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Atomic Structure and Electron Configuration

This document contains a summary of an exam on atomic structure and properties of elements and compounds. It includes 4 questions with multiple parts about: 1) Differences between the plum pudding model of the atom and modern atomic structure, and electron configurations of element R. 2) Calculating the concentration of sodium fluoride in toothpaste in parts per million, and determining safe and toxic amounts. 3) Properties of s-block metals including ionization energies, solubility, precipitation reactions and mass spectrometry. 4) Electron configurations of Cl- and Fe2+, and representing an oxidation-reduction reaction with state symbols.

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0% found this document useful (0 votes)

97 views81 pages

Atomic Structure and Electron Configuration

Uploaded by

successhustlerclub

We take content rights seriously. If you suspect this is your content, claim it here.

Available Formats

Download as RTF, PDF, TXT or read online on Scribd

You are on page 1/ 81

_______________________

Name:
_
3.1.1.3 Electron
configuration _______________________
Class:
_

_______________________
Date:
_

Time: 339 min.

Marks: 318 marks

Comments:

Page 1 of 81
Q1.
This question is about atomic structure.

In the nineteenth century JJ Thomson discovered the electron. He suggested that

negative electrons were found throughout an atom like ‘plums in a pudding of positive
charge’.

The diagram shows an atom of element R using the ‘plum pudding’ model.
An atom of R contains seven electrons.

(a) State two differences between the ‘plum pudding’ model and the model of atomic
structure used today.

1. _________________________________________________________________

___________________________________________________________________

2. _________________________________________________________________

___________________________________________________________________
(2)

(b) Deduce the full electron configuration of an atom of element R.

___________________________________________________________________
(1)

(c) Identify R and deduce the formula of the compound formed when R reacts with the
Group 2 metal in the same period as R.

___________________________________________________________________
(1)
(Total 4 marks)

Q2.
This question is about sodium fluoride (NaF).

Some toothpastes contain sodium fluoride.

The concentration of sodium fluoride can be expressed in parts per million (ppm).
1 ppm represents a concentration of 1 mg in every 1 kg of toothpaste.

(a) A 1.00 g sample of toothpaste was found to contain 2.88 × 10–5 mol of sodium
fluoride.

Calculate the concentration of sodium fluoride, in ppm, for the sample of toothpaste.
Give your answer to 3 significant figures.

Page 2 of 81
Concentration of sodium fluoride _______________________________ ppm
(4)

(b) Sodium fluoride is toxic in high concentrations.

Major health problems can occur if concentrations of sodium fluoride are greater
than 3.19 × 10–2 g per kilogram of body mass.

Deduce the maximum mass of sodium fluoride, in mg, that a 75.0 kg person could
swallow without reaching the toxic concentration.

Mass of sodium fluoride _______________________________ mg

(1)

(c) The concentration of sodium fluoride in a prescription toothpaste is 2800 ppm.

Use your answer to Question (b) to deduce the mass of toothpaste, in kg, that a
75.0 kg person could swallow without reaching the toxic concentration.

Mass of toothpaste _______________________________ kg

(1)

(d) Identify the diagram in the figure below that shows the correct relative sizes of the
ions in sodium fluoride.
Justify your answer.

Diagram __________

Justification _________________________________________________________

___________________________________________________________________

Page 3 of 81
(3)
(Total 9 marks)

Q3.
This question is about s-block metals.

(a) Give the full electron configuration for the calcium ion, Ca2+

___________________________________________________________________
(1)

(b) Explain why the second ionisation energy of calcium is lower than the second
ionisation energy of potassium.

___________________________________________________________________

___________________________________________________________________
(2)

___________________________________________________________________
(1)

(d) Give the formula of the hydroxide of the element in Group 2, from Mg to Ba, that is
least soluble in water.

___________________________________________________________________
(1)

(e) A student added 6 cm3 of 0.25 mol dm–3 barium chloride solution to 8 cm3 of 0.15
mol dm–3 sodium sulfate solution.
The student filtered off the precipitate and collected the filtrate.

Give an ionic equation for the formation of the precipitate.

Show by calculation which reagent is in excess.
Calculate the total volume of the other reagent which should be used by the student
so that the filtrate contains only one solute.

Ionic equation _______________________________________________________

Reagent in excess ___________________________________________________

Total volume of other reagent ___________________________________________

___________________________________________________________________

Page 4 of 81
___________________________________________________________________

___________________________________________________________________
(3)

(f) A sample of strontium has a relative atomic mass of 87.7 and consists of three
isotopes, 86Sr, 87Sr and 88Sr
In this sample, the ratio of abundances of the isotopes 86Sr :87Sr is 1:1

State why the isotopes of strontium have identical chemical properties.

Calculate the percentage abundance of the 88Sr isotope in this sample.

Why isotopes of strontium have identical chemical properties

___________________________________________________________________

Percentage abundance of 88Sr ____________________ %

(4)

(g) A time of flight (TOF) mass spectrum was obtained for a sample of barium that
contains the isotopes 136Ba, 137Ba and 138Ba

The sample of barium was ionised by electron impact.

Identify the ion with the longest time of flight.

___________________________________________________________________
(1)

(h) A 137Ba+ ion travels through the flight tube of a TOF mass spectrometer with a kinetic
energy of 3.65 × 10–16 J
This ion takes 2.71 × 10–5 s to reach the detector.

KE = mv2 where m = mass (kg) and v = speed (m s–1)

The Avogadro constant, L = 6.022 × 1023 mol–1

Calculate the length of the flight tube in metres.

Give your answer to the appropriate number of significant figures.

Page 5 of 81
Length of flight tube ____________________ m
(5)
(Total 18 marks)

Q4.
This question is about atomic structure.

(a) Write the full electron configuration for each of the following species.

Cl−________________________________________________________________

Fe2+_______________________________________________________________
(2)

(b) Write an equation, including state symbols, to represent the process that occurs
when the third ionisation energy of manganese is measured.

___________________________________________________________________

___________________________________________________________________
(1)

Explain your answer.

___________________________________________________________________

___________________________________________________________________
(3)

(d) A sample of nickel was analysed in a time of flight (TOF) mass spectrometer. The
sample was ionised by electron impact ionisation. The spectrum produced showed
three peaks with abundances as set out in the table.

m/z Abundance / %

58 61.0

60 29.1

61 9.9

Give the symbol, including mass number, of the ion that would reach the detector
first in the sample.

Page 6 of 81
Calculate the relative atomic mass of the nickel in the sample.

Give your answer to one decimal place.

Symbol of ion _______________________________________________________

Relative atomic mass _________________________________________________

(3)
(Total 9 marks)

Q5.
Which of these has the highest first ionisation energy?

A Na

B Al

C Si

D Cl
(Total 1 mark)

Q6.
A sample of titanium was ionised by electron impact in a time of flight (TOF) mass
spectrometer. Information from the mass spectrum about the isotopes of titanium in the
sample is shown in the table.

m/z 46 47 48 49

Abundance / % 9.1 7.8 74.6 8.5

(a) Calculate the relative atomic mass of titanium in this sample.

Give your answer to one decimal place.

Relative atomic mass of titanium in this sample ____________________

(2)

Page 7 of 81
(b) Write an equation, including state symbols, to show how an atom of titanium is
ionised by electron impact and give the m/z value of the ion that would reach the
detector first.

Equation ___________________________________________________________

m/z value ___________________________________________________________

(2)

(c) Calculate the mass, in kg, of one atom of 49Ti

The Avogadro constant L = 6.022 × 1023 mol−1

Mass ____________________ kg
(1)

(d) In a TOF mass spectrometer the time of flight, t, of an ion is shown by the equation

In this equation d is the length of the flight tube, m is the mass, in kg, of an ion and E
is the kinetic energy of the ions.

In this spectrometer, the kinetic energy of an ion in the flight tube is 1.013 × 10 −13 J

The time of flight of a 49Ti+ ion is 9.816 × 10−7 s

Calculate the time of flight of the 47Ti+ ion.

Give your answer to the appropriate number of significant figures.

Time of flight ____________________ s

(3)
(Total 8 marks)

Q7.
Element Q forms a sulfate with formula QSO4

Which of these could represent the electronic configuration of an atom of Q?

A [Ne]3s1

B [Ne]3s2

C [Ne]3s23p1

Page 8 of 81
D [Ne]3s13p2
(Total 1 mark)

Q8.
Bromine exists as two isotopes 79Br and 81Br, which are found in almost equal abundance.

Which of the statements is correct?

The first ionisation energy of 79Br is less than the first

A
ionisation energy of 81Br
The atomic radius of 79Br is less than the atomic radius
B
of 81Br
The mass spectrum of C3H7Br has two molecular ion
C
peaks at 122 and 124

D 79
Br is more reactive than 81Br
(Total 1 mark)

Q9.
This question is about the element iodine and its compounds.

(a) Iodine is in Group 7 of the Periodic Table.

Complete the electron configuration of an iodine atom.

[Kr]
________________________________________________________________
(1)

(b) Part of the structure of an iodine crystal is shown in the diagram.

Use your knowledge of structure and bonding to explain why the melting point of
iodine is low (113.5 °C) and why that of hydrogen iodide is very low (–50.8 °C).
(6)

(c) State why iodine does not conduct electricity.

___________________________________________________________________

Page 9 of 81
___________________________________________________________________
(1)

(d) Deduce an equation for the formation of hydrogen iodide from its elements.

___________________________________________________________________
(1)

(e) The triiodide ion is formed when an iodine molecule is bonded to an iodide ion.

What is the formula of ammonium triiodide?

Tick (✔) one box.

NH3I3

NH3I4

NH4I

NH4I3

(1)

(f) Draw the shape of the IF3 molecule and the shape of the IF4− ion.
Include any lone pairs of electrons that influence each shape.

(2)

(g) Deduce the oxidation state of iodine in the following species.

Ba(IO3)2
_____________________________________________________________

[H4IO6]−
_____________________________________________________________
(2)
(Total 14 marks)

Q10.
What is the electron configuration of Cu2+?

Page 10 of 81
A [Ar]3d94s2

B [Ar]3d104s1

C [Ar]3d9

D [Ar]3d10
(Total 1 mark)

Q11.
This question is about the elements in Group 2 and their compounds.

(a) Use the Periodic Table to deduce the full electron configuration of calcium.

___________________________________________________________________
(1)

(b) Write an ionic equation, with state symbols, to show the reaction of calcium with an
excess of water.

___________________________________________________________________
(1)

(c) State the role of water in the reaction with calcium.

___________________________________________________________________
(1)

(d) Write an equation to show the process that occurs when the first ionisation energy
of calcium is measured.

___________________________________________________________________
(1)

(e) State and explain the trend in the first ionisation energies of the elements in Group 2
from magnesium to barium.

Trend ______________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)
(Total 7 marks)

Page 11 of 81
Q12.
The table below shows some successive ionisation energy data for atoms of three
different elements X, Y and Z.

Elements X, Y and Z are Ca, Sc and V but not in that order.

First Second Third Fourth Fifth Sixth

X 648 1370 2870 4600 6280 12 400

Y 590 1150 4940 6480 8120 10 496

Z 632 1240 2390 7110 8870 10 720

(a) Which element is calcium?

Z
(1)

(b) Which element is vanadium?

Z
(1)

(c) Justify your choice of vanadium in part (b)

___________________________________________________________________

___________________________________________________________________
(1)

(d) An acidified solution of NH4VO3 reacts with zinc.

Explain how observations from this reaction show that vanadium exists in at least
two different oxidation states.

