Electronic Configuration & Filling Rules

Electronic Configuration of Atoms

The electronic configuration of an atom describes how its electrons are distributed among the various orbitals. Three fundamental rules govern this filling:

1. Aufbau Principle
2. Pauli Exclusion Principle
3. Hund's Rule of Maximum Multiplicity

1. Aufbau Principle

The word "Aufbau" is German for "building up." This principle states:

Electrons fill orbitals in order of increasing energy, starting from the lowest available energy orbital.

The filling order (from the $(n + l)$ rule): $$1s \to 2s \to 2p \to 3s \to 3p \to 4s \to 3d \to 4p \to 5s \to 4d \to 5p \to 6s \to 4f \to 5d \to 6p \to 7s \to 5f \to 6d$$

Memory Aid — The Diagonal Rule

Write the subshells in a grid and draw diagonal arrows:

1s
2s  2p
3s  3p  3d
4s  4p  4d  4f
5s  5p  5d  5f
6s  6p  6d
7s  7p

Follow the diagonals from top-right to bottom-left to get the filling order.

2. Pauli Exclusion Principle

Proposed by Wolfgang Pauli (1926):

No two electrons in an atom can have the same set of all four quantum numbers ($n, l, m_l, m_s$).

This means each orbital can hold a maximum of 2 electrons, and they must have opposite spins (↑↓).

Consequences:

• Max electrons in s-subshell ($l = 0$): $1 \times 2 = 2$
• Max electrons in p-subshell ($l = 1$): $3 \times 2 = 6$
• Max electrons in d-subshell ($l = 2$): $5 \times 2 = 10$
• Max electrons in f-subshell ($l = 3$): $7 \times 2 = 14$
• Max electrons in shell $n$: $2n^2$

3. Hund's Rule of Maximum Multiplicity

This rule deals with filling electrons in degenerate orbitals (orbitals of equal energy within the same subshell).

Pairing of electrons in orbitals of the same subshell does not take place until each orbital has one electron (singly occupied with parallel spins).

What This Means in Practice

For the p-subshell (3 orbitals of equal energy):

• 1 electron: ↑ _ _ (one orbital occupied)
• 2 electrons: ↑ ↑ _ (two orbitals, parallel spins)
• 3 electrons: ↑ ↑ ↑ (all three singly occupied — half-filled)
• 4 electrons: ↑↓ ↑ ↑ (pairing starts with the 4th electron)
• 5 electrons: ↑↓ ↑↓ ↑
• 6 electrons: ↑↓ ↑↓ ↑↓ (fully filled)

Why parallel spins? Electrons with the same spin in different orbitals have a lower electron-electron repulsion and stabilising exchange energy.

Example: Nitrogen ($Z = 7$)

Configuration: $1s^2 \, 2s^2 \, 2p^3$

Orbital diagram: [1s: ↑↓] [2s: ↑↓] [2p: ↑ | ↑ | ↑]

All three 2p orbitals are singly occupied with parallel spins — this is a half-filled configuration and is especially stable.

Electronic Configurations of Elements ($Z = 1$ to $30$)

