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A Level H2 Chemistry Periodic Table Quiz
Free A Level H2 Chemistry Periodic Table quiz, LongCat Exam version, with questions, answers, and A Level-style practice for Singapore students.
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A-Level Chemistry H2 Quiz - Periodic Table: Answer Key
Question 1 [5 marks]
(a) [2 marks]
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 notes:
- 1 mark for "remove one electron from a gaseous atom" (or equivalent).
- 1 mark for specifying "per mole" / one mole of gaseous atoms → one mole of gaseous ions.
- Common error: omitting "gaseous" — no mark for state.
(b) [3 marks]
Element Q is aluminium (Al).
Reasoning:
- There is a large jump between the 3rd IE (2745 kJ mol⁻¹) and the 4th IE (11577 kJ mol⁻¹). [1]
- This indicates that the first 3 electrons are removed from the outer shell, and the 4th electron is removed from an inner shell (closer to the nucleus, much more tightly held). [1]
- Therefore, Q has 3 valence electrons and is in Group 13. With relatively low first three IEs, it is aluminium in Period 3. [1]
Teaching note: A large jump in successive IEs signals moving from valence electrons to core electrons. The position of the jump tells you the group number (e.g., jump after 3rd IE → Group 13).
Question 2 [7 marks]
(a) [2 marks]
The atomic radius decreases across Period 3 from Na to Cl. [1] This is because the nuclear charge (number of protons) increases across the period, while electrons are added to the same shell (same principal quantum number). The increased nuclear attraction pulls the electron cloud closer to the nucleus. [1]
(b) [2 marks]
The outer electron in Al is in a 3p orbital, whereas in Mg the outer electrons are in the 3s orbital. [1] The 3p orbital is at a higher energy level (further from the nucleus) and is also slightly shielded by the 3s electrons, so the 3p electron is easier to remove. [1]
Common error: Students often say "Al has more electrons" without specifying the orbital type and shielding effect.
(c) [2 marks]
In phosphorus (P), the 3p orbitals are half-filled with one electron each (3p³), which is a stable arrangement. [1] In sulfur (S), one 3p orbital contains a pair of electrons. The electron-electron repulsion within this orbital makes it easier to remove one of the paired electrons. [1]
(d) [1 mark]
Argon does not form covalent bonds, so its atomic radius is measured as a van der Waals radius (not a covalent radius), which is significantly larger and not directly comparable to the covalent radii of the other elements.
Question 3 [4 marks]
(a)(i) [1 mark] 1s² 2s² 2p⁶ 3s² 3p⁴
(a)(ii) [1 mark] 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s² (or [Ar] 3d¹ 4s²)
(b) [2 marks]
Scandium is classified as a transition element because it can form at least one ion with an incomplete d subshell (Sc³⁺ has an empty 3d subshell, but Sc²⁺ has 3d¹; the definition requires the element to have an incomplete d subshell in one of its common oxidation states — Sc has [Ar] 3d¹ 4s² in its ground state). [1]
Sulfur is not a transition element because all its ions (e.g., S²⁻) have a complete d subshell or do not involve d orbitals at all — sulfur is a p-block element. [1]
Teaching note: The IUPAC definition of a transition element is an element that has an incomplete d subshell in any of its commonly occurring oxidation states. Scandium's ground state has 3d¹, satisfying this.
Question 4 [6 marks]
(a) [2 marks]
Helium (Z = 2) has the highest first ionisation energy. [1] This is because helium has a full 1s shell (1s²), a very stable configuration, and the electron is very close to the nucleus with no shielding from other electrons in the same shell, resulting in very strong nuclear attraction. [1]
(b) [2 marks]
From Z = 2 (He) to Z = 3 (Li), the first ionisation energy decreases sharply. [1] In Li, the outer electron is in the 2s orbital, which is further from the nucleus and is shielded by the inner 1s² electrons. This makes the outer electron easier to remove. [1]
(c) [2 marks]
From Z = 7 (N) to Z = 8 (O), the first ionisation energy decreases. [1] In oxygen, one of the 2p orbitals contains a pair of electrons. The electron-electron repulsion within this orbital makes it easier to remove one of the paired electrons compared to nitrogen, where each 2p orbital contains one electron (half-filled, stable). [1]
Question 5 [7 marks]
(a) [1 mark]
Ionic bonding.