___________________________________________________________________

Page 12 of 81
___________________________________________________________________
(2)

(e) The vanadium in 50.0 cm3 of a 0.800 mol dm−3 solution of NH4VO3 reacts with 506
cm3 of sulfur(IV) oxide gas measured at 20.0 °C and 98.0 kPa.

Use this information to calculate the oxidation state of the vanadium in the solution
after the reduction reaction with sulfur(IV) oxide.
Explain your working.
The gas constant R = 8.31 J K−1 mol−1.

Oxidation state = _______________

(6)
(Total 11 marks)

Q13.
Which change requires the largest amount of energy?

A He+(g) He2+(g) + e–

B Li(g) Li+(g) + e–

C Mg+(g) Mg2+(g) + e–

D N(g) N+(g) + e–
(Total 1 mark)

Q14.
This question is about electron configuration.

(a) Give the full electron configuration of an Al atom and of a Cr3+ ion.

Al atom ____________________________________________________________

Page 13 of 81
Cr3+ ion ____________________________________________________________
(2)

(b) Deduce the formula of the ion that has a charge of 2+ with the same electron
configuration as krypton.

___________________________________________________________________
(1)

(c) Deduce the formula of the compound that contains 2+ ions and 3− ions that both
have the same electron configuration as argon.

___________________________________________________________________
(1)
(Total 4 marks)

Q15.
This question is about Period 3 of the Periodic Table.

(a) Deduce which of Na+ and Mg2+ is the smaller ion.

Explain your answer.

Smaller ion _________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

(b) Write an equation to represent the process that occurs when the first ionisation
energy for sodium is measured.

___________________________________________________________________
(1)

Page 14 of 81
Complete the graph by plotting the approximate first ionisation energy values for
magnesium and sulfur.

Explain why the first ionisation energy of sulfur is different from that of phosphorus.

___________________________________________________________________

___________________________________________________________________
(4)
(Total 7 marks)

Q16.
Ions of two isotopes of iron are
53
Fe2+ 56
Fe2+

Which statement is correct?

Page 15 of 81
A The ions of both the isotopes have the electronic
configuration 1s22s22p63s23p64s23d6

B The ions of both the isotopes contains 26 neutrons

C 53
Fe2+ has fewer protons than 56Fe2+

D After acceleration to the same kinetic energy 56Fe2+

will move more slowly than 53Fe2+
(Total 1 mark)

Q17.
This question is about the periodicity of the Period 3 elements.

(a) State and explain the general trend in first ionisation energy across Period 3.

___________________________________________________________________

___________________________________________________________________
(4)

(b) Give one example of an element which deviates from the general trend in first
ionisation energy across Period 3.

Explain why this deviation occurs.

___________________________________________________________________

___________________________________________________________________
(3)

(c) The table shows successive ionisation energies of an element Y in Period 3.

Page 16 of 81
Ionisation number 1 2 3 4 5 6 7 8

Ionisation energy /
1000 2260 3390 4540 6990 8490 27 100 31 700
kJ mol–1

Identify element Y.

Explain your answer using data from the table.

___________________________________________________________________

___________________________________________________________________
(2)

(d) Identify the Period 3 element that has the highest melting point.

Explain your answer by reference to structure and bonding.

___________________________________________________________________

___________________________________________________________________
(4)
(Total 13 marks)

Q18.
This question is about the elements in Period 3 of the Periodic Table.

(a) State the element in Period 3 that has the highest melting point.
Explain your answer.

Element ____________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

Page 17 of 81
___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

(b) State the element in Period 3 that has the highest first ionisation energy.
Explain your answer.

Element ____________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

___________________________________________________________________
(1)

(d) Chlorine is a Period 3 element.

Chlorine forms the molecules ClF3 and CCl2

(i) Use your understanding of electron pair repulsion to draw the shape of ClF3
and the shape of CCl2
Include any lone pairs of electrons that influence the shape.

Shape of ClF3 Shape of CCl2

(2)

(ii) Name the shape of CCl2

______________________________________________________________
(1)

(iii) Write an equation to show the formation of one mole of ClF3 from its elements.

______________________________________________________________

Page 18 of 81
(1)
(Total 11 marks)

Q19.
(a) Table 1 shows some data about fundamental particles in an atom.
Table 1

Particle proton neutron electron

Mass / g 1.6725 × 10–24 1.6748 × 10–24 0.0009 × 10–24

(i) An atom of hydrogen can be represented as 1H

Use data from Table 1 to calculate the mass of this hydrogen atom.

______________________________________________________________
(1)

(ii) Which one of the following is a fundamental particle that would not be
deflected by an electric field?

A electron

B neutron

C proton

Write the correct letter, A, B or C, in the box.

(1)

(b) A naturally occurring sample of the element boron has a relative atomic mass of
10.8.
In this sample, boron exists as two isotopes, 10B and 11B

(i) Calculate the percentage abundance of 10B in this naturally occurring sample
of boron.

______________________________________________________________

______________________________________________________________
(2)

(ii) State, in terms of fundamental particles, why the isotopes 10B and 11B have
similar chemical reactions.

Page 19 of 81
______________________________________________________________

______________________________________________________________

______________________________________________________________
(1)

First Second Third Fourth Fifth

Ionisation energy / kJ mol–

1 799 2420 25 000 32 800

(1)

(d) Write an equation to show the process that occurs when the second ionisation
energy of boron is measured. Include state symbols in your equation.

___________________________________________________________________
(1)

(e) Explain why the second ionisation energy of boron is higher than the first ionisation
energy of boron.

___________________________________________________________________

___________________________________________________________________
(1)
(Total 8 marks)

Q20.
(a) Nickel is a metal with a high melting point.

(i) State the block in the Periodic Table that contains nickel.

______________________________________________________________
(1)

(ii) Explain, in terms of its structure and bonding, why nickel has a high melting
point.

______________________________________________________________

______________________________________________________________
(2)

(iii) Draw a labelled diagram to show the arrangement of particles in a crystal of

nickel.

Page 20 of 81
In your answer, include at least six particles of each type.

(2)

(iv) Explain why nickel is ductile (can be stretched into wires).

______________________________________________________________

______________________________________________________________
(1)

(b) Nickel forms the compound nickel(II) chloride (NiCl2).

(i) Give the full electron configuration of the Ni2+ ion.

______________________________________________________________
(1)

(ii) Balance the following equation to show how anhydrous nickel(II) chloride can
be obtained from the hydrated salt using SOCl2
Identify one substance that could react with both gaseous products.

......NiCl2.6H2O(s) + ...... SOCl2(g) ......NiCl2(s) + ......SO2(g) + ......HCl(g)

Substance ____________________________________________________
(2)
(Total 9 marks)

Q21.
Aluminium and thallium are elements in Group 3 of the Periodic Table.
Both elements form compounds and ions containing chlorine and bromine.

(a) Write an equation for the formation of aluminium chloride from its elements.

___________________________________________________________________
(1)

(b) An aluminium chloride molecule reacts with a chloride ion to form the AlCl4− ion.

Name the type of bond formed in this reaction. Explain how this type of bond is
formed in the AlCl4− ion.

Type of bond ________________________________________________________

Explanation _________________________________________________________

Page 21 of 81
___________________________________________________________________

___________________________________________________________________
(2)

Deduce the formula of the aluminium compound that has a relative molecular mass
of 267

___________________________________________________________________
(1)

(d) Deduce the name or formula of a compound that has the same number of atoms,
the same number of electrons and the same shape as the AlCl4− ion.

___________________________________________________________________
(1)

(e) Draw and name the shape of the TlBr52− ion.

Shape of the TlBr52− ion.

Name of shape ______________________________________________________

(2)

(f) (i) Draw the shape of the TlCl2+ ion.

(1)

(ii) Explain why the TlCl2+ ion has the shape that you have drawn in part (f)(i).

______________________________________________________________

______________________________________________________________
(1)

(g) Which one of the first, second or third ionisations of thallium produces an ion with
the electron configuration [Xe] 5d106s1?

Tick ( ) one box.

Page 22 of 81
First

Second

Third

(1)
(Total 10 marks)

Q22.
(a) Use your knowledge of electron configuration and ionisation energies to answer this
question. The following diagram shows the second ionisation energies of some
Period 3 elements.

(i) Draw an ‘X’ on the diagram to show the second ionisation energy of sulfur.
(1)

(ii) Write the full electron configuration of the Al2+ ion.

______________________________________________________________
(1)

(iii) Write an equation to show the process that occurs when the second ionisation
energy of aluminium is measured.

______________________________________________________________
(1)

(iv) Give one reason why the second ionisation energy of silicon is lower than the
second ionisation energy of aluminium.

______________________________________________________________

Page 23 of 81
______________________________________________________________
(1)

(b) Predict the element in Period 3 that has the highest second ionisation energy.
Give a reason for your answer.

Element ____________________________________________________________

Reason ____________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

First Second Third Fourth Fifth Sixth

Ionisation energy /
786 1580 3230 4360 16100 19800
kJ mol−1

Identify this element.

___________________________________________________________________
(1)

(d) Explain why the ionisation energy of every element is endothermic.

___________________________________________________________________

___________________________________________________________________
(1)
(Total 8 marks)

Q23.
The element rubidium exists as the isotopes 85Rb and 87Rb

(a) State the number of protons and the number of neutrons in an atom of the isotope
Rb
85

Number of protons ___________________________________________________

Number of neutrons __________________________________________________

(2)

(b) (i) Explain how the gaseous atoms of rubidium are ionised in a mass
spectrometer

______________________________________________________________

Page 24 of 81
______________________________________________________________

______________________________________________________________
(2)

(ii) Write an equation, including state symbols, to show the process that occurs
when the first ionisation energy of rubidium is measured.

______________________________________________________________
(1)

Element sodium potassium rubidium

First ionisation
494 418 402
energy / kJ mol–1

State one reason why the first ionisation energy of rubidium is lower than the first
ionisation energy of sodium.

___________________________________________________________________

___________________________________________________________________
(1)

(d) (i) State the block of elements in the Periodic Table that contains rubidium.

______________________________________________________________
(1)

(ii) Deduce the full electron configuration of a rubidium atom.

______________________________________________________________
(1)

(e) A sample of rubidium contains the isotopes 85Rb and 87Rb only.
The isotope 85Rb has an abundance 2.5 times greater than that of 87Rb

Calculate the relative atomic mass of rubidium in this sample.

Give your answer to one decimal place.

___________________________________________________________________

___________________________________________________________________
(3)

Page 25 of 81
(f) By reference to the relevant part of the mass spectrometer, explain how the
abundance of an isotope in a sample of rubidium is determined.

Name of relevant part _________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

(g) Predict whether an atom of 88Sr will have an atomic radius that is larger than, smaller
than or the same as the atomic radius of 87Rb. Explain your answer.

Atomic radius of 88Sr compared to 87Rb ___________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)
(Total 16 marks)

Q24.
The element nitrogen forms compounds with metals and non-metals.

(a) Nitrogen forms a nitride ion with the electron configuration 1s2 2s2 2p6
Write the formula of the nitride ion.

___________________________________________________________________
(1)

(b) An element forms an ion Q with a single negative charge that has the same electron
configuration as the nitride ion.
Identify the ion Q.

___________________________________________________________________
(1)

(d) Calcium nitride contains 81.1% by mass of the metal.

Page 26 of 81
Calculate the empirical formula of calcium nitride.
Show your working.

___________________________________________________________________

___________________________________________________________________
(3)

(e) Write an equation for the reaction between silicon and nitrogen to form silicon
nitride, Si3N4

___________________________________________________________________
(1)
(Total 7 marks)

Q25.
This question is about the first ionisation energies of some elements in the Periodic Table.

