonding pair to a different Br atom, so each atom leaves with one unpaired electron: Br₂ → 2Br•. This is homolytic fission, the initiation step of the chain reaction.
6 The diagram shows the electron movements in a propagation step of the reaction of methane with chlorine. Which products does the step shown form?
Answer and reasoning
CH₃Cl and a chloride ion — Arrow 3 is single-barbed, so it moves only ONE electron to the far Cl. That atom keeps one electron of the broken bond and is neutral with an unpaired electron, Cl•. A Cl⁻ ion would need a double-barbed arrow bringing the whole pair, and no ion forms in this radical step.
CH₃Cl and a Cl• radical — Arrows 1 and 2 each bring ONE electron into the space between C and the nearer Cl, forming the C–Cl bond of CH₃Cl. Arrow 3 moves the other electron of the Cl–Cl bond to the far Cl, which leaves with one unpaired electron: •CH₃ + Cl₂ → CH₃Cl + Cl•. The Cl• formed carries the chain on.
CH₃Cl and an HCl molecule — CH₃Cl and HCl are the products of the OVERALL equation, not of this step. No hydrogen atom takes part in the drawing: the arrows show only the C–Cl bond forming and the Cl–Cl bond breaking. HCl is made in the other propagation step, CH₄ + Cl• → •CH₃ + HCl.
CH₃Cl alone, with no radical — A student who thinks radicals are used up in propagation picks this. Count the atoms: Cl₂ brings two Cl atoms. Arrow 2 puts one bonding electron into the new C–Cl bond and arrow 3 sends the other to the far Cl, which leaves as Cl•. Each propagation step regenerates a radical, which is why one initiation event makes many CH₃Cl molecules.
That was your twenty minutes. Real practice on R3.3 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
DP Science Cafe