Element	$Z$	Electronic Configuration
H	1	$1s^1$
He	2	$1s^2$
Li	3	$[\text{He}] \, 2s^1$
Be	4	$[\text{He}] \, 2s^2$
B	5	$[\text{He}] \, 2s^2 \, 2p^1$
C	6	$[\text{He}] \, 2s^2 \, 2p^2$
N	7	$[\text{He}] \, 2s^2 \, 2p^3$
O	8	$[\text{He}] \, 2s^2 \, 2p^4$
F	9	$[\text{He}] \, 2s^2 \, 2p^5$
Ne	10	$[\text{He}] \, 2s^2 \, 2p^6$
Na	11	$[\text{Ne}] \, 3s^1$
Mg	12	$[\text{Ne}] \, 3s^2$
Al	13	$[\text{Ne}] \, 3s^2 \, 3p^1$
Si	14	$[\text{Ne}] \, 3s^2 \, 3p^2$
P	15	$[\text{Ne}] \, 3s^2 \, 3p^3$
S	16	$[\text{Ne}] \, 3s^2 \, 3p^4$
Cl	17	$[\text{Ne}] \, 3s^2 \, 3p^5$
Ar	18	$[\text{Ne}] \, 3s^2 \, 3p^6$
K	19	$[\text{Ar}] \, 4s^1$
Ca	20	$[\text{Ar}] \, 4s^2$
Sc	21	$[\text{Ar}] \, 3d^1 \, 4s^2$
Ti	22	$[\text{Ar}] \, 3d^2 \, 4s^2$
V	23	$[\text{Ar}] \, 3d^3 \, 4s^2$
Cr	24	$[\text{Ar}] \, 3d^5 \, 4s^1$ ⚠️
Mn	25	$[\text{Ar}] \, 3d^5 \, 4s^2$
Fe	26	$[\text{Ar}] \, 3d^6 \, 4s^2$
Co	27	$[\text{Ar}] \, 3d^7 \, 4s^2$
Ni	28	$[\text{Ar}] \, 3d^8 \, 4s^2$
Cu	29	$[\text{Ar}] \, 3d^{10} \, 4s^1$ ⚠️
Zn	30	$[\text{Ar}] \, 3d^{10} \, 4s^2$

[NEET Tip] Pay special attention to Cr and Cu — they are the most commonly tested exceptions!

Stability of Half-Filled and Fully-Filled Subshells

The electronic configurations of Chromium (Cr) and Copper (Cu) are exceptions to the Aufbau principle:

• Cr: expected $3d^4 \, 4s^2$ → actual $3d^5 \, 4s^1$
• Cu: expected $3d^9 \, 4s^2$ → actual $3d^{10} \, 4s^1$

Why? Because half-filled ($d^5$) and fully-filled ($d^{10}$) subshells have extra stability.

Reasons for Extra Stability

1. Symmetrical Distribution: Half-filled and fully-filled subshells have electrons symmetrically distributed among all orbitals. Symmetry leads to stability.

2. Exchange Energy: When two or more electrons with the same spin are present in degenerate orbitals, they can "exchange" positions. The energy released during these exchanges (exchange energy) stabilises the configuration.

The number of exchanges is maximum for half-filled and fully-filled configurations:

• $d^5$ (5 parallel electrons): $\binom{5}{2} = 10$ exchanges
• $d^4$ (4 parallel electrons): $\binom{4}{2} = 6$ exchanges

More exchanges → more exchange energy → greater stability.

Stable Configurations to Remember

• $p^3$ (half-filled p): N ($2p^3$), P ($3p^3$)
• $p^6$ (fully-filled p): Ne ($2p^6$), Ar ($3p^6$)
• $d^5$ (half-filled d): Cr ($3d^5$), Mn$^{2+}$ ($3d^5$), Fe$^{3+}$ ($3d^5$)
• $d^{10}$ (fully-filled d): Cu ($3d^{10}$), Zn ($3d^{10}$)

[JEE Tip] During ionisation of transition metals, 4s electrons are removed first (even though 4s was filled first). This is because after the 3d orbital is occupied, it becomes lower in energy than 4s. So Fe$^{2+}$ is $[\text{Ar}] \, 3d^6$ (not $3d^4 \, 4s^2$).

Solved Examples

Example 1: Writing Electronic Configuration

Write the electronic configuration of sulphur ($Z = 16$).

Solution: 16 electrons to fill in order: $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^4$

Condensed form: $[\text{Ne}] \, 3s^2 \, 3p^4$

Answer: $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^4$ or $[\text{Ne}] \, 3s^2 \, 3p^4$.

Example 2: Configuration of Chromium

Why is the electronic configuration of Cr ($Z = 24$) $[\text{Ar}] \, 3d^5 \, 4s^1$ and not $[\text{Ar}] \, 3d^4 \, 4s^2$?

Solution: The half-filled $3d^5$ configuration has extra stability due to:

1. Symmetrical distribution of electrons in all five d-orbitals
2. Maximum exchange energy ($\binom{5}{2} = 10$ exchanges vs $\binom{4}{2} = 6$ for $3d^4$)

The extra stability from $3d^5$ outweighs having 2 electrons in 4s.

