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  2. Chemistry
  3. Paper 1A: multiple choice

Paper 1A: multiple choice

How Paper 1A multiple-choice questions are built, a routine and time-saving strategies for answering them, the slips behind the wrong options, and thirteen worked questions.

  • IB Chemistry
  • Exam skills
  • SL and HL

17 min readLast reviewed 14 Oct 2026

Paper 1A is the multiple-choice part of Paper 1. Each question has four options, labelled A to D, and exactly one is correct. Every question is worth one mark and no marks are deducted for a wrong answer, so no question should be left blank. A calculator and a clean copy of the data booklet are available, but the numbers are chosen so that the arithmetic is short: most questions test whether the right chemistry is chosen, not whether a long calculation is done.

The questions run roughly in syllabus order, from the particulate nature of matter to organic reaction types, and they sample every part of the course. About three quarters are conceptual (bonding, trends, definitions, mechanisms, practical techniques) and about a quarter need some arithmetic. SL papers test SL content only; HL papers test both the SL content and the additional HL content, and some questions appear on both.

Key points

  • Each wrong option, a distractor, is the answer produced by one specific, common slip.

  • Deciding on your own answer before reading the options protects you from being led by them.

  • Combination tables, I/II/III statement sets and unevaluated expressions each have a quick method.

  • Units, coefficients, signs and state symbols are where most distractors are hidden.

  • Only the letter counts, so a reliable routine and a final check matter more than in a written paper.

How the questions are built

A typical stem is short, often a single line of 10 to 25 words, sometimes with an equation, a data line or a figure (an energy profile, a structure, a graph or a spectrum). Most stems are direct questions beginning "Which …?" or "What is …?". When a number is asked for, the unit sits in the stem ("What is the amount, in mol, of …?") and the options are bare numbers, usually in ascending order.

Format

What it looks like

How to approach it

Statements

Four sentences or phrases, such as four explanations of why a catalyst increases the rate

Decide in a few words what you would write, then find the option that says it

Combination table

Four rows that each pair two answers, such as "electron domain geometry / molecular geometry" or "oxidation state / charge"

Decide each column on its own; the row that matches both is the key

Formulas, expressions or orders

Four equations, four KK expressions, four bond-enthalpy sums, or four orders such as "Na<Mg<AlNa<Mg<Al"

Check one feature at a time: state symbols, coefficients, signs, which end of the order

Numbers

Four values, almost always in ascending order, or from negative to positive

Work out an answer or a rough size first; neighbouring options are often a factor of 2, 10 or a sign away

I, II and III

Three statements; options A "I and II only", B "I and III only", C "II and III only", D "I, II and III"

Judge each statement true or false separately, then pick the combination

The options are written so that each wrong one is reached by a real mistake: a mole ratio not applied, a sign reversed, a temperature left in degrees Celsius, the molecular formula given when the empirical formula was asked for. A student who makes that slip finds their answer among the options. "My answer is there, so it must be right" is therefore not a safe check.

Definition

A distractor is a wrong option in a multiple-choice question, designed so that a particular mistake in reasoning or calculation leads to it.

Stems also contain cue words that carry chemistry. "At the same temperature and pressure" means equal volumes of gases contain equal amounts. "Excess" tells you which reactant is limiting. "Aqueous", "dilute" and "concentrated" affect what happens at an electrode or how far an acid ionizes. "In the dark" or "in UV light" separates addition to an alkene from substitution in an alkane. "Assume complete combustion" fixes the products as carbon dioxide and water. A question that seems to lack information usually has the missing piece in one of these phrases.

A routine for every question

  1. Read the whole stem, including the last line. Underline what is asked for and any cue word. Many marks are lost by answering a slightly different question: the empirical formula instead of the molecular formula, the reverse activation energy instead of the forward one, the number of molecules instead of the number of atoms.

  2. Cover the options and predict. For a calculation, get a value or at least its size. For a conceptual question, say in a few words what the answer should be.

