JEE Mains Chemistry · Structure of Atom

Orbital Energies and Electronic Configuration

In a many-electron atom orbital energy follows the (n + l) rule; in a one-electron atom it depends on n alone. Filling orbitals in that order gives the configuration, with Cr and Cu as the exceptions.

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Why this matters

Twenty-two PYQs, nineteen of them multiple choice. Seven order the orbitals of a many-electron atom by the (n + l) rule; seven test one-electron atoms, where energy depends on n alone, and how an orbital's energy falls as Z rises; eight write configurations, count electrons by l or mₗ, or explain the extra stability of half-filled and filled subshells. Three ideas cover the page.

Concept 1 of 3: The (n + l) rule in many-electron atoms

In an atom with many electrons, inner electrons shield the outer ones. An s electron gets closer to the nucleus than a p or d electron of the same shell, so it is held more tightly. The

(n+l)

rule captures this: lower

n+l

means lower energy, and a tie goes to the lower

n

Definition

Lower $(n+l)$ $\Rightarrow$ lower energy.
Equal $(n+l)$ : the lower $n$ is lower in energy (3d below 4p; 3p below 4s).
Filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, …
Orbitals with the same $n$ and $l$ are degenerate. $m_l$ and $m_s$ do not change the energy without an external field.

Order of orbital energy

E\uparrow\ \text{with}\ (n+l);\quad \text{tie}\Rightarrow\text{lower } n \text{ first}

Worked example

Arrange 4f, 5d, 6s, 6p and 5p in order of increasing energy in a many-electron atom.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q27Moderate

Example 1 · Structure of Atom · Orbital Energies and Electronic Configuration

Arrange the following atomic orbitals of multi electron atoms in order of increasing energy. (A)

n = 3,\mathcal{l} = 2,\text{ }m = + 1

(B)

n = 4,\mathcal{l} = 0,m = 0

(C)

n = 6,\mathcal{l} = 1,\text{ }m = 0

(D)

n = 5,\mathcal{l} = 1,\text{ }m = + 1

(E)

n = 2,\mathcal{l} = 1,\text{ }m = + 1

Choose the correct answer from the options given below :

A tie goes to the lower n

3d and 4p both have

n+l=5

. 3d is lower because its

n

is smaller. Do not order a tie by

l

m does not enter

Orbitals given as

(n, l, m)

are ordered by

n

and

l

only. Two sets that differ only in

m_l

m_s

have the same energy.

Drill 6 more on the (n + l) rule in many-electron atoms

Concept 2 of 3: One-electron atoms and the effect of Z

With only one electron there is no shielding. The energy depends on

n

alone, as in Bohr's formula. So in hydrogen 3s, 3p and 3d have exactly the same energy, and 4s sits above 3d. Across atoms, a larger nuclear charge pulls every orbital lower.

Definition

One-electron species (H, $\mathrm{He^+}$ , $\mathrm{Li^{2+}}$ ): $E_n=-13.6\dfrac{Z^2}{n^2}\ \mathrm{eV}$ , independent of $l$ .
Hydrogen order: $1s<2s=2p<3s=3p=3d<4s=4p=4d=4f$ .
Shell $n$ of hydrogen has $n^2$ degenerate orbitals.
A jump between degenerate orbitals ( $2p_x\to2p_y$ ) has $\Delta E=0$ , so gives no line.
The same subshell falls in energy as $Z$ rises: 2s of Li lies below 2s of H.
Size grows with $n$ : $2p_x$ is smaller than $3p_x$ .

One-electron energy

E_n=-13.6\,\frac{Z^2}{n^2}\ \mathrm{eV}\quad(\text{no } l \text{ dependence})

Worked example

Arrange 3d, 4s, 3s and 2p of

\mathrm{He^+}

by increasing energy.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q104Moderate

Example 2 · Structure of Atom · Orbital Energies and Electronic Configuration

For hydrogen atom, the orbital/s with lowest energy is/are : (A) 4 s (B)

3p_{x}

(C)

3d_{x^{2}-y^{2}}

(D)

3{\text{ }d}_{z^{2}}

(E)

4p_{z}

Choose the correct answer from the options given below :

Hydrogen does not follow Aufbau

In H, 4s is above 3d and 2s equals 2p. The

(n+l)

order applies only to atoms with more than one electron.

Higher Z lowers the orbital

"Energies of orbitals in the same subshell increase with atomic number" is false. They decrease, because the nucleus pulls harder.

Drill 6 more on one-electron atoms and the effect of z

Concept 3 of 3: Configurations, exceptions and electron counts

Fill orbitals in

(n+l)

order, one electron per orbital with parallel spins before any pairing (Hund), and at most two per orbital (Pauli). Chromium and copper break the order to reach a half-filled or full 3d. Once the configuration is written, any count by

l

, by

m_l

, or by noble-gas match is just reading it.

Definition

Hund's rule: in degenerate orbitals, electrons spread out with parallel spins before pairing.
Exceptions: Cr $[\mathrm{Ar}]3d^54s^1$ , Cu $[\mathrm{Ar}]3d^{10}4s^1$ .
Half-filled and filled subshells are extra stable: symmetry, larger exchange energy (possible only among degenerate orbitals), smaller repulsion and shielding.
Electrons with a given $l$ : add all the s ( $l=0$ ), p ( $l=1$ ) or d ( $l=2$ ) electrons.
Each subshell has exactly one orbital with $m_l=0$ . A filled one needs two electrons.
Noble-gas configuration: the ion's electron count is 2, 10, 18, 36, 54 or 86 and it matches that gas's configuration.

Electrons in a subshell

N_{max}=2(2l+1)

Worked example

For copper (

Z=29

), how many electrons have

l=0

and how many have

l=2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q113Moderate

Example 3 · Structure of Atom · Orbital Energies and Electronic Configuration

Consider the ground state of chromium atom

(Z = 24)

. How many electrons are with Azimuthal quantum number

l= 1

and

l= 2

respectively?

Exchange energy needs degenerate orbitals

Extra stability comes from same-spin electrons exchanging among orbitals of equal energy. A statement placing them in non-degenerate orbitals is false.

Half-filled is not filled

When counting completely filled orbitals with

m_l=0

, a p orbital holding one electron does not count. In Ge (

4p^2

) the 4p electrons fill no orbital.

Drill 7 more on configurations, exceptions and electron counts

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 (3)

The (n + l) rule in many-electron atoms
Order of orbital energy
$E\uparrow\ \text{with}\ (n+l);\quad \text{tie}\Rightarrow\text{lower } n \text{ first}$
One-electron atoms and the effect of Z
One-electron energy
$E_n=-13.6\,\frac{Z^2}{n^2}\ \mathrm{eV}\quad(\text{no } l \text{ dependence})$
Configurations, exceptions and electron counts
Electrons in a subshell
$N_{max}=2(2l+1)$

Watch out for (6)

A tie goes to the lower n
→ The (n + l) rule in many-electron atoms
m does not enter
→ The (n + l) rule in many-electron atoms
Hydrogen does not follow Aufbau
→ One-electron atoms and the effect of Z
Higher Z lowers the orbital
→ One-electron atoms and the effect of Z
Exchange energy needs degenerate orbitals
→ Configurations, exceptions and electron counts
Half-filled is not filled
→ Configurations, exceptions and electron counts

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