Questions & Answers

Q1a(i).
Short Answer 1 mark

Diamond and graphite are both crystalline solids made from carbon atoms. Graphite conducts electricity. Diamond does not conduct electricity. Name the type of lattice structure present in the crystalline solids diamond and graphite.

Model Answer:
Giant covalent lattice
Explanation: Both diamond and graphite consist of a giant covalent network of carbon atoms.
Q1a(ii).
Short Answer 1 mark

Explain how graphite conducts electricity.

Model Answer:
Graphite has delocalised electrons that are free to move through the structure and carry charge.
Explanation: The mobile delocalised electrons allow electrical conduction.
Q1a(iii).
Short Answer 1 mark

Explain why diamond does not conduct electricity.

Model Answer:
Diamond has no delocalised electrons / all outer electrons are used in covalent bonds, so there are no mobile charge carriers.
Explanation: Without mobile electrons or ions, diamond cannot conduct electricity.
Q1b(i).
Short Answer 2 marks

Separate samples of phosphorus(V) chloride and silicon(IV) chloride are each added to an excess of cold water. Write an equation for each reaction.

Model Answer:
PCl5 + 4H2O -> H3PO4 + 5HCl
Model Answer:
SiCl4 + 2H2O -> SiO2 + 4HCl
Q1b(ii).
Short Answer 2 marks

Describe the appearance of phosphorus(V) chloride and silicon(IV) chloride at room temperature.

Model Answer:
phosphorus(V) chloride: white solid
Model Answer:
silicon(IV) chloride: colourless liquid
Q1b(iii).
Short Answer 2 marks

Compare the appearance of the mixtures produced when each reaction is complete.

Model Answer:
PCl5 gives a white mist / white fumes due to HCl gas; SiCl4 gives a white precipitate / white solid (silica) in a colourless solution.
Q1c(i).
Short Answer 1 mark

Some oxides are amphoteric. Describe what is meant by amphoteric.

Model Answer:
Able to react as both an acid and a base.
Q1c(ii).
Short Answer 1 mark

Identify the formula of a Period 3 oxide that is amphoteric.

Model Answer:
Al2O3
Q1d(i).
Short Answer 3 marks

Complete Table 1.1 by identifying the strongest force of attraction in each oxide that is broken during melting. Use the abbreviations i.d. = instantaneous dipole-induced dipole, p.d. = permanent dipole-permanent dipole, H = hydrogen bond, C = covalent bond, I = ionic bond.

Model Answer:
SO2: p.d.
Model Answer:
H2O: H
Model Answer:
SiO2: C
Model Answer:
Na2O: I
Model Answer:
MgO: I
Q1d(ii).
Short Answer 2 marks

A student suggests the following hypothesis. The stronger the covalent bond between atoms in non-metal oxides, the higher the melting point. Use Table 1.1 to deduce if this hypothesis is true or false or if there is not enough information to make a conclusion. Explain your answer.

Model Answer:
Not enough information to make a conclusion.
Explanation: Table 1.1 compares different types of bonding/structure, not just covalent bond strength in non-metal oxides. Melting point depends on more than covalent bond strength alone.
Q2a(i).
Short Answer 2 marks

Define relative atomic mass.

Model Answer:
The weighted mean mass of an atom of an element compared with 1/12 of the mass of an atom of carbon-12.
Q2a(ii).
Short Answer 2 marks

Table 2.1 shows the abundances of two of the isotopes present in the sample of iron. Use Table 2.1 to calculate the relative isotopic mass of the third isotope of iron in the sample. Show your working.

Model Answer:
56
Explanation: Using the weighted average: 55.8 = (53.9×6.0 + 55.9×91.9 + x×2.1)/100, giving x ≈ 56.
Q2b.
Short Answer 1 mark

Deduce the number of pairs of electrons in the 3d sub-shell in an iron(II) ion.

Model Answer:
2
Q2c.
Short Answer 1 mark

Sketch the shape of the lowest energy orbital in the shell with principal quantum number n = 2.

Model Answer:
A spherical 2s orbital.
Q2d.
Short Answer 2 marks

Complete Table 2.2 to show information about particles in one ion of 56Fe3+.

Model Answer:
electrons = 23
Model Answer:
protons = 26
Q2e.
Short Answer 2 marks

The atomic radius of iron is 1.26 × 10^-10 m. Suggest the change to the radius, if any, after an iron atom reacts to produce an Fe3+ ion. Explain your answer.

Model Answer:
The radius decreases.
Explanation: Electrons are lost to form Fe3+, reducing electron-electron repulsion and increasing the effective attraction between the nucleus and remaining electrons.
Q3a(i).
Short Answer 1 mark

C3H6 reacts with HBr(g) in an addition reaction. Define addition reaction.

Model Answer:
A reaction in which two molecules combine to form one product molecule.
Q3a(ii).
Structured 4 marks

Complete Fig. 3.1 to show the mechanism for the addition reaction between C3H6 and HBr to produce 2-bromopropane. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

Model Answer:
The C=C bond attacks Hδ+ of HBr; the H–Br bond breaks heterolytically to Br−; a secondary carbocation forms; Br− attacks the carbocation to form 2-bromopropane.
Q3a(iii).
Short Answer 2 marks

Explain why the major product of this reaction is 2-bromopropane rather than 1-bromopropane.

Model Answer:
The reaction proceeds via the more stable secondary carbocation intermediate, which is formed preferentially, so 2-bromopropane is the major product.
Q3b(i).
Short Answer 1 mark

H2(g) and nickel are added to alkene X. Fig. 3.2 shows how the concentration of X changes with time. Use Fig. 3.2 to identify the limiting reagent in this reaction. Explain your answer.

Model Answer:
Alkene X is the limiting reagent, because its concentration falls to zero.
Q3b(ii).
Short Answer 1 mark

Use Fig. 3.2 to describe how the gradient changes as the reaction proceeds. State what this shows about the rate during this reaction.

Model Answer:
The gradient becomes less steep with time and eventually becomes zero; the rate decreases and then becomes zero as the reaction finishes.
Q3c.
Short Answer 2 marks

Hydrocarbon Y contains two C=C groups. There are no other functional groups present. Y reacts with an excess of H2(g) to produce hexane, C6H14. Use Table 3.1 to calculate the enthalpy change per mole of C6H14 produced in this reaction.

Model Answer:
enthalpy change = -240 kJ mol^-1
Explanation: Breaking 2 C=C and 2 H-H; forming 2 C-C and 4 C-H gives ΔH = (2×610 + 2×436) - (2×350 + 4×410) = -240 kJ mol^-1.