Quantum Numbers

ChemistryAtomic StructureFor JEE aspirants

Modern quantum mechanics replaces Bohr's fixed orbits with orbitals - three-dimensional regions where an electron is most likely to be found. Every electron in an atom is uniquely described by four quantum numbers - $n, ℓ, m, s$ - which specify its shell, sub-shell (orbital shape), orientation, and spin. Filling of orbitals follows three rules (Aufbau, Pauli, Hund) which together determine an atom's electronic configuration - the key to periodic trends and chemical bonding.

Key Formulas - Quick Reference

Principal quantum number $n = 1, 2, 3, ...$ (shell/energy level)
Azimuthal quantum number $ℓ = 0, 1, 2, ..., (n - 1)$ (subshell shape)
Magnetic quantum number $m_{ℓ} = - ℓ, ..., 0, ..., + ℓ$ ( $2 ℓ + 1$ values)
Spin quantum number $s = + \frac{1}{2}$ or $- \frac{1}{2}$
Maximum electrons in nth shell $= 2 n^{2}$
Maximum electrons in subshell with quantum number $ℓ$ $= 2 (2 ℓ + 1)$
Number of orbitals in nth shell $= n^{2}$
Orbital labels: $ℓ = 0, 1, 2, 3 \to s, p, d, f$

1. The Four Quantum Numbers

Principal quantum number ( $n$ )

n

takes integer values

1, 2, 3, ...

It gives the main energy level (shell) - labelled K, L, M, N for

n = 1, 2, 3, 4

Determines the size and energy of the orbital (larger $n$ = larger, higher energy)
Maximum electrons in the nth shell $= 2 n^{2}$ : K $= 2$ , L $= 8$ , M $= 18$ , N $= 32$
Number of orbitals in nth shell $= n^{2}$

Azimuthal (angular momentum) quantum number ( $ℓ$ )

For a given

n

ℓ = 0, 1, 2, ..., (n - 1)

- so

n

possible values.

ℓ

determines the shape of the orbital (subshell).

$ℓ$	Subshell	Shape	Orbitals per subshell
0	s	Spherical	1
1	p	Dumbbell	3
2	d	Clover / donut	5
3	f	Complex (8 lobes)	7

Magnetic quantum number ( $m_{ℓ}$ )

For a given

ℓ

m_{ℓ} = - ℓ, - ℓ + 1, ..., 0, ..., ℓ - 1, + ℓ

(total

2 ℓ + 1

values).

m_{ℓ}

specifies the orientation of the orbital in space.

Examples: $s$ ( $ℓ = 0$ ) has 1 orientation. $p$ ( $ℓ = 1$ ) has 3 ( $p_{x}, p_{y}, p_{z}$ ). $d$ ( $ℓ = 2$ ) has 5.

Spin quantum number ( $s$ or $m_{s}$ )

s = + \frac{1}{2}

(spin up,

↑

) or

- \frac{1}{2}

(spin down,

↓

).
Represents the intrinsic angular momentum of the electron.

Solved Example 1

If the principal quantum number

n = 3

, what are the permitted values of

ℓ

and

m

Solution:

For $n = 3$ : $ℓ = 0, 1, 2$ (three subshells: 3s, 3p, 3d).

$m_{ℓ}$ values:

$ℓ = 0$ : $m = 0$ (1 value - the 3s orbital)
$ℓ = 1$ : $m = - 1, 0, + 1$ (3 values - three 3p orbitals)
$ℓ = 2$ : $m = - 2, - 1, 0, + 1, + 2$ (5 values - five 3d orbitals)

Total: $1 + 3 + 5 = 9 = 3^{2}$ orbitals, holding max $2 n^{2} = 18$ electrons. $✓$

Solved Example 2

In which orbital will the electron reside if it has the quantum numbers

n = 4, ℓ = 2, m = - 1, s = + 1/2

Solution:

$n = 4$ (4th shell), $ℓ = 2$ (d-subshell), $m = - 1$ (one specific d-orbital). Orbital: 4d.

