IB DP Chemistry SL Prediction Paper 2 for 2025 Exams
IB DP Chemistry SL Prediction Paper 2- April/May 2025 Exam
IB DP Chemistry SL Prediction Paper 2: Prepare for the IB exams with subject-specific Prediction questions, model answers. All topics covered.
Prepared by IB teachers: Access our IB DP Chemistry SL Prediction Paper 2 with model answer. Students: Practice with exam-style papers for IB DP Chemistry SL Exam
Question 1: Properties of Group I Elements and Sodium Hydride Reaction [8 marks]
This question examines the trends in physical properties of Group I elements, the definition and uses of radioactive isotopes, and the chemical behavior of sodium hydride in a redox reaction.
a(i) Question (a)(i) [2 marks] – Predicting Properties
(a)(i) Complete the table by predicting the boiling point of rubidium and the atomic radius of potassium.
▶️Answer/Explanation
(a)(i) Predicting Properties:
Boiling point of rubidium (Rb): Group I elements show a decreasing trend in boiling points down the group. Given Na (883°C), K (759°C), and Cs (671°C), Rb’s boiling point is between 671°C and 759°C. Predicted value: Any value between 675°C and 755°C (e.g., 700°C).
Atomic radius of potassium (K): Atomic radius increases down the group due to additional electron shells. Given Na (0.190 nm) and Rb (0.250 nm), K’s radius is between 0.190 nm and 0.250 nm. Predicted value: Any value between 0.195 nm and 0.245 nm (e.g., 0.227 nm).
a(ii) Question (a)(ii) [1 mark] – Melting Point Trend
(a)(ii) Describe the trend in the melting point of the Group I elements down the group.
▶️Answer/Explanation
(a)(ii) Melting Point Trend:
The melting point of Group I elements decreases down the group. This is due to weaker metallic bonding as the atomic radius increases, reducing the strength of the electrostatic forces between the delocalized electrons and the positive ions.
a(iii) Question (a)(iii) [2 marks] – Physical State of Potassium
(a)(iii) Deduce the physical state of potassium at 60°C. Explain your answer.
▶️Answer/Explanation
(a)(iii) Physical State of Potassium:
Physical state: Solid.
Explanation: The melting point of potassium is 63°C (from the table). At 60°C, the temperature is below the melting point, so potassium remains in the solid state.
b(i) Question (b)(i) [1 mark] – Proton Number
(b)(i) Define proton number.
▶️Answer/Explanation
(b)(i) Proton Number:
The proton number is the number of protons in the nucleus of an atom, which determines the element’s identity and its positive charge.
b(ii) Question (b)(ii) [1 mark] – Use of Radioactive Isotopes
(b)(ii) State one industrial use of radioactive isotopes.
▶️Answer/Explanation
(b)(ii) Use of Radioactive Isotopes:
Detecting leaks in pipelines (e.g., using gamma-emitting isotopes to trace fluid flow).
c(i) Question (c)(i) [2 marks] – Balancing Equation
(c)(i) Balance the equation for the reaction: \(\mathrm{Fe}_2\mathrm{O}_3 + 3\mathrm{NaH} \rightarrow \dots \mathrm{Fe} + \dots \mathrm{NaOH}\).
▶️Answer/Explanation
(c)(i) Balancing Equation:
Balance Fe: \(\mathrm{Fe}_2\mathrm{O}_3 \rightarrow 2\mathrm{Fe}\).
Balance Na and H: \(3\mathrm{NaH} \rightarrow 3\mathrm{NaOH}\).
Balance O: Left side has 3 O (from \(\mathrm{Fe}_2\mathrm{O}_3\)), right side has 3 O (from \(3\mathrm{NaOH}\)).
Balanced equation: \(\mathrm{Fe}_2\mathrm{O}_3 + 3\mathrm{NaH} \rightarrow 2\mathrm{Fe} + 3\mathrm{NaOH}\).
c(ii) Question (c)(ii) [1 mark] – Reduction Explanation
(c)(ii) Explain how this equation shows that iron(III) oxide is reduced.
▶️Answer/Explanation
(c)(ii) Reduction Explanation:
Iron(III) oxide (\(\mathrm{Fe}_2\mathrm{O}_3\)) is reduced because it loses oxygen atoms. In the reaction, \(\mathrm{Fe}_2\mathrm{O}_3\) is converted to elemental iron (\(\mathrm{Fe}\)), indicating the removal of oxygen, which is a characteristic of reduction.