Orbitals and Electron Configuration
Learn and apply Orbitals and Electron Configuration in the published Chemistry course sequence.
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The core idea
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Orbitals and Electron Configuration: Orientation
Electron configuration questions are workflow questions: count electrons, fill orbitals in the correct energy order, then check the total. This lesson focuses on the exam-safe order and the common traps (ions, especially 4s vs 3d).
Definitions (Must Know)
A. Orbital
An orbital is a region of space around the nucleus in which there is a high probability of finding an electron. It can hold up to two electrons with opposite spins.
B. Subshells (s, p, d)
A subshell is a set of orbitals of the same type within a shell.
- s subshell: 1 orbital → holds 2 electrons
- p subshell: 3 orbitals → holds 6 electrons
- d subshell: 5 orbitals → holds 10 electrons
C. Shell (principal quantum number, n)
A shell is an energy level labelled by n = 1, 2, 3, ….
Detailed Explanations
A. Writing a configuration (workflow)
- Count electrons (atomic number; adjust for charge if it is an ion).
- Fill orbitals using the energy order.
- Check the total electrons match what you started with.
If it is an ion, do the electron count first (e.g. Al³⁺ has 13-3 = 10 electrons).
Quick examples:
- O: 1s² 2s² 2p⁴
- Cl: 1s² 2s² 2p⁶ 3s² 3p⁵
B. Ions (what changes)
Use the charge to adjust the electron count:
-
Xⁿ⁺: the atom has lost n electrons → electrons = Z - n.
-
Xⁿ⁻: the atom has gained n electrons → electrons = Z + n.
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Anions: add electrons (more negative).
-
Main-group cations: remove electrons from the outer shell (highest n).
-
Transition-metal cations: remove 4s electrons before 3d (the most common exam trap).
Mini example: S²⁻ has 16 + 2 = 18 electrons, so its configuration ends at 3p⁶.
C. 4s vs 3d removal (the exam rule)
In neutral atoms, 4s fills before 3d. But when transition metals form cations, the 4s electrons are removed first because (once 3d is occupied) the 4s electrons are higher in energy and more exposed.
So if you see a transition-metal ion, remove 4s electrons before 3d electrons.
D. Chromium and copper exceptions
The simple filling order predicts [Ar]3d⁴ 4s² for chromium and [Ar]3d⁹ 4s² for copper, but their observed ground-state configurations are:
- Cr: [Ar]3d⁵ 4s¹
- Cu: [Ar]3d¹⁰ 4s¹
Apply the ion rule after writing the correct neutral-atom configuration:
- Cr³⁺: remove the 4s electron, then two 3d electrons → [Ar]3d³
- Cu²⁺: remove the 4s electron, then one 3d electron → [Ar]3d⁹
Worked Examples
Modelled example 1
Write the Configuration of an Aluminium Ion
Problem
Write the electron configuration of Al³⁺.
Study the worked solution
Write the neutral-atom configuration
Method
Place all 13 electrons into subshells in energy order.Reason
The ion configuration is derived from the neutral atom.Working
Al: 1s² 2s² 2p⁶ 3s² 3p¹Remove the outer electrons
Method
Remove one electron from 3p, then two from 3s.Reason
A 3 + ion has three fewer electrons, removed from the highest occupied shell.Working
3p¹ → 3p⁰ and 3s² → 3s⁰.State the ion configuration
Working
Al³⁺: 1s² 2s² 2p⁶
Guided practice 2
Write the Configuration of a Sulfide Ion
Problem
Write the electron configuration of S²⁻.
Try this before viewing the solution
Hints
Hint 1: start from sulfur
Hint 2: interpret the charge
View solution step by step
Write neutral sulfur
Method
Fill 16 electrons into the available subshells.Reason
This establishes where the added electrons go.Working
S: 1s² 2s² 2p⁶ 3s² 3p⁴Add two electrons
Method
Complete the 3p subshell.Reason
The 2- ion contains 18 electrons.Working
3p⁴ → 3p⁶State the ion configuration
Working
S²⁻: 1s² 2s² 2p⁶ 3s² 3p⁶
Common misconception 3
Correct Reversed Ion Electron Counts
Learner attempt
Asked for the electron configurations of Na⁺ and Cl⁻, a learner assigns 12 electrons to Na⁺ and 16 electrons to Cl⁻. Identify the first error and write both correct configurations.
Correct the charge rule
View solution step by step
Correct the sodium ion
Method
Subtract one electron from neutral sodium.Reason
A 1 + ion has one fewer electron than proton.Working
Na⁺ has 11-1 = 10 electrons: 1s² 2s² 2p⁶.Correct the chloride ion
Method
Add one electron to neutral chlorine.Reason
A 1- ion has one more electron than proton.Working
Cl⁻ has 17 + 1 = 18 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶.
Examiner practice 4
Compare Nitrogen and Oxygen
Problem
For the ground-state atoms N and O: (a) write their electron configurations; (b) state the number of unpaired electrons in each atom. [4 marks]
Try this before viewing the solution
View solution step by step
Write the nitrogen configuration
1 markMethod
Fill the subshells with nitrogen’s seven electrons.Reason
Nitrogen has seven electrons.Working
N: 1s² 2s² 2p³Write the oxygen configuration
1 markReason
Oxygen has eight electrons.Working
O: 1s² 2s² 2p⁴Count nitrogen's unpaired electrons
1 markMethod
Place the three 2p electrons singly into the three orbitals.Reason
Electrons occupy separate degenerate orbitals before pairing.Working
N has three unpaired electrons.Count oxygen's unpaired electrons
1 markMethod
Pair the fourth 2p electron in one orbital.Reason
The other two 2p orbitals remain singly occupied.Working
O has two unpaired electrons.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Credit each configuration and each unpaired-electron count separately.
Challenge 5
Infer Periodic Position from a Configuration
Problem
An atom has electron configuration 1s² 2s² 2p⁶ 3s² 3p⁴. Identify the element, its period and its group.
Try this before viewing the solution
Hints
Hint 1: identify the atom
Hint 2: read periodic position
View solution step by step
Identify the element
Method
Add the electron counts.Reason
A neutral atom has the same number of electrons and protons.Working
2 + 2 + 6 + 2 + 4 = 16, so the element is sulfur, S.Identify the period
Method
Find the highest occupied shell.Reason
The largest principal quantum number is 3.Working
Period 3.Identify the group
Method
Count the outer-shell s and p electrons.Reason
3s² 3p⁴ gives six valence electrons for a main-group element.Working
Group 16.
Mind Stretchers
Mind stretcher 1Extension
Write the electron configuration of Fe²⁺ and Fe³⁺ using noble gas notation.
Show Answer
Mark scheme:
- Fe is [Ar]3d⁶ 4s².
- Remove electrons from 4s before 3d:
- Fe²⁺: [Ar]3d⁶
- Fe³⁺: [Ar]3d⁵
Mind stretcher 2Extension
Write the electron configurations of Cr, Cr³⁺, Cu and Cu²⁺ using noble-gas notation.
Show Answer
Mark scheme:
- Cr: [Ar]3d⁵ 4s¹; Cr³⁺: [Ar]3d³.
- Cu: [Ar]3d¹⁰ 4s¹; Cu²⁺: [Ar]3d⁹.
- For each cation, remove 4s before 3d.