  3. Match and check. Find your prediction among the options. Then ask which slip produces each neighbouring option. If you can explain every distractor, you can be confident in the key.

  4. If stuck, eliminate. Cross out options with the wrong sign, an impossible value (a pH above 14, a negative amount), a broken trend or a state symbol that cannot be right. Choosing between two options is far better than guessing between four.

  5. Mark and move on. If a question resists you for more than a couple of minutes, choose the best remaining option, mark the question number, and return to it at the end.

Tip

Write on the question paper. A quick structure, a balanced equation, a mole ratio or a T and F beside each statement takes seconds and prevents most slips, which are errors of attention rather than gaps in knowledge.

Strategies for each type of question

Combination tables

A two-column table asks two questions at once, such as "What are the electron domain geometry and the molecular geometry of …?" or "What is the oxidation state of the metal and the overall charge of the complex?". Answer the column you are surest of first. Usually only two rows survive, and the second column then decides between them. Rows that are internally inconsistent, such as a trigonal planar electron domain geometry with a trigonal pyramidal molecular geometry, can be crossed out without any thought about the molecule.

Statements I, II and III

Judge each statement on its own and write T or F beside it before looking at the options. Because the four combinations are fixed, deciding that one statement is false removes two or three options at once: if statement III is false, only A ("I and II only") can be right, provided I and II are both true. Read each statement as a whole; a statement that is half right is false.

Expressions instead of values

Questions on bond enthalpies, Hess's law and equilibrium constants often give four unevaluated expressions, such as "[4(414)+2(498)]−[2(804)+4(463)][4(414)+2(498)]−[2(804)+4(463)]", so that the method is tested without the arithmetic. Check three things in order: the direction of the subtraction, the coefficients, and that every bond or species is counted. For bond enthalpies, ΔH=∑(bonds broken)−∑(bonds formed)ΔH=∑(bonds broken)−∑(bonds formed); for enthalpies of formation, the opposite order, products minus reactants, applies. For an equilibrium constant expression, products go on top and every concentration is raised to the power of its coefficient.

Common mistake

Using "products minus reactants" for bond enthalpies. That rule belongs to enthalpies of formation. With bond enthalpies the energy taken in to break the reactant bonds comes first: ΔH=∑(broken)−∑(formed)ΔH=∑(broken)−∑(formed). The reversed expression gives the right size with the wrong sign, and it is always one of the options.

Orders and trends

For "What is the order of increasing …?" questions, decide the two ends first: the substance with the lowest boiling point, the smallest radius or the weakest acid. That often leaves one option. Then check whether the question involves a known exception to the general trend, which is exactly what the distractors test: the electron configurations of chromium and copper, or the lower first ionization energies of the group 13 and group 16 elements at HL.

Short calculations

Calculation questions are designed to take a minute or less. Write the equation or the mole ratio, find the amount in moles, and scale. Then compare the size of your answer with the options: if they differ by factors of 2, the slip is usually a mole ratio or a coefficient; if they differ by factors of 10, it is a unit conversion (cm3cm3 to dm3dm3, gg to kgkg) or a power of ten in the Avogadro constant.

Tip

Two quick checks catch most numerical slips. For pH, every factor of 10 in [HX+][HX+] changes the pH by exactly one unit, so a hundredfold dilution of a strong acid raises the pH by 2. For gases, convert temperatures to kelvin before using any ratio: doubling a temperature from 27 ∘C27 ∘C means going to 327 ∘C327 ∘C (600 K), not to 54 ∘C54 ∘C.

Practical and data questions

One to three questions on each paper test practical work directly: which technique purifies a solid (recrystallization), how a reading should be recorded (to the precision of the instrument, so a balance reading to ±0.01 g±0.01 g is recorded as 5.00 g5.00 g, not 5 g5 g), what the percentage uncertainty of a reading is, which property can be monitored to follow a rate, or which graph shows a change in conductivity during a titration. The ideas are those of Paper 1B, asked in a single step.

