Periodicity Exam Practice | AQA A-Level Chemistry

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

(b) Explain why the first ionisation energy of aluminium is lower than that of magnesium. [2]

(c) Explain why the first ionisation energy of sulfur is lower than that of phosphorus. [3]

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(a)

General trend: First ionisation energy increases [1]
Nuclear charge (number of protons) increases [1]
Shielding remains constant (electrons are added to the same main shell) and atomic radius decreases, leading to stronger electrostatic attraction between the nucleus and outer electrons [1]

(b)

Aluminium outer electron is in a 3p subshell, whereas magnesium outer electron is in a 3s subshell [1]
The 3p subshell is higher in energy / further from the nucleus, and is shielded by the 3s electrons, making it easier to remove [1]

(c)

Phosphorus outer configuration is 3p3 (three singly occupied orbitals) while sulfur outer configuration is 3p4 (one paired orbital, two singly occupied) [1]
In sulfur, the outer electron is removed from a paired 3p orbital [1]
Repulsion between the two paired electrons in the same orbital makes it easier to remove an electron [1]

Examiner tip: State configurations where appropriate. When explaining deviations, reference the specific subshell (3s vs 3p for Mg vs Al) or orbital pairing (3p3 vs 3p4 for P vs S). Avoid vague terms like "sub-level" without naming it.

(a) Describe and explain the trend in melting point from sodium to aluminium. [3]

(b) Explain why silicon has a high melting point. [2]

(c) Compare the melting points of phosphorus (P4) and sulfur (S8), and explain the difference in terms of their structures and intermolecular forces. [4]

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(a)

Melting point increases from Na to Mg to Al [1]
Charge density increases / number of delocalised electrons per atom increases (from Na+ to Mg2+ to Al3+) and ionic radius decreases [1]
Stronger electrostatic attraction between the metal cations and the delocalised electrons, requiring more thermal energy to break the metallic bonds [1]

(b)

Silicon has a giant covalent (macromolecular) structure [1]
Contains many strong covalent bonds that require a very large amount of energy to break [1]

(c)

Sulfur (S8) has a higher melting point than phosphorus (P4) [1]
Both exist as simple molecular structures with weak van der Waals forces between molecules [1]
S8 molecules are larger and contain more electrons than P4 molecules [1]
Stronger van der Waals forces between S8 molecules require more energy to overcome [1]

Examiner tip: Ensure you state that van der Waals forces are intermolecular (between molecules), not intramolecular. Breaking covalent bonds in silicon is required for melting, but melting simple molecular elements only overcomes weak intermolecular forces, not covalent bonds.

(a) Define the term electronegativity. [2]

(b) State and explain the trend in electronegativity down Group 2. [3]

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(a)

The power / ability of an atom to attract [1]
The electron pair in a covalent bond [1]

(b)

Electronegativity decreases down the group [1]
Atomic radius increases and shielding increases (more inner shells) [1]
There is a weaker electrostatic attraction between the positive nucleus and the bonding pair of electrons [1]

Examiner tip: Memorise the definition of electronegativity word-for-word. A common error is defining it in terms of attracting a lone electron or forming a negative ion, which is electron affinity, not electronegativity.

3.2.1 Periodicity Exam Practice

📋 Structured Questions

Question 1: Period 3 Ionisation Energy Trends

Question 2: Structure and Melting Points across Period 3

Question 3: Electronegativity