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A Level H1 Chemistry Periodic Table Quiz

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Questions

A-Level Chemistry H1 Quiz - Periodic Table

Name: ________________________________________
Class: ________________________________________
Date: ________________________________________
Score: ____ / 50

Duration: 60 minutes
Total Marks: 50

Instructions:

Answer ALL questions.
Write your answers in the spaces provided.
Show all working for calculation questions.
The number of marks for each question is shown in brackets [ ].
You may use a calculator where appropriate.
A copy of the Periodic Table is provided separately.

Section A: Multiple Choice Questions (10 marks)

Questions 1–10 are multiple choice. Choose the ONE correct answer for each question.

1. Which of the following correctly describes the trend in atomic radius across Period 3 from Na to Ar?

A. Atomic radius increases because the number of electron shells increases.
B. Atomic radius decreases because the nuclear charge increases while the number of electron shells remains constant.
C. Atomic radius increases because electrons are added to the same shell.
D. Atomic radius remains constant because the number of protons equals the number of electrons.

[1]

2. An element X has the electronic configuration $1s^2\,2s^2\,2p^6\,3s^2\,3p^4$ . Which group and period does X belong to?

A. Group 4, Period 3
B. Group 6, Period 3
C. Group 16, Period 3
D. Group 14, Period 3

[1]

3. Which of the following elements has the highest first ionisation energy?

A. Na
B. Mg
C. Al
D. Si

[1]

4. Which statement best explains why the first ionisation energy of nitrogen is greater than that of oxygen?

A. Nitrogen has a smaller atomic radius than oxygen.
B. Nitrogen has a half-filled $2p$ subshell, which is relatively stable.
C. Oxygen has more protons than nitrogen.
D. Oxygen has a higher electronegativity than nitrogen.

[1]

5. Which oxide reacts with both hydrochloric acid and aqueous sodium hydroxide?

A. $Na_2O$
B. $Al_2O_3$
C. $SiO_2$
D. $P_4O_{10}$

[1]

6. Which of the following correctly describes the trend in electronegativity across Period 3?

A. Electronegativity decreases from Na to Cl.
B. Electronegativity increases from Na to Cl because the nuclear charge increases.
C. Electronegativity remains constant across a period.
D. Electronegativity increases from Na to Cl because atomic radius increases.

[1]

7. An element in Period 3 forms a chloride with the formula $XCl_3$ . Which group does X belong to?

A. Group 1
B. Group 2
C. Group 13
D. Group 15

[1]

8. Which of the following has the highest melting point?

A. $NaCl$
B. $MgO$
C. $AlCl_3$
D. $SiCl_4$

[1]

9. Which of the following statements about the elements in Group 17 is correct?

A. The boiling point decreases down the group.
B. The oxidising ability increases down the group.
C. The colour of the elements darkens down the group.
D. The bond dissociation energy increases down the group.

[1]

10. Which of the following ions has the largest ionic radius?

A. $O^{2-}$
B. $F^-$
C. $Na^+$
D. $Mg^{2+}$

[1]

Section B: Structured Questions (30 marks)

11. The table below shows the first ionisation energies (in $kJ\,mol^{-1}$ ) of the elements in Period 3.

Element	Na	Mg	Al	Si	P	S	Cl	Ar
1st IE / $kJ\,mol^{-1}$	496	738	578	789	1012	1000	1251	1521

(a) Define the term first ionisation energy.
[2]

(b) Explain why the first ionisation energy generally increases across Period 3 from Na to Ar.
[2]

(d) Explain why the first ionisation energy of S is lower than that of P.
[2]

[Total: 8]

12. The table below gives the atomic radii and boiling points of the Group 17 elements.

Element	Atomic radius / nm	Boiling point / °C
$F_2$	0.143	−188
$Cl_2$	0.199	−34
$Br_2$	0.228	59
$I_2$	0.267	184

(a) State and explain the trend in boiling point of the Group 17 elements.
[3]

