IB MYP Sciences – Atomic structure and bonding 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover atomic structure and bonding for IB Middle Years Programme Sciences, aligned to the International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014, which lists this topic for the MYP sciences on-screen examinations. MYP has no SL/HL split; the set suits MYP years 4 and 5.

MYP has no prescribed content list – schools design their own units – and your teacher will share the task-specific clarifications. Questions are labelled by criterion. Real MYP work is judged against criterion level descriptors (eight achievement levels, 1–8, in four bands), so the [1] points below are a revision aid, not IB marks.

Learn the content with the study guide first. See also the course hub and printable checklist.

Questions

1. (Criterion A) Give the relative mass, relative charge and location of a proton, a neutron and an electron. [3]

2. (Criterion A) An atom of potassium is written ³⁹₁₉K. State the number of protons, neutrons and electrons in this atom, and write its electron arrangement. [4]

3. (Criterion A)

(a) State the numbers of protons, neutrons and electrons in the ion ²⁴₁₂Mg²⁺. [2] (b) State the numbers of protons, neutrons and electrons in the ion ³²₁₆S²⁻. [2] (c) Write the electron arrangements of both ions and explain why these ions are stable. [2]

4. (Criterion A and C) A sample of gallium contains two isotopes: 60% gallium-69 and 40% gallium-71.

(a) Define the term isotopes. [1] (b) Calculate the relative atomic mass of gallium in this sample. Give your answer to 1 decimal place. [2] (c) Explain why gallium-69 and gallium-71 have the same chemical properties. [1]

5. (Criterion A)

(a) Magnesium reacts with oxygen to form magnesium oxide. Describe, in terms of electrons, how the ions in magnesium oxide form. You may use a dot-and-cross diagram. [3] (b) Write the formulas of calcium chloride, aluminium oxide and potassium sulfide. [3]

6. (Criterion A)

(a) State the number of shared (bonding) pairs and lone pairs on the nitrogen atom in ammonia, NH₃. [2] (b) Nitrogen gas contains N₂ molecules. Explain why the two nitrogen atoms share three pairs of electrons. [2] (c) Explain why ammonia has a low boiling point even though its covalent bonds are strong. [2]

7. (Criterion C) A student tested five solids, P to T (fictional samples).

Substance	Melting point / °C	Conducts as solid?	Conducts when molten?	Soluble in water?
P	801	No	Yes	Yes
Q	44	No	No	No
R	1710	No	No	No
S	660	Yes	Yes	No
T	2852	No	Yes	No

(a) Classify each substance as giant ionic, simple molecular, giant covalent or giant metallic. [5] (b) Justify your classification of R using two pieces of evidence from the table. [2]

8. (Criterion B) You are asked to investigate how the concentration of sodium chloride solution affects the electric current it carries.

(a) Write a testable hypothesis and explain it using scientific reasoning. [3] (b) Identify the independent variable, the dependent variable and two control variables. [4] (c) State one safety precaution for this investigation. [1]

9. (Criterion C) A class carried out the investigation in Question 8. Results (fictional data) are shown.

Concentration / g dm⁻³	Trial 1 / mA	Trial 2 / mA	Trial 3 / mA
5	42	44	43
10	81	79	83
15	118	122	120
20	161	159	124
25	196	200	198

(a) Identify the anomalous result and calculate the mean current for each concentration, leaving out the anomaly. [3] (b) Describe the trend in the data and explain it in terms of particles. [2] (c) Comment on the reliability of the data and suggest one improvement. [2]

10. (Criterion A)

(a) Describe the structure of a metal and the nature of metallic bonding. [2] (b) Explain why copper conducts electricity as a solid. [1] (c) Explain why steel, an alloy of iron, is harder than pure iron. [2]

11. (Criterion D – extended response) Diamond tips drills and saws that cut stone; industrial diamonds are mined or made synthetically at high temperature and pressure. Graphite is used as a dry lubricant and in electrodes. Discuss and evaluate these uses of carbon. In your answer:

• explain the properties of diamond and graphite in terms of their structure and bonding
• evaluate one advantage and one limitation of synthetic diamonds compared with mined diamonds for industrial cutting tools
• give a justified conclusion. [8]

12. (Criterion A and D) In the gold-foil experiment, alpha particles (which are positive) were fired at a very thin sheet of gold.

(a) Most alpha particles passed straight through; a very few bounced back. Explain what each observation shows about the structure of the atom. [2] (b) A gold atom is about 3.0 × 10⁻¹⁰ m across and its nucleus about 1.5 × 10⁻¹⁴ m across. Calculate how many times wider the atom is than the nucleus. [2]

Answers

1. Proton: mass 1, charge +1, nucleus [1]. Neutron: mass 1, charge 0, nucleus [1]. Electron: mass about 1/1840 (negligible), charge −1, shells [1]. [3] Examiner insight: Each particle’s point needs all three entries right, so “electron mass 0” loses the electron point.

2. Protons = 19 [1]. Neutrons = 39 − 19 = 20 [1]. Electrons = 19 (neutral atom) [1]. Arrangement 2,8,8,1 [1]. [4] Examiner insight: The 19th electron goes into the fourth shell; writing 2,8,9 is the most common way to drop the arrangement point.

3. (a) 12 protons and 12 neutrons [1]; 12 − 2 = 10 electrons [1]. (b) 16 protons and 16 neutrons [1]; 16 + 2 = 18 electrons [1]. (c) Mg²⁺ is 2,8 and S²⁻ is 2,8,8 [1]. Both have full outer shells, the same arrangement as a noble gas (neon and argon), which is stable [1]. Examiner insight: Proton numbers never change when an ion forms; changing them in (a) or (b) loses the point even if the electrons are right.

