ChemistryCBSE Class 11 · Structure of Atom

Quantum Numbers, Orbitals and Electronic Configuration Class 11

Updated 17 October 2026 · Fact-checked

The four quantum numbers n (n = 1, 2, …), l (0 to n − 1), mₗ (−l to +l) and mₛ (±½) specify an electron. A subshell l holds 2(2l + 1) electrons, a shell n holds 2n². Orbitals fill by increasing (n + l) (Aufbau), no two electrons share all four quantum numbers (Pauli), and degenerate orbitals fill singly first (Hund). An orbital has n − l − 1 radial nodes and l angular nodes.

Fill orbitals from the bottom up by increasing n + l: s has 1 box, p 3, d 5, and 4s fills before 3d.

Illustration drawn for QuestionBuddy.

Understand Quantum Numbers, Orbitals and Electronic Configuration

Quantum numbers.

Principal $n = 1, 2, 3, \dots$ : shell, size and energy.
Azimuthal $l = 0, 1, \dots, n - 1$ (s, p, d, f): subshell, shape.
Magnetic $m_{l} = - l, \dots, 0, \dots, + l$ : orientation; $2 l + 1$ orbitals per subshell.
Spin $m_{s} = \pm \frac{1}{2}$ .

A shell $n$ has $n^{2}$ orbitals and holds up to $2 n^{2}$ electrons; a subshell holds $2 (2 l + 1)$ electrons (s 2, p 6, d 10, f 14).

Shapes and nodes. s orbitals are spherical, p dumbbell-shaped along x, y or z, d mostly four-lobed. Number of radial (spherical) nodes $= n - l - 1$ ; angular nodes $= l$ ; total nodes $= n - 1$ . Probability density $∣ ψ ∣^{2}$ gives the chance of finding the electron; an orbital is a region of high probability, not an orbit.

Filling rules.

*Aufbau:* fill in order of increasing $(n + l)$ ; for equal $(n + l)$ , lower $n$ first: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, …
*Pauli exclusion:* no two electrons in an atom have the same four quantum numbers; an orbital holds at most two electrons of opposite spin.
*Hund's rule:* in degenerate orbitals, electrons occupy separate orbitals with parallel spins before pairing.

Stable configurations. Exactly half-filled and completely filled subshells are extra stable (symmetry and exchange energy), so Cr is $[Ar] 3 d^{5} 4 s^{1}$ and Cu is $[Ar] 3 d^{10} 4 s^{1}$ . When forming cations of transition elements, electrons are removed from the 4s orbital first, then 3d: Fe²⁺ is $[Ar] 3 d^{6}$ , Mn²⁺ is $[Ar] 3 d^{5}$ .

Unpaired electrons are counted from the orbital box diagram; substances with unpaired electrons are paramagnetic, those with none are diamagnetic. Exchange stabilisation is greater when more electrons with parallel spin are present (for $d^{5}$ , 10 exchange pairs).

Quantum Numbers, Orbitals and Electronic Configuration: formulas and key facts

Orbitals in a subshell: $2 l + 1$; s 1, p 3, d 5, f 7.
Electrons in a shell: $2 n^{2}$; Shell capacity.
Radial nodes: $n - l - 1$; Spherical nodes.
Angular nodes: $l$; Planar or conical.
Total nodes: $n - 1$; Radial + angular.
Aufbau energy order: $(n + l) increasing$; Lower $n$ for ties.

How to solve Quantum Numbers, Orbitals and Electronic Configuration questions

List the allowed quantum numbers shell by shell, fill the orbitals by the three rules, and then check for half-filled and filled exceptions.

1Write the electron count Z (subtract electrons for cations, add for anions).
2Fill in the order 1s 2s 2p 3s 3p 4s 3d 4p …
3Check the exceptions: Cr and Cu (and their analogues).
4For cations, remove electrons from the outermost shell (4s before 3d).
5Draw the orbital boxes to count unpaired electrons with Hund's rule.
6Use n − l − 1 and l for radial and angular nodes.

Quantum Numbers, Orbitals 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.

A school-locker rule

Suppose a locker bank has shells like the atom: shell n has n² lockers and each locker holds at most two students.

How many students can the n = 4 shell hold?

Try it first, then show the solutionSolution

Orbitals = n² = 16.
Electrons = 2 × 16 = 32.

Answer: 32 electrons.

Memory hook: The capacity of a shell is 2n².

