AS Chemistry · Cambridge 9701 · Atomic Structure

Ionisation Energies & Electron Configuration

A step-by-step method for sketching successive ionisation energy graphs — using tellurium as a worked example.

The question

Tellurium (Te) is in Group 16. Its electronic configuration is [Kr] 4d¹⁰ 5s² 5p⁴.
Sketch a graph to show the trend in the first seven ionisation energies of Te.

Strip away the noble-gas core and identify the outer electrons

The [Kr] core is fully filled and acts as an inner shield. Every electron you care about is outside that core. Write out only the outer subshells and count them.

[Kr] 4d¹⁰ 5s² 5p⁴

The n = 5 shell holds 6 electrons (5s² + 5p⁴). The 4d¹⁰ is in shell n = 4 — closer to the nucleus, acting as an inner shell relative to the 5s and 5p electrons.

The first 7 ionisations will remove: 4 electrons from 5p, 2 from 5s, and 1 from 4d.

Map each ionisation to the subshell it removes from

Electrons are always removed from the highest-energy (outermost) subshell first. For Te, the 5p subshell is outermost. Within 5p⁴, one orbital is paired — that paired electron is removed first.

5p⁴ →

↑↓

↑

(one paired, two singly filled)

IE	Electron removed from	Subshell	Approx. IE / kJ mol⁻¹
1st	5p⁴ — paired orbital (slight ease from repulsion)	5p	870
2nd	5p³	5p	1 045
3rd	5p²	5p	1 800
4th	5p¹ — last 5p electron	5p	2 190
5th	5s² — first 5s electron (new subshell)	5s	3 000
6th	5s¹ — last 5s electron	5s	3 610
7th	4d¹⁰ — first inner-shell electron (shell n = 4)	4d	5 800

Explain the general upward trend

Each successive removal leaves fewer electrons shielding the same nuclear charge. The remaining electrons feel a stronger effective nuclear attraction → more energy is needed to remove each one.

Within the same subshell (e.g. across IEs 1–4, all from 5p), the increase is gradual and relatively steady.

Identify where the jumps occur — and explain each one

A jump appears whenever you cross into a lower-energy subshell or a new (inner) principal quantum shell. That electron is held more tightly — closer to nucleus, better shielded from outside but not from within.

⚡

Smaller jump — between IE4 and IE5: moving from 5p → 5s. Both are in shell n = 5, but the 5s orbital penetrates closer to the nucleus and is lower in energy. A noticeable step up is visible.

🔺

Large jump — between IE6 and IE7: crossing from n = 5 (5s) into n = 4 (4d). This is a principal quantum shell boundary. The 4d electrons are far closer to the nucleus with much less shielding from outer electrons → very large jump. This is the signature that reveals Te belongs to Group 16 — 6 outer electrons before the big jump.

Step 5 — Sketch the graph

Three features earn the marks: a general upward trend, a step between IE4 and IE5, and a much larger jump between IE6 and IE7.

5p electrons (IE 1–4)

5s electrons (IE 5–6)

4d electrons (IE 7)

Exam technique — what markers look for

Count the electrons before the large (shell-crossing) jump to find the group number. Te has 6 electrons before the IE6→IE7 jump → Group 16.
Show seven discrete data points (crosses or dots), not a continuous curve — these are separate measured quantities.
The jump at IE7 must look much larger than the step at IE5. Relative heights matter on a sketch graph even if exact values are not required.
Label both axes: ionisation energy / kJ mol⁻¹ on the y-axis, successive ionisation on the x-axis, with 1–7 marked.
The 1st IE may be very slightly lower than IE2 (because IE1 removes from the paired 5p orbital where repulsion lowers the barrier). This is a bonus insight — only discuss it if the question explicitly asks about Group 16 pairing effects.