(b) Explain the trend in atomic radius down Group 17.
[2]

[Total: 7]

13. (a) Describe the trend in the nature of the oxides across Period 3 from $Na_2O$ to $P_4O_{10}$ . Include in your answer the nature (basic, amphoteric, or acidic) of each oxide and how it changes across the period.
[4]

(b) Write an equation to show the reaction of aluminium oxide with:
(i) hydrochloric acid
(ii) aqueous sodium hydroxide
[2]

[Total: 6]

14. The diagram below shows the variation of a property for the first 20 elements in the Periodic Table.

<image_placeholder> id: Q14-fig1 type: graph linked_question: Q14 description: A graph showing the trend in first ionisation energy (y-axis, in kJ mol^-1) against atomic number (x-axis, 1 to 20) for the first 20 elements. The graph shows a general increase from left to right across each period, with notable dips at Be/B, N/O, Mg/Al, and P/S. Key labelled points: He at the top right (highest IE ~2372), Li at the start of period 2 (~520), Ne near the end of period 2 (~2081), Na at the start of period 3 (~496), Ar near the end of period 3 (~1521), K at the start of period 4 (~419), Ca (~590), and Kr near the end of period 4 (~1351). labels: y-axis: First Ionisation Energy / kJ mol^-1; x-axis: Atomic Number (1–20); labelled peaks at He, Ne, Ar, Kr; labelled dips at B, O, Al, S values: He ≈ 2372, Ne ≈ 2081, Ar ≈ 1521, Kr ≈ 1351, Li ≈ 520, Na ≈ 496, K ≈ 419, B ≈ 801, Be ≈ 899, N ≈ 1402, O ≈ 1314, Mg ≈ 738, Al ≈ 578, P ≈ 1012, S ≈ 1000 must_show: General increasing trend across each period, group peaks at noble gases, dips at Group 13 (B, Al) and Group 16 (O, S) relative to preceding elements, clear axis labels and units </image_placeholder>

(a) Using the graph, state the element with the highest first ionisation energy. Explain your answer.
[2]

(b) Explain why there is a dip in first ionisation energy from Be to B.
[2]

(d) Explain why the first ionisation energy of K is lower than that of Na.
[2]

[Total: 8]

15. (a) Define the term electronegativity.
[1]

(b) Explain the trend in electronegativity across Period 3 from Na to Cl.
[2]

[Total: 5]

Section C: Data-Based Question (10 marks)

16. The following table provides information about four elements W, X, Y, and Z, which belong to Period 3 of the Periodic Table.

Element	Formula of oxide	Nature of oxide	Melting point of oxide / °C	Electrical conductivity of molten oxide
W	$W_2O$	Basic	1275	Good
X	$X_2O_3$	Amphoteric	2072	Good
Y	$YO_2$	Acidic	1610	Poor
Z	$Z_2O_7$	Acidic	−91	Poor

(a) Identify elements W, X, Y, and Z. Give a reason for each identification.
[4]

(b) Explain why the melting point of $X_2O_3$ is higher than that of $W_2O$ .
[2]

(d) Write an equation for the reaction of $X_2O_3$ with aqueous sodium hydroxide.
[2]

[Total: 10]

17. The table below shows the melting points and bonding types of the Period 3 chlorides.

Chloride	Melting point / °C	Bonding type
$NaCl$	801	Ionic
$MgCl_2$	714	Ionic
$AlCl_3$	192 (sublimes)	Covalent molecular
$SiCl_4$	−70	Covalent molecular
$PCl_3$	−93	Covalent molecular

(a) Explain why $NaCl$ has a higher melting point than $MgCl_2$ despite both being ionic.
[2]

(b) Explain why $AlCl_3$ has a much lower melting point than $MgCl_2$ .
[2]

[Total: 6]

18. (a) State the trend in atomic radius down Group 1 (alkali metals) and explain the trend.
[2]

(b) The table below shows the first ionisation energies of the Group 1 elements.

