Atomic Structure (A Level Sciences) - Worksheets, Questions and Revision

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A-Level · Chemistry

AC1 Atomic Structure

AQA 7405 · Calculator allowed · about 115 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question is about the sub-atomic particles found in atoms and ions, and the concept of isotopes.

(a)Complete the table below to show the relative mass and relative charge of a proton, a neutron and an electron.

Particle	Relative mass	Relative charge
Proton	1	+1
Neutron	...	...
Electron	...	...

(3)

(b)Define the term isotopes.(1)

(c)Chlorine exists naturally as two isotopes, 35Cl and 37Cl. Chlorine has atomic number 17. State the number of protons, neutrons and electrons in one atom of 37Cl.(3)

(d)35Cl and 37Cl are chemically identical to one another but can have slightly different physical properties, such as density. Explain why.(2)

(Total for Question 1 is 9 marks)

This question tests key definitions and facts about atomic structure. For each part, identify the correct option.

(a)Potassium-39 has atomic number 19. Which row correctly gives the number of protons and neutrons in one atom of potassium-39?(1)

A) 19 protons, 19 neutrons
B) 19 protons, 20 neutrons
C) 20 protons, 19 neutrons
D) 39 protons, 19 neutrons

(b)Which statement correctly defines the relative atomic mass of an element?(1)

A) The mass of one atom of the element compared with the mass of one atom of hydrogen-1.
B) The average mass of all the isotopes of the element compared with the mass of one atom of carbon-12.
C) The weighted mean mass of the isotopes of an element compared with 1/12 the mass of one atom of carbon-12.
D) The mass number of the most abundant isotope of the element.

(c)How many orbitals are there in total within a p sub-shell?(1)

A) 1
B) 2
C) 3
D) 5

(d)Which ionisation method is typically used in a modern time-of-flight mass spectrometer to ionise large or delicate molecules without breaking them apart?(1)

A) Electron impact ionisation
B) Electrospray ionisation
C) Thermal ionisation
D) Flame ionisation

(Total for Question 2 is 4 marks)

This question is about the numbers of sub-atomic particles present in ions.

(a)An ion has the formula 24Mg2+ (magnesium has atomic number 12). State the number of protons, neutrons and electrons present in this ion.(3)

(b)An ion has the formula 79Br- (bromine has atomic number 35). State the number of protons, neutrons and electrons present in this ion.(3)

(c)Explain, in terms of sub-atomic particles, why atoms are electrically neutral overall, but ions such as Mg2+ and Br- are not.(2)

(Total for Question 3 is 8 marks)

This question is about calculating relative atomic mass from isotopic abundance data.

(a)Bromine consists of two isotopes: 79Br (55.0% abundance) and 81Br (45.0% abundance). Calculate the relative atomic mass of bromine, giving your answer to 1 decimal place.(2)

(b)Neon consists of three isotopes: 20Ne (90.5% abundance), 21Ne (0.3% abundance) and 22Ne (9.2% abundance). Show that the relative atomic mass of neon is 20.2 (to 1 decimal place).(3)

(c)The relative atomic mass of chlorine is 35.5. Explain why this value is not a whole number, even though the mass number of every individual isotope of chlorine is always a whole number.(2)

(Total for Question 4 is 7 marks)

This question is about atomic orbitals and sub-shells.

(a)State the maximum number of electrons that can occupy: (i) a single atomic orbital, (ii) an s sub-shell, (iii) a p sub-shell.(3)

(b)State the number of orbitals within a p sub-shell and describe their shape and relative orientation to one another.(2)

(c)Nitrogen has the electron configuration 1s2 2s2 2p3. Complete a diagram of boxes to show how the three 2p electrons are arranged within the three 2p orbitals, and use Hund's rule to explain the arrangement you have shown.
2p: [ ] [ ] [ ](3)

(Total for Question 5 is 8 marks)

This question is about writing electron configurations, including for a transition metal and its ion.