Solved Example 3

Write all quantum numbers for one electron in each of: (a) 2p, (b) 4d.

Solution:

(a) 2p: $n = 2$ , $ℓ = 1$ , $m = - 1, 0, or + 1$ , $s = \pm \frac{1}{2}$
(b) 4d: $n = 4$ , $ℓ = 2$ , $m = - 2, - 1, 0, + 1, or + 2$ , $s = \pm \frac{1}{2}$

2. Shapes of Orbitals

s-orbital (spherical)

$s$ -orbitals are spherically symmetric. The 1s is smallest (densest at nucleus, decreasing outward). Higher s-orbitals (2s, 3s) are larger with additional radial nodes.

Figure 1: Shape of the s-orbital - spherically symmetric electron cloud, densest near the nucleus.

p-orbitals (dumbbell)

Three p-orbitals ( $p_{x}$ , $p_{y}$ , $p_{z}$ ) each have two lobes along an axis, separated by a nodal plane at the nucleus. Opposite lobes have opposite signs of the wave function $ψ$ .

Figure 2: Three p-orbitals (

p_{x}

p_{y}

p_{z}

) - dumbbell-shaped with a nodal plane at the nucleus. Opposite lobes have opposite signs of

ψ

d-orbitals (clover and donut)

Five d-orbitals: $d_{x y}$ , $d_{y z}$ , $d_{x z}$ , $d_{x^{2} - y^{2}}$ (four-lobe clovers), and $d_{z^{2}}$ (dumbbell + toroidal ring in the xy-plane).

Figure 3: Five d-orbitals - four clover-shaped (

d_{x y}

d_{y z}

d_{x z}

d_{x^{2} - y^{2}}

) and one dumbbell-plus-donut (

d_{z^{2}}

Solved Example 4

An electron has spin quantum number

+ 1/2

and magnetic quantum number

- 1

. It cannot occupy an s-orbital. Why?

Solution:

For an s-orbital, $ℓ = 0$ so $m_{ℓ}$ can only be $0$ . Since our electron has $m_{ℓ} = - 1$ , it must occupy an orbital with $ℓ \geq 1$ - i.e., p, d, or f - not s.

Solved Example 5

Which d-orbital has electron density in all the planes (xy, yz, xz)?

Solution:

The $d_{z^{2}}$ orbital has electron density above and below the xy-plane (lobes along z) and in the xy-plane (the donut). So it has density in all three planes. All other d-orbitals have nodes coinciding with at least one of these planes.

3. Rules for Filling Electrons

(a) Aufbau principle

Electrons fill orbitals in order of increasing energy. In multi-electron atoms, energy is determined by the $(n + ℓ)$ rule: orbital with lower

(n + ℓ)

is filled first. If

(n + ℓ)

is the same, the one with lower

n

is filled first.

Figure 4: Aufbau principle - diagonal rule for filling orbitals in order of increasing

(n + ℓ)

. Follow arrows top-to-bottom.

Orbital	$n$	$ℓ$	$n + ℓ$	Filling order
1s	1	0	1	1
2s	2	0	2	2
2p	2	1	3	3
3s	3	0	3	4
3p	3	1	4	5
4s	4	0	4	6 (lower n at same $n + ℓ$ )
3d	3	2	5	7
4p	4	1	5	8

Order: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p

(b) Pauli's exclusion principle

No two electrons in an atom can have all four quantum numbers identical. So each orbital (

n, ℓ, m

) holds a maximum of 2 electrons, and they must have opposite spins.

(c) Hund's rule of maximum multiplicity

Electrons fill degenerate orbitals (same energy) singly first, with parallel spins, before any pairing begins. This maximises the number of unpaired electrons and lowers energy due to exchange energy.

Figure 5: Hund's rule of maximum multiplicity - electrons occupy degenerate orbitals singly (with parallel spins) before pairing.

Exchange energy and half/fully-filled stability

When electrons in degenerate orbitals have parallel spins, they can "exchange" positions, releasing energy called exchange energy. Number of exchanges $= (2 N) = N (N - 1) /2$ where $N$ is the number of parallel-spin electrons.