Traps hidden in the distractors

The same slips appear across the whole course. Recognising them makes most distractors easy to reject.

Trap

Example of the distractor

Stoichiometry ignored

The 1 : 2 mole ratio in an equation not applied; two OHX−OHX− ions per Ca(OH)X2Ca(OH)X2​ forgotten; the number of atoms per molecule not counted

A power of ten

6.02×10246.02×1024 for 6.02×10236.02×1023; a volume left in cm3cm3 when dm3dm3 was needed

A reversed sign or subtraction

Bond enthalpy "formed minus broken"; the cell potential calculated the wrong way round; the reverse activation energy taken for the forward one

Right number, wrong question

The molecular formula offered when the empirical formula is asked for; the number of molecules when atoms are asked for

State symbols

NaCl(g)NaCl(g) in a lattice enthalpy equation; HX2O(g)HX2​O(g) where the standard state is liquid; deposition confused with condensation

Sublevels overlooked

[Ar] 4s23d4[Ar]4s23d4 for chromium; a p orbital said to hold six electrons (a p sublevel holds six)

Inverted proportionality

Volume taken as proportional to pressure; a temperature left in ∘C∘C in a gas-law ratio

Terms confused

Surface area and particle size; emission and absorption; Lewis and Brønsted–Lowry acids; "more collisions" in place of "more collisions with enough energy"

Bonds broken on boiling

Covalent bonds broken when a molecular substance boils; a giant covalent solid described as ionic

Exam tip

When your answer matches an option, look at the options either side of it. If one of them is exactly double, half, the reverse sign or a power of ten away from yours, check the step that would produce it. This takes ten seconds and catches most careless errors.

Worked multiple-choice questions

Each question is followed by its answer, the reasoning and the slip behind each distractor. Try each one before opening the solution. Questions 1 to 11 and 13 use SL content; question 12 is HL.

Question 1: counting atoms

Worked example

How many oxygen atoms are present in 0.200 mol0.200 mol of sulfuric acid, HX2SOX4HX2​SOX4​? (NA=6.02×1023 mol−1NA​=6.02×1023 mol−1)

  • A. 1.20×10231.20×1023

  • B. 4.82×10234.82×1023

  • C. 8.43×10238.43×1023

  • D. 4.82×10244.82×1024

Show the answer

Answer B. Each molecule contains four oxygen atoms, so the amount of oxygen atoms is 4×0.200=0.800 mol4×0.200=0.800 mol, and the number is 0.800×6.02×1023=4.82×10230.800×6.02×1023=4.82×1023.

Distractors. A is the number of molecules, not atoms. C counts all seven atoms in each molecule. D has the right digits and the wrong power of ten.

Question 2: a gas under new conditions

Worked example

A fixed mass of an ideal gas occupies 2.00 dm32.00 dm3 at 27 ∘C27 ∘C. The pressure is halved and the temperature is raised to 127 ∘C127 ∘C. What is the new volume, in dm3dm3?

  • A. 1.33

  • B. 4.00

  • C. 5.33

  • D. 18.8

Show the answer

Answer C. Use P1V1T1=P2V2T2T1​P1​V1​​=T2​P2​V2​​ with temperatures in kelvin (300 K and 400 K). Halving the pressure doubles the volume, and the temperature ratio multiplies it by 400300300400​:

V2=2.00×2×400300=5.33 dm3V2​=2.00×2×300400​=5.33 dm3

Distractors. A inverts the pressure ratio. B ignores the temperature change. D uses Celsius temperatures, 1272727127​, which is the classic trap: a ratio of Celsius temperatures has no physical meaning.

Question 3: a molecular shape

The options for this question are given as a combination table.

Electron domain geometry

Molecular geometry

A.

trigonal planar

bent

B.

tetrahedral

tetrahedral

C.

linear

linear

D.

tetrahedral

bent

Worked example

What are the electron domain geometry and the molecular geometry of oxygen difluoride, OFX2OFX2​?