(a)Write the full electron configuration of a chlorine atom (atomic number 17).(1)

(b)Write the full electron configuration of the Ca2+ ion (calcium has atomic number 20).(1)

(c)Write the full electron configuration of (i) an iron atom, Fe (atomic number 26), and (ii) the Fe3+ ion.(2)

(d)Explain why electrons are removed from the 4s sub-shell before the 3d sub-shell when a d-block atom such as iron forms a positive ion, even though the 4s sub-shell fills before the 3d sub-shell when building up the atom's electron configuration.(2)

(e)Complete a diagram of boxes to show the arrangement of electrons within the five 3d orbitals of the Fe3+ ion, and hence state the number of unpaired electrons in Fe3+.
3d: [ ] [ ] [ ] [ ] [ ](2)

(Total for Question 6 is 8 marks)

A time-of-flight (TOF) mass spectrometer is used to analyse a sample of magnesium. Singly charged Mg+ ions (relative mass 24, charge = 1.60 x 10^-19 C) are accelerated through a potential difference of 15.0 kV before travelling along a flight tube of length 2.00 m to the detector.
You may use: energy gained during acceleration, Ek = q x V, where q is the ion's charge and V is the accelerating voltage; kinetic energy, Ek = 1/2 x m x v²; time of flight, t = distance / v; 1 unified atomic mass unit = 1.66 x 10^-27 kg.

(a)State, in order, the four main stages a sample passes through in a TOF mass spectrometer, and describe what happens to the sample during the ionisation stage.(2)

(b)Explain why the entire flight tube of the mass spectrometer must be maintained under a high vacuum.(2)

(c)Calculate the velocity of the Mg+ ions as they leave the acceleration stage and enter the flight tube. Give your answer to 3 significant figures.(4)

(d)Hence calculate the time taken for the Mg+ ions to travel along the 2.00 m flight tube. Give your answer to 3 significant figures.(2)

(Total for Question 7 is 10 marks)

The table shows the first ionisation energies of the elements in Period 3, from sodium to argon.

Element	Na	Mg	Al	Si	P	S	Cl	Ar
First IE (kJ/mol)	496	738	578	786	1012	1000	1251	1521

(a)Describe and explain the general trend in first ionisation energy across Period 3, from sodium to argon.(3)

(b)Explain why the first ionisation energy of aluminium (578 kJ/mol) is lower than that of magnesium (738 kJ/mol), even though aluminium has a greater nuclear charge.(3)

(c)Explain why the first ionisation energy of sulfur (1000 kJ/mol) is lower than that of phosphorus (1012 kJ/mol), even though sulfur has a greater nuclear charge.(3)

(Total for Question 8 is 9 marks)

Element Y is in Period 2 of the Periodic Table. The table shows its first five successive ionisation energies.

Ionisation	IE1	IE2	IE3	IE4	IE5
Energy (kJ/mol)	801	2427	3660	25026	32827

(a)Explain why successive ionisation energies always increase for a given element.(2)

(b)Calculate the ratio of IE4 to IE3, and use it to show that there is a much larger increase between the removal of the 3rd and 4th electrons than between other consecutive electrons removed.(2)

(c)Use this information to deduce the group of the Periodic Table to which element Y belongs, explaining your reasoning.(2)

(d)Given that element Y is in Period 2, state its full electron configuration.(1)

(Total for Question 9 is 7 marks)

Successive ionisation energy data for a single element, and the pattern of first ionisation energies across a period, both provide evidence for the arrangement of electrons into shells and sub-shells. Discuss how this evidence supports the shell and sub-shell model of electron arrangement in atoms. You should refer to both successive ionisation energies within one element and the trend in first ionisation energy across a period.

(Total for Question 10 is 6 marks)

The mass spectrum of 2-chloropropane (C3H7Cl) shows two molecular ion peaks, at m/z = 78 and m/z = 80, in an approximate intensity ratio of 3:1.

(a)Explain the origin of the molecular ion peak at m/z = 80.(2)

(b)State what the approximate 3:1 ratio of peak intensities tells you about the relative natural abundance of the isotopes 35Cl and 37Cl.(2)

(c)Using Ar(C) = 12.0, Ar(H) = 1.0 and Ar(Cl) = 35.5, calculate the relative molecular mass, Mr, of 2-chloropropane, and explain why this value is not equal to either of the mass spectrum peaks at m/z = 78 or m/z = 80.(3)

(Total for Question 11 is 7 marks)

The first ionisation energy of hydrogen can be found from the minimum frequency of electromagnetic radiation needed to remove the single electron from a hydrogen atom in its ground state.
You may use: E = h f, where h = 6.63 x 10^-34 J s and f is frequency in Hz; Avogadro constant, NA = 6.02 x 10²³ mol^-1.

