2.3 Exam Practice
Exam-style practice questions on The Metallic Model
Section B: Data Analysis (Paper 1B Style)
Calculator and Data Booklet permitted. Show all working clearly.
Question 1: Melting Points of Period 3 Metals Explain
5 marksThe table shows the melting points and selected properties of the Period 3 metals.
| Element | Atomic radius / pm | Delocalised electrons | Melting point / °C |
|---|---|---|---|
| Na | 186 | 1 | 98 |
| Mg | 160 | 2 | 650 |
| Al | 143 | 3 | 660 |
(a) Describe and explain the trend in melting points from Na to Al. [3]
(b) The melting point of silicon (Si) is 1414 °C. Explain why silicon does not fit the metallic trend. [2]
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(a) Melting points increase from Na to Al [1]
The number of delocalised electrons increases from 1 to 3, and the ionic radius decreases from Na⁺ to Al³⁺ [1]
This results in a greater charge density on the cation and more delocalised electrons in the "sea", creating a stronger electrostatic attraction between cations and the delocalised electron sea [1]
(b) Silicon is not a metal; it is a metalloid with a covalent network (giant covalent) structure [1]
Its high melting point is due to strong covalent bonds throughout the lattice (like diamond), not metallic bonding [1]
Section C: Structured Questions (Paper 2 Style)
Show all working. State answers with appropriate significant figures and units.
Question 2: Properties of Metals Explain
4 marksCopper is used in electrical wiring and gold is used in jewellery.
(a) Explain why copper is a good electrical conductor using the metallic bonding model. [2]
(b) Explain why metals are malleable, unlike ionic compounds which are brittle. [2]
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(a) Copper has delocalised electrons throughout the metallic lattice [1]
When a potential difference is applied, these delocalised electrons can move freely through the structure, carrying charge [1]
(b) In metals, the metallic bonding is non-directional. When a force is applied, layers of cations can slide over each other without disrupting the bonding, because the delocalised electron sea adjusts and maintains attraction [1]
In ionic compounds, shifting layers brings ions of the same charge into alignment, causing electrostatic repulsion and the crystal shatters [1]
Question 3: Transition Metals and d-Electrons HL Explain
5 marksTransition metals such as iron (Fe), chromium (Cr), and tungsten (W) have particularly high melting points compared to s-block metals like sodium (Na) and potassium (K).
(a) Explain, using the concept of d-electron delocalization, why transition metals generally have higher melting points than s-block metals. [3]
(b) Tungsten has the highest melting point of any metal (3422 °C). Suggest why it is used as the filament in incandescent light bulbs. [1]
(c) Explain why electrical conductivity in metals decreases as temperature increases. [1]
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(a) In s-block metals, only the outermost s-electrons are delocalised (1 or 2 electrons per atom) [1]
Transition metals can also delocalise their d-electrons into the "electron sea", increasing the total number of delocalised electrons per atom [1]
More delocalised electrons and smaller ionic radii result in a stronger electrostatic attraction between the cations and the electron sea, giving higher melting points [1]
(b) Its extremely high melting point allows the filament to reach very high temperatures (>2000 °C) without melting, producing visible light through incandescence [1]
(c) At higher temperatures, cations vibrate more vigorously, scattering the delocalised electrons more frequently and reducing their net drift velocity / flow of charge [1]