2.4 Determining Electronic Configuration of Atoms and Ions | Atomic Structure | Chemistry 11

The electronic configuration of an atom or ion is determined by following a three-step process. In a neutral atom, the atomic number (Z), which equals the number of protons, is equal to the number of electrons.

Identify the neutral atom's electron configuration
- The number of electrons in a neutral atom corresponds to its atomic number, found on the Periodic Table.
- For example, a neutral fluorine atom (F) has an atomic number of 9, meaning it has 9 electrons.
Determine the charge of the ion
- Ions are atoms that have gained or lost electrons, resulting in a net electrical charge.
- Cations are positive ions that have lost one or more electrons.
- Anions are negative ions that have gained one or more electrons.
Write the electron configuration for the ion
- For cations, start with the neutral atom's configuration and remove electrons from the outermost shell (highest principal quantum number). The number of electrons removed equals the magnitude of the positive charge.
- For anions, start with the neutral atom's configuration and add electrons to the outermost shell. The number of electrons added equals the magnitude of the negative charge.

Principles Governing Electronic Configuration

The distribution of electrons in orbitals is governed by several key principles related to energy and repulsion. For detailed explanations of these principles, refer to Rules of Electronic Configuration→.

Inter-electron Repulsion

Electrons, being like-charged particles, repel each other. To minimize this repulsion, electrons prefer to occupy separate orbitals within the same subshell before pairing up. This minimizes the energy of the atom, leading to a more stable configuration.

Orbital Energy

The energy of an orbital is influenced by its distance from the nucleus and the shielding effect of other electrons. According to the Aufbau principle, electrons fill the lowest energy orbitals available first.

Spin Pairing Repulsion (Hund's Rule)

Electrons with the same spin repel each other more than electrons with opposite spins. Hund's rule states that every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Writing Detailed Electronic Configurations

A shorthand configuration like $3 p^{6}$ represents the full distribution of electrons across the degenerate p-orbitals. Each orbital can hold two electrons with opposite spins (one spin up, one spin down).

For example, $3 p^{6}$ fully expanded is: $3 p_{x}^{2}, 3 p_{y}^{2}, 3 p_{z}^{2}$ .
This can also be written showing spins: $3 p_{x}^{↑↓}, 3 p_{y}^{↑↓}, 3 p_{z}^{↑↓}$ .

For an element with atomic number 25 (Manganese, Mn), the detailed electronic configuration showing unpaired electrons in the d-orbitals is:

$1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{2} 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{2} 3 p_{z}^{2} 4 s^{2} 3 d_{x y}^{↑} 3 d_{y z}^{↑} 3 d_{z x}^{↑} 3 d_{x^{2} - y^{2}}^{↑} 3 d_{z^{2}}^{↑}$

The more common condensed form is:

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{5}$

Worked Examples

Example 1: Fluoride Ion ( $F^{-}$ )

Neutral Atom (Fluorine, F):
- Atomic Number (Z) = 9, so it has 9 electrons.
- Neutral Configuration: $1 s^{2} 2 s^{2} 2 p^{5}$
Ion Formation (Fluoride, $F^{-}$ ):
- The charge is -1, indicating the atom has gained one electron.
- Total electrons = $9 + 1 = 10$ electrons.
Ion Configuration:
- The added electron fills the partially filled 2p orbital.
- Final Configuration for $F^{-}$ : $1 s^{2} 2 s^{2} 2 p^{6}$

Example 2: Sodium Ion ( $N a^{+}$ )

Neutral Atom (Sodium, Na):
- Atomic Number (Z) = 11, so it has 11 electrons.
- Neutral Configuration: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$
Ion Formation (Sodium ion, $N a^{+}$ ):
- The charge is +1, indicating the atom has lost one electron.
- Total electrons = $11 - 1 = 10$ electrons.
Ion Configuration:
- The electron is removed from the outermost orbital, which is $3 s$ .
- Final Configuration for $N a^{+}$ : $1 s^{2} 2 s^{2} 2 p^{6}$

Example 3: Iron(II) Ion ( $F e^{2 +}$ )

Neutral Atom (Iron, Fe):
- Atomic Number (Z) = 26, so it has 26 electrons.
- Neutral Configuration: $[A r] 4 s^{2} 3 d^{6}$
Ion Formation ( $F e^{2 +}$ ):
- The charge is +2, indicating the atom has lost two electrons.
- Electrons are removed from the outermost $4 s$ orbital first (not from $3 d$ ).
Ion Configuration: $[A r] 3 d^{6}$

Key Rule for Transition Metal Cations: Always remove electrons from the highest principal quantum number shell first (i.e., $4 s$ before $3 d$ ), even though $3 d$ was filled after $4 s$ during the Aufbau process.