(a)The minimum frequency of radiation needed to ionise a single hydrogen atom is 3.29 x 10¹⁵ Hz. Calculate the energy, in J, needed to ionise one hydrogen atom.(2)

(b)Calculate the first ionisation energy of hydrogen in kJ/mol, using your answer to (a) and the Avogadro constant. Give your answer to 3 significant figures.(3)

(c)Suggest, in terms of quantum theory, why only photons with a frequency equal to or greater than this minimum value can ionise a hydrogen atom, while a very intense beam of lower-frequency photons cannot.(2)

(Total for Question 12 is 7 marks)

Mark scheme · AC1 Atomic Structure

Question 1

(a) B1 neutron: relative mass 1 AND relative charge 0
(a) B1 electron: relative mass 1/1840 (accept approx 0.00055 or 5.5 x 10^-4 or 'negligible')
(a) B1 electron: relative charge -1
(a) Answer: Neutron: mass 1, charge 0. Electron: mass 1/1840 (approx 0.00055), charge -1.
(b) B1 atoms of the same element (same number of protons/same atomic number) with different numbers of neutrons (different mass numbers) oe
(b) Answer: Atoms of the same element with the same number of protons but different numbers of neutrons.
(c) B1 protons = 17
(c) B1 neutrons = 20
(c) B1 electrons = 17
(c) Answer: Protons = 17, neutrons = 20, electrons = 17.
(d) B1 chemical properties depend on the number and arrangement of electrons, which is identical for both isotopes (same atomic number) oe
(d) B1 physical properties (e.g. density) depend on the mass of the atom, which differs because the isotopes have different numbers of neutrons (different mass numbers) oe
(d) Answer: Isotopes have identical electron arrangements so identical chemical properties, but different numbers of neutrons give different masses, so physical properties differ slightly.

Question 2

(a) B1 B
(a) Answer: B
(b) B1 C
(b) Answer: C
(c) B1 C
(c) Answer: C
(d) B1 B
(d) Answer: B

Question 3

(a) B1 protons = 12
(a) B1 neutrons = 12
(a) B1 electrons = 10
(a) Answer: Protons = 12, neutrons = 12, electrons = 10.
(b) B1 protons = 35
(b) B1 neutrons = 44
(b) B1 electrons = 36
(b) Answer: Protons = 35, neutrons = 44, electrons = 36.
(c) B1 atoms have equal numbers of protons and electrons, so the positive and negative charges cancel, giving no overall (net) charge oe
(c) B1 ions have unequal numbers of protons and electrons (electrons have been lost or gained), giving an overall positive or negative charge oe
(c) Answer: Atoms have equal numbers of protons and electrons so their charges cancel; ions have unequal numbers of protons and electrons, giving a net charge.

Question 4

(a) M1 (79 x 0.550) + (81 x 0.450) or equivalent correct method
(a) A1 79.9 cao
(a) Answer: 79.9
(b) M1 correct substitution: (20 x 0.905) + (21 x 0.003) + (22 x 0.092)
(b) M1 evaluates to 20.187
(b) A1 = 20.2 (1 dp), as required, cso
(b) Answer: 20.2 (cso, as given)
(c) B1 chlorine exists naturally as a mixture of isotopes (35Cl and 37Cl), each with an integer mass number oe
(c) B1 relative atomic mass is the weighted mean (average) of the isotopic masses, weighted by their relative abundance, so the result is not necessarily a whole number oe
(c) Answer: Chlorine is a mixture of isotopes; relative atomic mass is a weighted mean of these, so it need not be an integer.

Question 5

(a) B1 (i) 2
(a) B1 (ii) 2
(a) B1 (iii) 6
(a) Answer: Orbital: 2. s sub-shell: 2. p sub-shell: 6.
(b) B1 3 orbitals (px, py, pz)
(b) B1 each is dumbbell (figure-of-eight) shaped, and the three orbitals are oriented at right angles (90 degrees) to each other along the x, y and z axes oe
(b) Answer: 3 orbitals, each dumbbell-shaped, arranged at 90 degrees to one another along the x, y and z axes.
(c) M1 three separate 2p boxes each shown with a single (unpaired) arrow, all pointing the same direction (no pairing shown)
(c) B1 states Hund's rule: electrons occupy separate orbitals of equal energy (degenerate orbitals) singly, with parallel spins, before any pairing occurs oe
(c) B1 this arrangement minimises the repulsion between electrons, giving a more stable (lower energy) configuration oe
(c) Answer: 2p: one electron in each of the three 2p orbitals, all spins parallel (up, up, up).

