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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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?

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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

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?

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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 3

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?

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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 4

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.

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