IB DP Chemistry Structure 1.2: The nuclear atom – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set practises Structure 1.2, The nuclear atom, from the IB’s Diploma Programme Chemistry guide (first assessment in 2025; published February 2023). It tests Structure 1.2.1 and 1.2.2, for SL and HL, and Structure 1.2.3, which is HL only; questions on mass spectra are labelled “HL only”. They apply to the DP Chemistry course first assessed in 2025. The last two questions are extended problems drawing on more than one understanding.

Revise first with the nuclear atom study guide and the nuclear atom revision notes. Other topics are on the IB DP Chemistry hub, and the printable checklist helps you track each understanding. To spot gaps elsewhere, use the free diagnostics.

Where a question needs data, use the values given in it.

Questions

1. Copy and complete the table for these three species.

Species	Protons	Neutrons	Electrons
⁴⁵₂₁Sc³⁺
¹²⁷₅₃I⁻
²⁰⁷₈₂Pb²⁺

[3]

2. A species contains 34 protons, 46 neutrons and 36 electrons. Write its nuclear symbol, including its charge. [2]

3. This question is about the particles in an atom.

(a) Give the relative charge and the relative mass of a proton and of an electron. [2] (b) A hydrogen-1 atom has one proton and one electron. The mass of a proton is 1.673 × 10⁻²⁷ kg and the mass of an electron is 9.109 × 10⁻³¹ kg. Calculate the percentage of the atom’s mass that is due to its electron. [2]

4. ⁵⁸Ni and ⁶⁰Ni are isotopes of nickel.

(a) What is meant by the term isotopes? [1] (b) Explain why ⁵⁸Ni and ⁶⁰Ni have identical chemical properties. [2] (c) Suggest one physical property that differs between pure samples of the two isotopes. [1]

5. Thallium contains 29.5% ²⁰³Tl (relative isotopic mass 202.97) and 70.5% ²⁰⁵Tl (relative isotopic mass 204.97). Calculate the relative atomic mass of thallium, to two decimal places. [3]

6. Indium has two naturally occurring isotopes, ¹¹³In (relative isotopic mass 112.90) and ¹¹⁵In (relative isotopic mass 114.90). Its relative atomic mass is 114.82. Calculate the percentage abundance of each isotope. [4]

7. Ordinary water is ¹H₂¹⁶O. Heavy water is ²H₂¹⁶O, in which both hydrogen atoms are hydrogen-2.

(a) Using mass numbers, calculate the relative molecular mass of each molecule. [1] (b) Give the electron count of each molecule, and use it to explain why the two waters take part in the same chemical reactions. [2] (c) Suggest, with a reason, which liquid has the higher density. [1]

8. (HL only) The mass spectrum of a metal shows two peaks, both from 1+ ions. The peak at m/z 151 has a relative height of 91.6 and the peak at m/z 153 has a relative height of 100.0.

(a) State what each peak represents. [1] (b) Calculate the percentage abundance of the isotope with m/z 151. [1] (c) Calculate the relative atomic mass of the metal, to two decimal places. [2] (d) Identify the metal, using these relative atomic masses: Sm 150.36, Eu 151.96, Gd 157.25. [1]

9. (HL only) The mass spectrum of a sample of nickel gives these results for 1+ ions.

m/z	58	60	61	62	64
Relative abundance / %	68.1	26.2	1.2	3.6	0.9

(a) State the number of isotopes in the sample. [1] (b) Calculate the relative atomic mass of nickel from these data, to two decimal places. [2] (c) Explain why your answer to (b) lies much nearer 58 than 64. [1]

10. Rubidium has two isotopes: ⁸⁵Rb (relative isotopic mass 84.91) and ⁸⁷Rb (relative isotopic mass 86.91). The atomic number of rubidium is 37.