(a) Write an equation, including state symbols, to show the reaction that occurs when
the first ionisation energy of lithium is measured.

___________________________________________________________________
(1)

(b) State and explain the general trend in first ionisation energies for the Period 3
elements aluminium to argon.

Trend _____________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

State how selenium deviates from this general trend and explain your answer.

How selenium deviates from this trend ____________________________________

Explanation _________________________________________________________

___________________________________________________________________

Page 27 of 81
___________________________________________________________________
(3)

(d) Suggest why the first ionisation energy of krypton is lower than the first ionisation
energy of argon.

___________________________________________________________________

___________________________________________________________________
(1)

(e) The table below gives the successive ionisation energies of an element.

First Second Third Fourth Fifth

Ionisation energy / kJ mol–1 590 1150 4940 6480 8120

Deduce the group in the Periodic Table that contains this element.

___________________________________________________________________
(1)

(f) Identify the element that has a 5+ ion with an electron configuration of
1s2 2s2 2p6 3s2 3p6 3d10

___________________________________________________________________
(1)
(Total 10 marks)

Q26.
The following diagram shows the first ionisation energies of some Period 3 elements.

(a) Draw a cross on the diagram to show the first ionisation energy of aluminium.
(1)

(b) Write an equation to show the process that occurs when the first ionisation energy
of aluminium is measured.

Page 28 of 81
___________________________________________________________________
(2)

(c) State which of the first, second or third ionisations of aluminium would produce an
ion with the electron configuration 1s2 2s2 2p6 3s1

___________________________________________________________________
(1)

(d) Explain why the value of the first ionisation energy of sulfur is less than the value of
the first ionisation energy of phosphorus.

___________________________________________________________________

___________________________________________________________________
(2)

(e) Identify the element in Period 2 that has the highest first ionisation energy and give
its electron configuration.

Element ____________________________________________________________

Electron configuration _________________________________________________

(2)

(f) State the trend in first ionisation energies in Group 2 from beryllium to barium.
Explain your answer in terms of a suitable model of atomic structure.

Trend ______________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)
(Total 11 marks)

Q27.
Ionisation energies provide evidence for the arrangement of electrons in atoms.

(a) Complete the electron configuration of the Mg+ ion.

1s2 _______________________________________________________________
(1)

(b) (i) State the meaning of the term first ionisation energy.

______________________________________________________________

Page 29 of 81
______________________________________________________________

______________________________________________________________
(2)

(ii) Write an equation, including state symbols, to show the reaction that occurs
when the second ionisation energy of magnesium is measured.

______________________________________________________________
(1)

(iii) Explain why the second ionisation energy of magnesium is greater than the
first ionisation energy of magnesium.

______________________________________________________________

______________________________________________________________
(1)

(iv) Use your understanding of electron arrangement to complete the table by

suggesting a value for the third ionisation energy of magnesium.

First Second Third Fourth Fifth

Ionisation energies of
736 1450 10 500 13 629
magnesium / kJ mol–1

(1)

Trend ______________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

(d) State how the element sulfur deviates from the general trend in first ionisation
energies across Period 3. Explain your answer.

How sulfur deviates from the trend _______________________________________

___________________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

Page 30 of 81
___________________________________________________________________
(3)

(e) A general trend exists in the first ionisation energies of the Period 2 elements lithium
to fluorine. Identify one element which deviates from this general trend.

___________________________________________________________________
(1)
(Total 13 marks)

Q28.
(a) Complete the electronic configuration for the sodium ion, Na+
1s2 ________________________________________________________________
(1)

(b) (i) Write an equation, including state symbols, to represent the process for which
the energy change is the second ionisation energy of sodium.

______________________________________________________________
(2)

(ii) Explain why the second ionisation energy of sodium is greater than the second
ionisation energy of magnesium.

______________________________________________________________

______________________________________________________________
(3)

(iii) An element X in Period 3 of the Periodic Table has the following successive
ionisation energies.

First Second Third Fourth

Ionisation energies / kJ mol –1

577 1820 2740 11600

Deduce the identity of element X.

______________________________________________________________
(1)

Trend ______________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

Page 31 of 81
___________________________________________________________________
(3)

(d) Explain why sodium has a lower melting point than magnesium.

___________________________________________________________________

___________________________________________________________________
(3)

(e) Sodium reacts with ammonia to form the compound NaNH2 which contains the NH2–
ion. Draw the shape of the NH2– ion, including any lone pairs of electrons.
Name the shape made by the three atoms in the NH2– ion.

Shape of NH2–

Name of shape ______________________________________________________

(2)

(f) In terms of its electronic configuration, give one reason why neon does not form
compounds with sodium.

___________________________________________________________________
(1)
(Total 16 marks)

Q29.
(a) State the meaning of the term first ionisation energy of an atom.

___________________________________________________________________

___________________________________________________________________
(2)

(b) Complete the electron arrangement for the Mg2+ ion.

1s2 ________________________________________________________________
(1)

___________________________________________________________________

Page 32 of 81
(1)

(d) Write an equation to illustrate the process occurring when the second ionisation
energy of magnesium is measured.

___________________________________________________________________
(1)

(e) The Ne atom and the Mg2+ ion have the same number of electrons. Give two
reasons why the first ionisation energy of neon is lower than the third ionisation
energy of magnesium.

Reason 1 ___________________________________________________________

Reason 2 ___________________________________________________________
(2)

(f) There is a general trend in the first ionisation energies of the Period 3 elements, Na
– Ar

(i) State and explain this general trend.

Trend ________________________________________________________

Explanation ____________________________________________________

______________________________________________________________

(ii) Explain why the first ionisation energy of sulphur is lower than would be
predicted from the general trend.

______________________________________________________________

______________________________________________________________
(5)
(Total 12 marks)

Q30.
(a) When aluminium is added to an aqueous solution of copper(II) chloride, CuCl2,
copper metal and aluminium chloride, AlCl3, are formed. Write an equation to
represent this reaction.

___________________________________________________________________
(1)

(b) (i) State the general trend in the first ionisation energy of the Period 3 elements
from Na to Ar.

______________________________________________________________

(ii) State how, and explain why, the first ionisation energy of aluminium does not
follow this general trend.

Page 33 of 81
______________________________________________________________

______________________________________________________________

______________________________________________________________
(4)

(c) Give the equation, including state symbols, for the process which represents the
second ionisation energy of aluminium.

___________________________________________________________________
(1)

(d) State and explain the trend in the melting points of the Period 3 metals Na, Mg and
Al.

Trend _____________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)
(Total 9 marks)

Q31.
Photochromic glass contains silver ions and copper ions. A simplified version of a redox
equilibrium is shown below. In bright sunlight the high energy u.v. light causes silver atoms
to form and the glass darkens. When the intensity of the light is reduced the reaction is
reversed and the glass lightens.

Cu+(s) + Ag+(s) Cu2+(s) + Ag(s)

clear glass dark glass

Which one of the following is a correct electron arrangement?

A Cu+ is [Ar]3d94s1

B Cu is [Ar]3d104s2

C Cu2+ is [Ar]3d84s1

D Cu+ is [Ar]3d10

(Total 1 mark)

Q32.
Which one of the following statements is not correct?

A The first ionisation energy of iron is greater than its second ionisation energy.

Page 34 of 81
B The magnitude of the lattice enthalpy of magnesium oxide is greater than that of
barium oxide.

C The oxidation state of iron in [Fe(CN)6]3− is greater than the oxidation state of copper
in [CuCl2]−

D The boiling point of C3H8 is lower than that of CH3CH2OH

(Total 1 mark)

Q33.
The values of the first ionisation energies of neon, sodium and magnesium are 2080, 494
and 736 kJ mol–1, respectively.

(a) Explain the meaning of the term first ionisation of an atom.

___________________________________________________________________

___________________________________________________________________
(2)

(b) Write an equation to illustrate the process occurring when the second ionisation
energy of magnesium is measured.

___________________________________________________________________

___________________________________________________________________
(2)

___________________________________________________________________

___________________________________________________________________
(2)

(d) Explain why the value of the first ionisation energy of neon is higher than that of
sodium.

___________________________________________________________________

___________________________________________________________________
(2)
(Total 8 marks)

Q34.

Page 35 of 81
Lithium hydride, LiH, is an ionic compound containing the hydride ion, H–
The reaction between LiH and aluminium chloride, AlCl3, produces the ionic compound
LiAlH4

(a) Balance the equation below which represents the reaction between LiH and AlCl3

LiH + AlCl3 → LiAlH4 + LiCl

(1)

(b) Give the electronic configuration of the hydride ion, H–

___________________________________________________________________
(1)

Shape _____________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

(d) A bond in can be represented by H → Al

Name this type of bond and explain how it is formed.

Type of bond ________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)
(Total 8 marks)

Q35.
(a) One isotope of sodium has a relative mass of 23.

(i) Define, in terms of the fundamental particles present, the meaning of the term
isotopes.

______________________________________________________________

(ii) Explain why isotopes of the same element have the same chemical properties.

______________________________________________________________

Page 36 of 81
(iii) Calculate the mass, in grams, of a single atom of this isotope of sodium.
(The Avogadro constant, L, is 6.023 × 1023 mol–1)

______________________________________________________________

______________________________________________________________
(5)

(b) Give the electronic configuration, showing all sub-levels, for a sodium atom.

___________________________________________________________________
(1)

___________________________________________________________________

___________________________________________________________________
(1)

(d) An atom has half as many protons as an atom of 28Si and also has six fewer
neutrons than an atom of 28Si. Give the symbol, including the mass number and the
atomic number, of this atom.

___________________________________________________________________
(2)
(Total 9 marks)

Q36.
The diagram below shows the values of the first ionisation energies of some of the
elements in Period 3.

(a) On the above diagram, use crosses to mark the approximate positions of the values
of the first ionisation energies for the elements Na, P and S. Complete the diagram
by joining the crosses.
(3)

Page 37 of 81
(b) Explain the general increase in the values of the first ionisation energies of the
elements Na–Ar.

___________________________________________________________________

___________________________________________________________________
(3)

(c) In terms of the electron sub-levels involved, explain the position of aluminium and
the position of sulphur in the diagram.

Explanation for aluminium ______________________________________________

___________________________________________________________________

Explanation for sulphur ________________________________________________

___________________________________________________________________

___________________________________________________________________
(4)
(Total 10 marks)

Q37.
There is a general trend in the values of the first ionisation energies of the elements Na to
Ar. The first ionisation energies of the elements Al and S deviate from this trend.

(a) Write an equation, including state symbols, to represent the process for which the
energy change is the first ionisation energy of Na.

___________________________________________________________________
(2)

(b) State and explain the general trend in the values of the first ionisation energies of
the elements Na to Ar.

Trend _____________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

(c) State how, and explain why, the values of the first ionisation energies of the
elements Al and S deviate from the general trend.

How the values deviate from the trend ____________________________________

Page 38 of 81
Explanation for Al ____________________________________________________

___________________________________________________________________

Explanation for S ___________________________________________________

___________________________________________________________________
(5)
(Total 10 marks)

Q38.
Which one of the following is the electronic configuration of the strongest reducing agent?

A 1s2 2s2 2p5

B 1s2 2s2 2p6 3s2

C 1s2 2s2 2p6 3s2 3p5

D 1s2 2s2 2p6 3s2 3p6 4s2

(Total 1 mark)

Q39.
(a) State the meaning of the term electronegativity.

___________________________________________________________________

___________________________________________________________________
(2)

(b) State and explain the trend in electronegativity values across Period 3 from sodium
to chlorine.

