IB Chemistry HL Structure 2: Models of bonding and structure questions

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Question 1

Multiple choiceEasy1 mark

A calcium fluoride crystal has the empirical formula CaF₂. Which description of its ionic lattice is consistent with this formula?

Show answer and explanation

Answer: B. An ionic empirical formula gives the simplest whole-number ratio of ions in the lattice. In CaF₂, one Ca²⁺ balances the charge of two F⁻ ions, giving a ratio of 1:2.

A: reverses the subscripts, giving a ratio of 2:1.
C: assigns calcium a +1 charge, so the stated ratio is not charge-balanced.
D: uses equal numbers of ions, leaving the charges unbalanced.

Remember: ionic formulas give ratios of ions in a lattice, not discrete molecules.

An investigation compares the ionic solids magnesium oxide, MgO, and calcium oxide, CaO. Both have the same arrangement of ions in their lattices. The ionic radii for ions in this arrangement are shown below.

Ion	Ionic radius / pm
Mg²⁺	72
Ca²⁺	100
O²⁻	140

(a)

Explain why solid CaO does not conduct electricity but molten CaO does.

[2]

(b)

Use the data to predict which oxide has the larger magnitude of lattice enthalpy. Explain your prediction.

[2]

(c)

Hence, identify which oxide is expected to be less volatile. Explain your answer.

[2]

Question 3

Paper 2Hard11 marks

A controlled preparation is used to investigate the ionic composition of an iron sulfide solid. Each batch contains $12.0\times10^{-3}\ \mathrm{mol}$ of iron ions of unknown charge. Different amounts of sulfide ions, $\mathrm{S^{2-}}$ are added. Assume that the iron ions keep the same charge throughout, no redox occurs, and the only solid formed contains iron and sulfide ions.

Figure 1 shows the measured mass of dried solid for each amount of sulfide added. A preliminary report identifies the solid as iron(III) sulfide, $\mathrm{Fe_2S_3}$

(a)

Describe the pattern in the graph as more sulfide ions are added, and identify the ion that limits formation of the solid beyond the plateau.

[2]

(b)

Use Figure 1 to assess the preliminary report. Deduce the simplest formula and name of the solid.

[5]

(c)

Write balanced equations showing the electron transfer from a neutral iron atom and to a neutral sulfur atom for the ions in the solid identified in (b).

[2]

(d)

Explain the nature of the bonding between the ions in the solid identified in (b).

[2]

Question 4

Multiple choiceEasy1 mark

Four molecules are compared to assess how bond polarity and molecular geometry affect molecular polarity. Which molecule has a net dipole moment?

Show answer and explanation

Answer: A. A molecule has a net dipole when its bond dipoles do not cancel. H $_2$ O is bent, so its two O–H bond dipoles do not cancel.

B: Treats polar C=O bonds as sufficient for molecular polarity, overlooking cancellation in linear CO $_2$ C: Treats polar B–F bonds as sufficient, overlooking cancellation in trigonal-planar BF $_3$ D: Treats polar C–Cl bonds as sufficient, overlooking cancellation in tetrahedral CCl $_4$

Remember: molecular polarity depends on both bond polarity and geometry.

Question 5

Paper 2Medium6 marks

Figure 1 shows reference bond angles for methane, ammonia and water. Each central atom has four electron domains.

(a)

Use Figure 1 to state the bond angle in water and compare the bond angles in ammonia and water.

[2]

(b)

Explain the difference between the bond angles in ammonia and water in terms of the electron domains around their central atoms.

[2]

(c)

Ammonia reacts with a hydrogen ion to form an ammonium ion:

\ce{NH3 + H+ -> NH4+}

Identify the atom that donates the electron pair to $\ce{H+}$ and explain why the N–H bond formed is a coordination bond.

[2]

Question 6

Multiple choiceEasy1 mark

A piece of aluminium can be hammered into a thin sheet without fragmenting. Which feature of its metallic structure explains this malleability?

Show answer and explanation

Answer: C. Metallic bonding is the electrostatic attraction between lattice cations and delocalized electrons. When layers of cations shift during hammering, the electrons still attract the cations in their new positions, so the structure remains bonded.

