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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. Marks will be awarded for correct reasoning and method.
The number of marks for each question is shown in brackets [ ].
Clean numerical answers should be given to 2 or 3 significant figures unless otherwise stated.

Section A: Multiple Choice (Questions 1–5)

Each question carries 2 marks. Choose the ONE best answer.

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 effective nuclear charge increases.
C. Atomic radius increases because the shielding effect increases.
D. Atomic radius remains constant because the number of electron shells is the same.

2. An element X has the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^4$ . Which statement about element X is correct?

A. X is a metal in Group 16 of the Periodic Table.
B. X forms a basic oxide with formula $\text{XO}_2$ .
C. X has a more negative electron affinity than the element with configuration $1s^2 2s^2 2p^6 3s^2 3p^5$ .
D. X forms an ion with charge $2-$ and is in Group 16.

3. Which of the following oxides is amphoteric?

A. $\text{Na}_2\text{O}$
B. $\text{Al}_2\text{O}_3$
C. $\text{SO}_3$
D. $\text{Cl}_2\text{O}_7$

4. The first four ionisation energies ( $\text{kJ mol}^{-1}$ ) of an element Y are given below:

Ionisation energy	1st	2nd	3rd	4th
Value	590	1145	4912	6491

Which group of the Periodic Table does element Y most likely belong to?

A. Group 1
B. Group 2
C. Group 13
D. Group 14

5. Which of the following correctly explains why the melting point of the halogens increases down Group 17?

A. The bond strength of the covalent bond within the halogen molecule increases.
B. The van der Waals forces between molecules increase as the number of electrons increases.
C. The electronegativity of the halogens increases down the group.
D. The atomic radius decreases, allowing closer packing of molecules.

Section B: Short Answer & Structured Response (Questions 6–15)

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

(b) Write an equation, including state symbols, to represent the first ionisation energy of magnesium. [1]

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

Element	Na	Mg	Al	Si	P	S	Cl	Ar
1st I.E.	496	738	578	789	1012	1000	1251	1521

(a) Describe the general trend in first ionisation energy across Period 3. [1]

(b) Explain why aluminium has a lower first ionisation energy than magnesium. [2]

8. (a) State and explain the trend in electronegativity across Period 3 from sodium to chlorine. [3]

(b) Using your knowledge of electronegativity, predict and explain the type of bonding present in the compound aluminium chloride, $\text{AlCl}_3$ , given that aluminium has an electronegativity of 1.5 and chlorine has an electronegativity of 3.0. [2]

9. The table below shows the melting points of the Period 3 oxides.

Oxide	$\text{Na}_2\text{O}$	$\text{MgO}$	$\text{Al}_2\text{O}_3$	$\text{SiO}_2$	$\text{P}_4\text{O}_{10}$	$\text{SO}_3$
Melting point / °C	1275	2852	2072	1710	300	-10

(a) Explain why $\text{MgO}$ has a very high melting point. [3]

(b) Explain why $\text{SO}_3$ has a much lower melting point than $\text{SiO}_2$ . [3]

10. (a) Describe the trend in atomic radius down a group in the Periodic Table. [1]

(b) Explain this trend in terms of electronic structure and effective nuclear charge. [3]

11. Element Z is in Period 3 and forms an oxide with formula $\text{ZO}_2$ . This oxide reacts with aqueous sodium hydroxide to form a salt and water.

