AS Chemistry: Atomic Structure and Ionisation Energy — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Atomic Structure and Ionisation Energy revision notes

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Section A

1. Define first ionisation energy. [3]

2. State the three factors that determine the size of an ionisation energy. [3]

3. Write the electron configuration of (a) Fe²⁺ (b) Cr. [2]

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Section B

4. The first ionisation energies across period 3 generally increase, but there are two dips.

(a) Explain the general increase. [3]

(b) Explain why aluminium is lower than magnesium. [2]

(c) Explain why sulfur is lower than phosphorus. [2]

(d) State what these two dips demonstrate about atomic structure. [1]

5. The successive ionisation energies of an element (kJ mol⁻¹) are: 590, 1145, 4912, 6474, 8144.

(a) Identify the group, explaining your reasoning. [3]

(b) Explain why there is a very large jump between the second and third values. [2]

6. Explain why the first ionisation energy decreases down group 2. [3]

7. Fluorine has the electron configuration 1s²2s²2p⁵.

(a) Describe, using electrons-in-boxes notation, how the five 2p electrons are arranged across the three 2p orbitals. [2] (b) State the rule this arrangement follows, and explain why electrons arrange this way rather than pairing up in fewer orbitals first. [2]

8. Define the second ionisation energy of magnesium, giving the equation with state symbols. [3]

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Answers

1. The energy required to remove one mole of electrons [1] from one mole of gaseous atoms [1] to form one mole of gaseous 1+ ions [1].

2. Nuclear charge [1]; atomic radius / distance of the outer electron from the nucleus [1]; shielding by inner shells [1].

3. (a) [Ar]3d⁶ [1] — the 4s electrons are lost first. (b) [Ar]3d⁵4s¹ [1].

4. (a) Nuclear charge increases across the period [1] while shielding stays approximately constant [1], and the atomic radius decreases, so the outer electron is held more strongly [1].

(b) Aluminium’s outer electron is in a 3p sub-shell, which is higher in energy [1] and is slightly shielded by the 3s electrons, so less energy is needed to remove it [1].

(c) In sulfur, two electrons occupy the same 3p orbital [1] and repel each other, so one is more easily removed than an unpaired 3p electron in phosphorus [1].

(d) They provide direct evidence for the existence of sub-shells (and orbitals) within the main shells [1].

5. (a) Group 2 [1]. There is a large jump after the second ionisation energy [1], showing that two electrons are readily removed from the outer shell before an electron must be taken from a shell closer to the nucleus [1].

(b) The third electron is removed from a shell closer to the nucleus [1], which experiences less shielding and a stronger attraction, so far more energy is required [1].

6. Down the group the outer electron is in a shell further from the nucleus [1] with more shielding from inner shells [1], so the attraction between the nucleus and the outer electron is weaker and it is removed more easily [1].

7. (a) Each of the three 2p orbitals holds one unpaired electron first (arrows pointing the same way) [1], and only the fourth and fifth electrons cause one orbital to become doubly occupied, with the second electron in that orbital having the opposite spin [1]. (b) This follows Hund’s rule [1]. Electrons occupy separate orbitals singly before pairing up, because electrons carry the same (negative) charge and repel each other; spreading them across separate orbitals minimises this repulsion, making the configuration lower in energy than pairing electrons in fewer orbitals would be [1].

8. The energy required to remove one mole of electrons from one mole of gaseous Mg⁺ ions [1], forming one mole of gaseous Mg²⁺ ions [1]. Mg⁺(g) → Mg²⁺(g) + e⁻ [1].

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Where marks are usually lost

• Omitting “one mole” or “gaseous” from the definition.
• Explaining the Al dip by nuclear charge rather than sub-shell.
• Explaining the S dip by anything other than paired-electron repulsion.
• Writing Fe²⁺ as 3d⁴4s².
• Counting the jump position incorrectly when deducing the group.
• Forgetting that a second (or later) ionisation energy always starts from the ion already formed, not from the neutral atom — the equation must show the correct starting species.
• Filling 2p orbitals by pairing electrons up two at a time instead of spreading them singly across all three orbitals first (Hund’s rule).

Why the S dip and Hund’s rule are really the same idea

The sulfur-below-phosphorus dip in Period 3 (question 4c) and the electrons-in-boxes arrangement for fluorine (question 7) both rest on the same underlying physics: electrons repel each other more strongly when forced to share a single orbital than when spread across separate orbitals of the same sub-shell. That repulsion is what makes one of sulfur’s paired 3p electrons easier to remove than phosphorus’s unpaired one, and it is also exactly why Hund’s rule has electrons filling empty orbitals singly before any pairing occurs. Recognising this as one principle applied in two contexts, rather than two unrelated facts to memorise separately, makes both questions much more predictable. For the full explanation, including orbital shapes and free radicals, see the Atomic Structure and Ionisation Energy revision notes.