(b) [2 marks]
Both Na₂O and MgO are ionic compounds. Mg²⁺ has a higher charge than Na⁺, and Mg²⁺ has a smaller ionic radius than Na⁺. [1] The greater charge density of Mg²⁺ results in stronger electrostatic attraction between Mg²⁺ and O²⁻ ions, requiring more energy to break the lattice, hence a higher melting point. [1]
(c) [2 marks]
SiO₂ has a giant covalent (macromolecular) structure where each Si atom is covalently bonded to four O atoms in a tetrahedral arrangement, forming a 3D network. Breaking this structure requires breaking many strong covalent bonds. [1] P₄O₁₀ has a simple molecular structure with weak intermolecular forces (van der Waals forces) between molecules, which require little energy to overcome. [1]
(d) [2 marks]
SO₃ has a simple molecular structure. [1] The molecules are held together by weak van der Waals forces, which require very little energy to overcome, resulting in a low melting point. [1]
Question 6 [4 marks]
(a) [2 marks]
The relative atomic mass is the weighted average mass of one atom of an element relative to 1/12 the mass of a carbon-12 atom. [2]
Marking notes: 1 mark for "weighted average mass of atoms of the element"; 1 mark for "relative to 1/12 mass of one atom of carbon-12."
(b) [2 marks]
Let the percentage abundance of ³⁵Cl = x%. Then ³⁷Cl = (100 − x)%.
Therefore: ³⁵Cl = 75% and ³⁷Cl = 25%. [2]
Marking notes: 1 mark for correct setup; 1 mark for correct answer.
Question 7 [6 marks]
(a) [2 marks]
In solid NaCl, the ions (Na⁺ and Cl⁻) are held in fixed positions in the ionic lattice and cannot move freely, so they cannot carry charge. [1] In liquid (molten) NaCl, the ions are free to move and can carry electrical charge through the liquid, hence it conducts electricity. [1]
(b) [2 marks]
AlCl₃ is covalent (molecular) in the liquid state. [1] It does not contain free ions; the molecules are neutral, so there are no charge carriers to conduct electricity. [1]
(c) [2 marks]
SiCl₄ is covalent. [1] Silicon and chlorine have a small electronegativity difference, and SiCl₄ is a simple molecular compound with no ions present. It has a low melting point (−70 °C), consistent with weak intermolecular forces. [1]
Question 8 [4 marks]
(a) [2 marks]
The atomic radius increases down Group II. [1] This is because each successive element has an additional electron shell, so the outer electrons are further from the nucleus and more shielded from the nuclear charge. [1]
(b) [2 marks]
The first ionisation energy decreases down Group II. [1] The outer electrons are further from the nucleus and more shielded by inner electron shells, so the attractive force from the nucleus is weaker, making the electrons easier to remove. [1]
Question 9 [5 marks]
(a) [2 marks]
Magnesium: Very slow reaction / few bubbles of gas / the magnesium ribbon dissolves slowly. [1]
Calcium: Faster reaction / more vigorous bubbling / the calcium dissolves more quickly. [1]
(b) [1 mark]
(c) [2 marks]
Barium will react more vigorously than calcium. [1] This is because barium is below calcium in Group II, so its outer electrons are further from the nucleus and more shielded, resulting in a lower ionisation energy and greater reactivity. [1]
Question 10 [4 marks]
(a) [1 mark]
(b) [3 marks]
The thermal stability of Group II carbonates increases down the group because the cation size increases. [1] Ba²⁺ is larger than Mg²⁺, so it has a lower charge density. [1] The larger Ba²⁺ ion is less able to polarise the carbonate ion (CO₃²⁻), so the C–O bonds in the carbonate are less weakened, and more energy (higher temperature) is required to decompose it. [1]
Teaching note: This is a key application of polarising power (Fajan's rule). Smaller, highly charged cations distort the electron cloud of the anion more effectively, weakening the bonds within the anion.