(a) Identify the type of oxide $\text{ZO}_2$ is. [1]

(b) Write a balanced equation for the reaction of $\text{ZO}_2$ with aqueous $\text{NaOH}$ . [1]

12. (a) What is meant by the term amphoteric? [1]

(b) Using aluminium oxide as an example, write one equation to show its reaction with an acid and one equation to show its reaction with a base. [2]

With acid: __________________________________________________________________

With base: __________________________________________________________________

13. The table below shows some properties of the Period 3 chlorides.

Chloride	$\text{NaCl}$	$\text{MgCl}_2$	$\text{AlCl}_3$	$\text{SiCl}_4$
Melting point / °C	801	714	190 (sublimes)	-68
Electrical conductivity (molten)	Good	Good	Poor	None

(a) Explain why $\text{NaCl}$ and $\text{MgCl}_2$ conduct electricity in the molten state but $\text{SiCl}_4$ does not. [3]

(b) Molten $\text{AlCl}_3$ is a poor conductor of electricity. What does this suggest about the bonding in $\text{AlCl}_3$ ? [1]

14. (a) State the relationship between the position of an element in the Periodic Table and the charge of the ion it commonly forms. Give one example. [2]

(b) Explain why the ionic radius of $\text{Na}^+$ is smaller than the atomic radius of $\text{Na}$ . [2]

15. The graph below shows the second ionisation energies of the first 20 elements.

<image_placeholder> id: Q15-fig1 type: graph linked_question: Q15 description: A graph of second ionisation energy (y-axis, in kJ mol⁻¹) against atomic number (x-axis, 1 to 20). The graph shows a generally increasing trend from H to He, drop at Li, gradual increase to Be, drop at B, increase to C, increase to N, drop at O, increase to F, increase to Ne, sharp drop at Na, gradual increase to Mg, drop at Al, increase to Si, increase to P, drop at S, increase to Cl, increase to Ar, sharp drop at K. Key features: sharp drops after noble gas configurations (He→Li, Ne→Na, Ar→K), and drops after Group 2 (Be→B, Mg→Al) and after Group 15 (N→O, P→S). labels: y-axis: "Second Ionisation Energy / kJ mol⁻¹", x-axis: "Atomic Number", title: "Second Ionisation Energy of Elements 1–20" values: Key data points: Li has a very high second IE (~7298 kJ/mol), Be has a relatively high second IE (~1757 kJ/mol), B has a lower second IE (~2427 kJ/mol), N has a high second IE (~2856 kJ/mol), O has a lower second IE (~3388 kJ/mol), Na has a very high second IE (~4562 kJ/mol), Mg has a relatively high second IE (~1451 kJ/mol), Al has a lower second IE (~1817 kJ/mol), P has a high second IE (~1903 kJ/mol), S has a lower second IE (~2252 kJ/mol), K has a very high second IE (~3052 kJ/mol), Ca has a relatively high second IE (~1145 kJ/mol). must_show: Clear drops at Li (very high), Na (very high), K (very high) — indicating removal of an electron from a noble gas core. Also drops from Group 2 to Group 13 (Be→B, Mg→Al) and Group 15 to Group 16 (N→O, P→S) indicating electron removal from a new subshell or paired orbital.

(a) Explain why the second ionisation energy of lithium is much higher than that of beryllium. [3]

(b) Explain why the second ionisation energy of oxygen is lower than that of nitrogen. [2]

Section C: Extended Response & Data Analysis (Questions 16–20)

16. The following table gives some properties of the Period 3 elements.

Property	Na	Mg	Al	Si	P	S	Cl
Atomic radius / nm	0.186	0.160	0.143	0.117	0.110	0.104	0.099
1st I.E. / $\text{kJ mol}^{-1}$	496	738	578	789	1012	1000	1251
Electronegativity	0.9	1.2	1.5	1.8	2.1	2.5	3.0
Electrical conductivity of element	Good	Good	Good	Semiconductor	None	None	None

(a) Describe and explain the trend in atomic radius across Period 3. [3]

(b) Silicon is described as a semiconductor. Explain, in terms of its structure and bonding, why silicon has this property. [2]

(c) Using the data in the table, explain the relationship between electronegativity and first ionisation energy across Period 3. [2]

17. Chlorine, bromine, and iodine are Group 17 elements.

(a) Describe the trend in boiling point down Group 17 and explain this trend in terms of structure and intermolecular forces. [4]