Question 6

(a) B1 1s2 2s2 2p6 3s2 3p5
(a) Answer: 1s2 2s2 2p6 3s2 3p5
(b) B1 1s2 2s2 2p6 3s2 3p6
(b) Answer: 1s2 2s2 2p6 3s2 3p6
(c) B1 (i) Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2
(c) B1 (ii) Fe3+: 1s2 2s2 2p6 3s2 3p6 3d5
(c) Answer: Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2. Fe3+: 1s2 2s2 2p6 3s2 3p6 3d5.
(d) B1 once the 3d sub-shell begins to fill, the 3d electrons drop to a lower energy than the 4s electrons (3d becomes lower in energy than 4s in the resulting atom/ion) oe
(d) B1 electrons are always removed from the highest energy (outermost) sub-shell first, which is now 4s, not 3d oe
(d) Answer: Once 3d starts filling it drops below 4s in energy, so 4s (now the higher-energy, outermost sub-shell) loses its electrons first.
(e) M1 five separate 3d boxes each shown with a single (unpaired) arrow, all parallel spins (half-filled sub-shell, per Hund's rule)
(e) A1 5 unpaired electrons
(e) Answer: 5 unpaired electrons (one in each of the five 3d orbitals).

Question 7

(a) B1 correct order of the four stages: ionisation, acceleration, ion drift (flight tube), detection
(a) B1 describes ionisation, e.g. the sample is vaporised and ionised, for example by electrospray ionisation, in which molecules gain a proton (H+) to form positive ions such as [M+H]+ oe
(a) Answer: Ionisation, acceleration, ion drift, detection. In ionisation the sample is vaporised and converted into positive ions, e.g. by electrospray ionisation.
(b) B1 prevents the (positive) ions colliding with air/gas molecules oe
(b) B1 such collisions would deflect/scatter the ions or cause them to lose energy, giving inaccurate flight times and an unreliable spectrum oe
(b) Answer: A vacuum prevents ions colliding with air molecules, which would otherwise deflect them and give inaccurate flight times.
(c) M1 Ek = qV = 1.60 x 10^-19 x 1.50 x 10⁴ = 2.40 x 10^-15 J
(c) M1 mass of ion in kg = 24 x 1.66 x 10^-27 = 3.98 x 10^-26 kg
(c) M1 rearranges Ek = 1/2 m v² to v = √2 Ek / m and substitutes values
(c) A1 v = awrt 3.47 x 10⁵ m/s
(c) Answer: v = 3.47 x 10⁵ m/s
(d) M1 t = d / v = 2.00 / (their answer to (c)) (ft/ecf from (c))
(d) A1 t = awrt 5.76 x 10^-6 s
(d) Answer: t = 5.76 x 10^-6 s

Question 8

(a) B1 general trend is an increase in first ionisation energy from Na to Ar (with two dips, at Al and at S)
(a) B1 nuclear charge (number of protons) increases across the period while the number of occupied shells, and therefore shielding by inner shells, stays approximately constant oe
(a) B1 atomic radius decreases across the period, so the outer electrons are held more strongly, meaning more energy is needed to remove one oe
(a) Answer: First IE generally increases Na to Ar because nuclear charge increases with roughly constant shielding, so atomic radius falls and outer electrons are held more strongly.
(b) B1 in magnesium the outer electron removed is from the full 3s sub-shell (3s2), which has extra stability oe
(b) B1 in aluminium the outer electron removed is from the 3p sub-shell, which is at a higher energy than the 3s sub-shell (and/or experiences extra shielding from the full 3s2 sub-shell) oe
(b) B1 so the 3p electron in aluminium is easier to remove than a 3s electron in magnesium, despite aluminium's higher nuclear charge oe
(b) Answer: Mg's outer electron is in a stable, full 3s sub-shell; Al's outer electron is in the higher-energy 3p sub-shell, so it is easier to remove despite Al's greater nuclear charge.
(c) B1 phosphorus (3p3) has one electron in each of the three 3p orbitals (unpaired), which minimises electron-electron repulsion oe
(c) B1 sulfur (3p4) has one 3p orbital containing a pair of electrons oe
(c) B1 extra repulsion between the two paired electrons in the same orbital makes one of them easier to remove, lowering sulfur's first ionisation energy despite its greater nuclear charge oe
(c) Answer: P's 3p3 electrons are all unpaired (low repulsion); S's 3p4 has a pair in one orbital, and the extra repulsion makes an electron easier to remove.

