AS Chemistry: Atomic Structure and Ionisation Energy — Revision Notes

Condensed for the final weeks. For the full explanation, use the Atomic Structure: Orbitals and Ionisation Energy study guide.

Sub-shells and orbitals

Sub-shell	Orbitals	Max electrons
s	1	2
p	3	6
d	5	10
f (beyond 9701 scope — 1.3.2 assesses only s, p, d)	7	14

Each orbital holds a maximum of two electrons with opposite spins.

Filling order (9701 scope, up to Kr): 1s 2s 2p 3s 3p 4s 3d 4p (beyond this — 5s, 4d… — is beyond the 9701 scope, which assesses only H to Kr per 1.3/1.4).

4s fills before 3d because it is slightly lower in energy — but 4s is emptied first on ionisation. Both halves of that sentence are examined.

Worked example. Iron’s full configuration is 1s²2s²2p⁶3s²3p⁶3d⁶4s² — written as [Ar] 3d⁶4s², with 4s shown last even though it filled first. When Fe²⁺ forms, the two 4s electrons leave first, giving [Ar] 3d⁶ — not the wrong answer [Ar] 3d⁴4s². Fe³⁺ then loses one 3d electron: [Ar] 3d⁵.

The two exceptions: chromium 3d⁵ 4s¹ and copper 3d¹⁰ 4s¹ — a half-filled or full d sub-shell is more stable.

Orbital shapes and Hund’s rule

An s orbital is spherical, centred on the nucleus. A p orbital is shaped like two lobes either side of the nucleus (a dumbbell, or figure-of-eight), and each p sub-shell has three such orbitals pointing along different axes.

Electrons-in-boxes notation shows each orbital as a box and each electron as an arrow, arrow direction representing spin. Hund’s rule: electrons occupy separate orbitals within a sub-shell, spins unpaired, before any orbital is doubly occupied — this minimises repulsion between electrons of like charge. For fluorine’s 2p⁵: each of the three 2p orbitals gets one arrow first, and only the fourth and fifth electrons force one orbital to hold a second, oppositely-spinning arrow. A species with one or more unpaired electrons is a free radical — the same idea reappears later for free-radical substitution in organic chemistry.

Ionisation energy

The energy required to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions.

X(g)  ->  X+(g)  +  e-

A complete answer should include the (g) state symbols — a full, precise equation includes them even though the underlying chemistry is unaffected by their absence.

The three controlling factors

1. Nuclear charge — more protons, stronger attraction, higher IE.
2. Atomic radius — further out, weaker attraction, lower IE.
3. Shielding — more inner shells, weaker effective attraction, lower IE.

Any explanation should cite all three.

Trends

Across a period: IE increases — nuclear charge rises while shielding stays roughly constant and radius decreases.

Down a group: IE decreases — radius and shielding both increase, outweighing the greater nuclear charge.

The two Period 3 dips — learn these precisely

Dip	Explanation
Al < Mg (Group 13 dips below Group 2)	Al’s outer electron is in a 3p orbital, higher in energy and more shielded than Mg’s 3s
S < P (Group 16 dips below Group 15)	In sulfur, two electrons share one 3p orbital and repel each other, so one is more easily removed

Explaining Al with “more shielding by inner shells” is the classic wrong answer — it is the 3p/3s energy difference.

Successive ionisation energies

A large jump occurs when an electron is removed from a shell closer to the nucleus. The position of the jump gives the group.

Element X:  590, 1150, 4940, 6480 kJ/mol
                   ^^^^ jump after the 2nd

Two outer electrons  ->  GROUP 2

Exam traps

• Omitting (g) state symbols in the equation.
• Removing 3d before 4s on ionisation.
• Explaining the Al dip with inner-shell shielding.
• Saying “the nucleus is bigger” rather than “greater nuclear charge”.
• Ionisation energy is always endothermic and therefore positive.
• Successive IEs always increase — the question is where the big jump falls.
• Writing “energy released” instead of “energy required” for ionisation energy — it is never exothermic.
• Drawing all five arrows in one 2p box instead of spreading them across three orbitals first.

Self-test

1. Define first ionisation energy and write the equation for sodium.
2. Give the electron configuration of chromium and explain the anomaly.
3. Why is aluminium’s first IE lower than magnesium’s?
4. Why is sulfur’s lower than phosphorus’s?
5. Successive IEs are 738, 1451, 7733, 10 540. Which group is the element in?
6. Sketch the shape of a p orbital and state how many orbitals make up a p sub-shell.
7. Draw the electrons-in-boxes for fluorine’s 2p⁵ and explain which rule you have applied.

Answers: 1. The energy required to remove one electron from each atom in one mole of gaseous atoms: Na(g) → Na⁺(g) + e⁻. 2. 1s²2s²2p⁶3s²3p⁶3d⁵4s¹ — a half-filled 3d sub-shell is more stable than 3d⁴4s². 3. Its outer electron occupies a 3p orbital, which is higher in energy and better shielded than the 3s orbital, so it is removed more easily. 4. Sulfur has two electrons paired in one 3p orbital; their mutual repulsion makes one easier to remove. 5. The big jump is between the 2nd and 3rd, so there are two outer electrons — Group 2. 6. Two lobes either side of the nucleus (a dumbbell/figure-of-eight shape); a p sub-shell has three orbitals. 7. Three separate arrows, one per 2p orbital, pointing the same way, then a fourth arrow doubling up (opposite spin) in one orbital and a fifth doubling up in a second — this is Hund’s rule, filling orbitals singly before pairing.