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

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

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Multiple choice Written

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Question 1

Multiple choiceMedium1 mark

An ionic solid is made from X²⁺ and Y⁻ ions in a three-dimensional lattice. Which description gives both its empirical formula and its electrical conductivity in the solid and molten states?

Show answer and explanation

Answer: C. One X²⁺ requires two Y⁻ ions to give an electrically neutral formula, XY₂. In the solid lattice, ions are fixed in position and cannot carry charge through the solid. When molten, the ions are mobile and conduct electricity.
A: assumes a 1:1 ion ratio, so the formula does not balance charge.
B: reverses the conductivity behaviour: ions are fixed in the solid but mobile when molten.
D: assumes ions cannot conduct even when molten, overlooking their mobility.
Remember: ionic compounds conduct when their ions are mobile.

Question 2

Paper 2Medium6 marks

Aluminium oxide is a hard material used in abrasive surfaces. An aluminium atom has electron configuration 2,8,3, and an oxygen atom has electron configuration 2,6. In forming ions in this compound, each atom attains a full outer energy level.

(a)

Deduce the ion formed by an aluminium atom, stating the number of electrons transferred.

[2]

(b)

Deduce the ion formed by an oxygen atom, stating the number of electrons transferred.

[2]

(c)

Use your answers to (a) and (b) to determine the simplest formula of aluminium oxide. Show how the charges determine the ratio of ions.

[2]

Question 3

Multiple choiceMedium1 mark

For molecules whose central atoms have four electron domains, the graph shows mean bond angle against the number of lone pairs on the central atom.

What is the average decrease in mean bond angle per added lone pair from 0 to 2 lone pairs, and the molecular geometry at 2 lone pairs?

Show answer and explanation

Answer: A. The average decrease is (109.5° − 104.5°) / (2 − 0) = 2.5° per added lone pair. With four electron domains and two lone pairs, the molecular geometry is bent.

B: subtracts the endpoint angles but does not divide by the two added lone pairs, giving 5.0° per lone pair.
C: assigns the trigonal-pyramidal geometry associated with one lone pair rather than the bent geometry for two.
D: combines the undivided angle change with the one-lone-pair geometry.

Remember: molecular geometry describes the arrangement of atoms, not lone pairs.

Question 4

Paper 2Medium4 marks

Figure 1 shows Lewis representations of ammonia, $\ce{NH3}$ and methane, $\ce{CH4}$ Each line represents a single covalent bond. The arrangements shown do not represent the three-dimensional molecular shapes.

(a)

State the electron-domain geometry around the central atom in each molecule.

[1]

(b)

Compare the molecular geometries of $\ce{NH3}$ and $\ce{CH4}$

[1]

(c)

The H–N–H bond angle in $\ce{NH3}$ is smaller than the H–C–H bond angle in $\ce{CH4}$ Explain why.

[2]

Question 5

Multiple choiceMedium1 mark

In the metallic model, sodium forms Na⁺ ions and magnesium forms Mg²⁺ ions; Mg²⁺ has a smaller radius than Na⁺. Which row predicts the metal with stronger metallic bonding and the metal with the higher melting point, respectively?

Show answer and explanation

Answer: C. Mg²⁺ has a greater charge and smaller radius than Na⁺, so it attracts delocalized electrons more strongly. Stronger metallic bonding requires more energy to overcome during melting, so Mg has the higher melting point.

A: treats the larger radius of Na⁺ as strengthening the attraction and carries that reversal through to melting point.
B: identifies Mg as more strongly bonded but reverses the relationship between bonding strength and melting point.
D: assigns the higher melting point to Mg but incorrectly assigns stronger metallic bonding to Na.

Remember: greater cation charge and smaller radius strengthen metallic bonding; stronger bonding generally corresponds to a higher melting point.

Question 6

Paper 2Hard12 marks

Three metals are compared using the simplified metallic model. Figure 1 shows formula-unit representations of their chlorides. The chloride-ion charges are shown, but the metal-ion charges are omitted. Table 1 gives the cation radii adopted for the model and the melting points of the pure metals at standard pressure. The students use melting point as evidence of the relative strength of metallic bonding.

Metal	Cation radius adopted in the model / pm	Melting point / °C
K	138	63.4
Na	102	97.8
Mg	72	650

(a)

Use Figure 1 to deduce the charge of the metal ion in each of the three compounds.

[3]

(b)

A student wants to examine the effect of cation radius on melting point while keeping cation charge constant. Identify the appropriate pair of metals, calculate the difference in their cation radii and melting points, and use the results to explain which metal has stronger metallic bonding according to the model.

[4]

(c)

Compare magnesium with sodium to explain their difference in melting points. Refer to both cation charge and radius.

[2]

(d)

A student concludes that the rise in melting point from sodium to magnesium is caused only by the increase in cation charge. Assess whether this conclusion is justified, using evidence from Figure 1 and Table 1.

[3]

Question 7

Multiple choiceMedium1 mark

Two batches of the same copper–zinc metallic material have the compositions shown. Use $A_\mathrm{r}(\mathrm{Cu})=64.0$ and $A_\mathrm{r}(\mathrm{Zn})=65.0$ Which option gives the amount ratios $n(\mathrm{Cu}):n(\mathrm{Zn})$ and the correct classification?

Batch	Mass of Cu / g	Mass of Zn / g
I	6.40	6.50
II	12.8	6.50

Show answer and explanation

Answer: C. For an element, $M$ its molar mass in $\mathrm{g\,mol^{-1}}$ has the same numerical value as $A_\mathrm{r}$ use $n=m/M$

[1] Batch I gives $6.40/64.0 : 6.50/65.0 = 0.100:0.100 = 1:1$
[1] Batch II gives $12.8/64.0 : 6.50/65.0 = 0.200:0.100 = 2:1$
[1] The differing ratios are consistent with a mixture, which does not have a fixed ratio of elements.

Total: [3]

A: calculates both ratios correctly but classifies the material as a compound, which has a fixed element ratio.
B: reverses the Cu:Zn ratio for batch II.
D: swaps the ratios for batches I and II.

Remember: alloys are mixtures, not compounds with a fixed element ratio.

Question 8

Paper 2Medium6 marks

Brass is an alloy of copper and zinc. It is generally harder than pure copper. Figure 1 shows schematic representations of two brass samples and the structure of polymer P.

(a)

Use Figure 1(a) to explain why brass is classified as a mixture rather than a compound.

[2]

(b)

Using Figure 1(a), explain how the structure of brass can account for its greater hardness than pure copper. Include the role of metallic bonding.

[2]

(c)

For polymer P in Figure 1(b), name the monomer and state the structural feature that identifies P as a polymer.

[2]

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