(b) Explain why hydrogen chloride is a gas at room temperature while sodium chloride is a solid. [3]

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

Ionisation energy	1st	2nd	3rd	4th	5th	6th	7th
Value	496	4562	6912	9543	13353	16610	20114

(a) Identify the group to which element Q belongs. Explain your reasoning using the data. [3]

(b) Write the full electronic configuration of element Q. [1]

19. Consider the following oxides of Period 3 elements: $\text{Na}_2\text{O}$ , $\text{MgO}$ , $\text{Al}_2\text{O}_3$ , $\text{SiO}_2$ , $\text{P}_4\text{O}_{10}$ , and $\text{SO}_3$ .

(a) Classify each oxide as basic, acidic, or amphoteric. [3]

$\text{Na}_2\text{O}$ : ___________________________

$\text{MgO}$ : ___________________________

$\text{Al}_2\text{O}_3$ : ___________________________

$\text{SiO}_2$ : ___________________________

$\text{P}_4\text{O}_{10}$ : ___________________________

$\text{SO}_3$ : ___________________________

(b) Write balanced equations for the reaction of $\text{Na}_2\text{O}$ with water and $\text{P}_4\text{O}_{10}$ with water. [2]

$\text{Na}_2\text{O}$ with water: _______________________________________________________

$\text{P}_4\text{O}_{10}$ with water: _______________________________________________________

(c) Explain, in terms of bonding and structure, why $\text{SiO}_2$ has a much higher melting point than $\text{P}_4\text{O}_{10}$ . [3]

20. An element R is in Period 4, Group 2 of the Periodic Table.

(a) Write the full electronic configuration of R. [1]

(b) Predict and explain the trend in the following properties of the Group 2 elements from Be to Ba: (i) Atomic radius [2]

(ii) First ionisation energy [2]

(d) A solution of the hydroxide of R has a pH greater than 7. Explain why, with reference to the dissociation of the hydroxide in water. [2]

END OF QUIZ

Answers

A-Level Chemistry H1 Quiz - Periodic Table

Answer Key & Teaching Notes

Choosing A: The covalent bond strength within the molecule does not determine melting point; intermolecular forces do.
Choosing C: Electronegativity actually decreases down the group.

Section B: Short Answer & Structured Response

Question 6

(a) [2 marks]

Answer: The 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 1+ ions.

Marking: 1 mark for "remove electron from gaseous atom"; 1 mark for correct specification of "one mole of electrons from one mole of atoms forming one mole of unipositive gaseous ions."

(b) [1 mark]

Answer: $\text{Mg}(g) \rightarrow \text{Mg}^+(g) + e^-$

Marking: Correct species and state symbols. Must show gaseous state and the electron as a product.

(c) [2 marks]

Explanation:

Mg has a higher nuclear charge (12 protons) than Na (11 protons). [1]
Both have the same shielding from inner electrons (core electrons: $1s^2 2s^2 2p^6$ ), so the effective nuclear charge experienced by the outer electron is greater in Mg. [1]
The outer electron in Mg is held more tightly, requiring more energy to remove.

Common Mistakes: Students may say "more electron shells" or "more electrons" without specifying effective nuclear charge and shielding.

Question 7

(a) [1 mark]

Answer: The first ionisation energy generally increases across Period 3 from Na to Ar.

(b) [2 marks]

Explanation:

In Mg, the outermost electron is in the 3s orbital. [1]
In Al, the outermost electron is in the 3p orbital, which is at a higher energy level (slightly further from the nucleus) and is shielded by the 3s electrons. [1]
Therefore, the 3p electron in Al is easier to remove, giving a lower ionisation energy.

(c) [2 marks]

Explanation:

In P, the 3p orbitals contain three unpaired electrons (one in each orbital). [1]
In S, one of the 3p orbitals contains a pair of electrons. The electron-electron repulsion within the paired orbital makes it easier to remove one of the paired electrons. [1]

Common Mistakes: Students may confuse the explanation for Al vs Mg with that of S vs P. The Al vs Mg drop is due to subshell energy; the S vs P drop is due to electron pairing.

