ChemistryISC Class 11 · Structure of Atom

Quantum Numbers, Orbitals and Electronic Configuration

Q: Why does 4s fill before 3d?

By the (n + l) rule 4s (4) has a lower value than 3d (5), so it has lower energy in a many-electron atom and fills first.

Q: Why do we say the d⁵ and d¹⁰ configurations are stable?

Symmetry and exchange energy: electrons with parallel spins in degenerate orbitals can exchange positions, which lowers the energy.

Q: Is an orbital the same as an orbit?

No. An orbit is a fixed path in the Bohr model. An orbital is a region of space where the probability of finding the electron is high.

Updated 17 October 2026 · Fact-checked

Four quantum numbers define an electron: n (shell), l = 0 to n − 1 (subshell), m = −l to +l (orbital) and s = ±½ (spin). Electrons fill orbitals by the Aufbau principle (n + l rule), the Pauli exclusion principle and Hund's rule, giving configurations such as Fe: [Ar] 3d⁶ 4s².

Notice 4s fills first and holds a pair with opposite spins, while 3d electrons sit singly with parallel spins before the sixth pairs up.

Illustration drawn for QuestionBuddy.

Understand Quantum Numbers and Electronic Configuration

Quantum numbers. The principal quantum number n = 1, 2, 3... fixes the size and energy of the shell (K, L, M...). The azimuthal quantum number l = 0, 1, ..., n − 1 fixes the shape of the subshell (s, p, d, f for l = 0, 1, 2, 3). The magnetic quantum number m = −l, ..., 0, ..., +l fixes the orientation; a subshell has (2l + 1) orbitals. The spin quantum number s = +½ or −½.

Counting. The shell n has n² orbitals and holds 2n² electrons. Subshells s, p, d, f have 1, 3, 5, 7 orbitals and hold 2, 6, 10, 14 electrons.

Shapes and nodes. s orbitals are spherical, p orbitals are dumb-bell shaped along x, y, z, and d orbitals are mostly double dumb-bells (four lobes). The number of radial nodes = n − l − 1, the number of angular nodes = l, and the total number of nodes = n − 1.

Filling rules. Aufbau: orbitals fill in increasing order of (n + l); for equal (n + l), the lower n fills first: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s. Pauli: no two electrons in an atom have the same set of four quantum numbers, so an orbital holds at most two electrons of opposite spin. Hund: in degenerate orbitals electrons occupy singly with parallel spins before pairing.

Stability and exceptions. Half-filled and fully filled subshells are specially stable (extra exchange energy), so Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹. In forming cations the electrons leave the outermost shell first (4s before 3d).

Magnetic behaviour. Spin-only magnetic moment μ = √(n(n + 2)) BM, where n is the number of unpaired electrons.

Quantum Numbers and Electronic Configuration: formulas and key facts

Allowed values: $l = 0 t o n - 1∣ m = - l t o + l ∣ s = \pm ½$; Orbitals in a subshell = 2l + 1.
Capacity: $O r bi t a l s in s h e l l = n^{2} ∣ E l ec t r o n s in s h e l l = 2 n^{2}$; Subshell: 2(2l + 1) electrons.
Nodes: $R a d ia l = n - l - 1∣ A n g u l a r = l ∣ T o t a l = n - 1$; Radial nodes are spherical shells.
Magnetic moment: $μ = \sqrt (n (n + 2)) B M$; n = number of unpaired electrons.

Quantum Numbers and Electronic Configuration in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Why iron is magnetic

Suppose a student writes the configuration of iron (Z = 26) and its ion Fe³⁺ to explain its magnetic behaviour.

Find the number of unpaired electrons and the spin-only moment of Fe and Fe³⁺.

Try it first, then show the solutionSolution

Fe: [Ar] 3d⁶ 4s². The 3d⁶ part has one paired set and four unpaired electrons, so n = 4.
Fe³⁺: lose 4s² and one 3d electron: [Ar] 3d⁵, with five unpaired electrons.
μ(Fe) = √(4 × 6) = 4.90 BM; μ(Fe³⁺) = √(5 × 7) = 5.92 BM.

Answer: Fe: 4 unpaired, 4.90 BM; Fe³⁺: 5 unpaired, 5.92 BM.

Memory hook: Cations lose the outer 4s electrons before 3d.

Tricks to remember Quantum Numbers and Electronic Configuration

Short memory aids. Say each one aloud once, then use it on a question the same day.

