JEE Mains Chemistry · Structure of Atom

Quantum Numbers and Electron Counting

Four quantum numbers, n, l, mₗ and mₛ, label every electron in an atom. Their allowed values decide how many orbitals and electrons a shell or subshell can hold.

All Structure of Atom notes JEE Mains Chemistry strategy

Why this matters

Sixteen PYQs, ten of them multiple choice. Eleven test the allowed values of n, l, mₗ and mₛ, count the orbitals or electrons that share some quantum numbers, or ask for the orbital angular momentum; five name the four quantum numbers of one particular electron. Two ideas cover the page.

Concept 1 of 2: Allowed values and counting orbitals and electrons

Each quantum number is limited by the one before it.

n

picks the shell,

l

runs from 0 to

n-1

, and

m_l

runs from

-l

+l

. Each orbital holds two electrons of opposite spin. To count electrons that share some quantum numbers, list the subshells that allow them, count the orbitals, then double (or not, if

m_s

is fixed).

Definition

$n=1,2,3,\ldots$ ; $l=0,1,\ldots,n-1$ (s, p, d, f).
$m_l=-l,\ldots,0,\ldots,+l$ : $2l+1$ orbitals in a subshell.
$m_s=+\tfrac12$ or $-\tfrac12$ .
Shell $n$ : $n^2$ orbitals, $2n^2$ electrons. Subshell: $2(2l+1)$ electrons.
A fixed $m_l$ occurs once in every subshell with $l\ge|m_l|$ .
All four fixed: exactly one electron (Pauli exclusion principle).
Orbital angular momentum $L=\sqrt{l(l+1)}\,\dfrac{h}{2\pi}$ . It depends on $l$ only, and is zero for every s orbital.

Orbital angular momentum

L=\sqrt{l(l+1)}\,\frac{h}{2\pi}

Worked example

How many electrons in an atom can have

n=4

and

m_l=+2

? How many of these can also have

m_s=+\tfrac12

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q36Moderate

Example 1 · Structure of Atom · Quantum Numbers and Electron Counting

Given, (A)

n = 5,m_{l}= - 1

(B)

n = 3,l = 2,m_{l}= - 1,m_{s}= +\frac{1}{2}

The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are :

l stops at n − 1

l=n

is never allowed. So

n=3, l=3

and

n=2, l=2

are invalid sets, whatever

m_l

is.

Angular momentum uses l, not n

L=\sqrt{l(l+1)}\,h/2\pi

. A 2s and a 3s electron both have

L=0

; a 2p electron has

\sqrt2\,h/2\pi

\sqrt6

belongs to

l=2

Drill 10 more on allowed values and counting orbitals and electrons

Concept 2 of 2: Quantum numbers of a particular electron

To name an electron's quantum numbers, write the configuration and find the subshell the electron sits in. The subshell gives

n

and

l

. An s electron always has

m_l=0

. For ions, remember that electrons leave from the highest

n

first, which is not always the last subshell filled.

Definition

Subshell $\to$ $n$ and $l$ : 4s is $n=4, l=0$ ; 3d is $n=3, l=2$ .
The valence electron of an alkali metal is $ns^1$ : $(n,0,0,\pm\tfrac12)$ .
Cations lose electrons from the highest $n$ first: 4s leaves before 3d.
"In accordance with the Aufbau principle" means strict filling order, even for Cr and Cu.

Worked example

Find the azimuthal quantum number of the outermost electron of

\mathrm{In^+}

(

Z=49

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q28Moderate

Example 2 · Structure of Atom · Quantum Numbers and Electron Counting

Which of the following is correct set of 4 quantum number of

19^{\text{th}}

electron in Chromium (Atomic number

= 24

) in accordance with Aufbau principle?

Ions lose the highest-n electrons first

\mathrm{Ga^+}

[\mathrm{Ar}]3d^{10}4s^2

: the 4p electron left, so the valence electron has

l=0

, not

l=1

"By Aufbau" ignores the exceptions

Cr is really

3d^54s^1

, but a question that says "in accordance with Aufbau" wants the strict order, where the 19th electron goes into 4s.

Drill 4 more on quantum numbers of a particular electron

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Allowed values and counting orbitals and electrons
Orbital angular momentum
$L=\sqrt{l(l+1)}\,\frac{h}{2\pi}$

Watch out for (4)

l stops at n − 1
→ Allowed values and counting orbitals and electrons
Angular momentum uses l, not n
→ Allowed values and counting orbitals and electrons
Ions lose the highest-n electrons first
→ Quantum numbers of a particular electron
"By Aufbau" ignores the exceptions
→ Quantum numbers of a particular electron

Test yourself on Structure of Atom

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