Question 8

(a) [3 marks]

Answer:

Electronegativity increases across Period 3 from Na to Cl. [1]
This is because the nuclear charge increases while the number of inner electron shells remains the same. [1]
The effective nuclear charge increases, so the bonding electrons are attracted more strongly by the atom. [1]

(b) [2 marks]

Explanation:

The electronegativity difference is $3.0 - 1.5 = 1.5$ . [1]
Since the difference is relatively small and aluminium is a metal near the metal–non-metal boundary, $\text{AlCl}_3$ is covalent (or predominantly covalent). It exists as a simple molecular compound (dimer $\text{Al}_2\text{Cl}_6$ in vapour state) rather than an ionic lattice. [1]

Common Mistakes: Students may assume all metal–non-metal compounds are ionic. The electronegativity difference and the position of Al near the metalloid boundary must be considered.

Question 9

(a) [3 marks]

Answer:

$\text{MgO}$ has a giant ionic lattice structure. [1]
It consists of $\text{Mg}^{2+}$ and $\text{O}^{2-}$ ions held together by strong electrostatic forces of attraction. [1]
A large amount of energy is needed to overcome these strong ionic bonds, resulting in a very high melting point. [1]

(b) [3 marks]

Answer:

$\text{SiO}_2$ has a giant covalent (macromolecular) structure where each Si atom is bonded to four O atoms in a tetrahedral arrangement, forming a 3D network. [1]
$\text{SO}_3$ exists as simple covalent molecules ( $\text{SO}_3$ molecules) held together by weak van der Waals forces. [1]
The strong covalent bonds throughout the $\text{SiO}_2$ lattice require much more energy to break than the weak intermolecular forces between $\text{SO}_3$ molecules. [1]

Common Mistakes: Students may say "ionic bonds" for $\text{SiO}_2$ ; it is covalent. Also, students may not distinguish between breaking covalent bonds (in $\text{SiO}_2$ ) and overcoming intermolecular forces (in $\text{SO}_3$ ).

Question 10

(a) [1 mark]

Answer: Atomic radius increases down a group.

(b) [3 marks]

Explanation:

Going down a group, the number of electron shells increases. [1]
The outermost electrons are further from the nucleus. [1]
Although the nuclear charge increases, the additional inner electron shells provide more shielding, so the effective nuclear charge does not increase enough to compensate for the greater distance. [1]

Question 11

(a) [1 mark]

Answer: Acidic oxide.

(b) [1 mark]

Answer: $\text{ZO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZO}_2 + \text{H}_2\text{O}$

(Or a specific equation if Z is identified as Si: $\text{SiO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SiO}_3 + \text{H}_2\text{O}$ )

(c) [2 marks]

Answer: Z is silicon (Si). [1] Silicon is a metalloid in Period 3, Group 14, and its oxide $\text{SiO}_2$ is weakly acidic, reacting with NaOH to form sodium silicate and water. [1]

Alternative acceptable answer: Z could be sulfur (S), but $\text{SO}_2$ is the more common oxide of sulfur that reacts with NaOH. Given the formula $\text{ZO}_2$ , silicon is the best fit.

Question 12

(a) [1 mark]

Answer: An amphoteric substance is one that can react with both acids and bases.