Trend ______________________________________________________________

Explanation _________________________________________________________
(3)

(c) What is meant by the term first ionisation energy?

___________________________________________________________________

___________________________________________________________________
(2)

(d) The diagram below shows the variation in first ionisation energy across Period 3.

Page 39 of 81
(i) What is the maximum number of electrons that can be accommodated in an s
sub-level?

______________________________________________________________

(ii) What evidence from the diagram supports your answer to part (d)(i)?

______________________________________________________________

(iii) What evidence from the diagram supports the fact that the 3p sub-level is
higher in energy than the 3s?

______________________________________________________________

(iv) What evidence from the diagram supports the fact that no more than three
unpaired electrons can be accommodated in the 3p sub-level?

______________________________________________________________

______________________________________________________________
(5)
(Total 12 marks)

Q40.
Which one of the following atoms has only two unpaired electrons in its ground (lowest
energy) state?

A helium

B beryllium

Page 40 of 81
C nitrogen

D oxygen
(Total 1 mark)

Q41.
Which one of the following does not have a pair of s electrons in its highest filled electron
energy sub-level?

A H−

B Mg

C P3+

D Ar
(Total 1 mark)

Q42.
Which one of the following explains why boron has a lower first ionisation energy than
beryllium?

A A boron atom is smaller than a beryllium atom.

B In beryllium all the electrons are paired in full sub-shells.

C A beryllium atom has fewer protons than a boron atom.

D In boron the 2p electron occupies a higher energy level than a 2s electron.

(Total 1 mark)

Q43.
Which one of the following ionisations requires less energy than the first ionisation energy
of oxygen?

A S(g) → S+(g) + e−

B O+(g) → O2+(g) + e−

C N(g) → N+(g) + e−

D F(g) → F+(g) + e−

(Total 1 mark)

Q44.
Which atom has an incomplete sub-shell?

A Be

B Ca

C Ge

Page 41 of 81
D Zn
(Total 1 mark)

Q45.
In which one of the following pairs is the first ionisation energy of element Y greater than
that of element X?

electronic
electronic configuration
configuration
of element X
of element Y

A 1s1 ls2

B 1s2 2s2 ls22s2 2p1

C 1s2 2s22p3 ls22s22p4

D 1s2 2s22p6 ls22s22p6 3s1

(Total 1 mark)

Q46.
Which one of the following lists the first ionisation energies (in kJ mol−1) of the elements
Mg, Al, Si, P and S in this order?

A 577 786 1060 1000 1260

B 736 577 786 1060 1000

C 786 1060 1000 1260 1520

D 1060 1000 1260 1520 418

(Total 1 mark)

Q47.
Which one of the following is the electronic configuration of an element with a maximum
oxidation state of +5?

A 1s2 2s2 2p5

B 1s2 2s2 2p6 3s2 3p1

C 1s2 2s2 2p6 3s2 3p3

D 1s2 2s2 2p6 3s2 3p6 3d7 4s2

(Total 1 mark)

Page 42 of 81
Mark schemes

Q1.
(a) Assume current model unless otherwise stated.

Statement about the nucleus:

(Central) nucleus contains protons and neutrons.
Allow “protons and neutrons are in the centre of the atom”
1

Statement about electrons

Electrons are now arranged in energy levels/shells/orbitals
Ignore “mostly empty space”
Ignore electrons surround / orbit nucleus
Allow additional statement about neutrons but must be
separate from statement about nucleus to score
e.g.
no neutrons in plum pudding / neutrons now recognised
1

(b) 1s22s22p3

Ignore commas, capitals and subscripts

Allow 1s22s22px12py12pz1
1

Formula Be3N2
Accept Be3R2 only if stated R = nitrogen
Accept N2Be3
1
[4]

Q2.
(a) Mr NaF = 42(.0)
Incorrect Mr loses M1 & M4
1

Mass NaF in 1 g = 2.88 × 10–5 × 42.0 (= 1.210 (1.2096) × 10–3 g)

Mass NaF in 1 kg = 1.210 (1.2096) g

M3 = M2 × 1000 (g)
Units, if given, must match answer
1

(Mass in mg = 1210 (1209.6) mg)

Concentration of NaF = 1210 (ppm)

Allow 1.21 × 103 ppm
1

Page 43 of 81
(b) Toxic mass = 3.19 × 10–2 × 75 × 1000
= 2390 mg
Allow 2393
1

(c) Mass of toothpaste needed =

= 0.854 kg
Mark consequential to (b)
(b) ÷ 2800 (to at least 2 sig fig)
Allow 0.85 - 0.86 kg
1

(d) B
If not B, allow M2 only
If blank, read on.
1

Both Na+ and F– same electron arrangement (1s22s22p6) or isoelectronic

Electronegativity, molecules or IMF = CE, M1 only
1

Sodium (ion) has more protons so attracts (outer) electrons closer / Sodium (ion)
has more protons so stronger attractions for (outer) electrons
Ignore shielding, higher charge density, atomic radius
If reference to fluorine rather than fluoride, then penalise 1
mark only
1
[9]

Q3.
(a) 1s2 2s2 2p6 3s2 3p6 (4s0)
1

(b) M1 In Ca(+) (outer) electron(s) is further from nucleus

Or Ca(+) loses electron from a higher (energy) orbital

Or Ca(+) loses electron from a 4(s) orbital or 4th energy level or 4th energy
shell and K(+) loses electron from a 3(p) orbital or 3rd energy level or 3rd
energy shell
Must be comparative
Allow converse arguments
1

M2 More shielding (in Ca+)

(d) Mg(OH)2
1

(e) Ba2+ + SO42– → BaSO4

Ignore state symbols

Page 44 of 81
1

n BaCl2 (6/1000 × 0.25) = 1.5 × 10–3 and n Na2SO4 = (8/1000 × 0.15) = 1.2 × 10–3

and BaCl2 /barium chloride in excess

Working required or 3 × 10–4 of BaCl2
1

10 cm3 (of 0.15 mol dm–3 sodium sulfate)

or 0.01dm3
1

(f) M1 Same electronic configuration / same number of electrons (in outer shell) / all
have 37 electrons (1)
Ignore protons and neutrons unless incorrect numbers
Not just electrons determine chemical properties
1

M2 = 87.7
Alternative M2:

M3 x = 10% (or × = 0.1)

M3 y = 8
1

M4 (% abundance of 88 isotope is 100 – 2x10) = 80(.0)%

M4 % of 88 isotope is 100 – 10y = 80(.0) %
Allow other alternative methods
1

(g) 138
Ba+

(h) M1 = 2.275 × 10–25 (kg)

Calculation of m in kg
If not converted to kg, max 4
If not divided by L lose M1 and M5, max 3
1

M2 v2 = = = 3.2088 × 109

For re-arrangement
1

M3 v= (v = 5.6646 × 104)
For expression with square root

Page 45 of 81
1

M4 v = d/t or d = vt or with numbers

M5 d = (5.6646 × 104 × 2.71 × 10–5) = 1.53 - 1.54 (m)

M5 must be to 3sf
If not converted to kg, answer = 0.0485-0.0486 (3sf). This
scores 4 marks
1

Alternative method

M1 = 2.275 × 10–25

M1 Calculation of m in kg
1

M2 v = d/t
M2, M3 and M4 are for algebraic expressions or correct
expressions with numbers
1

M3 d2 =
1

M4 d= (= √ (3.65 × 10–16 × 2 × (2.71 × 10–5)2 / 2.275 × 10–25))

M5 d = 1.53 – 1.54 (m)

M5 must be to 3sf
1
[18]

Q4.
(a) Cl− 1s22s22p63s23p6
1
2+
Fe 1s 2s 2p 3s 3p 3d
2 2 6 2 6 6

1
If [Ne] or [Ar] used then Max 1if both correct
Ignore 4s0
Allow subscripts

(b) Mn2+ (g) ⟶ Mn3+ (g) + e−

1
States symbols are required
Allow Mn2+ (g) − e− ⟶ Mn3+ (g)
Negative charge needed on electron

Page 46 of 81
1
(Outer) electron in (3)p sublevel / orbital
Not just level or shell
1
Higher in energy / further from the nucleus
so easier to remove OWTTE
Both required for M3
1
Ignore shielding

(d) 58
Ni+

M1 needs mass and charge – allow subscripts

1
Ar= [(58 × 61.0) + (60 × 29.1) + (61 × 9.9)] / 100
1
Ar= 58.9 must be to 1dp
1
[9]

Q5.
D
[1]

Q6.

(a)
1

= 47.8
Correct answer scores 2 marks.
Allow alternative methods.
Allow 1dp or more.
Ignore units
1

(b) Ti(g) → Ti+(g) +e−

or Ti(g) + e−→ Ti+(g) +2e−

or Ti(g) − e−→ Ti+(g)

State symbols essential
Allow electrons without − charge shown.
1

46
1

(c) 8.1(37) × 10−26

(d) M1 is for re-arranging the equation

Page 47 of 81
Allow t α square root of m
1

d = 1.5(47)
This scores 2 marks
Allow this expression for M2

= 9.6(14) × 10−7

Correct answer scores 3 marks.

1
[8]

Q7.
B
[1]

Q8.
C
[1]

Q9.
(a) [Kr] 5s2 4d105p5
1

(b) This question is marked using levels of response. Refer to the Mark Scheme
Instructions for Examiners for guidance on how to mark this question.

Level 3
All stages are covered and the explanation of each stage is correct and complete.

Answer communicates the whole explanation coherently and shows a logical

progression from stage 1 to stage 2 and then stage 3.
5-6 marks

Level 2
All stages are covered but the explanation of each stage may be incomplete or may
contain inaccuracies OR two stages are covered and the explanations are generally
correct and virtually complete.

Answer is mainly coherent and shows a progression through the stages. Some

Page 48 of 81
steps in each stage may be out of order and incomplete.
3-4 marks

Level 1
Two stages are covered but the explanation of each stage may be incomplete or
may contain inaccuracies, OR only one stage is covered but the explanation is
generally correct and virtually complete.