Question 9

(a) B1 after each electron is removed, the resulting ion has one fewer electron but the same number of protons, so there is less repulsion between the remaining electrons and a greater effective nuclear attraction per electron oe
(a) B1 more energy is therefore required to remove each successive electron oe
(a) Answer: Each successive ion has one fewer electron but the same nuclear charge, so remaining electrons are attracted more strongly, needing more energy to remove.
(b) M1 IE4 / IE3 = 25026 / 3660 = awrt 6.8
(b) A1 compares with other ratios, e.g. IE2/IE1 approx 3.0 and IE3/IE2 approx 1.5, both much smaller, confirming a genuinely large jump between IE3 and IE4 oe
(b) Answer: IE4/IE3 = 6.84, much larger than IE2/IE1 (approx 3.03) or IE3/IE2 (approx 1.51), confirming a large jump.
(c) M1 the large jump between IE3 and IE4 shows that the 4th electron removed comes from an inner shell (closer to the nucleus, with less shielding) than the first three electrons oe
(c) A1 this means Y has 3 electrons in its outer shell, so Y is in Group 3 (Group 13)
(c) Answer: Group 3 (Group 13).
(d) B1 1s2 2s2 2p1 (ft from (c): Period 2, Group 3 = boron)
(d) Answer: 1s2 2s2 2p1 (boron, Z = 5)

Question 10

Level 3 (5-6): A detailed, coherent and logically structured discussion that correctly uses both successive ionisation energy data (large jumps between shells) and the periodic trend in first ionisation energy (including sub-shell anomalies) to support the existence of shells and sub-shells. Accurate scientific terminology is used throughout.
Level 2 (3-4): Some relevant scientific points relating to either shells or sub-shells are made, with some structure and reference to ionisation energy evidence, but the discussion is not fully developed, may focus mainly on one type of evidence, or contains minor inaccuracies.
Level 1 (1-2): Basic, isolated relevant statements are made with little structure or use of data. Limited or inaccurate use of scientific terminology.
Level 0 (0): No relevant content, or the answer does not relate to the question.
Indicative content:
- Within one element, successive ionisation energies increase steadily because each remaining electron experiences a greater effective nuclear attraction as electrons are removed (less repulsion, same nuclear charge).
- A very large jump between two successive ionisation energies occurs when the next electron removed comes from a shell closer to the nucleus (an inner shell), which is less shielded and more strongly attracted.
- The position of this large jump indicates the number of electrons in the outer shell, and therefore the group of the element; this provides direct evidence for the existence of discrete electron shells.
- Within a shell, smaller step changes in successive ionisation energies can reflect the existence of sub-shells, for example a somewhat bigger increase when electrons start being removed from an s sub-shell after a p sub-shell is emptied.
- The general trend of increasing first ionisation energy across a period occurs because nuclear charge increases while shielding remains roughly constant, so atomic radius decreases and outer electrons are held more strongly.
- Two anomalies interrupt this general increase: a decrease from Group 2 to Group 3 (e.g. Mg to Al), and a decrease from Group 5 to Group 6 (e.g. P to S).
- The Group 2 to Group 3 dip occurs because the outer electron moves from a stable, full s sub-shell into a higher energy p sub-shell, which is easier to remove despite the increased nuclear charge; this is evidence for the s and p sub-shells having different energies.
- The Group 5 to Group 6 dip occurs because a p orbital becomes occupied by a pair of electrons, and the extra repulsion between paired electrons in the same orbital makes one easier to remove; this is evidence for the existence of individual orbitals within a sub-shell.
- Together, the repeating (periodic) pattern of these features in every period provides strong, consistent evidence for a shell and sub-shell model of electron arrangement, rather than electrons being arranged randomly or singly around the nucleus.

Question 11

(a) B1 this peak arises from molecules containing the heavier chlorine isotope, 37Cl, in place of 35Cl oe
(a) B1 the C3H7 part of the molecule has mass 43, so 43 + 37 = 80, two mass units higher than the 78 peak (43 + 35)
(a) Answer: The m/z = 80 peak is from molecules containing 37Cl instead of 35Cl (43 + 37 = 80).
(b) B1 35Cl is approximately three times as abundant as 37Cl oe
(b) B1 i.e. natural chlorine is approximately 75% 35Cl and 25% 37Cl
(b) Answer: 35Cl is about 3 times as abundant as 37Cl (approximately 75% : 25%).
(c) M1 Mr = (3 x 12.0) + (7 x 1.0) + 35.5
(c) A1 Mr = 78.5
(c) B1 Mr uses the weighted-average Ar of chlorine (accounting for both isotopes' abundances), whereas each mass spectrum peak corresponds to molecules containing one specific integer-mass isotope, so neither integer peak equals the non-integer average Mr oe
(c) Answer: Mr = 78.5. Neither peak equals this because each peak is from molecules with one specific chlorine isotope, while Mr uses the weighted-average Ar.