Element	Li	Na	K	Rb	Cs
1st IE / $kJ\,mol^{-1}$	520	496	419	403	376

Explain the trend in first ionisation energy down Group 1.
[2]

[Total: 6]

19. (a) Explain why the second ionisation energy of sodium is much higher than its first ionisation energy.
[2]

(b) The table below shows the successive ionisation energies (in $kJ\,mol^{-1}$ ) of an element Q.

Ionisation	1st	2nd	3rd	4th	5th	6th
IE / $kJ\,mol^{-1}$	578	1817	2745	11578	14831	18378

(i) Identify the group to which element Q belongs. Explain your answer.
[2]

(ii) Identify element Q.
[1]

[Total: 5]

20. The table below provides data on the physical properties of four Period 3 elements.

Element	Melting point / °C	Electrical conductivity (solid)	Oxide formula
A	98	Good	$A_2O$
B	650	Good	$BO$
C	1410	Poor	$C_2O_3$
D	1610	Poor	$DO_2$

(a) Identify elements A, B, C, and D. Give a reason for each identification.
[4]

(b) Explain why element C has a high melting point but does not conduct electricity in the solid state.
[2]

[Total: 8]

End of Quiz

Answers

A-Level Chemistry H1 Quiz - Periodic Table

Answer Key

Section A: Multiple Choice Questions

1. B [1]
Explanation: Across Period 3, electrons are added to the same principal energy level (n = 3) while protons are added to the nucleus. The increasing nuclear charge pulls the electron cloud closer, decreasing atomic radius. The number of electron shells remains constant across a period, so option A is incorrect.

2. C [1]
Explanation: The electronic configuration $1s^2\,2s^2\,2p^6\,3s^2\,3p^4$ has 6 electrons in the outermost shell (3s²3p⁴), placing it in Group 16. The highest principal quantum number is 3, so it is in Period 3. This is sulfur (S).

3. D [1]
Explanation: Across Period 3, first ionisation energy generally increases due to increasing nuclear charge and decreasing atomic radius. Among the options, Si has the highest nuclear charge and smallest atomic radius, so it has the highest first ionisation energy. (Note: P would be even higher, but it is not listed.)

4. B [1]
Explanation: Nitrogen has the configuration $1s^2\,2s^2\,2p^3$ , with a half-filled $2p$ subshell. This half-filled configuration is relatively stable, so more energy is required to remove an electron. Oxygen ( $2p^4$ ) has one paired electron in a $2p$ orbital, which experiences greater electron-electron repulsion, making it easier to remove.

5. B [1]
Explanation: $Al_2O_3$ is amphoteric, meaning it reacts with both acids and bases. $Na_2O$ is basic and reacts only with acids. $SiO_2$ and $P_4O_{10}$ are acidic and react only with bases.

6. B [1]
Explanation: Electronegativity increases across a period because the nuclear charge increases while the atomic radius decreases. The bonding electrons are held more strongly by the nucleus, increasing the atom's ability to attract bonding electrons.

7. C [1]
Explanation: The formula $XCl_3$ indicates that element X has a valency of 3 (since Cl has a valency of 1). In Period 3, the element with a valency of 3 is aluminium (Al), which belongs to Group 13.

8. B [1]
Explanation: Melting point of ionic compounds depends on the strength of the electrostatic attraction between ions. $MgO$ has $Mg^{2+}$ and $O^{2-}$ ions with higher charges and smaller ionic radii compared to $NaCl$ ( $Na^+$ and $Cl^-$ ), resulting in stronger ionic bonds and a higher melting point. $AlCl_3$ and $SiCl_4$ are covalent molecular compounds with much lower melting points.

9. C [1]
Explanation: Down Group 17, the colour of the elements darkens: $F_2$ is pale yellow, $Cl_2$ is greenish-yellow, $Br_2$ is reddish-brown, and $I_2$ is dark purple/black. Boiling point increases down the group (A is wrong). Oxidising ability decreases down the group (B is wrong). Bond dissociation energy decreases down the group (D is wrong).

