Assessor’s
use only
QUESTION ONE:
(a)
Complete the table below by:

drawing the Lewis structure (electron dot diagram) for each of the following
molecules.

drawing the 3-D shape to show the shape of each molecule.
Appendix 1
Molecule
Lewis diagram
3-D shape
H
NH3
N
H
H
SO2
O
COCl2
C
Cl
1
Cl
(b)
The following table shows the Lewis diagrams for the molecules H2S and SO3.
Molecule
H2 S
Lewis
S
diagram
H
SO3
O
H
S
O
O
Discuss the shapes and bond angles of these two molecules.
In your answer you should include:

the shape of hydrogen sulfide, H2S and sulfur trioxide, SO3.

the factors that determine the shape of each molecule.

the bond angle of H2S and SO3.

an explanation of the bond angles of H2S and SO3.
Step 1 – state the shape and bond angle:
H2S is a bent (or angular) shape with a 109o bond angle
Step 2 – outline the numbers of electron pairs/clouds around the central
atom and the subsequent repulsion
There are a total of 4 electron clouds around the central sulfur atom.
These clouds are repelling each other, pointing to the 4 corners of a
tetrahedron. As a consequence the angle between these pairs, and hence
the bond angle is 109.5o.
Step 3 – use numbers of bonding vs non-bonding electron clouds to
justify the shape
There are 2 bonding clouds vs 2 non-bonding clouds of electrons. The
non-bonding clouds influence the shape, but are not part of the shape.
Thus the molecule is bent in shape.
SO3 has a trigonal planar shape with a 120o bond angle.
There is a total of 3 clouds around the central sulfur atom. The clouds a
repelling each other, pointing to the 3 corners of a triangle. As a
consequence the angle between electron clouds is 120o.
All three clouds are bonding clouds, so all three are part of the shape. Thus
the molecule has a trigonal planar shape.
2
(c)
The three dimensional shapes of two molecules are shown below.
F
F
C
F
F
N
F
F
F
Circle the word that describes the polarity of each of the molecules CF4 and NF3.
CF4
polar
non-polar
NF3
polar
non-polar
For each molecule, justify your choice.
Step 1 – outline the polarity of the C-F bond, with reasoning related
to electronegativity differences
The C-F bond is polar/a dipole, as fluorine is more electronegative
than carbon.
Step 2 – comment on the symmetry of the molecule and whether
these dipoles cancel out
The molecule is tetrahedral in shape, with the polar C-F bonds (the
dipoles) symmetrically (evenly) distributed around the central carbon.
As a result these dipoles cancel out.
Step 3 – restate the polarity of the molecule as a consequence of what
has been written
Thus the CF4 molecules in non-polar
The N-F bond is polar as fluorine has a greater electronegativity that
nitrogen. However, the (trigonal pyramidal) is not very symmetrical OR
the N-F bonds are not symmetrically (evenly) distributed around the
nitrogen. As a result the dipoles of the N-F bonds do not cancel out.
Thus the molecule is polar.
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QUESTION TWO:
(a)
Complete the table below by stating the type of particle and the bonding (attractive
force) between the particles for each of the substances.
(b)
Substance
Type of particle
Bonding between particles
Copper, Cu
Atoms
Metallic
Copper sulfate,
CuSO4
Ions
Ionic
Contrast both the malleability and solubility in water for both copper, Cu, and copper
sulfate, CuSO4, using your knowledge of structure and bonding.
Step 1 – Outline the structure and bonding of a substance
Copper consists of a 3D lattice of metal atoms, from which the
valance electrons have become delocalised. The positive nuclei of the
atoms are attracted the delocalised electrons, forming the metallic
bond.
Step 2 – Relate this structure to one of the properties
Copper is malleable, so when the metal bends the metallic bonds are
not broken. This is because the attractions involved mobile electrons.
Thus as the atoms move, the electrons can also move, preventing the
attractions from breaking.
OR
AS the electrons are delocalised, they are considered to be nondirectional, so don’t break as the metal is deformed.
Step 3 – Relate the structure to the second property
The metallic bond is very strong. For a substance to dissolve the
attractions between its particles and water molecules needs to stronger
than between its own particles. Attractions between water molecules
and copper atom are relatively weak, so copper does not dissolve.
4
Copper sulfate consists of a 3D lattice of ions. The oppositely charged
ions (Cu2+ and SO42-) are attracted to one another. The attraction between
the ions is called an ionic bond.
As the ions are in a rigid lattice (i.e. no particles are free to move), the