Show the answer

Answer D. Oxygen has two bonding pairs (to the two fluorine atoms) and two lone pairs: four electron domains, arranged tetrahedrally. The molecular geometry describes the positions of the atoms only, so with two of the four positions occupied by lone pairs the molecule is bent (about 103∘103∘).

Distractors. A counts only one of the lone pairs. B counts the lone pairs in the molecular geometry as well. C ignores the lone pairs altogether, as if two bonds always gave a linear molecule.

Question 4: an order of boiling points

Worked example

Propane, CHX3CHX2CHX3CHX3​CHX2​CHX3​, methoxymethane, CHX3OCHX3CHX3​OCHX3​, and ethanol, CHX3CHX2OHCHX3​CHX2​OH, have similar molar masses. What is the order of increasing boiling point?

  • A. methoxymethane < propane < ethanol

  • B. propane < ethanol < methoxymethane

  • C. propane < methoxymethane < ethanol

  • D. ethanol < methoxymethane < propane

Show the answer

Answer C. With similar molar masses, the London forces are similar, and the other intermolecular forces decide the order. Propane is non-polar, with London forces only. Methoxymethane is polar, so it also has dipole–dipole forces, but it has no hydrogen atom bonded to oxygen and cannot form hydrogen bonds between its molecules. Ethanol forms hydrogen bonds. The boiling points are about −42−42, −24−24 and 78 ∘C78 ∘C.

Distractors. A ignores the dipole–dipole forces in methoxymethane. B treats methoxymethane as hydrogen-bonded because it contains oxygen and hydrogen. D is the order reversed.

Question 5: a bond-enthalpy expression

Worked example

Which expression gives the enthalpy change, in kJkJ, for the hydrogenation of ethene, CX2HX4(g)+HX2(g)→CX2HX6(g)CX2​HX4​(g)+HX2​(g)​CX2​HX6​(g)? Bond enthalpies / kJ mol−1kJ mol−1: C=CC=C 614, H−HH−H 436, C−CC−C 346, C−HC−H 414.

  • A. (614+436)−(346+2(414))(614+436)−(346+2(414))

  • B. (346+2(414))−(614+436)(346+2(414))−(614+436)

  • C. (614+436)−(346+414)(614+436)−(346+414)

  • D. (614+436)−346(614+436)−346

Show the answer

Answer A. The four C–H bonds of ethene are still present in ethane, so they cancel. The bonds broken are one C=CC=C and one H−HH−H; the bonds formed are one C−CC−C and two new C−HC−H. ΔH=∑(broken)−∑(formed)=1050−1174=−124 kJΔH=∑(broken)−∑(formed)=1050−1174=−124 kJ.

Distractors. B is formed minus broken, which reverses the sign. C counts only one new C–H bond. D forgets the new C–H bonds entirely.

Question 6: reading an energy profile

Energy profile of an exothermic reaction: potential energy against reaction coordinate. Arrow P rises from the reactant level to the peak, Q rises from the product level to the peak, R points down from the reactant level to the product level and S points up from the product level to the reactant…Energy profile of an exothermic reaction: potential energy against reaction coordinate. Arrow P rises from the reactant level to the peak, Q rises from the product level to the peak, R points down from the reactant level to the product level and S points up from the product level to the reactant…
Energy profile for worked question 6. The dashed lines mark the energy of the reactants and of the transition state at the peak.

Worked example

The energy profile is for an exothermic reaction. Which arrow represents the activation energy of the reverse reaction?

  • A. P

  • B. R

  • C. S

  • D. Q

Show the answer

Answer D. The reverse reaction starts from the products, so its activation energy is the rise from the product level to the transition state: arrow Q. It equals the forward activation energy plus the magnitude of ΔHΔH.

Distractors. P is the activation energy of the forward reaction. R is ΔHΔH for the forward reaction (negative, exothermic). S is ΔHΔH for the reverse reaction, which is positive.

