Periodic Table, Group 2 &
The Halogens
AS & A Level
Model Answers 1
Level
A Level
Subject
Chemistry
Exam Board
OCR
Module
Periodic Table & Energy
Topic
Periodic Table, Group 2 & The Halogens
Paper
AS & A Level
Booklet
Model Answers 1
Time allowed:
63 minutes
Score:
/47
Percentage:
/100
Grade Boundaries:
A*
A
B
C
D
E
>85%
73%
60%
47%
34%
21%
Dr. Asher Rana
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Question 1
This question is about the attraction between particles.
(a) State how and explain why the attraction between nuclei and outermost electrons in gaseous
atoms varies across Period 3.
[2]
The attraction between the nuclei and outermost electrons increases across the period. This is
because:
•
the number of protons increases (= nuclear charge increases)
•
the outer electrons are in the same shell, so they experience similar shielding /OR/
•
the atomic radius decreases (because the outer electrons are attracted more strongly as
the nuclei contain a greater number of protons)
A diagram to show the trends in atomic size across the periodic table:
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(b) The table shows the boiling points of ammonia, fluorine and bromine
Boiling point / °C
ammonia, NH3
– 33
fluorine, F2
– 188
59
bromine, Br2
Explain the different boiling points of NH3, F2 and Br2.
Include the names of any relevant forces and particles.
In your answer you should use appropriate technical terms, spelled correctly.
.
NH3
It has hydrogen bonding between molecules. A hydrogen
bond forms between a lone pair of electrons in an
electronegative atom such as oxygen (also F and N) and
the Hδ+ attached to one of those electronegative atoms in
a neighbouring molecule:
F2 and Br2
They both have van der Waals’ forces between molecules
Comparisons
•
The van der Waals’ forces in Br2 are stronger than in F2 because bromine has more
electrons than fluorine.
•
The van der Waals’ forces in Br2 are greater than hydrogen bonding in NH3, but the
hydrogen bonding in NH3 is stronger than van der Waals’ forces in F2
Hydrogen
Bonding
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Permanent
Dipoles
Induced
Dipoles
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[5 ]
[Total 7 Marks ]
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Question 2
Group 2 elements react with halogens.
(a) Describe and explain the trend in reactivity of Group 2 elements with chlorine as the group is
descended.
In your answer you should use appropriate technical terms, spelled correctly.
[5]
The reactivity of group 2 elements with chlorine increases as the group is descended because:
•
the atomic radius increases
•
there are more shells, therefore more shielding
•
so the outermost electrons experience less attraction to the nucleus (even though the
nuclear charge increases, the increased shielding outweighs this factor)
Therefore, it is easier to remove the outer electrons (or in other words: the ionisation energy
decreases).
A diagram to show the trends in atomic radii in the periodic table:
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(b) A student was provided with an aqueous solution of calcium iodide.
The student carried out a chemical test to show that the solution contained iodide ions. In this
test, a precipitation reaction took place.
(i)
State the reagent that the student would need to add to the solution of calcium iodide. [1]
aqueous AgNO3 (silver nitrate)
(ii)
What observation would show that the solution contained iodide ions?
[1]
yellow precipitate
(iii)
Write an ionic equation, including state symbols, for the reaction that took place.
[1]
Ag+ (aq) + I- (aq) → AgI (s)
(iv)
The student is provided with an aqueous solution of calcium bromide that is contaminated
with calcium iodide.
The student carries out the same chemical test but this time needs to add a second
reagent to show that iodide ions are present.
State the second reagent that the student would need to add.
Concentrated NH3 (aq)
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[1]
Explanation for part b.
•
AgNO3 (aq) + X- (aq) → AgX (s) + NO3- (aq)
•
The solid AgX has a different colour depending on the halide ion (X-)
•
However, in the real experiment, it is hard to differentiate between the silver halide colours,
therefore ammonia solution is added to confirm the precipitate, using the following
observations:
X-
AgX
Colour of AgX (s) adding dilute NH3
Cl -
AgCl
white
dissolves,
colourless solution
Br-
AgBr
cream
insoluble
I-
AgI
pale yellow
insoluble
adding concentrated
NH3
dissolves, colourless
solution
dissolves, colourless
solution
insoluble
[Total 9 Marks]
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Question 3
Periodicity is a repeating pattern across different periods.
(a) First ionisation energy shows a trend across Period 2.
The first ionisation energies of lithium, carbon and fluorine are shown in Table 5.1 below.
Element
Lithium
Carbon
Fluorine
First ionisation energy / kJ mol–1
520
1086
1681
Table 5.1
(i)
Explain the trend across Period 2 shown in Table 5.1.
In your answer, you should use appropriate technical terms, spelled correctly.
Across the period,
•
the number of protons increases, therefore there is greater nuclear charge
•
since they are all in the same period, the outermost electrons experience similar
shielding
•
therefore, the outer electrons are attracted more strongly to the nucleus
The general trend across a period: As the atomic number increases (i.e. number of protons
increases), the first ionisation energy increases:
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[3]
(ii)
Solid carbon exists in two forms, diamond and graphite.