4. (a) Atoms of the same element with the same number of protons but different numbers of neutrons [1]. (b) Aᵣ = (69 × 60 + 71 × 40) ÷ 100 [1] = (4140 + 2840) ÷ 100 = 69.8 [1]. (c) They have the same electron arrangement (same number of electrons), and chemical reactions depend on electrons [1]. Examiner insight: Show the weighted-mean line; a bare 69.8 leaves no method to credit if the arithmetic slips.

5. (a) Magnesium (2,8,2) loses its two outer electrons [1]. Oxygen (2,6) gains those two electrons [1]. This forms Mg²⁺ (2,8) and O²⁻ (2,8), which attract strongly because they have opposite charges [1]. (b) CaCl₂ [1]; Al₂O₃ [1]; K₂S [1]. Examiner insight: You need the direction of transfer (magnesium to oxygen); “they share electrons” describes covalent bonding and earns nothing.

6. (a) 3 bonding pairs [1] and 1 lone pair [1]. (b) Each nitrogen atom (2,5) needs three more electrons for a full outer shell [1], so each shares three electrons: a triple bond giving 8 outer electrons each [1]. (c) Ammonia is a simple molecular substance; boiling only overcomes the weak forces between molecules [1]. Little energy is needed, and the strong covalent N-H bonds are not broken [1]. Examiner insight: In (c), “between molecules” earns the point; “the bonds are weak” implies covalent bonds break and is wrong.

7. (a) P: giant ionic [1]. Q: simple molecular [1]. R: giant covalent [1]. S: giant metallic [1]. T: giant ionic [1]. (b) R has a very high melting point (1710 °C), so strong bonds throughout a giant structure must be broken [1]. It does not conduct as a solid or when molten, so it has no free ions or delocalised electrons – ruling out ionic and metallic [1]. Examiner insight: T is insoluble but still ionic; justify ionic labels with “conducts only when molten”, not solubility.

8. (a) Hypothesis: if the concentration of sodium chloride solution increases, the current will increase [1], because there will be more ions per unit volume [1] to move and carry charge between the electrodes [1]. (b) Independent: concentration of the solution [1]. Dependent: current, in mA, measured with an ammeter [1]. Controls, any two: supply voltage [1]; electrode distance, material or area, or temperature [1]. (c) Use a low voltage and wear eye protection when handling solutions [1]. Examiner insight: A hypothesis with no reason earns only the prediction point; the reasoning must name moving ions.

9. (a) Anomaly: 124 mA at 20 g dm⁻³ [1]. Means: 5 → 43, 10 → 81, 15 → 120 [1]; 20 → 160, 25 → 198 mA [1]. (b) Current increases with concentration, roughly in proportion (about 8 mA per g dm⁻³) [1], because more ions per unit volume carry charge [1]. (c) Apart from the anomaly, repeats agree within 4 mA, so the data are reliable [1]. Improvement: repeat the 20 g dm⁻³ reading, or add more concentrations [1]. Examiner insight: Including 124 gives a mean of 148 mA and loses the accuracy point; say which value you excluded.

10. (a) A lattice of positive metal ions [1] surrounded by a sea of delocalised electrons; the metallic bond is the strong attraction between the positive ions and the delocalised electrons [1]. (b) Delocalised electrons can move through the structure and carry charge [1]. (c) In steel, atoms of a different size (such as carbon) distort the regular layers of iron ions [1], so the layers cannot slide over each other as easily [1]. Examiner insight: In (b), writing “free ions” loses the point – metals conduct by electrons.

11. Indicative points:

• Diamond: each carbon forms four strong covalent bonds in a giant rigid network [1], so it is extremely hard and cuts stone [1].
• Its very high melting point means frictional heat does not damage the tip [1].
• Graphite: three bonds per carbon in layers held by weak forces, so layers slide – a dry lubricant [1].
• One delocalised electron per carbon, so graphite conducts – used for electrodes [1].
• Advantage of synthetic diamonds: controlled size and quality, and no mining, which disturbs land and ecosystems [1].
• Limitation: very high temperature and pressure mean large energy use [1].
• Justified conclusion, for example: synthetic diamonds suit cutting tools because their hardness comes from the same bonding, provided their energy use is weighed against the impact of mining [1]. [8] Examiner insight: The final point needs a conclusion that follows from your own evaluation, including the limitation you gave.

12. (a) Most passed straight through, so the atom is mostly empty space [1]. A few bounced back, so the positive charge and most of the mass are concentrated in a tiny nucleus that repels the alpha particles [1]. (b) Ratio = (3.0 × 10⁻¹⁰) ÷ (1.5 × 10⁻¹⁴) [1] = 2.0 × 10⁴ (20 000 times) [1]. Examiner insight: In (a), leaving out that the nucleus is positive drops the second point.

Where marks are usually lost

• Giving the electron’s relative mass as 0 rather than negligible or about 1/1840.
• Writing 2,8,9 for potassium or 2,8,10 for calcium instead of starting the fourth shell.
• Subtracting electrons for a negative ion (S²⁻ has 18 electrons, not 14).
• Unbalanced ionic formulas, or charges written inside the formula.
• Saying boiling breaks covalent bonds in simple molecules.
• Classifying from one piece of data only; use melting point and conductivity.
• Means that include an obvious anomaly without comment.
• Extended responses that describe uses but never evaluate or conclude.

Next steps

• Atomic structure and bonding revision notes
• Atomic structure and bonding study guide
• Investigation skills eAssessment preparation and criteria in practice questions
• IB MYP Sciences course hub
• Printable MYP Sciences checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014.