Half-filled and fully-filled subshells are extra stable because they maximise exchange energy. This explains anomalous configurations of Cr and Cu (see below).

Aufbau exceptions (Cr and Cu)

Element	Expected	Actual	Reason
Cr (Z=24)	[Ar] 3d $^{4}$ 4s $^{2}$	[Ar] 3d $^{5}$ 4s $^{1}$	Half-filled 3d + half-filled 4s more stable
Cu (Z=29)	[Ar] 3d $^{9}$ 4s $^{2}$	[Ar] 3d $^{10}$ 4s $^{1}$	Fully-filled 3d + half-filled 4s more stable

Solved Example 6

Write the electronic configuration of nitrogen (Z = 7).

Solution:

Fill in Aufbau order: 1s $^{2}$ 2s $^{2}$ 2p $^{3}$ . By Hund, the three 2p electrons occupy $p_{x}^{1} p_{y}^{1} p_{z}^{1}$ singly with parallel spins.

Solved Example 7

Write electronic configurations of (a) Ne (Z=10), (b) Ar (Z=18), (c) Fe (Z=26), (d) Cu (Z=29).

Solution:

(a) Ne: 1s $^{2}$ 2s $^{2}$ 2p $^{6}$
(b) Ar: 1s $^{2}$ 2s $^{2}$ 2p $^{6}$ 3s $^{2}$ 3p $^{6}$
(c) Fe: [Ar] 3d $^{6}$ 4s $^{2}$
(d) Cu: [Ar] 3d $^{10}$ 4s $^{1}$ (exception - fully-filled 3d)

Solved Example 8

Which metal has the highest exchange energy for exchange stabilisation: Cr, Mn, Fe, or Cu?

Solution:

Exchange energy $\propto N (N - 1) /2$ . Count unpaired electrons in d-subshell:

Cr [Ar]3d $^{5}$ 4s $^{1}$ : 5 unpaired in 3d, exchanges = $5 \times 4/2 = 10$
Mn [Ar]3d $^{5}$ 4s $^{2}$ : 5 unpaired in 3d, exchanges = 10
Fe [Ar]3d $^{6}$ 4s $^{2}$ : 4 unpaired
Cu [Ar]3d $^{10}$ 4s $^{1}$ : 0 unpaired in 3d

Answer: Cr and Mn (tied). Cr also gains exchange stabilisation from its 4s $^{1}$ electron.

Solved Example 9

The total spin resulting from a d

^{7}

configuration is: (A) 1 (B) 2 (C) 5/2 (D) 3/2

Solution:

d $^{7}$ : In 5 d-orbitals, first fill 5 singly (Hund), then pair 2. So 3 unpaired + 2 paired. Unpaired = 3.
Total spin $S = 3 \times \frac{1}{2} = 3/2$ . Answer: (D).

Solved Example 10

The electronic configuration 1s

^{2}

^{2}

^{6}

^{2}

^{6}

^{5}

^{1}

represents:
(A) Ground state (B) Excited state (C) Cation (D) Anion

Solution:

Total electrons = 24 = Chromium. This IS the ground state (Cr uses the half-filled 3d $^{5}$ 4s $^{1}$ configuration). Answer: (A).

Solved Example 11

Write electronic configurations of (a) Fe

^{2 +}

, (b) Fe

^{3 +}

, (c) Cu

^{+}

, (d) Cr

^{3 +}

Solution:

Rule for ions: remove electrons first from the highest n (outermost shell), so 4s before 3d for transition metals.

Fe (26): [Ar] 3d $^{6}$ 4s $^{2}$ → Fe $^{2 +}$ : [Ar] 3d $^{6}$
Fe $^{3 +}$ : [Ar] 3d $^{5}$ (half-filled - extra stable)
Cu (29): [Ar] 3d $^{10}$ 4s $^{1}$ → Cu $^{+}$ : [Ar] 3d $^{10}$
Cr (24): [Ar] 3d $^{5}$ 4s $^{1}$ → Cr $^{3 +}$ : [Ar] 3d $^{3}$

Solved Example 12

Explain how to arrange 3 electrons in the p-orbitals using Hund's rule.