Answer includes some isolated statements, but these are not presented in a logical
order or show confused reasoning.
1-2 marks

Level 0
Insufficient correct chemistry to warrant a mark.
0 marks
Indicative Chemistry content
Stage 1
I2 is molecular.
HI is molecular.
Stage 2
IMF hold the molecules together.
There are weak IMF forces hence the melting point is low in
both substances.
I2 bigger molecule than HI so I2 has more electrons.
Stage 3
Therefore stronger van der Waals between molecules in I2
that need more energy to break causing the melting point to
be higher.
HI also shows permanent dipole-dipole attraction between
molecules but these forces are less than the vdW forces in
iodine.
6

(c) No delocalised electrons or ions

(d) ⟶ HI
Allow multiples
1

(e) NH4I3
1

(f)
Allow any shape with 3 bond pairs and 2 lone pairs
1

Page 49 of 81
Allow any shape with 4 bond pairs and 2 lone pairs (e.g. lone
pairs in equatorial positions)
1

(g) +5
1

+7
1
[14]

Q10.
C
[1]

Q11.
(a) 1s22s22p63s23p64s2
Allow correct numbers that are not superscripted
1

(b) Ca(s)+ 2H2O(l) Ca2+(aq) + 2OH–(aq) + H2(g)

State symbols essential
1

(c) Oxidising agent

(d) Ca(g) Ca+(g) + e–

State symbols essential

Allow ‘e’ without the negative sign
1

(e) Decrease
If answer to ‘trend’ is not ‘decrease’, then chemical error = 0 /
3
1

Ions get bigger / more (energy) shells

Allow atoms instead of ions
1

Weaker attraction of ion to lost electron

1
[7]

Page 50 of 81
Q12.
(a) Y
1

(b) X
1

(c) Jump in trend of ionisation energies after removal of fifth electron

Fits with an element with 5 outer electrons (4s23d3) like V

(d) Explanation: Two different colours of solution are observed

Because each colour is due to vanadium in a different oxidation state

(e) Stage 1: mole calculations in either order

Moles of vanadium = 50.0 × 0.800 / 1000 = 4.00 × 10–2

Extended response
Maximum of 5 marks for answers which do not show a
sustained line of reasoning which is coherent, relevant,
substantiated and logically structured.
1

Moles of SO2 = pV / RT = (98 000 × 506 × 10–6 ) / (8.31 × 293)

= 2.04 × 10–2

Stage 2: moles of electrons added to NH4VO3

When SO2 (sulfur(IV) oxide) acts as a reducing agent, it is oxidised to

sulfate(VI) ions so this is a two electron change
1

Moles of electrons released when SO2 is oxidised = 2.04 × 10–2 × 2

= 4.08 × 10–2

Stage 3: conclusion

But in NH4VO3 vanadium is in oxidation state 5

4.00 × 10–2 mol vanadium has gained 4.08 × 10–2 mol of electrons
therefore 1 mol vanadium has gained 4.08 × 10–2 / 4.00 × 10 – 2 = 1 mol
of electrons to the nearest integer, so new oxidation state is 5 – 1 = 4
1
[11]

Q13.

Page 51 of 81
A
[1]

Q14.
(a) 1s22s22p63s23p1
1

1s22s22p63s23p63d3

1
If noble gas core used correctly in both then scores 1
Allow subscripts and capitals
Ignore 4s0
(b) Sr2+

Ignore name and correct proton/mass number

Allow Sr+2
1

Q15.
(a) Mg(2+) or Magnesium
Na+ CE=0
1

Because Mg2+ has more protons

AND
With the same shielding/screening/electron
arrangement/number of electrons (or isoelectronic)
Allow larger/stronger nuclear charge
Ignore atomic radius
1

(b) Na(g) → Na+(g) + e−

1 for correct species and gas phase

Allow e without charge
Allow Na(g) – e− → Na+(g)
Na(g) + e− → Na+(g) + 2e−
1

(c) Mg between 600-800

S between 800-1040
If S not lower than P on graph then M1 only
If no plots on graph must state S below P to access M3 & M4
1

Page 52 of 81
e− paired in (3)p orbital in S (owtte)
Allow (3)p subshell/sublevel provided pair mentioned
1

Paired e− repel (so less energy needed to remove)

1
[7]

Q16.
D
[1]

Q17.
(a) General increase
If not increase then CE
1
Greater nuclear charge / more protons
1
Same shielding / electrons added to same shell
Allow similar
1
Stronger attraction (from nucleus) for outer electron(s)
Allow electron in outer shell
1

(b) Aluminium / Al (lower than Mg)

CE if not Al or S
1
(Outer) electron in (3)p orbital / sub-shell (level)
If 2p or 4p orbital lose M2 and M3
1
(3p) higher in energy
Allow more shielded or weaker nuclear attraction
M3 is dependent on M2
1
or
Sulfur / S (lower than P)
(Outer) electrons in (3)p orbital begin to pair
Repel
If 2p or 4p orbital lose M2 and M3
Allow 2 electrons in (3)p
M3 is dependent on M2

(d) Silicon
CE if not Si
1
Giant covalent structure / macromolecule

Page 53 of 81
1
Covalent (bonds)
Giant covalent scores M2 and M3
1
Many / strong (covalent bonds) or
(covalent bonds) need lots of energy to break
CE for M2-M4 if molecules / metallic / ionic / IMFs mentioned
1
[13]

Q18.
(a) Silicon / Si
If not silicon then CE = 0 / 3
1

covalent (bonds)
M3 dependent on correct M2
1

Strong or many of the (covalent) bonds need to be broken / needs a lot of

energy to break the (covalent) bonds
Ignore hard to break
1

(b) Argon / Ar
If not argon then CE = 0 / 3. But if Kr chosen, lose M1 and
allow M2+M3
1

Large(st) number of protons / large(st) nuclear charge

Ignore smallest atomic radius
1

Same amount of shielding / same number of shells / same number of energy

levels
Allow similar shielding
1

(d) (i)

Or any structure with 3 bonds and 2 lone pairs

Ignore any angles shown
1

Page 54 of 81
Or a structure with 2 bonds and 1 lone pair
1

(ii) Bent / v shape

Ignore non-linear, angular and triangular
Apply list principle
1

(iii) Cl2 + F2 ClF3

No multiples
Ignore state symbols
1
[11]

Q19.
(a) (i) 1.6734 × 10−24 (g)
Only.

1.6734 × 10−27 kg
Not 1.67 × 10−24 (g).
1

(ii) B
1

(b) (i) = 10.8

OR ratio 10:11 = 1:4 OR 20:80 etc

Allow idea that there are 5 × 0.2 divisions between 10 and
11.
1

abundance of 10B is 20(%)

= 10.8

10x + 1100 − 11x = 1080

∴ x = 1100 − 1080 = 20%

Correct answer scores M1 and M2.
1

(ii) Same number of electrons (in outer shell or orbital)

Ignore electrons determine chemical properties.

Page 55 of 81
Same electronic configuration / arrangement
Ignore protons unless wrong.
1

(c) Range between 3500 and 10 000 kJ mol−1

(d) B+(g) B2+(g) + e(−)

B+(g) − e(−) B2+(g)

B+(g) + e(−) B2+(g) + 2e(−)

Ignore state symbol on electron even if wrong.

(e) Electron being removed from a positive ion (therefore needs more energy) /
electron being removed is closer to the nucleus
Must imply removal of an electron.
Allow electron removed from a + particle / species or from a
2+ ion.
Not electron removed from a higher / lower energy level /
shell.
Not electron removed from a higher energy sub-level /
orbital.
Ignore electron removed from a lower energy sub-level /
orbital.
Ignore ‘more protons than electrons’.
Not ‘greater nuclear charge’.
Ignore ‘greater effective nuclear charge’.
Ignore shielding.
1
[8]

Q20.
(a) (i) d (block) OR D (block)
Ignore transition metals / series.
Do not allow any numbers in the answer.
1

(ii) Contains positive (metal) ions or protons or nuclei and delocalised /

mobile / free / sea of electrons
Ignore atoms.
1

Strong attraction between them or strong metallic bonds

Allow ‘needs a lot of energy to break / overcome’ instead of
‘strong’.
If strong attraction between incorrect particles, then CE = 0 /
2.
If molecules / intermolecular forces / covalent bonding / ionic
bonding mentioned then CE=0.
1

Page 56 of 81
(iii)

M1 is for regular arrangement of atoms / ions (min 6 metal

particles).
M2 for + sign in each metal atom / ion.
Allow 2+ sign.
2

(iv) Layers / planes / sheets of atoms or ions can slide over one another
QoL.
1

(b) (i) 1s2 2s2 2p6 3s2 3p6 3d8 (4s0)

Only.
1

(ii) NiCl2.6H2O + 6 SOCl2 NiCl2 + 6 SO2 + 12 HCl

Allow multiples.
1

NaOH / NH3 / CaCO3 / CaO

Allow any name or formula of alkali or base.
Allow water.
1
[9]

Q21.
(a) Al + 1.5Cl → AlCl3
2

Accept multiples.
Also 2Al + 3Cl2 → Al2Cl6
Ignore state symbols.
1

(b) Coordinate / dative (covalent)

If wrong CE=0/2 if covalent mark on.
1

Electron pair on Cl donated to Al(Cl )

−
3

QoL
Lone pair from Cl not just Cl
−

Penalise wrong species.

(c) Al2Cl6 or AlBr3

Allow Br3Al or Cl6Al2
Upper and lower case letters must be as shown.
Not 2AlCl3
1

(d) SiCl4 / silicon tetrachloride

Accept silicon(4) chloride or silicon(IV) chloride.

Page 57 of 81
Upper and lower case letters must be as shown.
Not silicon chloride.
1

(e)

Accept shape containing 5 bonds and no lone pairs from Tl

to each of 5 Br atoms.
Ignore charge.
1

Trigonal bipyramid(al)
1

(f) (i) Cl — Tl — Cl

Accept this linear structure only with no lone pair on Tl

(ii) (Two) bonds (pairs of electrons) repel equally / (electrons in) the bonds
repel to be as far apart as possible
Dependent on linear structure in (f)(i).
Do not allow electrons / electron pairs repel alone.
1

(g) Second
1
[10]

Q22.
(a) (i) Higher than P
1

(ii) 1s 2s 2p 3s
2 2 6 1

Allow any order

(iii) Al (g) + e
+ (−)
Al +(g) + 2e
2 (−)

OR
Al (g)
+
Al (g) + e
2+ (−)

OR
Al (g) − e
+ (−)
Al (g)
2+

(iv) Electron in Si (removed from) (3)p orbital / electron (removed) from

higher energy orbital or sub-shell / electron in silicon is more shielded

Page 58 of 81
Accept converse arguments relating to Al
Penalise incorrect p-orbital
1

(b) Sodium / Na
Allow Na +

Electron (removed) from the 2 shell / 2p (orbital)

M2 is dependent on M1
Allow electron from shell nearer the nucleus (so more
attraction)
1

(d) Heat or energy needed to overcome the attraction between the (negative)
electron and the (positive) nucleus or protons
Not breaking bonds
QoL

Or words to that effect eg electron promoted to higher energy level (infinity) so

energy must be supplied
1
[8]

Q23.
(a) 37
These answers only.
Allow answers in words.
1

48
Ignore any sum(s) shown to work out the answers.
1

(b) (i) Electron gun / high speed/high energy electrons

Not just electrons.
Not highly charged electrons.
1

Knock out electron(s)

Remove an electron.
1

(ii) Rb(g) → Rb+(g) + e(–)

OR
Rb(g) + e(–) → Rb+(g) + 2e(–)
OR
Rb(g) - e(–) → Rb+(g)
Ignore state symbols for electron.
1

Page 59 of 81
(c) Rb is a bigger (atom) / e further from nucleus / electron lost from a higher
energy level/ More shielding in Rb / less attraction of nucleus in Rb for outer
electron / more shells
Answer should refer to Rb not Rb molecule
If converse stated it must be obvious it refers to Na
Answer should be comparative.
1

(d) (i) s / block s / group s

Only
1

(ii) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1

Allow 3d10 before 4s2

Allow in any order.
1

(e) (85 × 2.5) + 87 ×1

3.5
M1 is for top line
1
1

= 85.6
Only
1

(58 × 5) + 87 ×2
7
M185Rb 71.4% and 87Rb 28.6%
M2 divide by 100
1
1

85.6
M3 = 85.6
1

(f) Detector
Mark independently
Allow detection (plate).
1

Current / digital pulses / electrical signal related to abundance

Not electrical charge.
1

(g) Smaller
Chemical error if not smaller, CE = 0/3
If blank mark on.
1

Bigger nuclear charge / more protons in Sr

Page 60 of 81
Not bigger nucleus.
1

Similar/same shielding
QWC
(Outer) electron entering same shell/sub shell/orbital/same
number of shells.
Do not allow incorrect orbital.
1
[16]

Q24.
(a) N3- / N–3
1

(b) F–/ fluoride

Ignore fluorine/F
Penalise Fl
1

(c) Li3N / NLi3

(d)
M1 for correct fractions
1

(=2.02 = 1.35)