(b) [2 marks]

With acid: $\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}$ [1]

With base: $\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{Na}[\text{Al(OH)}_4]$ [1]

(Alternative for base: $\text{Al}_2\text{O}_3 + 2\text{NaOH} \rightarrow 2\text{NaAlO}_2 + \text{H}_2\text{O}$ )

Question 13

(a) [3 marks]

Explanation:

$\text{NaCl}$ and $\text{MgCl}_2$ are ionic compounds. In the molten state, the ions are free to move and carry charge, so they conduct electricity. [1]
$\text{SiCl}_4$ is a covalent (simple molecular) compound. It consists of neutral molecules with no mobile ions or electrons. [1]
Therefore, $\text{SiCl}_4$ does not conduct electricity in any state. [1]

(b) [1 mark]

Answer: This suggests that $\text{AlCl}_3$ is predominantly covalent (or a poor ionic conductor), existing as molecules rather than as a fully ionic lattice. In the molten state, it exists as $\text{Al}_2\text{Cl}_6$ dimers with few free ions.

Question 14

(a) [2 marks]

Answer: For main group elements, the charge of the ion formed is related to the group number. Group 1 elements form 1+ ions, Group 2 elements form 2+ ions, Group 13 elements form 3+ ions, Group 16 elements form 2− ions, and Group 17 elements form 1− ions. [1]

Example: Sodium (Group 1) forms $\text{Na}^+$ ; oxygen (Group 16) forms $\text{O}^{2-}$ . [1]

(b) [2 marks]

Explanation:

$\text{Na}^+$ has lost its outer 3s electron, so it has one fewer electron shell (now only n = 2). [1]
The remaining electrons experience a greater effective nuclear charge per electron and are pulled closer to the nucleus, making the ionic radius smaller than the atomic radius. [1]

Question 15

(a) [3 marks]

Explanation:

Li has the configuration $1s^2 2s^1$ . After losing one electron, $\text{Li}^+$ has the configuration $1s^2$ (a stable noble gas configuration). [1]
The second ionisation energy of lithium involves removing an electron from the 1s orbital, which is much closer to the nucleus and experiences very little shielding. [1]
Be has the configuration $1s^2 2s^2$ . The second ionisation removes a 2s electron, which is in a higher energy level and is shielded by the 1s electrons. [1]
Therefore, the second IE of Li is much higher than that of Be.

(b) [2 marks]

Explanation:

Nitrogen has the configuration $1s^2 2s^2 2p^3$ , with one electron in each 2p orbital (all unpaired). [1]
Oxygen has the configuration $1s^2 2s^2 2p^4$ , with one paired set of electrons in a 2p orbital. The electron-electron repulsion in the paired orbital makes it easier to remove an electron, so the second IE of O is lower than that of N. [1]

Note on graph: The graph should show very high second IE values for Li, Na, and K (removing an electron from a noble gas core), and drops from Group 2→13 and Group 15→16.

Section C: Extended Response & Data Analysis

Question 16

(a) [3 marks]

Answer:

Atomic radius decreases across Period 3. [1]
The number of protons (nuclear charge) increases while electrons are added to the same principal energy level (n = 3). [1]
The shielding effect from inner electrons remains approximately constant, so the effective nuclear charge increases, pulling the electron cloud closer to the nucleus. [1]

(b) [2 marks]

Answer:

Silicon has a giant covalent (diamond-like) structure where each Si atom is covalently bonded to four other Si atoms in a tetrahedral arrangement. [1]
At low temperatures, electrons are held in covalent bonds and cannot conduct. At higher temperatures, some electrons gain enough energy to move into the conduction band, allowing limited conductivity. This intermediate behaviour makes it a semiconductor. [1]

(c) [2 marks]

Answer:

Both electronegativity and first ionisation energy generally increase across Period 3. [1]
This is because both properties depend on the effective nuclear charge: a higher effective nuclear charge means the atom attracts bonding electrons more strongly (higher electronegativity) and holds its own electrons more tightly (higher ionisation energy). [1]

Question 17

(a) [4 marks]

Answer:

The boiling point increases down Group 17 (from $\text{F}_2$ to $\text{I}_2$ ). [1]
Halogens exist as diatomic molecules ( $\text{F}_2$ , $\text{Cl}_2$ , $\text{Br}_2$ , $\text{I}_2$ ) held together by weak van der Waals (London dispersion) forces between molecules. [1]
Going down the group, the number of electrons in the molecule increases. [1]
More electrons mean stronger temporary dipoles and hence stronger van der Waals forces, requiring more energy to overcome, so the boiling point increases. [1]