10. A [1]
Explanation: All four species are isoelectronic (10 electrons each). For isoelectronic species, ionic radius decreases as nuclear charge increases. $O^{2-}$ has the fewest protons (8), so the electrons are held least tightly, giving it the largest ionic radius. Order: $O^{2-} > F^- > Na^+ > Mg^{2+}$ .

Section B: Structured Questions

11.

(a) First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous unipositive ions. [2]
Marking: 1 mark for "remove one electron from each gaseous atom" or equivalent; 1 mark for correct state symbols and mole quantities.
Common error: Omitting "gaseous" or not specifying "one mole of" — both are required for full marks.

(b) Across Period 3, the nuclear charge (number of protons) increases while electrons are added to the same principal energy level (n = 3). The increasing nuclear charge attracts the outer electrons more strongly, so more energy is required to remove an electron. The atomic radius also decreases, bringing the outer electrons closer to the nucleus. [2]
Marking: 1 mark for increasing nuclear charge; 1 mark for electrons in same shell / decreasing atomic radius.

(c) Mg has the configuration $[Ne]\,3s^2$ (fully filled 3s subshell), which is relatively stable. Al has the configuration $[Ne]\,3s^2\,3p^1$ . The $3p$ electron in Al is at a higher energy level than the $3s$ electrons and is also shielded by the $3s^2$ electrons, so it is easier to remove. [2]
Marking: 1 mark for identifying that the electron removed from Al is from a $3p$ orbital (higher energy); 1 mark for explaining that the $3p$ electron is shielded by $3s^2$ electrons / further from nucleus.

(d) P has the configuration $[Ne]\,3s^2\,3p^3$ with a half-filled $3p$ subshell, which is relatively stable. S has the configuration $[Ne]\,3s^2\,3p^4$ . In S, one of the $3p$ orbitals contains a paired electron. The electron-electron repulsion within this paired orbital makes it easier to remove one of these electrons. [2]
Marking: 1 mark for identifying half-filled $3p$ subshell stability in P; 1 mark for electron-electron repulsion in paired $3p$ orbital in S.

12.

(a) The boiling point increases down Group 17. As the molecules get larger (from $F_2$ to $I_2$ ), the number of electrons increases, leading to stronger instantaneous dipole-induced dipole (van der Waals / London dispersion) forces between molecules. More energy is required to overcome these stronger intermolecular forces, so the boiling point increases. [3]
Marking: 1 mark for stating the trend (boiling point increases down the group); 1 mark for increasing number of electrons / increasing molecular size; 1 mark for stronger van der Waals / London dispersion forces.

(b) Going down Group 17, each successive element has an additional electron shell. The outer electrons are further from the nucleus and are more shielded by inner shell electrons, so the atomic radius increases. [2]
Marking: 1 mark for additional electron shells; 1 mark for increased shielding / greater distance from nucleus.

(c) The boiling point of $At_2$ would be expected to be higher than that of $I_2$ (above 184 °C). Astatine is below iodine in Group 17, so $At_2$ molecules are larger with more electrons, resulting in even stronger van der Waals forces. More energy is needed to overcome these forces. [2]
Marking: 1 mark for predicting boiling point above 184 °C; 1 mark for explanation in terms of larger molecule / more electrons / stronger van der Waals forces.

13.

(a) Across Period 3, the nature of the oxides changes from basic to amphoteric to acidic:

$Na_2O$ and $MgO$ are basic oxides (react with acids only).
$Al_2O_3$ is amphoteric (reacts with both acids and bases).
$SiO_2$ , $P_4O_{10}$ , and $SO_3$ are acidic oxides (react with bases only).