Question 11 [5 marks]
(a) [1 mark]
Magnesium sulfate is more soluble than barium sulfate.
(b) [1 mark]
Barium hydroxide is more soluble than magnesium hydroxide.
(c) [3 marks]
Reagent: Aqueous sodium sulfate (or dilute sulfuric acid). [1]
Observations:
- Magnesium chloride: No visible change / no precipitate. [1]
- Barium chloride: White precipitate forms. [1]
Alternative acceptable answer: Aqueous sodium hydroxide — MgCl₂ gives a white precipitate (Mg(OH)₂), BaCl₂ gives no precipitate (Ba(OH)₂ is soluble enough in dilute solution). Full marks for correct reagent with correct observations.
Question 12 [6 marks]
(a) [2 marks]
With potassium bromide: [1]
With potassium iodide: [1]
(b) [2 marks]
With potassium bromide: The solution turns orange/brown. [1]
With potassium iodide: The solution turns brown (or purple in organic solvent). [1]
(c) [2 marks]
The oxidising ability of the halogens decreases down Group VII. [1] This is because the atomic radius increases down the group, so the outer shell is further from the nucleus and more shielded. The ability to attract and gain an electron (to form X⁻) decreases, so the oxidising power decreases. [1]
Question 13 [6 marks]
(a) [1 mark]
A white precipitate forms.
(b) [1 mark]
The precipitate dissolves in dilute ammonia.
(c) [1 mark]
(d) [3 marks]
Sodium bromide: A cream precipitate (AgBr) forms. The precipitate does not dissolve in dilute ammonia (it dissolves only in concentrated ammonia). [1]
Sodium iodide: A yellow precipitate (AgI) forms. The precipitate does not dissolve in dilute or concentrated ammonia. [1]
Explanation: The solubility of silver halides in ammonia decreases from AgCl to AgI. AgCl is the most soluble in dilute ammonia because the Ag⁺ ion forms a soluble complex ion [Ag(NH₃)₂]⁺ with ammonia. AgBr is less soluble and requires concentrated ammonia. AgI is so insoluble that it does not dissolve even in concentrated ammonia. [1]
Question 14 [5 marks]
(a) [3 marks]
From Figure 2, E°(Cl₂/Cl⁻) = +1.36 V and E°(Br₂/Br⁻) = +1.07 V. [1] Since E°(Cl₂/Cl⁻) > E°(Br₂/Br⁻), chlorine is a stronger oxidising agent than bromine and can oxidise bromide ions to bromine (the reaction is spontaneous because E°cell > 0). [1] However, E°(F₂/F⁻) = +2.87 V, which is more positive than E°(Cl₂/Cl⁻), so fluorine is a stronger oxidising agent than chlorine. Chlorine cannot oxidise fluoride ions because the reaction would have a negative E°cell (non-spontaneous). [1]
(b) [2 marks]
[1 mark for correct formula; 1 mark for correct answer with unit]
Question 15 [4 marks]
(a) [1 mark]
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ (or [Ar] 3d¹⁰ 4s¹)
(b) [1 mark]
Copper is a transition element because it has an incomplete d subshell in its common oxidation state Cu²⁺ (3d⁹). [1]
(c) [2 marks]
In a free Cu atom, all five 3d orbitals have the same energy (they are degenerate). [1] In a compound, the ligands cause the 3d orbitals to split into two sets of different energies. Electrons can absorb visible light to move (d-d transition) between these split d orbitals. The wavelength absorbed depends on the energy gap, and the complementary colour is observed. [1]
Question 16 [4 marks]
(a) [2 marks]