Nodes: n − 1 in total
Radial n − l − 1, angular l; they always sum to n − 1.
Use it when: Node counts for 2s, 3p, 4d.
Half-filled is happy
d⁵ and d¹⁰ are extra stable, so Cr and Cu "borrow" a 4s electron.
Use it when: Configurations of Cr and Cu.

Quantum Numbers and Electronic Configuration: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Orbitals in a shell · 1 mark · MCQBoard level

The number of orbitals in the shell with n = 3 is:
(a) 3
(b) 6
(c) 9
(d) 18

Try it first, then show the solutionSolution

n² = 9 orbitals (one 3s, three 3p, five 3d).

Answer: (c) 9

Example 2: Allowed quantum numbers · 3 marks · Short answerBoard level

List the possible values of l and m for n = 3 and find the number of orbitals and the maximum number of electrons in this shell.

Try it first, then show the solutionSolution

l = 0, 1, 2 (s, p, d). (1 mark)
m: 0 for l = 0; −1, 0, +1 for l = 1; −2 to +2 for l = 2, giving 1 + 3 + 5 = 9 orbitals. (1 mark)
Maximum electrons = 2 × 9 = 18. (1 mark)

Answer: l = 0, 1, 2; 9 orbitals; 18 electrons.

Example 3: Validity of quantum number sets · 4 marks · Short answerHarder board level

State which sets (n, l, m, s) are not permitted, with reasons: (a) 2, 2, 0, +½; (b) 3, 1, −1, −½; (c) 1, 0, 1, +½; (d) 4, 3, +3, +½.

Try it first, then show the solutionSolution

(a) Not permitted: l must be less than n (l ≤ 1 for n = 2). (1 mark)
(b) Permitted: l = 1 < 3, and |m| = 1 ≤ l. (1 mark)
(c) Not permitted: m can only be 0 when l = 0. (1 mark)
(d) Permitted: l = 3 < 4 and m = +3 is within −3 to +3. (1 mark)

Answer: (a) and (c) are not permitted; (b) and (d) are permitted.

Example 4: Configurations and magnetism · 4 marks · Short answerHarder board level

Write the electronic configurations of Cr (Z = 24), Cu (Z = 29), Fe (Z = 26) and Fe³⁺, state the number of unpaired electrons in Fe and Fe³⁺ and their spin-only magnetic moments.

Try it first, then show the solutionSolution

Cr: [Ar] 3d⁵ 4s¹; Cu: [Ar] 3d¹⁰ 4s¹ (half-filled and full-filled stability). (1 mark)
Fe: [Ar] 3d⁶ 4s², 4 unpaired electrons; Fe³⁺: [Ar] 3d⁵, 5 unpaired electrons. (1 mark)
μ(Fe) = √(4 × 6) = √24 = 4.90 BM. (1 mark)
μ(Fe³⁺) = √(5 × 7) = √35 = 5.92 BM. (1 mark)

Answer: Fe: 4 unpaired, 4.90 BM; Fe³⁺: 5 unpaired, 5.92 BM.

Example 5: Chlorine and nodes · 5 marks · Long answerHOTS / competency

Write the configuration of chlorine (Z = 17) and Cl⁻. Find the number of electrons with l = 1 in Cl, the number of occupied orbitals and the number of unpaired electrons. Find the radial, angular and total nodes in 3p, 4d and 2s orbitals.

Try it first, then show the solutionSolution

Cl: 1s² 2s² 2p⁶ 3s² 3p⁵; Cl⁻: 1s² 2s² 2p⁶ 3s² 3p⁶ (same as argon). (1 mark)
Electrons with l = 1: 2p⁶ + 3p⁵ = 11. (1 mark)
Orbitals occupied: 1s, 2s, 2p (3), 3s, 3p (3) = 9; the 3p⁵ part leaves 1 unpaired electron. (1 mark)
3p: radial 3 − 1 − 1 = 1, angular 1, total 2. 4d: radial 1, angular 2, total 3. (1 mark)
2s: radial 2 − 0 − 1 = 1, angular 0, total 1. (1 mark)

Answer: Cl: 1s² 2s² 2p⁶ 3s² 3p⁵; 11 electrons with l = 1; 9 occupied orbitals; 1 unpaired; nodes: 3p (1, 1, 2), 4d (1, 2, 3), 2s (1, 0, 1).

Where students lose marks in Quantum Numbers and Electronic Configuration

The slips examiners see most often, and what to write instead.