This trend occurs because the elements become more non-metallic across the period. The oxides change from ionic (basic) to covalent (acidic) in character. The electronegativity of the elements increases, so the oxides become more electron-withdrawing and behave as acidic oxides. [4]
Marking: 1 mark for identifying basic nature of $Na_2O$ / $MgO$ ; 1 mark for identifying amphoteric nature of $Al_2O_3$ ; 1 mark for identifying acidic nature of $SiO_2$ / $P_4O_{10}$ / $SO_3$ ; 1 mark for explanation linking to metallic/non-metallic character or ionic/covalent nature.

(b)
(i) $Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$ [1]
(ii) $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]$ (or $Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$ ) [1]
Marking: 1 mark each for correct balanced equation. Accept either product for (ii): sodium aluminate or sodium tetrahydroxoaluminate.

14.

(a) Helium (He) has the highest first ionisation energy. It is a noble gas with a complete outer shell ( $1s^2$ ) and the smallest atomic radius among the first 20 elements. The outer electrons experience the strongest effective nuclear charge and are held most tightly, so the most energy is required to remove an electron. [2]
Marking: 1 mark for identifying He; 1 mark for explanation in terms of complete outer shell / smallest atomic radius / strongest nuclear attraction.

(b) Be has the configuration $1s^2\,2s^2$ with a fully filled $2s$ subshell, which is relatively stable. B has the configuration $1s^2\,2s^2\,2p^1$ . The $2p$ electron in B is at a higher energy level than the $2s$ electrons and is also shielded by the $2s^2$ electrons, so it is easier to remove. [2]
Marking: 1 mark for identifying that the electron removed from B is from a $2p$ orbital (higher energy); 1 mark for explaining that the $2p$ electron is shielded by $2s^2$ electrons / further from nucleus.

(c) N has the configuration $1s^2\,2s^2\,2p^3$ with a half-filled $2p$ subshell, which is relatively stable. O has the configuration $1s^2\,2s^2\,2p^4$ . In O, one of the $2p$ orbitals contains a paired electron. The electron-electron repulsion within this paired orbital makes it easier to remove one of these electrons. [2]
Marking: 1 mark for identifying half-filled $2p$ subshell stability in N; 1 mark for electron-electron repulsion in paired $2p$ orbital in O.

(d) K is below Na in Group 1. K has more electron shells than Na, so the outer electron is further from the nucleus and experiences more shielding from inner shell electrons. The effective nuclear charge on the outer electron is weaker, so less energy is required to remove it. [2]
Marking: 1 mark for more electron shells / greater distance from nucleus; 1 mark for increased shielding / weaker effective nuclear charge.

15.

(a) Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond towards itself. [1]
Marking: 1 mark for correct definition. Must include "bonding pair" or "shared pair" and "attract towards itself."

(b) Across Period 3, electronegativity increases from Na to Cl. The nuclear charge increases while electrons are added to the same principal energy level. The atomic radius decreases, so the bonding electrons are closer to the nucleus and experience a stronger attraction. The atom's ability to attract bonding electrons therefore increases. [2]
Marking: 1 mark for increasing nuclear charge / decreasing atomic radius; 1 mark for stronger attraction of bonding electrons.

(c) Down Group 1, electronegativity decreases from Li to Cs. Each successive element has an additional electron shell, so the atomic radius increases. The bonding electrons are further from the nucleus and are more shielded by inner shell electrons. The effective nuclear charge on the bonding electrons is weaker, so the atom's ability to attract bonding electrons decreases. [2]
Marking: 1 mark for increasing atomic radius / more electron shells; 1 mark for increased shielding / weaker attraction of bonding electrons.

Section C: Data-Based Question

16.

(a)

W = Sodium (Na): $W_2O$ is a basic oxide with formula $Na_2O$ , consistent with a Group 1 element. [1]
X = Aluminium (Al): $X_2O_3$ is an amphoteric oxide with formula $Al_2O_3$ , consistent with a Group 13 element. [1]
Y = Silicon (Si): $YO_2$ is an acidic oxide with formula $SiO_2$ , consistent with a Group 14 element. [1]
Z = Chlorine (Cl): $Z_2O_7$ is an acidic oxide with formula $Cl_2O_7$ , consistent with a Group 17 element. [1]
Marking: 1 mark each for correct identification with valid reason.