(a) Deduce the proton, neutron and electron counts for a ⁸⁷Rb⁺ ion. [1] (b) A natural sample of rubidium has a relative atomic mass of 85.47. Calculate the percentage abundance of ⁸⁵Rb in this sample. [3] (c) A laboratory prepares a sample that is 50.0% of each isotope. Calculate its relative atomic mass, to two decimal places. [2] (d) Rubidium chloride made from the prepared sample and from the natural sample react in the same way with silver nitrate solution. Explain why. [2]

11. (HL only) The mass spectrum of a transition metal shows peaks for 1+ ions at the m/z values below. The heights are given relative to the tallest peak.

m/z	90	91	92	94	96
Relative height	100.0	21.8	33.3	33.8	5.4

(a) Calculate the percentage abundance of the isotope with m/z 90, to one decimal place. [2] (b) Calculate the relative atomic mass of the metal, to two decimal places. [3] (c) Identify the metal, using these relative atomic masses: Y 88.91, Zr 91.22, Nb 92.91. Explain why your Ar is not exactly equal to the value you matched. [2] (d) The isotope with m/z 94 forms an ion with a 4+ charge. Deduce how many protons, neutrons and electrons this ion has. [2]

Answers

1. ⁴⁵₂₁Sc³⁺: 21, 24, 18 [1] ¹²⁷₅₃I⁻: 53, 74, 54 [1] ²⁰⁷₈₂Pb²⁺: 82, 125, 80 [1] Examiner insight: One slip in a row, such as 52 electrons for the iodide ion, loses that row’s mark here; write neutrons = A − Z and electrons = Z − charge beside the table so you can check each entry.

2. A = 34 + 46 = 80 and Z = 34, so the element is selenium [1]; charge = 34 − 36 = −2, giving ⁸⁰₃₄Se²⁻ [1] Examiner insight: A symbol without the charge, or with 2+ instead of 2−, describes a different species; check that more electrons than protons means a negative ion.

3. (a) Proton: relative mass 1, relative charge +1 [1]; electron: relative mass negligible (about 1/1836), relative charge −1 [1] (b) Percentage = 9.109 × 10⁻³¹ / (1.673 × 10⁻²⁷ + 9.109 × 10⁻³¹) × 100 [1] = 0.0544% [1] Examiner insight: In (b) the denominator is the whole atom (proton plus electron); dividing by the proton’s mass alone gives a slightly different value, so write out the full set-up before you calculate.

4. (a) Atoms with the same proton number (same element) but different neutron numbers [1] (b) Both have 28 protons, so the neutral atoms have the same number of electrons [1]; chemical reactions depend on electrons, so the two isotopes react in the same way [1] (c) Any one of: density, melting point, boiling point (the ⁶⁰Ni sample is denser because its atoms are heavier) [1] Examiner insight: “Different number of particles” is too vague for (a); name the neutron, and say “same element” or “same number of protons”.

5. Ar = (202.97 × 29.5 + 204.97 × 70.5) / 100 [1] = (5987.615 + 14450.385) / 100 = 20438.000 / 100 [1] = 204.38 [1] Examiner insight: Keep every digit until the last line; rounding the products early can move the second decimal place, and the question asks for two decimal places.

6. Let the fraction of ¹¹³In be x: 112.90x + 114.90(1 − x) = 114.82 [1] 114.90 − 2.00x = 114.82, so 2.00x = 0.08 [1] x = 0.040, so ¹¹³In = 4.0% [1] ¹¹⁵In = 100 − 4.0 = 96.0% [1] Examiner insight: Set up the equation in terms of one unknown; two answers that do not add to 100% show the setting-up step went wrong.

7. (a) ¹H₂¹⁶O: 18; ²H₂¹⁶O: 20 [1] (b) Each molecule has 10 electrons [1]; the isotopes differ only in neutrons, and the electrons, which take part in reactions, are the same, so the chemical reactions are the same [1] (c) Heavy water: its molecules are heavier but about the same size, so a given volume has more mass [1] Examiner insight: For (b) the link to electrons carries the explanation mark; “they are both water” restates the question rather than explaining it.