Table: Electronic Configuration of the First 18 Elements

Element	Atomic Number (Z)	Electronic Configuration
Hydrogen	1	$1 s^{1}$
Helium	2	$1 s^{2}$
Lithium	3	$1 s^{2} 2 s^{1}$
Beryllium	4	$1 s^{2} 2 s^{2}$
Boron	5	$1 s^{2} 2 s^{2} 2 p^{1}$
Carbon	6	$1 s^{2} 2 s^{2} 2 p^{2}$
Nitrogen	7	$1 s^{2} 2 s^{2} 2 p^{3}$
Oxygen	8	$1 s^{2} 2 s^{2} 2 p^{4}$
Fluorine	9	$1 s^{2} 2 s^{2} 2 p^{5}$
Neon	10	$1 s^{2} 2 s^{2} 2 p^{6}$
Sodium	11	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$
Magnesium	12	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$
Aluminum	13	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$
Silicon	14	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2}$
Phosphorus	15	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}$
Sulfur	16	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}$
Chlorine	17	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$
Argon	18	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$

Concept Assessment Exercise 2.5 — Worked Answers

Q1: Write down the electronic configuration of the following elements:

a. $^{21} Sc$
b. $^{24} Cr$
c. $^{25} Mn$
d. $^{30} Zn$

A1:

a. Scandium (Sc, Z=21): $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{1}$ or $[Ar] 4 s^{2} 3 d^{1}$
b. Chromium (Cr, Z=24): $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{5}$ or $[Ar] 4 s^{1} 3 d^{5}$ (Anomalous: half-filled $3 d^{5}$ is extra stable, so one $4 s$ electron is promoted to $3 d$ )
c. Manganese (Mn, Z=25): $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{5}$ or $[Ar] 4 s^{2} 3 d^{5}$
d. Zinc (Zn, Z=30): $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10}$ or $[Ar] 4 s^{2} 3 d^{10}$

Worked Examples

Neutral Atom (Fluorine, F):

Atomic Number (Z) = 9, so it has 9 electrons.
Neutral Configuration: $1 s^{2} 2 s^{2} 2 p^{5}$

Ion Formation (Fluoride, $F^{-}$ ):

The charge is -1, indicating the atom has gained one electron.
Total electrons = $9 + 1 = 10$ electrons.

Ion Configuration:

The added electron fills the partially filled 2p orbital.
Final Configuration for $F^{-}$ : $1 s^{2} 2 s^{2} 2 p^{6}$

Neutral Atom (Sodium, Na):

Atomic Number (Z) = 11, so it has 11 electrons.
Neutral Configuration: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$

Ion Formation (Sodium ion, $N a^{+}$ ):

The charge is +1, indicating the atom has lost one electron.
Total electrons = $11 - 1 = 10$ electrons.

Ion Configuration:

The electron is removed from the outermost orbital, which is $3 s$ .
Final Configuration for $N a^{+}$ : $1 s^{2} 2 s^{2} 2 p^{6}$

Neutral Atom (Iron, Fe):

Atomic Number (Z) = 26, so it has 26 electrons.
Neutral Configuration: $[A r] 4 s^{2} 3 d^{6}$

Ion Formation ( $F e^{2 +}$ ):

The charge is +2, indicating the atom has lost two electrons.
Electrons are removed from the outermost $4 s$ orbital first (not from $3 d$ ).

Ion Configuration: $[A r] 3 d^{6}$

Element	Atomic Number (Z)	Electronic Configuration
Hydrogen	1	$1 s^{1}$
Helium	2	$1 s^{2}$
Lithium	3	$1 s^{2} 2 s^{1}$
Beryllium	4	$1 s^{2} 2 s^{2}$
Boron	5	$1 s^{2} 2 s^{2} 2 p^{1}$
Carbon	6	$1 s^{2} 2 s^{2} 2 p^{2}$
Nitrogen	7	$1 s^{2} 2 s^{2} 2 p^{3}$
Oxygen	8	$1 s^{2} 2 s^{2} 2 p^{4}$
Fluorine	9	$1 s^{2} 2 s^{2} 2 p^{5}$
Neon	10	$1 s^{2} 2 s^{2} 2 p^{6}$
Sodium	11	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$
Magnesium	12	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$
Aluminum	13	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$
Silicon	14	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2}$
Phosphorus	15	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}$
Sulfur	16	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}$
Chlorine	17	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$
Argon	18	$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$

Concept Assessment Exercise 2.5 — Worked Answers

Q1: Write down the electronic configuration of the following elements:

^{21} Sc

^{24} Cr

^{25} Mn

^{30} Zn

A1:

a. Scandium (Sc, Z=21):

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{1}

[Ar] 4 s^{2} 3 d^{1}

b. Chromium (Cr, Z=24):

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{5}

[Ar] 4 s^{1} 3 d^{5}

(Anomalous: half-filled $3 d^{5}$ is extra stable, so one $4 s$ electron is promoted to $3 d$ )

c. Manganese (Mn, Z=25):

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{5}

[Ar] 4 s^{2} 3 d^{5}

d. Zinc (Zn, Z=30):

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10}

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