1.5 1 or 3:2
M2 for correct ratio
1

Ca3N2
If Ca3N2 shown and with no working award 3 marks
If Ca3N2 obtained by using atomic numbers then lose M1
1

(e) 3 Si + 2 N2 → Si3N4
Accept multiples
1
[7]

Q25.
(a) Li(g) → Li+(g) + e-(g)

Li(g) - e-(g) → Li+(g)

Li(g) + e-(g) → Li+(g) + 2e-

One mark for balanced equation with state symbols

Charge and state on electron need not be shown
1

Page 61 of 81
(b) Increases
If trend wrong then CE = 0/3 for (b). If blank mark on.
1

Increasing nuclear charge / increasing no of protons

Ignore effective with regard to nuclear charge
1

Same or similar shielding / same no of shells / electron

(taken) from same (sub)shell / electron closer to the
nucleus / smaller atomic radius
1

Paired electrons in a (4) p orbital

If incorrect p orbital then M2 = 0
1

(Paired electrons) repel

If shared pair of electrons M2 + M3 = 0
1

(d) Kr is a bigger atom / has more shells / more shielding

in Kr / electron removed further from nucleus/ electron
removed from a higher (principal or main) energy level
CE if molecule mentioned
Must be comparative answer
QWC
1

(e) 2 / two / II
1

(f) Arsenic / As
1
[10]

Q26.
(a) Cross between the Na cross and the Mg cross
1

(b) Al(g) → Al+(g) + e–

Al(g) – e– → Al+(g)
Al(g) + e– → Al+(g) + 2e–
One mark for state symbols consequential on getting
equation correct.
Electron does not have to have the – sign on it
Ignore (g) if put as state symbol with e– but penalise state
symbol mark if other state symbols on e–
2

Page 62 of 81
1

(d) Paired electrons in (3)p orbital

Penalise wrong number
If paired electrons repel allow M2
1

repel
1

(e) Neon/Ne
No consequential marking from wrong element
1

1s22s22p6/[He}2s22p6

Allow capital s and p

Allow subscript numbers
1

(f) Decreases
CE if wrong
1

Atomic radius increases/electron removed further from nucleus

or nuclear charge/electron in higher energy level/Atoms
get larger/more shells
Accept more repulsion between more electrons for M2
Mark is for distance from nucleus
Must be comparative answers from M2 and M3
CE M2 and M3 if mention molecules
Not more sub-shells
1

As group is descended more shielding

1
[11]

Q27.
(a) 2s22p63s1
1s2 can be rewritten
Allow 2s22px22py22pz23s1
Allow subscripts and capitals
1

(b) (i) Energy/enthalpy (needed) to remove one mole of electrons

from one mole of atoms/compounds/molecules/elements
1

Energy to form one mole of positive ions from one mole of atoms

Energy/enthalpy to remove one electron from one atom

Page 63 of 81
In the gaseous state (to form 1 mol of gaseous ions)
Energy given out loses M1
M2 is dependent on a reasonable attempt at M1
Energy needed for this change
X(g) → X+(g) + e(–) = 2 marks
This equation alone scores one mark
1

(ii) Mg+(g) → Mg2+(g) + e(–)

Mg+(g) + e(–) → Mg2+(g) + 2e(–)
Mg+(g) – e(–) → Mg2+(g)
Do not penalise MG
Not equation with X
1

(iii) Electron being removed from a positive ion (therefore need more
energy)/electron being removed is closer to the nucleus/Mg+
smaller (than Mg)/Mg+ more positive than Mg
Allow from a + particle/species
Not electron from a higher energy level/or higher sub-level
More protons = 0
1

(iv) Range from 5000 to 9000 kJ mol–1

Bigger nuclear charge (from Na to Cl)/more protons

QWC
1

electron (taken) from same (sub)shell/similar or same shielding/

electron closer to the nucleus/smaller atomic radius
If no shielding = 0
Smaller ionic radius = 0
1

(d) Lower
If not lower CE = 0/3
If blank mark on
Allow does not increase
1

Two/pair of electrons in (3)p orbital or implied

Not 2p
1

repel (each other)

M3 dependent upon a reasonable attempt at M2
1

(e) Boron/B or oxygen/O/O2

Page 64 of 81
1
[13]

Q28.
(a) 2s2 2p6;
If ignored the 1s2 given and written 1s22s22p6 mark as correct
Allow capitals and subscripts
1

(b) (i) Na+(g) → Na2+ (g) + e(–);

One mark for equation and one mark for state symbols

Na+(g) + e(–) → Na2+ (g) + 2e(–);

M2 dependent on M1
Allow Na+(g) – e(–) → Na(g)
Allow X+(g) → X2+ (g) + e = 1 mark
2

(ii) Na(2+) requires loss of e– from a 2(p) orbital or 2nd energy level or
2nd shell and Mg(2+) requires loss of e– from a 3(s) orbital or 3rd
energy level or 3rd shell / Na(2+) loses e from a lower (energy)
orbital/ or vice versa;
Not from 3p
1

Less shielding (in Na);

Or vice versa for Mg
1

e(–) closer to nucleus/ more attraction (of electron to nucleus) (in Na);
M3 needs to be comparative
1

(iii) Aluminium /Al;

Increasing nuclear charge/ increasing number of protons;

Electrons in same shell or level/ same shielding/ similar shielding;

(d) Answer refers to Na;

Allow converse answers relating to Mg.

Na fewer protons/smaller nuclear charge/ fewer delocalised electrons;

Allow Mg is 2+ and Na is +.
If vdw CE = 0.
1

Na is a bigger ion/ atom;

Page 65 of 81
1

Smaller attraction between nucleus and delocalised electrons;

If mentioned that charge density of Mg2+ is greater then allow
first 2 marks.
(ie charge / size / attraction).
M3 allow weaker metallic bonding.
1

(e) (Bent) shape showing 2 lone pairs + 2N−H bond pairs;

Atoms must be labelled.
Lone pairs can be with or without lobes.
1

Bent / v shape/ triangular;

Not tetrahedral.
Allow non-linear.
Bent-linear = contradiction.
1

(f) Ne has full sub-levels/ can’t get any more electrons in the sub-levels/
Ne has full shells;
Not 2s2 2p6 alone.
Not stable electron configuration.
1
[16]

Q29.
(a) enthalpy/energy change/required when an electron is removed/
knocked out / displaced/ to form a uni-positive ion
(ignore ‘minimum’ energy)
1

from a gaseous atom

(could get M2 from a correct equation here)
(accept ‘Enthalpy/energy change for the process...’
followed by an appropriate equation, for both marks)
(accept molar definitions)
1

(b) 1s2 2s22p6

(accept capitals and subscripts)

(c) ‘s’ block

(not a specific ‘s’ orbital – e.g. 2s)
1

(d) Mg+(g) → Mg2+(g) + e– or

Mg+(g) + e– → Mg2+(g) + 2e– or

Mg+(g) – e– → Mg2+(g)
1

Page 66 of 81
(e) Mg2+ ion smaller than Ne atom / Mg2+ e– closer to nucleus
(Not ‘atomic’ radius fo Mg2+)
1

Mg2+ has more protons than Ne / higher nuclear charge or

e– is removed from a charged Mg2+ion / neutral neon atom
(accept converse arguments)
(If used ‘It’ or Mg/magnesium/Mg3+ etc. & 2 correct reasons,
allow (1))
1

(f) (i) trend: increases

(if ‘decreases’, CE = 0/3)
1

Expln: more protons / increased proton number /

increased nuclear charge
(NOT increased atomic number)
1

same shell / same shielding / smaller size

(ii) QoL reference to the e– pair in the 3p sub-level

(penalise if wrong shell, e.g. ‘2p’, quoted)
1

repulsion between the e–in this e–pair

(if not stated, ‘e– pair’ must be clearly implied)
(mark M4 and M5 separately)
1
[12]

Q30.
(a) 2Al + 3CuCl2 → 2AlCl3 + 3Cu;
(accept multiples/fractions)

2Al+ 3Cu2+ → 2Al3+ + 3Cu;

(b) (i) increases;

(ii) lower than expected / lower than Mg /

less energy needed to ionise; e– removed from (3)p sub-level;

(‘e– removed’ may be implied)

of higher energy / further away from nucleus / shielded by 3s e–s;

Page 67 of 81
(c) Al+(g) → Al2+(g) + e–;
1

(d) trend: increases;

more protons / higher charge on cation / more delocalised e– / smaller

atomic/ionic radius;
stronger attraction between (cat)ions and delocalised/free/mobile e–

stronger metallic bonding;

1
[9]

Q31.
D
[1]

Q32.
A
[1]

Q33.
(a) Enthalpy change/required when an electron is removed/knocked
out/displaced (Ignore ‘minimum’ energy)
1

From a gaseous atom

(could get this mark from equation)
1

(b) Mg+(g) → Mg2+(g) + e– Equation

Or Mg+(g) + e– → Mg2+(g) + 2e– State symbols (Tied to M1)

(c) Increased/stronger nuclear charge or more protons

Smaller atom or electrons enter the same shell or

same/similar shielding
1

(d) Electron removed from a shell of lower energy or smaller

atom or e– nearer
1

nucleus or e– removed from 2p rather than from 3s

Less shielding
(Do not accept ‘e– from inner shell’)

Page 68 of 81
1
[8]

Q34.
(a) 4LiH + AlCl3 → LiAlH4 + 3LiCl
1

(b) H – = 1s2 or 1s2

(c) Tetrahedral or diagram

(Not distorted tetrahedral)
1

(Equal) repulsion
1

between four bonding pairs / bonds

(Not repulsion between H atoms loses M2 and M3)
(Not ‘separate as far as possible’)
(‘4’ may be inferred from a correct diagram)
1

(d) Dative (covalent) or coordinate

Lone pair or non-bonding pair of electron or both e–

QoL Donated from H– to Al or shared between H and Al

(tied to M2)
(Not ‘from H atom’) (Not ‘to Al ion’) (Not ‘e–s transferred’)
1
[8]

Q35.
(a) (i) Atoms with the same number of protons / proton number (1)
NOT same atomic number

with different numbers of neutrons (1)

NOT different mass number / fewer neutrons

(ii) Chemical properties depend on the number or amount of

(outer) electrons (1) OR, isotopes have the same electron
configuration / same number of e–

(iii) 23/6.023 × 1023 (1)

CE = 0 if inverted or multiplied

tied to M1 3.8(2) × 10–23 [2-5 sig figs] (1)

(b) 1s2 2s2 2p6 3s1 (1)

Page 69 of 81
accept subscripted figures
1

(c) Highest energy e– / outer e–s / last e– in (3)d sub-shell (1)

OR d sub-shell being filled / is incomplete
OR highest energy sub-shell is (3)d
NOT transition element / e– configuration ends at 3d
Q of L
1

(d) N correct symbol (1)

allow

Mass number = 15 AND atomic number = 7 (1)

2
[9]

Q36.
(a)

(b) Increased nuclear charge / proton number (1)

NOT increased atomic number

Electrons enter same shell / energy level OR atoms get smaller

OR same shielding (1)

Stronger attraction between nucleus and (outer) electrons (1)

Q of L
3

(c) Explanation for aluminium: (third) electron in (3)p sub-shell (1)

Sub-shell further away from nucleus OR of higher energy (1)
OR extra shielding from (3)s

Explanation for sulphur: Pair of electrons in (3)p orbital (1)

Repulsion between electrons (1)
tied to reference to e– pair in M3

Page 70 of 81
Penalise ‘2p’ once only
4
[10]