Question 7: rate and amount of product

Worked example

Calcium carbonate chips react with an excess of dilute hydrochloric acid. Which change increases the initial rate of reaction but does not change the total volume of carbon dioxide produced?

  • A. Using the same mass of calcium carbonate as a powder

  • B. Using a larger mass of the same chips

  • C. Adding more acid of the same concentration

  • D. Carrying out the reaction at a lower temperature

Show the answer

Answer A. The acid is in excess, so the calcium carbonate is the limiting reactant and fixes the amount of carbon dioxide. A powder of the same mass has a much larger surface area, so collisions with acid particles are more frequent and the rate increases, but the amount of calcium carbonate, and therefore of carbon dioxide, is unchanged.

Distractors. B increases the volume of gas as well as the rate. C leaves the concentration, and so the rate, unchanged. D decreases the rate.

Question 8: the equilibrium constant

Worked example

For the equilibrium 2 SOX2(g)+OX2(g)⇌2 SOX3(g)2SOX2​(g)+OX2​(g)​2SOX3​(g), ΔH=−196 kJΔH=−196 kJ. Which change increases the value of the equilibrium constant, KK?

  • A. Adding a catalyst

  • B. Increasing the pressure

  • C. Decreasing the temperature

  • D. Adding more oxygen

Show the answer

Answer C. The value of KK depends only on temperature. The forward reaction is exothermic, so lowering the temperature shifts the equilibrium to the right and increases KK.

Distractors. A catalyst speeds up the forward and reverse reactions equally and changes neither the position of equilibrium nor KK. Increasing the pressure and adding oxygen both shift the position of equilibrium to the right, but KK stays the same: the concentrations adjust until the ratio again equals the same KK.

Common mistake

Treating "the position of equilibrium moves to the right" and "KK increases" as the same thing. A change in concentration or pressure moves the position of equilibrium without changing KK. Only a change in temperature changes the value of KK.

Question 9: diluting a strong acid

Worked example

10.0 cm310.0 cm3 of hydrochloric acid with a pH of 2.00 is diluted with water to a total volume of 1.00 dm31.00 dm3. What is the pH of the diluted solution?

  • A. 1.00

  • B. 2.00

  • C. 3.00

  • D. 4.00

Show the answer

Answer D. The volume increases from 10.0 cm310.0 cm3 to 1000 cm31000 cm3, a factor of 100, so [HX+][HX+] falls from 1.00×10−21.00×10−2 to 1.00×10−4 mol dm−31.00×10−4 mol dm−3 and the pH rises by 2 units.

Distractors. A lowers the pH, as if dilution made the acid stronger. B assumes the pH does not change. C treats a hundredfold dilution as a tenfold one.

Question 10: recognising reduction

Worked example

In which change is nitrogen reduced?

  • A. NHX3→NHX4X+NHX3​​NHX4​X+

  • B. NOX3X−→NOX2NOX3​X−​NOX2​

  • C. NO→NOX2NO​NOX2​

  • D. NOX2→NOX3X−NOX2​​NOX3​X−

Show the answer

Answer B. Reduction is a decrease in oxidation state. In NOX3X−NOX3​X− nitrogen is +5+5; in NOX2NOX2​ it is +4+4.

Distractors. In A nitrogen is −3−3 on both sides: adding HX+HX+ is an acid–base change, not a redox change. In C the oxidation state rises from +2+2 to +4+4, and in D from +4+4 to +5+5: both are oxidations.

Question 11: recording a measurement

Worked example

A student weighs a sample on a balance with an uncertainty of ±0.01 g±0.01 g and records its mass as 2.50 g2.50 g. Which statements are correct?

  • I. The percentage uncertainty in the mass is 0.4 %.

  • II. Repeating the weighing on the same balance reduces the uncertainty of each reading.

  • III. The mass is recorded to three significant figures.