Solution:

Place one electron in each of $p_{x}, p_{y}, p_{z}$ with parallel spins (all $↑$ or all $↓$ ). Only after all three are singly occupied would a fourth begin to pair. This maximises exchange energy and gives the most stable arrangement.

Solved Example 13

Write the electronic configuration of Cr (Z = 24) explaining why it deviates from the expected Aufbau order.

Solution:

Expected: [Ar] 3d $^{4}$ 4s $^{2}$ . Actual: [Ar] 3d $^{5}$ 4s $^{1}$ .
Reason: shifting one electron from 4s to 3d gives a half-filled 3d $^{5}$ configuration. Both 3d $^{5}$ (5 unpaired) and 4s $^{1}$ (1 unpaired) subshells are half-filled - which is extra stable due to maximised exchange energy and symmetric electron distribution.

Solved Example 14

Why is potassium's configuration [Ar] 4s

^{1}

and not [Ar] 3d

^{1}

Solution:

Apply the $(n + ℓ)$ rule: 4s has $n + ℓ = 4 + 0 = 4$ ; 3d has $n + ℓ = 3 + 2 = 5$ . Since 4s has lower $(n + ℓ)$ , it fills first. So K: [Ar] 4s $^{1}$ .

Solved Example 15

What is the maximum number of electrons in the 3rd shell?

Solution:

Max electrons in nth shell $= 2 n^{2} = 2 (3)^{2} = 18$ .
Breakdown: 3s $^{2}$ (2) + 3p $^{6}$ (6) + 3d $^{10}$ (10) = 18. $✓$

Solved Example 16

Maximum number of electrons in a single d-orbital?

Solution:

A single orbital can hold 2 electrons (opposite spins) - Pauli's exclusion. The d-subshell (5 orbitals) holds 10.

Solved Example 17

Calculate the total number of d-electrons in molybdenum (Z = 42).

Solution:

Mo: [Kr] 4d $^{5}$ 5s $^{1}$ (analogous to Cr - half-filled 4d exception).
Total d-electrons = 3d $^{10}$ (from [Kr]) + 4d $^{5}$ = 15.

Aside on Zeeman/Stark: When atoms are placed in a strong magnetic field, spectral lines split - the Zeeman effect - because the

m_{ℓ}

sublevels (originally degenerate) split in energy. In an electric field, similar splitting occurs - the Stark effect. Both are natural in the quantum-mechanical picture but were unexplained by Bohr.

Common Mistakes to Avoid

Watch out

$m_{ℓ}$ range depends on $ℓ$ , not $n$ . For $ℓ = 2$ , $m$ has 5 values, regardless of $n$ .
Do not skip 3d after 4s in cations. For transition-metal ions, remove 4s electrons first, then 3d.
Hund's rule ≠ Pauli. Pauli says orbitals hold max 2 with opposite spins. Hund says fill singly first with parallel spins.
Anomalous configurations - only Cr and Cu at JEE level. Other exceptions (Nb, Mo, Ru, Rh, Pd, Ag, Pt, Au) are advanced and rarely asked.
$n + ℓ$ rule: tie-breaker is lower $n$ . Not higher. So 4s (n=4) fills before 3d (n=3) at $n + ℓ = 4$ vs 5.
Do not write configuration by shell order (K, L, M, N). Write by subshell energy (Aufbau order). E.g., Fe is written [Ar] 3d $^{6}$ 4s $^{2}$ , not [Ar] 4s $^{2}$ 3d $^{6}$ (either is acceptable, but be consistent).

Frequently Asked Questions

What are the four quantum numbers and what do they tell us?

The four quantum numbers describe every electron uniquely: (1) Principal ( $n$ ) - shell/size, values $1, 2, 3, ...$ ; (2) Azimuthal ( $ℓ$ ) - subshell/shape, values $0$ to $n - 1$ ; (3) Magnetic ( $m_{ℓ}$ ) - orientation, values $- ℓ$ to $+ ℓ$ ; (4) Spin ( $m_{s}$ ) - electron spin, $+ 1/2$ or $- 1/2$ .

