iGCSE Chemistry 0620 Extended - 12.5 Identification of ions and gases- Exam Style Questions Paper 4- New Syllabus
Question
Manganese is the element with atomic number 25 in the Periodic Table.
Calcium is the element with atomic number 20 in the Periodic Table.
(a) Complete Table 5.1 to show the number of protons, neutrons and electrons in the \(^{55}\)Mn atom and the \(^{42}\)Ca\(^{2+}\) ion.
(b) Manganese forms several oxides. The formulae of some of these oxides are shown.
MnO
Mn\(_2\)O\(_3\)
Mn\(_3\)O\(_4\)
MnO\(_2\)
Mn\(_2\)O\(_7\)
(i) Suggest why manganese is expected to form coloured oxides.
(ii) State which other property of manganese is shown by the formation of several oxides.
(iii) State the formula of manganese(II) oxide.
(c) Mn\(_3\)O\(_4\) is found in an ore of manganese. Manganese metal can be extracted from Mn\(_3\)O\(_4\) using aluminium as the reducing agent.
(i) Define the term reducing agent.
(ii) Complete the symbol equation by inserting the formula of the missing product and balancing the equation.
\[ \text{….} \, \text{Mn}_3\text{O}_4 + …. \, \text{Al} \rightarrow \text{……} + …. \, \text{Mn} \]
(d) MnO\(_2\) reacts with dilute hydrochloric acid as shown in the equation.
\[ \text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2 \]
(i) Calculate the volume of chlorine gas formed, in cm\(^3\), at r.t.p. when excess MnO\(_2\) reacts with 50.0 cm\(^3\) of 0.200 mol/dm\(^3\) HCl.
(ii) Describe a test for chlorine gas.
(iii) Explain, in terms of collision theory, why decreasing the temperature decreases the rate of this reaction.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 2.2 — Atomic structure and the Periodic Table (Part (a))
• Topic 8.4 — Transition elements (Parts (b)(i), (b)(ii))
• Topic 6.4 — Redox (Parts (b)(iii), (c)(i))
• Topic 3.3 — The mole and the Avogadro constant (Parts (d)(i), (d)(ii))
• Topic 12.5 — Identification of ions and gases (Part (d)(ii))
▶️ Answer/Explanation
(a)
For \(^{55}\)Mn: protons = atomic number = 25; neutrons = mass number – protons = 55-25 = 30; electrons = protons = 25 (neutral atom).
For \(^{42}\)Ca\(^{2+}\): protons = atomic number = 20; neutrons = 42-20 = 22; electrons = 20-2 = 18 (2+ ion).
(b)(i) (Mn is a) transition element
Transition metal compounds are often colored due to the presence of partially filled d-orbitals which allow electron transitions that absorb visible light.
(b)(ii) variable oxidation state
The different oxides show manganese in different oxidation states (+2 in MnO, +3 in Mn\(_2\)O\(_3\), etc.), demonstrating variable oxidation states.
(b)(iii) MnO
The (II) indicates manganese has a +2 oxidation state in this oxide.
(c)(i) a substance that reduces another substance and is itself oxidised
A reducing agent donates electrons to another substance, causing that substance to be reduced while the reducing agent itself is oxidized.
(c)(ii) \[ 3\text{Mn}_3\text{O}_4 + 8\text{Al} \rightarrow 4\text{Al}_2\text{O}_3 + 9\text{Mn} \]
The thermite reaction produces aluminum oxide and manganese metal. Balancing: 3 Mn\(_3\)O\(_4\) provides 9 Mn atoms, requiring 8 Al to form 4 Al\(_2\)O\(_3\).
(d)(i) 60.0 cm\(^3\)
Steps:
1. Moles HCl = 50.0 cm\(^3\) × 0.200 mol/dm\(^3\) ÷ 1000 = 0.0100 mol
2. From equation, 4 HCl produce 1 Cl\(_2\), so moles Cl\(_2\) = 0.0100 ÷ 4 = 0.00250 mol
3. Volume at r.t.p. = 0.00250 mol × 24000 cm\(^3\)/mol = 60.0 cm\(^3\)
(d)(ii) (damp) litmus (paper) and is bleached/goes white
Chlorine is a powerful oxidizing agent that bleaches the color from litmus paper.
(d)(iii) 1. Kinetic energy of particles decreases
2. Frequency of collisions between particles decreases
3. Lower percentage/proportion/fraction of collisions/particles have energy greater than/equal to activation energy
At lower temperatures, particles move slower with less energy, resulting in fewer effective collisions that can overcome the activation energy barrier.