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IB Chemistry – Structure 2: Bonding and Structure

Ionic, covalent and metallic bonding, VSEPR, intermolecular forces, hybridisation (HL).

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12 questions  ·  41 marks  ·  answers included  ·  IB Chemistry SL & HL (2023+ syllabus)
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This worksheet covers:

  • This worksheet covers ionic
  • Covalent and metallic bonding
  • VSEPR shapes
  • Intermolecular forces
  • (HL only, marked) hybridisation

No calculator is needed; questions marked (HL) are for Higher Level students only.

Questions

Section A – Bonding Models

1

Define ionic bonding and covalent bonding, and state the type of elements that typically form each.

Show answer
Ionic bonding: the electrostatic attraction between oppositely charged ions, typically formed between a metal and a non-metal (large electronegativity difference). Covalent bonding: the electrostatic attraction between a shared pair (or pairs) of electrons and the nuclei of the bonded atoms, typically formed between non-metal atoms (small or no electronegativity difference).
[4]
2

Explain, in terms of structure and bonding, why sodium chloride has a high melting point but does not conduct electricity when solid.

Show answer
NaCl is a giant ionic lattice held together by strong electrostatic forces of attraction between oppositely charged ions acting in all directions, so a large amount of energy is needed to overcome these forces and separate the ions, giving a high melting point. In the solid, the ions are fixed in place in the lattice and cannot move to carry charge, so it does not conduct (it conducts only when molten or dissolved in water, when the ions are free to move).
[3]
3

Metallic bonding in magnesium is stronger than in sodium. Explain this observation in terms of bonding.

Show answer
$\text{Mg}^{2+}$ ions have a greater positive charge and a smaller ionic radius than $\text{Na}^+$ ions, and each Mg atom contributes two delocalised electrons to the “sea” of electrons compared with one for Na. This results in stronger electrostatic attraction between the metal cations and the delocalised electrons, so metallic bonding (and melting point) is stronger in magnesium.
[3]
4

Use VSEPR theory to predict the shape and bond angle of:(a) $\text{CH}_4$(b) $\text{NH}_3$(c) $\text{H}_2\text{O}$

Show answer
(a) 4 bonding pairs, 0 lone pairs: tetrahedral, $109.5^{\circ}$.(b) 3 bonding pairs, 1 lone pair: trigonal pyramidal, approximately $107^{\circ}$.(c) 2 bonding pairs, 2 lone pairs: bent (angular), approximately $104.5^{\circ}$.
[6]

Section B – Intermolecular Forces

Exam-style
5

List the three types of intermolecular force covered in the IB course, and state which is/are present in a sample of pure liquid methanol, $\text{CH}_3\text{OH}$.

Show answer
London (dispersion) forces, dipole-dipole forces, and hydrogen bonding. Methanol contains an O–H bond (O is small and highly electronegative), so it experiences all three: London forces, dipole-dipole forces, and (dominantly) hydrogen bonding between molecules.
[4]
6

Explain why $\text{NH}_3$ (boiling point $-33^{\circ}\text{C}$) has a much higher boiling point than $\text{PH}_3$ (boiling point $-87^{\circ}\text{C}$), even though the molecules have a similar shape and P is below N in Group 15.

Show answer
$\text{NH}_3$ can form hydrogen bonds between molecules, since it has an N–H bond with N being small and highly electronegative. $\text{PH}_3$ cannot form hydrogen bonds because P is larger and much less electronegative, so it only has weaker dipole-dipole and London forces. Since hydrogen bonds are stronger than dipole-dipole/London forces, more energy is needed to separate $\text{NH}_3$ molecules, giving it the higher boiling point.
[3]
7

Explain why the boiling points of the noble gases (Group 18) increase down the group.

Show answer
Going down the group, atoms have more electrons and larger, more polarisable electron clouds, so the strength of the London (dispersion) forces between atoms increases. More energy is therefore required to overcome these intermolecular forces, so boiling point increases down the group.
[3]
8

Predict, with reasoning, which has the higher boiling point: $\text{CO}_2$ or $\text{SiO}_2$.

Show answer
$\text{SiO}_2$ has the much higher boiling point. $\text{CO}_2$ is a simple molecular covalent substance, held in the solid/liquid state by only weak London forces between molecules. $\text{SiO}_2$ is a giant covalent (macromolecular) lattice in which strong covalent bonds extend throughout the whole structure, so far more energy is needed to break these bonds and melt/boil it.
[3]

Section C – HL Extension: Hybridisation and Advanced Bonding

HL only – challenge
9

(HL) State the hybridisation of the carbon atom(s) in:(a) $\text{CH}_4$(b) $\text{C}_2\text{H}_4$ (ethene)(c) $\text{C}_2\text{H}_2$ (ethyne)

Show answer
(a) $sp^3$.(b) $sp^2$.(c) $sp$.
[3]
10

(HL) Explain, in terms of orbitals, how a sigma ($\sigma$) bond and a pi ($\pi$) bond are formed, and state how many of each are present in the triple bond of $\text{N}_2$.

Show answer
A $\sigma$ bond forms by the head-on (axial) overlap of two atomic orbitals directly along the internuclear axis, giving electron density concentrated between the two nuclei. A $\pi$ bond forms by the sideways (lateral) overlap of two parallel, unhybridised p-orbitals, giving electron density above and below the internuclear axis. The triple bond in $\text{N}_2$ consists of 1 $\sigma$ bond and 2 $\pi$ bonds.
[4]
11

(HL) Explain why the bond angle in $\text{NH}_3$ ($107^{\circ}$) is slightly less than the bond angle in $\text{CH}_4$ ($109.5^{\circ}$).

Show answer
$\text{NH}_3$ has one lone pair in addition to three bonding pairs on nitrogen, whereas $\text{CH}_4$ has four equivalent bonding pairs and no lone pairs on carbon. A lone pair occupies more space and repels adjacent bonding pairs more strongly than a bonding pair does (lone pair–bond pair repulsion is greater than bond pair–bond pair repulsion), which compresses the H–N–H angle slightly below the ideal tetrahedral angle.
[2]
12

(HL) Deduce the shape and hybridisation of the central atom in $\text{SF}_6$.

Show answer
Sulfur has 6 bonding pairs and 0 lone pairs, giving an octahedral shape (all F–S–F angles $90^{\circ}$). The hybridisation of the central S atom is $sp^3d^2$.
[3]

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