Q37.
(a) Na(g) → Na+(g) + e–
OR Na(g) + e– → Na+(g) + 2e–
(-) on electron not essential
equation (1)
state symbols (1)
Ignore state symbols on electrons
2

(b) Trend : Increases (1)

Explanation : Increased nuclear charge or proton number (1)

Stronger attraction (between nucleus and (outer) e–) (1)
Trend wrong
Allow M2 only if M3 correct (con)
3

(c) How values deviate from trend: (both values) too low (1)
Explanation for Al: e– removed from (3) p (1)
e– or orbital is higher in energy or better
shielded than (3)s
or p electron is shielded by 3s electrons (1)
Allow e– is further away

Mark independently

Explanation for S: e– removed from (3)p electron pair (1)

repulsion between paired e– (reduces energy required) (1)
Mark separately
If deviation wrong allow M2 and M4
If M3 and / or M5 right (con)
If used ‘d’ rather than ‘p’ orbital - lose M2 + M4 but may get
M3, M5 (explanation marks)
5
[10]

Q38.
D
[1]

Q39.
(a) Ability (or power) of an atom to attract electron density
(or electrons or -ve charge) (1)
in a covalent bond (1)
or shared pair
If remove an electron lose first mark
2

(b) Trend: increases (1)

Page 71 of 81
Explanation: nuclear charge (number of protons) increases (1)
electrons in same shell (1)
OR similar shielding
OR atoms similar size or smaller
OR 1 mol of e-
3

(c) Heat / enthalpy / energy for removal of one electron (1)

from a gaseous atom (1)
can score in an equation
must have first mark to score the second
2

(d) (i) 2 (1)

(ii) Two elements (or Na / Mg) before the drop (in energy) to Al (1)

(iii) ionisation energy of Al < that for Mg (1)

(iv) fall in energy from P to S (1)

or discontinuity in trend

From Al to P there are 3 additional electrons (1)

or three elements
For second mark idea of block of 3 elements
5
[12]

Q40.
D
[1]

Q41.
D
[1]

Q42.
D
[1]

Q43.
A
[1]

Q44.
C
[1]

Q45.

Page 72 of 81
A
[1]

Q46.
B
[1]

Q47.
C
[1]

Page 73 of 81
Examiner reports

Q1.
(a) Most students understood that electrons are now arranged in energy levels,
although many didn’t make reference to sub-atomic particles in their answer.
Students did not always make a comparison and it was not always clear which
model they were referring to in their responses.

(b) Almost all students (91.3%) were able to write the correct electron configuration in
terms of shells and sub-shells, although a few students made no reference to sub-
shells.

(c) Many students were able to identify that the two elements were nitrogen and
beryllium but many were then unable to write the correct formula of the compound
formed between them.

Q2.
(a) Almost all students were able to give the correct Mr for sodium fluoride but many
became confused by the different units involved in this question. Students should be
advised that they should explain their working for each step in a calculation. When
only numbers were provided in an answer, it was not clear what students were
calculating in each step.

(b) A good number of students (39.9%) were able to determine the maximum mass of
sodium fluoride that a 75.0 kg person could swallow without reaching a toxic
concentration, although a few struggled with the conversion to mg.

(c) Many students completed this calculation with the figures upside-down and so were
unable to score. A surprisingly high proportion of students (16%) made no attempt
whatsoever at this question.

(d) Many students were unable to identify the correct relative sizes – many stated that
both ions were the same size because they had the same number of electrons,
without considering the impact of the protons. Other students did not appreciate that
both ions were isoelectronic.

Q3.
(a) The electron configuration was well known (88.9% correct).

(b) Only the best students gained both marks here (19%); many students focused on
whether the electron was being removed from a full or partially full orbital instead of
considering why the attraction between the nucleus and the electron might be
different.

(d) The majority of students (76.3%) answered this correctly, although there were quite
a number who gave the incorrect formula.

(e) The ionic equation was generally correct and many students could also calculate
that barium chloride was in excess. Just under a third of students managed to score
all three marks.

Page 74 of 81
(f) Most students could state why isotopes have the same chemical properties. In the
calculation, many students gave an algebraic equation with two unknowns and could
not proceed any further. The students who gave a correct algebraic expression with
one unknown could work through the calculation and generally gained full marks.
Some students answered the question by trial and error and this was also given full
credit. Pleasingly, 43.7% of students gained all four marks available.

(g) Most students deduced that the mass number of the ion would be 138 but less than
30% of students could give the correct formula for the ion including its charge.

(h) Some students answered this part very well (with 36.6% scoring full marks), but
most students scored partial marks or no marks; common errors included failure to
use the Avogadro constant, not converting mass to kg, and incorrect rearrangement
of expressions.

Q4.
(a) This was surprisingly poorly answered. Many students were unable to write the
electronic configuration for Fe2+.

(b) Part (b) was generally well answered, but a significant number of answers had
missing state symbols, despite the direct instruction to include them.

(c) Part (c) was quite well answered.

(d) The symbol including mass number was often incorrect. The calculation of A r was
straight- forward and done well by many students.

Q5.
69% of students scored this mark.

Q6.
(a) This was generally well answered, though a few students added the abundances
incorrectly.

(b) Most students gave the correct equation, but some used the wrong state symbols or
omitted them. The identity of the isotope was well known but some students lost a
mark by using the symbol for the atom rather than the m/z value.

(d) Only the best students gained all three marks. All marks were available, even for
those students who had not answered part (c), or who had the wrong answer for it.
Some students were unable to re-arrange the equation. Students who appreciated
that the distance travelled was common to both ions usually went on to get the
correct answer, but a large number of students scored zero or one mark.

Q14.
Aspects of this question were well done but significant numbers were unable to write the
electronic configuration of the Cr3+ ion. part (b) was answered well but the unfamiliar
formula required in (c) proved tricky for the majority.

Q15.

Page 75 of 81
Most of this question was answered correctly but many students failed to appreciate that
the ions compared in part (a) had the same number of electrons therefore 1 mark was
scored rather than 2. It was pleasing to see over half the students scoring 3 or more
marks on part (c)

Q17.
In part (a) the trend in ionisation energy was explained well with many scoring three or
four marks. Many students did not read the question carefully and explained the
anomalies in the trend. Those who lost a mark usually omitted the key word ‘outer’ in
respect of the electron being removed. In part (b) many students correctly identified and
explained an element that deviated from the general trend. In part (c) most were able to
recognise the large jump after the sixth ionisation energy and deduce the element as
sulfur. Most students knew that silicon had the highest melting point and went on to score
full marks, although aluminium was a common incorrect answer. The quality of language
was not always sufficiently clear and some used chemical terms incorrectly such as
covalent bonds between molecules or strong van der Waals forces between atoms.

Q18.
The students who selected the correct element in part (a) generally went on to score full
marks. However a large number thought that argon had the highest melting point and
therefore did not score any marks. In part (b) the majority of students selected argon and
gave an explanation of their choice. If the answer was correct the most common problem
was missing the shielding factor necessary for the third mark. Part (c) was answered
correctly by most students. The shape of ClF₃ was found easier to draw than CCl₂ in part
(d)(i) and the name of the shape in (d)(ii) was answered correctly by many students. The
most common error in part (d)(iii) was to write an equation to produce two moles rather
than one mole as stated in the question.

Q19.
The overall performance in part (a)(i) was disappointing because many students did not
know the subatomic particles in the hydrogen atom. The mark in part (a)(ii) was scored by
most students. Many students scored both marks in part (b)(i) but several quoted an
equation involving x and y but could not progress beyond that. The majority of students
answered part (b)(ii) correctly and part (c) was also generally well done. In part (d), many
students used Br instead of B in their equation and some either wrote the wrong ionisation
energy or failed to add state symbols. Many answers to part (e) were superficial; many
students simply referred to a different proton / electron ratio or the removal of an electron
from a different energy level.

Q20.
This question was generally quite well answered. In part (a)(i), most students knew the
correct block in the Periodic Table although some seemed to think that the name
‘Transition Metals’ was enough to identify the d block. In part (a)(ii), the biggest issue was
the omission of ‘positive’ when describing ions and ‘delocalised’ when describing the
electrons involved in the bonding. The diagrams in part (a)(iii) generally scored highly but
some were untidy and failed to show a regular arrangement of ions. Surprisingly, many
students did not refer to layers of atoms / ions sliding in part (a)(iv) and therefore did not
score the mark. Answers to part (b)(i) were disappointing and showed that students could
not write the electron configuration of ions in the d block. Balancing the equation proved
difficult but a good number of students did this well and many students could suggest a
suitable substance in part (b)(ii).

Page 76 of 81
Q21.
The equation in part (a) was given correctly by only a few students. Far too many students
could not give the correct formula for aluminium chloride. The type of bonding in part (b)
was well answered but many students thought that the pair of electrons originated from
the Cl atom or from the Al atom. In part (e), the majority of students gave a correct
structure and formula. Incorrect answers were due to the insertion of lone pairs of
electrons. The linear shape in part (f) was generally answered well but the explanation of
the shape often did not refer to repulsion between the bonding electron pairs. The majority
of students scored the mark in part (g), realising that the outer electrons are removed first
during successive ionisations.

Q22.
The concept of second ionisation energy discriminated well. Good students could apply
their understanding of ionisation energy and gave good answers to this question. Weaker
students found much of the question difficult.

In part (a)(i), many students thought incorrectly that the cross was below phosphorus. The
electron configuration in part (a)(ii) was well answered and the main error in part (a)(iii)
was omitting state symbols. Answers to part (a)(iv) often failed to mention the electron
being removed and there were several confused responses with inappropriate use of shell
/ energy level rather than sub-shell / orbital. A large number of students thought that argon
had the highest second ionisation energy and therefore could not score the rest of the
marks in part (b). The students who did quote sodium often found the explanation difficult.
In part (c), various elements were quoted but it was evident that some students did not
know which was Period 3. In part (d), many students simply stated the meaning of
endothermic but did not explain why the ionisation was endothermic. Some students had
the idea that a bond was broken but failed to extend this to an attraction between the
nucleus and the electrons.

Q23.
Part (a) was done very well. In part (b)(i), many students stated that a positive ion was
formed but failed to say how. The equation in part (b)(ii) was well done with only a few
omitting the state symbols. In part (c), a number of students failed to score the mark since
they referred to a rubidium molecule, showing a lack of understanding of the metals given.
In part (d)(ii) there were many students who filled the 4d rather than the 5s orbitals and
some gave an abbreviated electron structure even though the question asked for the full
electron structure. Many students found part (e) difficult although it was pleasing to see
some gain full marks. Part (f) was not well understood with many students confusing
charge and current. The answers to part (g) were generally good although a few students
did not give complete explanations and lost the last mark.

Q24.
A lot of correct answers were seen in part (a) although many students found this difficult
and answers giving incorrect charges ranging from –1 to +4 were seen. Some students
lost marks by trying to write an equation rather than just stating the ion. In part (b), many
students did not refer to an ion in their answer. Many students did not appreciate the –ide
ending in part (c) and produced formulae such as LiNO3. Many of those who realised that
the formula contained only Li and N gave incorrect formulae such as LiN, LiN3, Li4N etc. In
the empirical formula calculation in part (d), there were many students who could get no
further than the 2.02 :1.35 ratio and weaker students divided the percentage of nitrogen
by 28 rather than by 14. A small number of students used incorrect fractions by dividing
the relative atomic mass by the percentage and getting Ca2N3. The equation in part (e)
proved demanding with many students writing atomic nitrogen, N and molecular silicon eg

Page 77 of 81
Si2, Si8. A few incorrect symbols for silicon were seen ie S rather than Si.