Hotel rooms and Hund's rule

Suppose three guests arrive for three identical rooms (like the three 2p orbitals) and each prefers a room alone.

How are three p electrons arranged?

Try it first, then show the solutionSolution

Hund's rule: each orbital gets one electron with parallel spin before any pairing.
So p³ has 3 unpaired electrons.

Answer: Three unpaired, parallel spins.

Memory hook: Fill singly first.

Tricks to remember Quantum Numbers, Orbitals and Electronic Configuration

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

n² and 2n²
Orbitals n², electrons 2n².
Use it when: Shell capacity.
Nodes: n−1
Radial n − l − 1 plus angular l.
Use it when: Node counts.
4s out first
In cations, remove 4s electrons before 3d.
Use it when: Transition-metal ions.
Half and full
d⁵ and d¹⁰ are extra stable.
Use it when: Cr and Cu.

Quantum Numbers, Orbitals and Electronic Configuration: solved examples

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

Example 1: Orbitals of the third shell · 2 marks · Very short answerBoard level

List the subshells of the n = 3 shell, the number of orbitals in each and the maximum number of electrons in the shell.

Try it first, then show the solutionSolution

l = 0, 1, 2: 3s, 3p, 3d with 1, 3 and 5 orbitals. (1)
Orbitals = 9; maximum electrons = 2 × 9 = 18. (1)

Answer: 3s, 3p, 3d; 9 orbitals; 18 electrons.

Example 2: Copper and its ion · 3 marks · Short answerBoard level

Write the electronic configuration of Cu (Z = 29) and Cu²⁺, give the number of unpaired electrons in each and say whether they are paramagnetic.

Try it first, then show the solutionSolution

Cu: [Ar] 3d¹⁰ 4s¹ (full d subshell is more stable). One unpaired electron (4s). (1)
Cu²⁺: remove the 4s electron and one 3d electron: [Ar] 3d⁹. (1)
3d⁹ has one unpaired electron, so both Cu and Cu²⁺ are paramagnetic. (1)

Answer: Cu [Ar]3d¹⁰4s¹; Cu²⁺ [Ar]3d⁹; one unpaired electron each; paramagnetic.

Example 3: Nodes in orbitals · 3 marks · Short answerHarder board level

Find the number of radial nodes, angular nodes and total nodes in (a) 4d and (b) 3p orbitals.

Try it first, then show the solutionSolution

(a) 4d: n = 4, l = 2; radial n − l − 1 = 1; angular l = 2; total n − 1 = 3. (1)
(b) 3p: n = 3, l = 1; radial 3 − 1 − 1 = 1; angular 1; total 2. (1)
Each total equals n − 1. (1)

Answer: (a) 1, 2, 3; (b) 1, 1, 2.

Example 4: Electrons of iron · 4 marks · Long answerHarder board level

Iron has Z = 26. Write its configuration and find (a) the number of electrons with n = 3, (b) the number with l = 2, (c) the number with l = 1, and (d) the number of unpaired electrons. Write the configuration of Fe²⁺ and Fe³⁺.

Try it first, then show the solutionSolution

Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s². (1)
(a) 3s² 3p⁶ 3d⁶ gives 14; (b) 3d⁶ gives 6; (c) 2p⁶ and 3p⁶ give 12. (1)
(d) 3d⁶ has one paired and four unpaired electrons: 4. (1)
Fe²⁺: [Ar]3d⁶; Fe³⁺: [Ar]3d⁵ (4s electrons are lost first). (1)

Answer: (a) 14; (b) 6; (c) 12; (d) 4; Fe²⁺ [Ar]3d⁶, Fe³⁺ [Ar]3d⁵.

Example 5: Manganese and its ions · 4 marks · Long answerHOTS / competency

Manganese has Z = 25. (a) Write the configurations of Mn, Mn²⁺ and Mn⁷⁺ and the number of unpaired electrons in each. (b) Explain, using (n + l), why 4s fills before 3d yet is lost first. (c) How many exchange pairs exist among the five parallel-spin electrons of Mn²⁺?

Try it first, then show the solutionSolution

(a) Mn: [Ar]3d⁵4s² (5 unpaired); Mn²⁺: [Ar]3d⁵ (5 unpaired); Mn⁷⁺: [Ar] (0 unpaired). (1)
(b) 4s has (n + l) = 4 and 3d has 5, so 4s fills first. (1)
Once 3d is occupied, its electrons are held closer to the nucleus and 4s becomes the higher-energy, outer orbital, so 4s electrons are lost first. (1)
(c) Five parallel electrons give C(5, 2) = 10 exchange pairs, which is why d⁵ is extra stable. (1)

Answer: (a) 5, 5, 0 unpaired; (b) lower (n + l) fills first, but 4s is outermost and lost first; (c) 10.