Answer: Half-filled $3d^5$ is more stable due to symmetry and exchange energy.

Example 3: Configuration of an Ion

Write the electronic configuration of $\text{Fe}^{2+}$ ($Z = 26$).

Solution: Fe: $[\text{Ar}] \, 3d^6 \, 4s^2$ (26 electrons)

$\text{Fe}^{2+}$: remove 2 electrons — from 4s first (not 3d)

$\text{Fe}^{2+}$: $[\text{Ar}] \, 3d^6$ (24 electrons)

Answer: $[\text{Ar}] \, 3d^6$.

Example 4: Number of Unpaired Electrons

How many unpaired electrons are present in nitrogen ($Z = 7$)?

Solution: N: $1s^2 \, 2s^2 \, 2p^3$

Orbital diagram for 2p: ↑ | ↑ | ↑ (Hund's rule — all singly occupied)

Answer: 3 unpaired electrons.

Example 5: Unpaired Electrons in $\text{Fe}^{3+}$

Calculate the number of unpaired electrons in $\text{Fe}^{3+}$.

Solution: Fe: $[\text{Ar}] \, 3d^6 \, 4s^2$ $\text{Fe}^{3+}$: remove 3 electrons (2 from 4s, 1 from 3d) → $[\text{Ar}] \, 3d^5$

$3d^5$: ↑ | ↑ | ↑ | ↑ | ↑ (half-filled, all unpaired)

Answer: 5 unpaired electrons.

Example 6: Configuration of Cu$^+$ and Cu$^{2+}$

Write the electronic configurations of $\text{Cu}^+$ and $\text{Cu}^{2+}$.

Solution: Cu ($Z = 29$): $[\text{Ar}] \, 3d^{10} \, 4s^1$

$\text{Cu}^+$: remove 1 electron from 4s → $[\text{Ar}] \, 3d^{10}$ (fully-filled d — very stable!)

$\text{Cu}^{2+}$: remove 2 electrons (1 from 4s, 1 from 3d) → $[\text{Ar}] \, 3d^9$

Answer: $\text{Cu}^+$: $[\text{Ar}] \, 3d^{10}$; $\text{Cu}^{2+}$: $[\text{Ar}] \, 3d^9$.

Example 7: Identifying the Element

An element has the electronic configuration $[\text{Ar}] \, 3d^3 \, 4s^2$. Identify the element and state its atomic number.

Solution: $[\text{Ar}]$ accounts for 18 electrons. Additional: $3d^3 (3) + 4s^2 (2) = 5$ Total electrons $= 18 + 5 = 23$ → Element is Vanadium (V)

Answer: Vanadium, $Z = 23$.

Example 8: Applying Hund's Rule

Draw the orbital diagram for oxygen ($Z = 8$) and determine the number of unpaired electrons.

Solution: O: $1s^2 \, 2s^2 \, 2p^4$

Orbital diagram:

• 1s: ↑↓
• 2s: ↑↓
• 2p: ↑↓ | ↑ | ↑ (Hund's rule: 3 electrons go in singly, 4th pairs up)

Answer: 2 unpaired electrons.

Example 9: Exchange Energy Calculation

Calculate the number of electron exchanges for a $d^3$ configuration.

Solution: In $d^3$, there are 3 electrons with the same spin in different orbitals.

Number of exchanges $= \binom{3}{2} = \frac{3!}{2! \times 1!} = 3$

The three exchanges are: $e_1 \leftrightarrow e_2$, $e_1 \leftrightarrow e_3$, $e_2 \leftrightarrow e_3$.

Answer: 3 exchanges.

Example 10: Configuration Beyond $Z = 30$

Write the electronic configuration of krypton ($Z = 36$).

Solution: After Zn ($Z = 30$): $[\text{Ar}] \, 3d^{10} \, 4s^2$

6 more electrons fill the 4p subshell: $$\text{Kr}: [\text{Ar}] \, 3d^{10} \, 4s^2 \, 4p^6$$

Expanded form: $$1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^6$$

Answer: $[\text{Ar}] \, 3d^{10} \, 4s^2 \, 4p^6$.