IB Chemistry SLTopic 3 — Classifying the ElementsPaper 1 & 2Core skill~12 min read
Electron Configuration and Periodicity
Position gives you the electron configuration. The electron configuration gives you the position. Once you can travel in both directions you can be handed a string like [Ar]3d104s24p3 and name the element without a periodic table in front of you — which is exactly what some exam questions are testing.
📘 What you need to know
Electron configuration shows how electrons are spread across shells, subshells and orbitals.
Subshells fill in order of increasing energy — the Aufbau principle. Crucially, 4s fills before 3d.
Subshell capacities: s holds 2, p holds 6, d holds 10, f holds 14.
The period number is the highest principal quantum number in the configuration.
The group comes from the number of valence electrons; the block comes from the subshell being filled last.
Shorthand notation replaces the inner electrons with the previous noble gas in square brackets, e.g. [Ne]3s1 for sodium.
When a metal forms a positive ion, electrons are removed from the outermost shell first — for transition metals that means 4s before 3d.
The filling order
Electrons go into the lowest available energy level first. That sounds obvious, and it is, until you reach the fourth shell — because the 4s subshell is lower in energy than the 3d subshell, so it fills first even though its shell number is higher.
The gap between 4s and 3d is tiny. That near-equality is why transition metals can lose different numbers of electrons and end up with several stable oxidation states.
🧩 Writing a configuration from scratch
Find the atomic number — that is how many electrons a neutral atom has.
Fill in energy order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p.
Respect the capacities: 2, 6, 10, 14 for s, p, d, f.
Stop when the electrons run out, then check the superscripts add up to the atomic number.
Rewrite in numerical order if you like — 3d is usually written before 4s even though 4s filled first. Both are accepted.
Always add up the superscripts at the end. It takes three seconds and catches almost every slip you can make in this topic.
Shorthand notation
Writing out all 36 electrons of krypton every time is a waste of your exam. Instead, replace the inner electrons with the previous noble gas in square brackets and write only what comes after.
The same atom, two ways
Br: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 = [Ar] 3d10 4s2 4p5
The bracket stands for a complete, unreactive core of electrons that takes no part in chemistry. Everything outside the bracket is what actually matters.
Going backwards: configuration to position
There is a fourth clue hiding in plain sight: the superscripts must total the atomic number, so the configuration names the element outright without any of the other reasoning.
Configurations of ions
To make a positive ion you take electrons away, and they leave from the outermost shell first. For a main-group metal this is straightforward: magnesium is [Ne]3s2, so Mg2+ is simply [Ne].
Transition metals catch people out. Iron fills 4s before 3d, but once the 4s is occupied it becomes the outer shell, so electrons leave from 4s first when the ion forms. Fe is [Ar]3d64s2, and Fe2+ is [Ar]3d6 — not [Ar]3d44s2.
Last in, first out does not apply here. The 4s fills first but empties first as well. Say “electrons are removed from the outermost shell” and you will always be right.
Worked examples
WORKED EXAMPLE
Write the full and shorthand electron configurations of sulfur, and of the sulfide ion S2−.
Step 1: how many electrons
Sulfur has atomic number 16, so a neutral atom has 16 electrons.
Step 2: fill in energy order1s² 2s² 2p⁶ 3s² 3p⁴
Check: 2 + 2 + 6 + 2 + 4 = 16 ✓
Step 3: shorthand and the ion
Previous noble gas is neon, so [Ne] 3s² 3p⁴
S2− has gained two electrons, filling the 3p.
S2− is 1s² 2s² 2p⁶ 3s² 3p⁶, or [Ar]the sulfide ion is isoelectronic with argon — same electrons, different nucleus
WORKED EXAMPLE
Identify the element with the configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3, and state its period, group and block.
Step 1: add the superscripts2+2+6+2+6+2+10+3 = 33Step 2: period and block
Highest shell number is 4, so period 4. The configuration ends in 4p, so p-block.
Step 3: group
Valence electrons are 4s² and 4p³, giving 5. In the p-block that means group 15.
Arsenic — period 4, group 15, p-blockthe 3d electrons are not valence electrons here; they sit in an inner shell
WORKED EXAMPLE
An element has the shorthand configuration [Kr] 5s2 4d10 5p5. Deduce its identity and predict one chemical property.
Step 1: count electronsKr = 36, then 2 + 10 + 5 = 17 more, total 53Step 2: place it
Highest shell is 5, ends in p, valence electrons 2 + 5 = 7.
Period 5, group 17, p-blockStep 3: predict
Group 17 means one electron short of a full shell, so it will gain one electron to form a 1− ion.
Iodine — a halogen, forms I− and exists as I2 moleculesthis is the whole point of the periodic table: the configuration predicted the chemistry
💡 Exam tip
Write configurations with no gaps in the filling order. A missing 3s2 in the middle is an easy mark to drop.
Both 3d-before-4s and 4s-before-3d orderings are accepted for the written configuration. Be consistent within one answer.
For ions, state which electrons you removed and from where. That reasoning is often worth a mark by itself.
Adding the superscripts to check against the atomic number takes seconds and catches most errors.
If asked why elements in a group behave similarly, name the same number of valence electrons in the same type of subshell.
Use square brackets properly: [Ne]3s1, not Ne3s1.
⚠ Common mix-up
Filling 3d before 4s. The 4s subshell is lower in energy and fills first.
Removing 3d electrons first when forming a transition metal ion. The 4s electrons go first because they are outermost.
Counting d electrons as valence electrons for a p-block element. In arsenic the 3d10 is buried in an inner shell.
Forgetting to change the electron count for an ion. A 2+ ion has two fewer electrons than the atom.
Confusing subshell with shell. Shell 3 contains the 3s, 3p and 3d subshells.
Writing 2p8 or 3d12. Capacities are fixed at 2, 6, 10 and 14.
Up next: Trends Across the Periodic Table — now that you can find the electrons, we look at how strongly the nucleus holds on to them, and how that one question explains size, ionisation energy and electronegativity all at once.
Want this explained one-to-one?
Book a free session with an experienced IB Chemistry tutor and get your trickiest topics made simple.