When concentrated HCl is added, the Cl⁻ ligands replace some of the H₂O ligands around Cu²⁺, forming the [CuCl₄]²⁻ complex ion (tetrachlorocuprate(II)). [1] The different ligand field strength of Cl⁻ compared to H₂O changes the energy gap between the split d orbitals, causing a different wavelength of light to be absorbed. The solution appears green (a mixture of blue from [Cu(H₂O)₆]²⁺ and yellow from [CuCl₄]²⁻). [1]
(b) [1 mark]
(c) [1 mark]
[Cu(NH₃)₄(H₂O)₂]²⁺ (tetraamminecopper(II) ion)
Question 17 [6 marks]
(a) [2 marks]
Fe²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ (or [Ar] 3d⁶) [1]
Fe³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ (or [Ar] 3d⁵) [1]
(b) [2 marks]
Fe³⁺ has a 3d⁵ configuration, which is a half-filled d subshell. [1] A half-filled d subshell is particularly stable due to maximum exchange energy and symmetrical distribution of electrons. Therefore, Fe²⁺ (3d⁶) readily loses one electron to achieve this stable half-filled configuration. [1]
(c) [2 marks]
Add aqueous NaOH to separate samples of each solution. [1] Fe²⁺(aq): A green precipitate of Fe(OH)₂ forms (which slowly turns brown on exposure to air due to oxidation). [½] Fe³⁺(aq): A brown/rust-coloured precipitate of Fe(OH)₃ forms. [½]
Marking note: Award 1 mark for correct reagent and 1 mark for correct observations for both ions.
Question 18 [5 marks]
(a) [2 marks]
The 3d and 4s orbitals are very close in energy. [1] A half-filled d subshell (3d⁵) provides extra stability due to maximum exchange energy and symmetrical electron distribution. One electron from the 4s orbital is promoted to the 3d orbital to achieve this stable half-filled configuration. [1]
(b) [1 mark]
+2, +3, +6 (or Cr²⁺, Cr³⁺, CrO₄²⁻/Cr₂O₇²⁻)
(c) [2 marks]
Cr³⁺ has a 3d³ configuration. In an octahedral field, this gives a t₂g³ configuration (all three electrons in the lower-energy t₂g set), which is a stable half-filled t₂g level. [1] Cr²⁺ has a 3d⁴ configuration (t₂g³ eg¹), which is less stable. Cr²⁺ is also a strong reducing agent and is easily oxidised to Cr³⁺ in aqueous solution. [1]
Question 19 [4 marks]
(a) [2 marks]
The large jump between the 4th and 5th IE indicates that the first 4 electrons are removed from the outer shells (4s and 3d), while the 5th electron is removed from an inner shell (3p). [1] The 5th electron is much closer to the nucleus and experiences much less shielding, so significantly more energy is required to remove it. [1]
(b) [2 marks]
The most common oxidation state of titanium is +4. [1] This is because the large jump occurs after the 4th IE, meaning 4 electrons can be relatively easily removed (from the 4s² and 3d² orbitals), but removing a 5th electron requires much more energy. Therefore, Ti⁴⁺ is the most stable and common ion. [1]
Question 20 [6 marks]
(a)(i) [1 mark] +3
(a)(ii) [1 mark] +5 (O is −2 each, so V + 2(−2) = +1 → V = +5... wait: VO₂⁺: V + 2(−2) = +1, so V = +5)
(b) [2 marks]
Yellow: +5 (VO₂⁺) [½] Blue: +4 (VO²⁺) [½] Green: +3 (V³⁺) [½] Violet: +2 (V²⁺) [½]
(c) [2 marks]
Marking notes:
- 1 mark for correct species and stoichiometry.
- 1 mark for correct number of electrons and charge balance.
- Check: Left side charge: +1 + 4 − 1 = +4. Right side: +3. Hmm, let me recalculate.
Corrected equation:
Verification:
- V: +5 in VO₂⁺ → +3 in V³⁺ (gain of 2 electrons) ✓
- Left charge: +1 + 4 − 2 = +3; Right charge: +3 ✓
[1 mark for correct half-equation; 1 mark for correct balancing of charge and atoms]
END OF ANSWER KEY