A: Treats electrons as localized pairs linking particular cations, rather than as delocalized electrons. B: Assigns movement to cations passing through stationary electrons. D: Reverses the electrostatic interaction: like-charged cations repel rather than attract.

Remember: In metallic malleability, layers can shift while attraction to the delocalized electrons is maintained.

Question 7

Paper 2Hard8 marks

A materials team is screening pure chromium, iron and copper for a component. The specification requires a melting point above 1400 °C; among metals that meet this limit, the electrical conductivity at 20 °C should be as high as possible. Approximate values are shown.

Metal	Melting point / °C	Electrical conductivity at 20 °C / S m⁻¹
Chromium	1907	7.9 × 10⁶
Iron	1538	1.00 × 10⁷
Copper	1085	5.96 × 10⁷

(a)

Describe the metallic model for a transition element, including the role of d-electrons.

[2]

(b)

Explain how this model accounts for electrical conductivity.

[2]

(c)

A trainee claims: “The metal with the greatest electrical conductivity must have the strongest metallic bonding and the highest melting point.” Assess this claim using the data and the metallic model, and identify the metal that meets the specification.

[4]

Question 8

Paper 2Hard8 marks

A chemistry student uses the metallic model to explain properties of lithium and to assess claims about Group 1 metals. Figure 1 shows a simplified model of lithium before and during shear. The table gives first ionization energies.

Element	First ionization energy / kJ mol⁻¹
Li	520
Na	496
K	419

A student argues: “The decrease in first ionization energy from Li to K explains the increase in Group 1 reactivity. It also proves that metallic bonding becomes stronger, so potassium must conduct electricity better than lithium.”

(a)

State the nature of the metallic bond in terms of the particles involved.

[1]

(b)

Use Figure 1 to explain how the metallic model accounts for malleability.

[2]

(c)

Explain how the metallic model accounts for electrical conductivity.

[1]

(d)

Evaluate both inferences made by the student, using Figure 1 and the table.

[4]

Question 9

Multiple choiceEasy1 mark

The electronegativity values, χ, of the elements in four binary compounds are shown.

Element	Mg	O	Na	Cl	Ca	F	C
χ	1.2	3.5	0.9	3.0	1.0	4.0	2.5

Which compound is predicted to have the greatest ionic contribution to its bonding?

Show answer and explanation

Answer: C [1]. A larger electronegativity difference indicates a greater ionic contribution. The differences are MgO: |3.5 − 1.2| = 2.3; NaCl: |3.0 − 0.9| = 2.1; CaF₂: |4.0 − 1.0| = 3.0; CF₄: |4.0 − 2.5| = 1.5. CaF₂ has the largest difference.

A: MgO has a substantial difference, but 2.3 is less than 3.0.
B: NaCl is ionic, but its difference of 2.1 is less than 3.0.
D: A high mean electronegativity alone does not indicate a greater covalent contribution; CF₄ is less ionic here because its electronegativity difference is smaller.

Remember: electronegativity difference indicates the ionic contribution to bonding.

Total: [1]

Question 10

Paper 2Hard11 marks

A furnace-lining manufacturer is comparing three binary compounds for a component that must remain solid at 1500 °C and limit electrical conduction. Use the electronegativity values in the Data Booklet for Mg, O, Si and C; for the third compound, use χ(Ni) = 1.9 and χ(Al) = 1.5.

In the qualitative bonding triangle, increasing the electronegativity difference, Δχ, moves a compound towards ionic bonding. When Δχ is small, a higher mean electronegativity favours covalent character, while a lower mean electronegativity favours metallic character.

Tests on the three samples gave the following results:

Compound	Remained solid at 1500 °C	Electrical conductivity in the test
MgO	Yes	Below the detection limit
SiC	Yes	Measurable
NiAl	Yes	High

(a)

For MgO and SiC, calculate Δχ and the mean electronegativity, χ, to two decimal places.

[4]

(b)

Using your results from (a) and the electronegativities given for NiAl, state which bonding contribution is greatest for each compound: ionic, covalent or metallic.

[3]

(c)

An engineer claims: “SiC must be a better electrical insulator than MgO because its smaller electronegativity difference places it closer to covalent bonding; covalent-dominant compounds do not conduct electricity.” Evaluate this claim and recommend one of the three compounds for the furnace lining, using the bonding model and the test results.

[4]

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