Explain why it is unnecessary to refer to carbon as either diamond or graphite in
Table 5.1
[1]
The first ionisation energy of an element is the energy required to remove 1 mol of electrons
from 1 mol of gaseous atoms, to form 1 mol of gaseous 1+ ions.
Therefore, the structure of the carbon (whether graphite or diamond) is irrelevant, as the
atoms are in the gaseous state when ionised.
(b) Lithium, carbon (in the form of diamond) and fluorine have very different melting points.
These differences in melting points are the result of different types of structure and different
forces or bonds between the particles in the structures.
Part of the table below has been filled in.
Complete the table below.
[6]
[Total 10 Marks]
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Question 4
Chemists can use the Periodic Table to predict the behaviour of elements.
(a) What is the name of the term used to describe the repeating patterns in the Periodic Table?
[1]
Periodicity
(b) Melting points show a trend across a period.
The table below shows the melting points of three elements in Period 3 of the Periodic
Table.
element
aluminium
silicon
phosphorus
melting point / °C
660
1410
44
Explain the trend shown in terms of bonding and structure.
In your answer, you should use appropriate technical terms spelled correctly.
•
Al has metallic bonding and a giant lattice structure:
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[6]
•
Si has covalent bonding and a giant lattice structure:
•
P has induced dipole/van der Waals’ forces between molecules and a simple covalent
structure:
•
Metallic bonds and covalent bonds are stronger than van der Waals’ forces (note you
are NOT breaking covalent bonds when you melt phosphorus but you are when you
melt silicon).
•
Therefore, more energy is needed to overcome bonds in Al and Si than the forces in P
TIP: When a question asks you to explain something in terms of bonding and structure, always check to see
that you have:
Described the bonding in all substances
Described the structure of all substances
Compare how they affect the physical property in the question (e.g. melting point)
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(c) Scientists use ‘sketch graphs’ to show trends.
(i)
Draw a sketch graph to show the general trend in ionisation energy across Period 3.
[1]
Increasing straight line OR curve from Na to Ar
(ii)
Draw a sketch graph to show the general trend in atomic radius across Period 3.
Decreasing straight line OR curve from Na to Ar
[1]
[Total 9 Marks]
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Question 5
This question is about halogens.
(a) Bromine is used to extract iodine from a solution containing iodide ions.
(i)
Write an ionic equation for the reaction.
[1]
Br2 + 2I– → I2 + 2Br–
Iodine is oxidised by bromine, which is more reactive.
(ii)
Explain why iodine is less reactive than bromine.
[3]
Iodine has a larger atomic radius
Iodine has greater shielding / more shells
Iodine has weaker / less nuclear attraction (on electron gained than bromine)
Reactivity in the Group VII elements decreases as you move down the group for the
reasons stated above. Iodine sits below bromine on the Periodic Table and therefore is
less reactive.
(b) Iodine can be used for the small-scale purification of drinking water.
(i)
[3]
Iodine reacts with water as shown below.
I2 + H2O
HI + HIO
Using oxidation numbers, explain why this reaction is a disproportionation.
Disproportionation
Oxidation AND reduction of same element/iodine
OR
Iodine has been oxidised and Iodine has been reduced
Disproportionation reactions involve oxidation and reduction of the same species in a
reaction.
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Oxidation from 0 to +1 in HIO
Reduction from 0 to –1 in HI
Focusing on the iodine in the equation:
●
I2 on the LHS undergoes reduction to produce HI, where the oxidation number of the
iodine goes from zero (elemental state) to -1.
●
I2 on the LHS also undergoes oxidation to produce the HIO, its oxidation number goes
from zero (elemental state) to +1.
(ii)
Chlorine is used to purify water on a large scale.
State one disadvantage of using chlorine for the purification of drinking water.
[1]
Chlorine is toxic / poisonous OR forms halogenated hydrocarbons OR forms carcinogens /
toxic compounds
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(c) Hydrogen reacts with chlorine to form hydrogen chloride, HCl:
H2(g) + Cl2(g)
∆H = –184 kJ mol–1
2HCl(g)
Table 3.1 shows bond enthalpies.
Bond
Bond Enthalpy/ kJ mol–1
H–H
+436
Cl–Cl
+243
Table 3.1
Calculate the bond enthalpy for the H–Cl bond from the information above.
[2]
Step
Working out
1.Calculate the bond enthalpy of 2 x HCl
ΔH= +436 + 243 + 184 = 863 (kJ mol-1)
bonds (product formed in the equation).
2. Divide by 2 to calculate the H-Cl bond
enthalpy for just one HCl molecule.
ΔH =
863
2
= 431.5 (kJ mol-1)
(d) ‘Enthalpy change of vaporisation’ is the enthalpy change when one mole of a substance
changes from a liquid to a gas at its boiling point.
(i)
Write an equation, including state symbols, to represent the enthalpy change of
vaporisation of bromine.
Br2(l) → Br2(g)
Vaporisation is the change in state from a liquid to a gas.
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[1]
(ii)
Suggest whether the enthalpy change of vaporisation of bromine is exothermic or
endothermic.
Explain your answer.
[1]
Endothermic AND energy required to overcome induced dipole–dipole forces/London
forces
ΔHvap is +ve and the process is endothermic for the vaporisation of bromine due to the
intermolecular forces present which need to be overcome.
(Total 12 marks)
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