Q25.
Candidates generally attempted part (a) well although there were some candidates who
failed to score the mark since the state symbols were missing or, due to unclear writing, it
was impossible to distinguish if the state symbol was an ‘s’ or ‘g’. Part (b) was well
answered by most candidates. In order to give a convincing answer in part (c), candidates
needed to understand what an orbital is. Many incorrect answers showed that candidates
confused principal energy levels and orbitals. Many answers stated that electrons repel
and did not refer to repulsion of the paired electrons in the relevant p orbital. There was
also confusion between paired electrons repelling and the repulsion of paired electrons
that is used to explain molecular shapes. Most candidates answered part (d) well. Parts
(e) and (f) discriminated quite well and a range of answers were seen.

Q26.
Part (a) was answered well. The equation in part (b) was generally well known although
many candidates lost the second mark for omitting the state symbols or having the
electron on the wrong side of the equation. Part (c) was answered correctly by a good
number of candidates. The explanation in part (d) was variable with much confusion over
paired electrons and lone pairs of electrons. There was much inaccuracy in the use of
terminology in candidates' answers. Answers to part (e) were variable and answers
included almost every element in period 2. Part (f) was well done by a majority of
candidates.

Q27.
Part (a) was answered well by most candidates with some writing the electron
configuration of Mg rather than the ion. Some candidates wrote subscripts for the
numbers of electrons even though the 1s2 was given. In part (b)(i) many candidates could
not give a precise definition. The commonest errors were to mix up moles and atoms or to
miss out the gaseous state. Many candidates missed the state symbols from the equation
in part (b)(ii) and some gave equations showing the removal of 2 electrons. Part (b)(iii)
was not well answered. There was a good attempt at part (b)(iv). In part (c) the trend was
generally well known as was the increasing nuclear charge. Imprecise technical language
cost several candidates a mark -”a bigger nucleus” is not the same as “a nucleus with a
bigger charge”. Part (d) was well answered with most candidates knowing the deviation
and the explanation. Marks were lost for not stating that the paired electrons were in the p
orbital and some thought they were in the 2p orbital. There were many correct answers to
part (e).

Q28.
The electronic configuration in part (a) was generally well known. In part (b)(i) the removal
of 2 electrons from the neutral atom was a common incorrect answer and in (b)(ii) many
candidates missed the effect of change of shielding. Often candidates referred to s and p
orbitals with no reference to the number of the principal energy level. Part (b)(iii) was
disappointing and showed that many candidates find it difficult to apply concepts to
different situations. Part (c) was generally well known although many candidates did not
score the second explanation mark. In part (d) many candidates did not realise that
metallic bonding, or an explanation of it, was required in the answer. Many thought that
van der Waals forces were involved or mentioned losing electrons causing ionisation. In
part (e) there were many unclear diagrams and many candidates had the wrong number
of lone pairs of electrons. Answers to the shape of the molecule were often incorrect or
contradictory e.g. bent-linear. Part (f) was well answered.

Page 78 of 81
Q29.
Candidates tended to score a little better on this question than on question 3 but, overall,
the question was not well answered. Part (a) was generally well done, although
incomplete answers were quite common. The most common error was to omit reference
to gaseous atoms.

Parts (b) and (c) were often well answered; however, in part (d) incorrect ionic charges
and missing or incorrect state symbols were quite often seen.

Part (e) was poorly done. Many candidates referred to difference in the number of electron
shells present in the two species despite being told in the question that they had the same
number of electrons. Some candidates attributed the difference to a special stability
enjoyed by neon as a consequence of it having a full outer electron shell. Clearly, the
majority of candidates failed to appreciate the significance of the two species having the
same number of electrons. Many answers referred to ”It’, magnesium or Mg, rather than
correctly identifying the species being discussed.

In part (f), candidates are asked to demonstrate their understanding of a trend required in
the specification. While part (f)(i) was often competently dealt with, answers in part (f)(ii)
frequently lacked sufficient specific detail to be awarded credit. Often, candidates failed to
identify the electron pair from which from which the first electron would be removed, or to
explain the relatively low first ionisation energy of sulphur in terms of repulsion between
the electrons in this electron pair. Vague phrases such as ‘electron pairs repel’, or a
reference to a more stable half-filled shell being formed, earned no credit.

Q30.
While less able, (or less well prepared,) candidates struggled with this question, the
majority coped quite well. The equation in part (a) was usually correct, although it was not
unusual to see Al2, Al3+ or Cu2 in the equation, or for the equation to be incorrectly
balanced. Most candidates correctly identified an increase in ionisation energy across the
group; however, the explanations in part (b)(ii) were frequently vague and unconvincing. In
particular, statements that the electron in aluminium as easier to remove, rather than
making direct reference to the lower ionisation energy of aluminium, failing to specify the
3p electron as being the first electron to be removed from aluminium or vaguely referring
to extra shielding, were quite common. In part (c) some candidates did not start their
equation with Al+ ions and many more either omitted the state symbols completely or used
the wrong one. The majority of candidates correctly identified a rising trend in the melting
point of the three metals in part (d) but the explanations offered were diverse and
frequently incomplete or wrong. Those candidates who did correctly attribute the trend to
an increase in proton number or to the number of delocalised electrons often proceeded
no further, or simply referred, without explanation, to an increase in metallic bonding.
Many explanations were poor, and references to molecules, ionic bonding, van der Waals'
forces and even hydrogen bonding were frequently seen.

Q33.
This question was, in general, poorly answered. In part (a), the definition was not well
known, element was frequently used in place of atom, and the requirement for gaseous
conditions was often omitted. In part (b) the equation was frequently correct; however,
equations showing Mg going to Mg2+ ions were not unusual. Again, the gaseous state
symbol was often omitted. In part (c), some candidates based their argument on the
stability of the full shell of the Na+ ion and the lower stability of the incomplete outer shell
of the Mg+ ion. Other candidates suggested that sodium atoms wanted to lose an electron.
These arguments earned no credit. In part (d) similar incorrect arguments to those used in
part (c) were frequently seen, as well as vague references to differences in charge density

Page 79 of 81
and comparisons of metallic bonding strengths.

Q34.
This question was, in general, also poorly answered. Of those candidates who attempted
part (a), most were successful; however, a surprisingly large number of candidates made
no attempt to balance the equation. Whilst, in part (b), most candidates gave the correct
electronic arrangement for the hydride ion; 1s0 and 1s12s1 were quite common errors.
Candidates encountered many problems in deducing the shape of the ion, with
trigonal planar, square planar and distorted tetrahedral being the most common errors.
Some candidates drew diagrams to illustrate their answer, many of which showed a lone
pair on the Al atom. The explanation was frequently incomplete, or incorrectly referred to
repulsions between hydrogen atoms or hydride ions, rather than between four bonding
electron pairs. In part (d), many candidates were able to identify the bonding as
dative/coordinate but then simply described, in general terms, the formation of such a
bond. References to the donation of a lone pair from the hydride ion to the aluminium
atom were relatively rare.

Q35.
Many good answers were seen here and high marks were quite common. However,
candidates frequently ignored the instruction in (a)(i) to define isotopes in terms of
fundamental particles, and offered a definition based wholly, or in part, on mass number
and atomic number. In (a)(ii), the link between chemical properties and electron
configuration was well known but many candidates attempted to establish a link between
proton, neutron or mass numbers and chemical properties. The calculation in (a)(iii) was
often poorly done, the most common errors being to invert the expression or to multiply
the mass number by L. Part (b) was generally well answered, although a few candidates
gave the electron arrangement showing 23 electrons. Similarly, part (c) was generally well
answered, however, some candidates referred to electrons being in the d block, while a
small minority offered nonsense explanations. Most candidates correctly identified
nitrogen as the element in part (d), but was a common error.

Q36.
Many candidates correctly positioned the crosses in the diagram, but errors were not
uncommon. In some cases, the cross for sodium was omitted. Other common errors were
to position the cross for sodium at the same level as that for aluminium, to place the cross
for phosphorus below that for silicon and to place the cross for sulphur level either level
with that for phosphorus, or level with that of silicon. Given that the questions in parts (b)
and (c) are frequently asked, and so should be an essential part of a revision programme,
the answers offered by many candidates were disappointing. The better candidates were
well prepared and so scored well on this question; others lacked much of the basic
knowledge required in this section of the specification. In part (b), many candidates
attempted to explain the deviations from the general increase in ionisation energy across
the period, rather than explaining the general increase itself, as required by the question,
and so failed to earn the marks allocated to part (b). Where the explanation offered in part
(b) helped to clarify an answer to part (c), which did require an explanation of the
deviations from the general trend, credit for the marks allocated to part (c) was given.
While some clear, concise and accurate explanations were seen in part (b), many
attempts were vague. All that was required was for a link to be established between the
increased nuclear charge of smaller atoms having no change in shielding and an increase
in the attraction between the nucleus and the outer electrons. Many explanations lacked
precision and failed to establish such a link. The phrase effective nuclear charge was
incorrectly used as an explanation by some candidates. Many explanations in part (c)
similarly were incomplete and lacked precision. Many candidates correctly identified the

Page 80 of 81
outermost electron in an aluminium atom as being in the 3p sub-level. The second mark,
for explaining why this electron is more easily lost was less frequently earned. Some
candidates successfully explained the second point, but omitted to state the location of the
outermost electron. Some candidates referred to an increase in the shielding of the outer
electron but omitted to identify the electrons responsible for this increase. In some cases,
candidates simply stated that aluminium was in the 3s with no reference at all being made
to electrons. While some candidates clearly explained the drop in ionisation energy
between phosphorus and sulphur, many others vaguely attributed this deviation from the
general trend to mutual repulsion, or to repulsion between electrons, without identifying
the location, or the paired nature, of the electrons involved. Some vague references to the
increased stability of the electron arrangement present were also seen.

Q37.
Although many candidates gave a correct equation for this process, errors were not
uncommon. Some candidates lost the mark for the state symbols as examiners, who
strove to differentiate between (g) and (s), were unable to clearly identify the squiggles
used in the equation. Other candidates omitted the state symbols completely. Other errors
seen in this equation included the use of X, Z or M to represent the metal, the use of
positrons, single electrons being shown on both sides of the equation, or the equation
being reversed. Part (b) was quite well done although unacceptable phrases such as an
increase in the nuclear force or core charge were not uncommon. Many candidates
struggled with part (c). Many knew that the first ionisation energy of both aluminium and
sulphur are lower than the trend would suggest but the explanations offered were
frequently vague, and the specific identification of the 3p1 electron, in aluminium, and the
3p4 electron pair, in sulphur, were often omitted.
Many candidates were aware that the outer electron in aluminium is of higher energy,
better shielded or further removed from the nucleus but it was often not clear that a
reference to mutual repulsion referred to the 3p paired electrons in sulphur. More often,
explanations referred to repulsion between electron pairs, or simply to repulsion between
electrons. Some candidates attempted to explain the deviation using vague references to
full or half-full shells/sub-shells having more stability.

Q39.
This question also discriminated very effectively. Answers to parts (a) and (b) were often
correct but answers to part (c) lacked precision and in many cases did not refer to removal
of an electron from a gaseous atom. Parts (d)(ii), (iii) and (iv) were not answered well.
Many candidates did not appear to have read or perhaps understood the question. They
were expected to compose their answers by showing how the statements were supported
by evidence from the diagram. Most candidates just explained the variations in ionisation
energies across the period.

Page 81 of 81

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