(b) $X_2O_3$ ( $Al_2O_3$ ) has $Al^{3+}$ and $O^{2-}$ ions with higher charges than $W_2O$ ( $Na_2O$ ) which has $Na^+$ and $O^{2-}$ ions. The greater charge density in $Al_2O_3$ results in stronger electrostatic attraction between the ions, requiring more energy to overcome the ionic bonds, hence a higher melting point. [2]
Marking: 1 mark for identifying higher ionic charges in $Al_2O_3$ ; 1 mark for stronger electrostatic attraction / stronger ionic bonds.

(c) $Z_2O_7$ ( $Cl_2O_7$ ) is a covalent molecular compound. In the molten state, it consists of discrete molecules with no free-moving ions or electrons to carry charge. Therefore, it does not conduct electricity. [2]
Marking: 1 mark for identifying covalent molecular structure; 1 mark for absence of free-moving ions / charge carriers.

(d) $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]$
(or $Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$ ) [2]
Marking: 1 mark for correct reactants; 1 mark for correct products and balancing.

17.

(a) Both $NaCl$ and $MgCl_2$ are ionic compounds. $NaCl$ has $Na^+$ and $Cl^-$ ions, while $MgCl_2$ has $Mg^{2+}$ and $Cl^-$ ions. Although $Mg^{2+}$ has a higher charge, the $Mg^{2+}$ ion is smaller than $Na^+$ , and the $Cl^-$ ions in $MgCl_2$ are not as efficiently packed. However, the key factor is that $NaCl$ has a 1:1 ratio of ions with optimal packing in the lattice, while $MgCl_2$ has a different crystal structure (layer lattice) with weaker interlayer forces. Alternatively, the higher charge on $Mg^{2+}$ polarises the $Cl^-$ ions more, introducing some covalent character which weakens the ionic bonding slightly. [2]
Marking: 1 mark for identifying both as ionic; 1 mark for explanation involving lattice structure / polarisation / packing efficiency.

(b) $MgCl_2$ is an ionic compound with strong electrostatic forces between $Mg^{2+}$ and $Cl^-$ ions throughout the lattice, requiring a lot of energy to overcome. $AlCl_3$ is a covalent molecular compound (exists as $Al_2Cl_6$ dimers) with weak van der Waals forces between molecules, requiring much less energy to overcome. [2]
Marking: 1 mark for identifying $MgCl_2$ as ionic and $AlCl_3$ as covalent molecular; 1 mark for comparing strength of ionic bonds vs van der Waals forces.

(c) Both $SiCl_4$ and $PCl_3$ are covalent molecular compounds. $SiCl_4$ has a higher melting point than $PCl_3$ because $SiCl_4$ is a larger molecule with more electrons, resulting in stronger van der Waals forces between molecules. More energy is required to overcome these stronger intermolecular forces. [2]
Marking: 1 mark for identifying both as covalent molecular; 1 mark for larger molecular size / more electrons / stronger van der Waals forces in $SiCl_4$ .

18.

(a) The atomic radius increases down Group 1. Each successive element has an additional electron shell. The outer electrons are further from the nucleus and are more shielded by inner shell electrons, so the atomic radius increases. [2]
Marking: 1 mark for stating the trend (atomic radius increases); 1 mark for additional shells / increased shielding / greater distance from nucleus.

(b) The first ionisation energy decreases down Group 1. As we go down the group, each element has more electron shells. The outer electron is further from the nucleus and experiences more shielding from inner shell electrons. The effective nuclear charge on the outer electron is weaker, so less energy is required to remove it. [2]
Marking: 1 mark for stating the trend (first ionisation energy decreases); 1 mark for more shells / increased shielding / weaker effective nuclear charge.