Mistake: Writing 3d before 4s for ions
Removing 3d electrons first from Fe.
Do this: The outermost (4s) electrons leave first.
Usually costs: 1 to 2 marks.
Mistake: Allowed values of m
Using m greater than l.
Do this: m runs from −l to +l.
Usually costs: 1 mark.
Mistake: Missing Cr and Cu exceptions
Writing 3d⁴ 4s² for Cr.
Do this: Half-filled and fully filled d subshells are stable: Cr 3d⁵ 4s¹, Cu 3d¹⁰ 4s¹.
Usually costs: 2 marks.

Exam tips

Use a box diagram when asked for unpaired electrons.
State the rule (Aufbau, Pauli, Hund) you apply.
Check that the sum of electrons equals Z (or Z minus the charge).
Give the magnetic moment with units BM.

Practice questions on Quantum Numbers and Electronic Configuration

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
According to the Pauli exclusion principle:
(a) electrons fill orbitals singly first
(b) no two electrons in an atom can have the same four quantum numbers
(c) orbitals fill in order of increasing energy
(d) the position and momentum cannot both be known
Hint
It limits an orbital to two electrons of opposite spin.
Answer
(b) no two electrons in an atom can have the same four quantum numbers
Q2 · 2 marks · Short answerBoard level
Write the electronic configuration of sodium (Z = 11) and of Na⁺.
Hint
Remove the outermost electron for the cation.
Answer
Na: 1s² 2s² 2p⁶ 3s¹; Na⁺: 1s² 2s² 2p⁶ (isoelectronic with neon).
Q3 · 3 marks · Short answerHarder board level
How many electrons in the shell n = 4 can have l = 2? How many electrons in the n = 3 shell can have m = +1?
Hint
Count orbitals first.
Answer
l = 2 means the 4d subshell: 5 orbitals, 10 electrons. For n = 3, m = +1 occurs for l = 1 and l = 2: 2 orbitals, 4 electrons.
Q4 · 3 marks · Short answerHarder board level
Arrange 3d, 4s, 4p, 5s, 4d and 5p in order of increasing energy using the (n + l) rule.
Hint
Equal (n + l): lower n first.
Answer
4s (4) < 3d (5) < 4p (5) < 5s (5) < 4d (6) < 5p (6). Within each group of equal n + l, the lower n comes first.
Q5 · 3 marks · Short answerHarder board level
Find the spin-only magnetic moments of Cr³⁺ (Z = 24) and Mn²⁺ (Z = 25).
Hint
Write the configuration of each ion first.
Answer
Cr³⁺: [Ar] 3d³, 3 unpaired: √15 = 3.87 BM. Mn²⁺: [Ar] 3d⁵, 5 unpaired: √35 = 5.92 BM.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student studies bromine (Z = 35).
(i) Write its configuration. (1 mark)
(ii) State the number of valence electrons and the number of unpaired electrons. (1 mark)
(iii) State its period and group. (1 mark)
(iv) Give possible quantum numbers of the last electron added (taking m in the order −1, 0, +1 for filling). (1 mark)
Hint
Fill 4s before 3d and 4p.
Answer
(i) [Ar] 3d¹⁰ 4s² 4p⁵. (ii) 7 valence electrons (4s² 4p⁵); 1 unpaired electron. (iii) Period 4, group 17. (iv) Filling m in the order −1, 0, +1, the first three 4p electrons enter m = −1, 0, +1 with s = +½. The fourth pairs in m = −1 and the fifth (last) pairs in m = 0. So the last electron has n = 4, l = 1, m = 0, s = −½ (the unpaired electron is in the m = +1 orbital).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Possible values of l for n = 3?Show answer
0, 1, 2.
2.Number of orbitals in a subshell?Show answer
2l + 1.
3.Number of radial nodes?Show answer
n − l − 1.
4.Configuration of Cu?Show answer
[Ar] 3d¹⁰ 4s¹.

Quantum Numbers and Electronic Configuration: frequently asked questions

Why does 4s fill before 3d?

By the (n + l) rule 4s (4) has a lower value than 3d (5), so it has lower energy in a many-electron atom and fills first.

Why do we say the d⁵ and d¹⁰ configurations are stable?

Symmetry and exchange energy: electrons with parallel spins in degenerate orbitals can exchange positions, which lowers the energy.

Is an orbital the same as an orbit?

No. An orbit is a fixed path in the Bohr model. An orbital is a region of space where the probability of finding the electron is high.