IB Chemistry SLTopic 3 — Classifying the ElementsPaper 1 & 2Core skill~11 min read
Electron Configuration and Periodicity
This is a two-way street, and both directions are worth marks. Give me an element’s position and I’ll write its electron configuration. Give me a configuration and I’ll tell you exactly which element it is. The method never changes, which makes this one of the most reliable questions on the paper.
📚 What you need to know
An electron configuration says which shell, which sub-shell and how many electrons — for example 3p5.
Sub-shells fill from lowest energy upwards, and 4s fills before 3d.
Configurations are written in numerical order, so 3d appears before 4s on the page.
Shorthand configurations replace the inner electrons with the previous noble gas in square brackets.
From a configuration: period = highest shell number, group = valence electrons, block = last sub-shell written.
When forming positive ions, electrons leave the outermost shell first — so 4s empties before 3d.
How a configuration is written
Every term has three pieces of information packed into it. Read them left to right:
Anatomy of a term
3p5 = shell 3 · sub-shell p · 5 electrons in it
Sub-shells hold a fixed number of electrons: s holds 2, p holds 6, d holds 10, f holds 14. Fill them from the lowest energy upwards and you never have to guess.
Sub-shells fill from the bottom up. The 4s slots in below 3d, which is why potassium and calcium fill 4s before the d block begins.
The one place students trip up is the 4s and 3d swap. The 4s sub-shell is lower in energy than 3d, so it fills first. But when you write the configuration out you put it in numerical order, which means 3d gets written before 4s. Filling order and writing order are two different things.
Fill in energy order, write in number order. It feels contradictory the first time, but both conventions exist for good reasons and examiners accept only the numerical order on the page.
Full and shorthand configurations
Writing out every electron for a heavy element gets tedious, so we use the previous noble gas as a shortcut. Its symbol in square brackets stands for all of its electrons.
WORKED EXAMPLE
Write the full and shorthand electron configurations of chlorine (Z = 17) and of zinc (Z = 30).
Chlorine — 17 electrons, fill from the bottom2 + 2 + 6 + 2 + 5 = 17 ✓1s² 2s² 2p⁶ 3s² 3p⁵Previous noble gas is neon (10 electrons), so shorthand = [Ne] 3s² 3p⁵Zinc — 30 electrons, and 4s fills before 3d2 + 2 + 6 + 2 + 6 + 10 + 2 = 30 ✓1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²Shorthand = [Ar] 3d¹⁰ 4s². Note 3d written before 4s.
Configurations for ions
To make a positive ion, take electrons away from the outermost shell first. For a d-block metal that means the 4s electrons leave before the 3d ones, even though 4s filled first. To make a negative ion, just add electrons into the next available space.
WORKED EXAMPLE
Write the electron configurations of Ca2+ and O2–.
Ca is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²Ca²⁺ has lost 2 electrons — taken from the outermost shell, the 4s.Ca²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ (= [Ar])O is 1s² 2s² 2p⁴O²⁻ has gained 2 electrons, which go into the 2p.O²⁻ = 1s² 2s² 2p⁶ (= [Ne])Both ions now match a noble gas — that is exactly why they form.
Going the other way: configuration to element
This is where periodicity earns its name. Because the table is a map of electron configurations, three quick checks locate any element.
Three quick checks turn any s- or p-block configuration into a position, and a position into a name.
🧩 Finding the element from a configuration
Period = the highest shell number that appears. In 3s2 3p5 that is 3.
Group = add up the valence electrons (the s and p electrons in that highest shell). 2 + 5 = 7, so Group 17.
Block = the last sub-shell written. Ends in p, so p-block — and the superscript tells you how far along.
Read the position off the table. Period 3, Group 17 is chlorine.
WORKED EXAMPLE
Identify the element with the shorthand configuration [Ar] 3d6 4s2.
Count the electrons18 (from Ar) + 6 + 2 = 26Locate itHighest shell number is 4 → Period 4. Last sub-shell being filled is 3d → d-block, 6th along.Z = 26, so the element is iron
WORKED EXAMPLE
An element is in Period 4 and Group 15. Deduce its outer electron configuration.
Period 4 → outer shell is n = 4Group 15 → 5 valence electrons, split 2 in s and 3 in pouter configuration = 4s² 4p³Full: [Ar] 3d¹⁰ 4s² 4p³ — the element is arsenic. Don’t forget the filled 3d.
Sense check: your electron count must equal the atomic number for an atom, or the atomic number minus the charge for an ion. If it doesn’t, something is wrong before you go any further.
💡 Exam tip
Always total your superscripts and check against Z. It catches almost every slip.
For p-block elements, the superscript on the final term tells you how many places along that block the element is.
Learn the noble gases (He, Ne, Ar, Kr, Xe, Rn) and their electron counts — shorthand configurations depend on them.
Write configurations in numerical order, even where that isn’t the filling order.
⚠️ Common mix-up
Filling order is not writing order. 4s fills first; 3d is written first.
Positive ions lose 4s before 3d, not the other way round.
Don’t count d electrons as valence electrons when working out the group of an s- or p-block element.
The superscript is a number of electrons, not a charge. 2p6 does not mean 6+.
Period comes from the highest shell number, not from the number of terms you have written.
Up next: Trends Across the Periodic Table — where all this arranging finally pays off, and one idea explains five different trends.
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