Where students lose marks in Quantum Numbers, Orbitals and Electronic Configuration

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

Mistake: Removing 3d electrons before 4s when forming transition-metal cations.
Following the filling order.
Do this: Remove 4s electrons first.
Usually costs: The ion mark.
Mistake: Writing Cr as 3d⁴4s².
Missing the exception.
Do this: Cr is 3d⁵4s¹; Cu is 3d¹⁰4s¹.
Usually costs: 1 mark.
Mistake: Pairing electrons in p orbitals before every orbital has one.
Ignoring Hund's rule.
Do this: Fill singly with parallel spins first.
Usually costs: The unpaired-electron count.
Mistake: Allowing l = n.
Quantum number error.
Do this: l runs from 0 to n − 1.
Usually costs: The whole question.

Exam tips

State the Aufbau, Pauli and Hund rules in the working lines.
Draw the orbital box diagram when asked for unpaired electrons.
Show (n − l − 1) and l for nodes explicitly.
Give configurations in the abbreviated noble-gas core form.
Remember the exceptions Cr and Cu and their ions.

Practice questions on Quantum Numbers, Orbitals 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
The maximum number of electrons in the f subshell is
(a) 6 (b) 10 (c) 14 (d) 18
Hint
2(2l + 1) with l = 3.
Answer
(c) 14.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The electronic configuration of chromium is [Ar]3d⁵4s¹ and not [Ar]3d⁴4s².
Reason (R): A half-filled d subshell has extra stability because of symmetry and greater exchange energy.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Half-filled d.
Answer
(a).
Q3 · 3 marks · Short answerHarder board level
Find the number of unpaired electrons in the ions Ti³⁺, V³⁺, Cr³⁺, Mn²⁺ and Fe²⁺ (Z = 22, 23, 24, 25, 26).
Hint
Remove 4s then 3d electrons.
Answer
Ti³⁺ 3d¹: 1; V³⁺ 3d²: 2; Cr³⁺ 3d³: 3; Mn²⁺ 3d⁵: 5; Fe²⁺ 3d⁶: 4.
Q4 · 3 marks · Short answerHarder board level
An orbital has n = 5 and l = 3. Name it and give the number of orbitals in the subshell, the maximum number of electrons, the radial nodes and the angular nodes.
Hint
Use 2l + 1 and n − l − 1.
Answer
5f; 7 orbitals; 14 electrons; radial nodes 5 − 3 − 1 = 1; angular nodes 3.
Q5 · 4 marks · Long answerHOTS / competency
A neutral element has the cation X²⁺ with the configuration [Ar]3d³. (a) Find its atomic number and write the ground-state configuration of the atom. (b) Give the four quantum numbers of the two electrons lost in forming X²⁺. (c) Write the number of unpaired electrons in X²⁺.
Hint
The 2+ ion lost its 4s² electrons.
Answer
(a) Electrons in X²⁺ = 18 + 3 = 21, so Z = 23 (vanadium): [Ar]3d³4s². (b) (4, 0, 0, +½) and (4, 0, 0, −½). (c) 3d³ has three unpaired electrons.

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.Allowed values of l?Show answer
0 to n − 1.
2.Allowed values of mₗ?Show answer
−l to +l.
3.Electrons in a shell?Show answer
2n².
4.Radial nodes?Show answer
n − l − 1.
5.Pauli exclusion principle?Show answer
No two electrons have all four quantum numbers equal.
6.Hund's rule?Show answer
Degenerate orbitals fill singly with parallel spins first.

Quantum Numbers, Orbitals and Electronic Configuration: frequently asked questions

Why does 4s fill before 3d?

Its (n + l) value of 4 is lower than 3d's value of 5, so it has lower energy in a neutral multi-electron atom.

What is an orbital?

A region of space around the nucleus, described by a wavefunction ψ, where the probability of finding the electron is high.

What is the difference between an orbit and an orbital?

An orbit is a definite circular path (Bohr); an orbital is a probability distribution (quantum mechanics).

Why are half-filled and full subshells stable?

They have symmetric distribution of electrons and large exchange energy.

How many orbitals in the third shell?

Nine: one 3s, three 3p and five 3d.