  • A. I and II only

  • B. I and III only

  • C. II and III only

  • D. I, II and III

Show the answer

Answer B. I is true: 0.012.50×100=0.4 %2.500.01​×100=0.4 %. III is true: the final zero is significant, because the balance reads to two decimal places. II is false: the uncertainty of a single reading is set by the instrument. Repeating measurements reduces the effect of random errors on the mean, but each reading is still ±0.01 g±0.01 g.

Question 12: a Lewis acid (HL)

Worked example

Which species is a Lewis acid but not a Brønsted–Lowry acid?

  • A. BFX3BFX3​

  • B. HClHCl

  • C. NHX4X+NHX4​X+

  • D. HX2OHX2​O

Show the answer

Answer A. Boron in BFX3BFX3​ has only six electrons in its outer shell and can accept an electron pair, so BFX3BFX3​ is a Lewis acid. It has no hydrogen atom to donate as a proton, so it cannot be a Brønsted–Lowry acid.

Distractors. HClHCl, NHX4X+NHX4​X+ and HX2OHX2​O can all donate a proton, so each can act as a Brønsted–Lowry acid.

Question 13: structural isomers

Worked example

Which compound is a structural isomer of butan-1-ol, CHX3CHX2CHX2CHX2OHCHX3​CHX2​CHX2​CHX2​OH?

  • A. butanal

  • B. butanoic acid

  • C. 2-methylpropan-2-ol

  • D. but-3-en-1-ol

Show the answer

Answer C. Structural isomers have the same molecular formula and different structures. Butan-1-ol is CX4HX10OCX4​HX10​O, and so is 2-methylpropan-2-ol, (CHX3)X3COH(CHX3​)X3​COH.

Distractors. Butanal and but-3-en-1-ol are both CX4HX8OCX4​HX8​O: two hydrogen atoms fewer. Butanoic acid is CX4HX8OX2CX4​HX8​OX2​. Each has four carbon atoms, which is the feature the distractors rely on; the molecular formula must be checked in full.

Using the whole paper well

Paper 1A and Paper 1B are taken together in one sitting, and you decide how to divide the time between them. Many students answer the multiple-choice questions first, because they are quick and build confidence, and then move to the data-based questions with the remaining time. Whichever order you choose, decide it before the exam and keep an eye on the clock.

  • Keep a steady pace. Most questions take well under two minutes. A question that is taking much longer is better left, with a provisional answer marked, than allowed to eat into the rest of the paper.

  • Answer every question. There is no penalty for a wrong answer, so a blank is a mark given away. When unsure, eliminate what you can and choose from what is left.

  • Change an answer only for a reason. A second reading that finds a definite error (a missed coefficient, a misread unit) is worth acting on; a vague feeling is not. First answers reached by sound reasoning are more often right.

  • Ignore patterns in the letters. Keys are spread roughly evenly over A to D, but runs of the same letter happen. Choose each answer on its own merits.

  • Check the answer sheet. Make sure each answer is entered against the right question number, especially after skipping a question.

Exam tip

Use the data booklet during Paper 1A as freely as in the written papers. Questions on electronegativity, bond enthalpies, ionization energies, electrode potentials, indicators and spectroscopic data assume that you look the values up rather than remember them.

Self-check

  • To do:

    I can recognise the five common formats of a Paper 1A question and the quickest approach to each.

  • To do:

    I can predict an answer before reading the options and then explain the slip behind each distractor.

  • To do:

    I can solve a combination-table question by deciding one column at a time.

  • To do:

    I can judge statements I, II and III independently and use the fixed option set to eliminate quickly.

  • To do:

    I can check a bond-enthalpy, Hess or equilibrium expression for direction, coefficients and completeness.

  • To do:

    I can spot the common traps: mole ratios, powers of ten, reversed signs, state symbols, °C for K.

  • To do:

    I can distinguish between a change in the position of equilibrium and a change in the value of K.

  • To do:

    I can answer short practical questions on precision, uncertainty and the choice of technique.

  • To do:

    I can pace the paper, mark questions to return to, and leave no question unanswered.

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