ionic bond is directional. Hence, when CuSO4 is deformed, ions of the
same charge are forced next to each other, leading to repulsion, and the
ionic bonds break. So CuSO4 is not malleable.
Water molecules are polar, and can form strong attractions to ions. When
CuSO4 is mixed with water, the slightly positive Hδ+ attracts the SO42- and
the slightly negative Oδ- attracts the Cu2+. So the ions are attracted out of
the lattice and CuSO4 is soluble.
(c)
Carbon dioxide, CO2, and silicon dioxide, SiO2, are both group 14 oxides.
(i)
Discuss the difference in structure and bonding between carbon dioxide and
silicon dioxide.
Describe the structure and bonding in each of the substances given
CO2 consists of molecules joined by weak intermolecular forces. An
individual molecule consists of one C atom bonded covalently to two
oxygen atoms.
SiO2 consists of silicon and oxygen atoms connected with covalent
bonds termed a covalent network (3D lattice).
(ii)
Use your answer above to justify the difference in melting point between
carbon dioxide, CO2 and silicon dioxide, SiO2.
Step 1 – state the melting point of each substance
CO2 has a low melting point; it is a gas at room temperature.
Whereas SiO2 has high melting point, making it a solid at room
temperature.
Step 2 – use the structure outlined above to justify these properties
When CO2 is melted, the intermolecular (van der Waals) forces
are broken. These forces are weak, only a small amount of heat
energy is needed to break the forces. So CO2’s melting point is
low.
When SiO2 is melted, the covalent bonds need to be broken. As
these bonds are strong, a large amount of heat energy is needed to
break the bonds. So SiO2’s melting point is high.
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QUESTION THREE:
Below is a drawing of two water molecules. Each individual molecule is drawn inside a box
for clarity.
O
H
(a)
H
H
O
H
With reference to the diagram above, describe the bonds that are broken when water
boils.
Answer by referring to a feature of the diagram
The dotted lines in the diagram are broken
(b)
In a well-known laboratory reaction, water vapour can be formed by burning
hydrogen in oxygen gas (air). This is shown in the equation 1 below.
Equation 1:
2H2 (g) + O2 (g)  2H2O (g)
(i) State which bonds are broken and which are formed during this reaction.
Bonds broken: H-H bonds and O=O (from the H2 and O2 molecules)
Bonds formed: O-H bonds (in the H2O molecules)
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(ii) Calculate the change in enthalpy for equation 1. You will need to use the values
in
the table below. (The values are true only for molecules in the equation above.)
molecule
Energy required to break 1 mol of bonds
inside molecule/ kJ
hydrogen gas
436
oxygen gas
498
water as a gas
464
Step 1 – calculate the energy required to break the bonds in the reactants;
care when counting the number of bonds
Bonds broken:
2 x H-H = 2 x 436 = 872
1 x O=O = 1 x 498 = 498
1370
Step 2 – calculate the total energy of the bonds formed in the product – you
need to know the structure of the product here i.e. H-O-H, so 2 O-H bonds
per molecule, and there are 2 of them
Bonds formed:
4 x O-H = 4 x 464 = 1856
Step 3 – calculate the change in enthalpy using:
Enthalpy change = bond broken – bond formed
= 1370 – 1856
= -486 kJ mol-1
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(c)
Another common reaction producing water is that of the combustion of
methane. This is represented in equation 2 below.
Equation 2:
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
ΔrH = -889 kJ mol-1
(i) The Ea for this reaction is 120 kJ mol-1 under particular conditions.
Sketch the reaction profile diagram in the box below for equation 2.
Your diagram need not be to scale.
Your diagram should have the following parts labelled (with their values where
relevant):

Reactants

Products

Enthalpy change

Activation energy
Reactants
CH4 + 2O2
Enthalpy
change =
889 kJ mol-1
Activation energy = 120 kJ mol1
Products
CO2 + 2H2O
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(d)
Calculate the energy released if 10.0 g of water is produced.
M (H2O) = 18.0 g mol-1
Step 1 – calculate number of moles of water formed (n(H2O))
n(H2O) = mass/molar mass
= 10g/18.0 gmol-1
= 0.556 mol
Step 2 – calculate the energy released when 1 mol of water is formed
From equation, 890 kJ for 2 mol of H2O
890/2 = 445 kJ for 1 mol of H2O
Step 3 – calculate energy released for no. of moles of H2O from step 1
ΔH = 0.556 x 445 = 247 kJmol-1
Its exothermic, so ΔH = -247 kJmol-1
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