8. (a) Each peak is one isotope of the metal (its m/z equals the isotope’s mass for a 1+ ion) [1] (b) 91.6 / (91.6 + 100.0) × 100 = 47.8% [1] (c) Ar = (151 × 91.6 + 153 × 100.0) / 191.6 [1] = 152.04 [1] (d) 152.04 is closest to 151.96, so the metal is europium [1] Examiner insight: Dividing by 100 in (c) instead of the total height 191.6 gives an Ar above 290, which is impossible for these peaks; check that your Ar lies between the two m/z values.

9. (a) 5 [1] (b) Ar = (58 × 68.1 + 60 × 26.2 + 61 × 1.2 + 62 × 3.6 + 64 × 0.9) / 100 [1] = 58.76 [1] (c) ⁵⁸Ni makes up most of the sample (68.1%), so it dominates the weighted mean [1] Examiner insight: A plain average of the five m/z values (61.0) ignores abundance; a weighted mean needs each mass multiplied by its own abundance before dividing.

10. (a) 37 protons, 87 − 37 = 50 neutrons, 37 − 1 = 36 electrons: 37, 50, 36 [1] (b) 84.91x + 86.91(1 − x) = 85.47 [1] 86.91 − 2.00x = 85.47, so x = 1.44 / 2.00 = 0.720 [1] ⁸⁵Rb = 72.0% [1] (c) Ar = (84.91 × 50.0 + 86.91 × 50.0) / 100 [1] = 85.91 [1] (d) Both samples contain rubidium atoms with 37 protons, so the Rb⁺ ions have the same electron arrangement [1]; the reaction with silver nitrate depends on the ions present, not on the neutron number, so it is the same [1] Examiner insight: In (d) a mark depends on naming protons or electrons as the reason; saying “isotopes have the same chemical properties” only repeats the stem.

11. (a) Total height = 100.0 + 21.8 + 33.3 + 33.8 + 5.4 = 194.3 [1] 100.0 / 194.3 × 100 = 51.5% [1] (b) Ar = (90 × 100.0 + 91 × 21.8 + 92 × 33.3 + 94 × 33.8 + 96 × 5.4) / 194.3 [1] = 17743.0 / 194.3 [1] = 91.32 [1] (c) 91.32 is closest to 91.22, so the metal is zirconium [1]; the calculation uses whole-number m/z values rather than exact isotopic masses, so it differs slightly [1] (d) Zirconium has Z = 40: 40 protons and 94 − 40 = 54 neutrons [1]; 40 − 4 = 36 electrons [1] Examiner insight: When heights are measured against the tallest peak, the total is not 100; showing the total of 194.3 makes the rest of your working easy to follow.

Where marks are usually lost

• Writing electrons = Z + charge for positive ions, which adds electrons instead of removing them.
• Changing A or Z when an ion forms; only the electron count changes.
• Using the mass of the most abundant isotope, or a plain average of m/z values, as Ar.
• (HL only) Dividing by 100 when the abundances are relative heights that do not total 100.
• Rounding intermediate products, so a two-decimal-place answer comes out wrong.
• Reverse calculations where the two percentages do not add to 100%.
• Linking isotopes’ identical chemistry to “the same mass” instead of to the same electrons.
• Giving a physical difference between isotopes without saying it comes from the different mass.
• (HL only) Describing how a mass spectrometer works when asked to interpret a spectrum.
• (HL only) Identifying the element from the tallest peak alone instead of from the calculated Ar.

Next steps

• Nuclear atom revision notes
• Nuclear atom study guide
• Structure 1 practice questions, for mixed questions across the whole of Structure 1
• IB DP Chemistry hub
• Printable IB DP Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Structure 1.2, The nuclear atom: International Baccalaureate Organization, Chemistry guide, Diploma Programme, first assessment 2025, published February 2023.