(b) [3 marks]

Answer:

$\text{HCl}$ is a simple covalent molecule with weak van der Waals forces between molecules. Little energy is needed to separate the molecules, so it is a gas at room temperature. [1]
$\text{NaCl}$ has a giant ionic lattice structure with strong electrostatic forces of attraction between $\text{Na}^+$ and $\text{Cl}^-$ ions. [1]
A large amount of energy is needed to overcome these strong ionic bonds, so $\text{NaCl}$ is a solid at room temperature. [1]

Question 18

(a) [3 marks]

Answer:

Element Q is in Group 1. [1]
There is a very large jump between the 1st IE (496) and the 2nd IE (4562). [1]
This indicates that the first electron is easy to remove (from the outer shell), but the second electron must be removed from a stable, inner noble gas configuration. This is characteristic of Group 1 elements with one valence electron. [1]

(b) [1 mark]

Answer: $1s^2 2s^2 2p^6 3s^1$ (sodium, Na)

(c) [1 mark]

Answer: $\text{Na}^{2+}(g) \rightarrow \text{Na}^{3+}(g) + e^-$

Note: The third ionisation energy involves removing an electron from $\text{Na}^{2+}$ , which has the configuration $1s^2 2s^2 2p^5$ . This requires removing an electron from the inner 2p subshell, which is why the 3rd IE is very high.

Question 19

(a) [3 marks]

Answers:

$\text{Na}_2\text{O}$ : Basic [0.5]
$\text{MgO}$ : Basic [0.5]
$\text{Al}_2\text{O}_3$ : Amphoteric [0.5]
$\text{SiO}_2$ : Acidic (weakly acidic) [0.5]
$\text{P}_4\text{O}_{10}$ : Acidic [0.5]
$\text{SO}_3$ : Acidic [0.5]

Marking: 0.5 marks each. Award 0.5 for each correct classification.

(b) [2 marks]

$\text{Na}_2\text{O}$ with water: $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$ [1]

$\text{P}_4\text{O}_{10}$ with water: $\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4$ [1]

(c) [3 marks]

Answer:

$\text{SiO}_2$ has a giant covalent (macromolecular) structure where each Si atom is bonded to four O atoms in a tetrahedral network. [1]
$\text{P}_4\text{O}_{10}$ exists as simple covalent molecules held together by weak van der Waals forces. [1]
Melting $\text{SiO}_2$ requires breaking strong covalent bonds throughout the lattice, while melting $\text{P}_4\text{O}_{10}$ only requires overcoming weak intermolecular forces. Hence $\text{SiO}_2$ has a much higher melting point. [1]

Question 20

(a) [1 mark]

Answer: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$ (calcium, Ca)

(b) [4 marks]

(i) Atomic radius increases down Group 2. [1] This is because each successive element has an additional electron shell, placing the outermost electrons further from the nucleus. The increased shielding from inner shells outweighs the increased nuclear charge. [1]

(ii) First ionisation energy decreases down Group 2. [1] The outermost electrons are further from the nucleus and experience more shielding from inner electrons, so the effective nuclear charge experienced by the outer electrons is less. Less energy is required to remove the outer electron. [1]

(c) [1 mark]

Answer: $\text{Ca}(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{H}_2(g)$

(d) [2 marks]

Answer:

Calcium hydroxide, $\text{Ca(OH)}_2$ , is a base that dissociates in water to produce hydroxide ions: $\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-$ [1]
The presence of $\text{OH}^-$ ions makes the solution alkaline, giving a pH greater than 7. [1]

Mark Summary

Section	Questions	Marks
A: MCQ	1–5	10
B: Short Answer/Structured	6–15	24
C: Extended Response	16–20	16
Total	20 questions	50