What is the shape of s, p, d, and f orbitals?

s-orbitals are spherically symmetric. p-orbitals have a dumbbell shape (two lobes with a nodal plane at the nucleus) - three of them, along x, y, z. d-orbitals have four-lobe clover shapes (four of them) plus one dumbbell + donut ( $d_{z^{2}}$ ) - five in total. f-orbitals have complex 8-lobe patterns - seven in total.

How does the Aufbau principle determine electron filling order?

Electrons fill orbitals in order of increasing energy. Energy of a multi-electron orbital is given by $(n + ℓ)$ : lower $(n + ℓ)$ fills first. If two orbitals have the same $(n + ℓ)$ , the one with lower $n$ fills first. This gives the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, ...

What is Hund's rule of maximum multiplicity?

In a set of degenerate (same-energy) orbitals, electrons occupy separate orbitals with parallel spins first, and only after each is singly occupied do they begin to pair. This maximises the number of unpaired electrons and thereby the exchange energy, which lowers total energy.

Why are Cr and Cu exceptions to the Aufbau principle?

The expected configurations [Ar]3d $^{4}$ 4s $^{2}$ (Cr) and [Ar]3d $^{9}$ 4s $^{2}$ (Cu) rearrange to [Ar]3d $^{5}$ 4s $^{1}$ (Cr) and [Ar]3d $^{10}$ 4s $^{1}$ (Cu). Half-filled ( $d^{5}$ ) and fully-filled ( $d^{10}$ ) subshells are extra stable due to maximum exchange energy and symmetric electron distribution.

What is the maximum number of electrons in the nth shell?

$2 n^{2}$ . So K (n=1) holds 2, L (n=2) holds 8, M (n=3) holds 18, N (n=4) holds 32. This follows from the number of orbitals ( $n^{2}$ ) with 2 electrons each (Pauli).

What is Pauli's exclusion principle?

No two electrons in an atom can have all four quantum numbers identical. In practice this means each orbital (fixed $n, ℓ, m_{ℓ}$ ) can hold at most 2 electrons, and they must have opposite spins.

How do you write the electronic configuration of ions?

First write the neutral atom's configuration. For cations, remove electrons from the outermost shell (highest $n$ ) first. So for transition metals, remove 4s before 3d. Example: Fe [Ar]3d $^{6}$ 4s $^{2}$ ; Fe $^{2 +}$ : [Ar]3d $^{6}$ ; Fe $^{3 +}$ : [Ar]3d $^{5}$ .

Previous year questions on Quantum Numbers

22 questions from past papers, each with a step-by-step solution.

Show all 22 questions

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1. The Four Quantum Numbers

Principal quantum number (n)

Azimuthal (angular momentum) quantum number (ℓ)

Magnetic quantum number (mℓ​)

Spin quantum number (s or ms​)

2. Shapes of Orbitals

s-orbital (spherical)

p-orbitals (dumbbell)

d-orbitals (clover and donut)

3. Rules for Filling Electrons

(a) Aufbau principle

(b) Pauli's exclusion principle

(c) Hund's rule of maximum multiplicity

Exchange energy and half/fully-filled stability

Aufbau exceptions (Cr and Cu)

Common Mistakes to Avoid

Frequently Asked Questions

What are the four quantum numbers and what do they tell us?

What is the shape of s, p, d, and f orbitals?

How does the Aufbau principle determine electron filling order?

What is Hund's rule of maximum multiplicity?

Why are Cr and Cu exceptions to the Aufbau principle?

What is the maximum number of electrons in the nth shell?

What is Pauli's exclusion principle?

How do you write the electronic configuration of ions?

Previous year questions on Quantum Numbers

Ready to master Atomic Structure?

Principal quantum number ( $n$ )

Azimuthal (angular momentum) quantum number ( $ℓ$ )

Magnetic quantum number ( $m_{ℓ}$ )

Spin quantum number ( $s$ or $m_{s}$ )