(c) The first ionisation energy of Fr would be expected to be lower than that of Cs (below 376 $kJ\,mol^{-1}$ ). Francium is below caesium in Group 1, so it has even more electron shells. The outer electron is even further from the nucleus and experiences even more shielding, making it even easier to remove. [2]
Marking: 1 mark for predicting first ionisation energy below 376 $kJ\,mol^{-1}$ ; 1 mark for explanation in terms of more shells / greater shielding / weaker nuclear attraction.

19.

(a) After the first electron is removed from Na, the resulting $Na^+$ ion has the electronic configuration $1s^2\,2s^2\,2p^6$ , which is the stable noble gas configuration of neon. Removing a second electron would require breaking into this stable, complete outer shell. Additionally, the second electron is removed from the $2p$ orbital which is closer to the nucleus and experiences less shielding, so it is held much more strongly. [2]
Marking: 1 mark for identifying the stable noble gas configuration of $Na^+$ ; 1 mark for second electron being closer to nucleus / less shielded / held more strongly.

(b)
(i) Group 13. There is a large jump in ionisation energy between the 3rd and 4th ionisation energies (2745 to 11578 $kJ\,mol^{-1}$ ). This indicates that the first 3 electrons are removed from the outer shell, while the 4th electron is removed from an inner shell. Therefore, the element has 3 electrons in its outermost shell, placing it in Group 13. [2]
Marking: 1 mark for identifying the large jump between 3rd and 4th IE; 1 mark for concluding 3 outer electrons / Group 13.

(ii) Aluminium (Al). The first ionisation energy of 578 $kJ\,mol^{-1}$ matches that of aluminium from the Period 3 data. The electronic configuration of Al is $1s^2\,2s^2\,2p^6\,3s^2\,3p^1$ , which has 3 electrons in the outer shell (3s²3p¹), consistent with the large jump after the 3rd ionisation. [1]
Marking: 1 mark for identifying Al with correct reasoning.

20.

(a)

A = Sodium (Na): $A_2O$ is a basic oxide with formula $Na_2O$ , consistent with a Group 1 element. Low melting point and good electrical conductivity indicate a metallic element. [1]
B = Magnesium (Mg): $BO$ is a basic oxide with formula $MgO$ , consistent with a Group 2 element. Good electrical conductivity indicates a metallic element. [1]
C = Aluminium (Al): $C_2O_3$ is an amphoteric oxide with formula $Al_2O_3$ , consistent with a Group 13 element. High melting point but poor electrical conductivity in solid state suggests a giant covalent or ionic structure with no free electrons. [1]
D = Silicon (Si): $DO_2$ is an acidic oxide with formula $SiO_2$ , consistent with a Group 14 element. High melting point and poor electrical conductivity indicate a giant covalent (macromolecular) structure. [1]
Marking: 1 mark each for correct identification with valid reason.

(b) Element C (Al) has a high melting point because $Al_2O_3$ has a giant ionic lattice with strong electrostatic forces between $Al^{3+}$ and $O^{2-}$ ions, requiring a lot of energy to overcome. However, in the solid state, the ions are fixed in position and cannot move to carry charge, so it does not conduct electricity. (Note: Al metal itself conducts electricity, but the question refers to the oxide $C_2O_3$ based on the table.) [2]
Marking: 1 mark for giant ionic lattice / strong electrostatic forces; 1 mark for ions fixed in position / no free-moving charge carriers in solid state.

(c) Element A (Na) is a metal with a sea of delocalised electrons that are free to move and carry charge, so it conducts electricity in the solid state. Element D (Si) is a covalent macromolecular solid (or metalloid) where all valence electrons are involved in covalent bonds and there are no free-moving charge carriers, so it does not conduct electricity in the solid state. [2]
Marking: 1 mark for delocalised electrons in A / metallic bonding; 1 mark for no free charge carriers in D / all electrons in covalent bonds.

End of Answer Key