Electronic Configuration of Atoms

Electronic Configuration of Atoms:

The distribution of electrons into orbitals of an atom is called its electronic configuration. If one keeps in mind the basic rules which govern the filling of different atomic orbitals, the electronic configurations of different atoms can be written very easily.

Example:

Write the electronic configuration for nitrogen which has 7 electrons.

We know that

1s

1 orbital

(2 electrons)

2s 2p

1 orbital 3 orbitals

(2 electrons) (2 electrons each)

1s^1	1s^2
2s^1	2s^2	2p^1	2p^2	2p^3	2p^4	2p^5	2p^6

Each orbital contains 2 electrons each. So 1s and 2s get filled completely and 2p gets partially filled with 3 electrons and 3 places are empty.

We say,

1s and 2s are filled and 2p is half-filled.

Usually orbitals with lower principal quantum numbers have lower energies, but that isn't the case when you compare the energies of the 3d orbitals and the energies of the 4s orbitals. In this case, the 4s orbitals are lower in energy than the 3d orbitals even though they have higher principal quantum numbers.

To check this let us look at Potassium atom which has 19 electrons.

1s^1

1s^2

2s^1

2s^2

2p^1

2p^2

2p^3

2p^4

2p^5

2p^6

3s^1

3s^2

3p^1

3p^2

3p^3

3p^4

3p^5

3p^6

3d^1

3d^2

3d^3

3d^4

3d^5

3d^6

3d^7

3d^8

3d^9

3d^10

4s^1

4s^2

4p^1

4p^2

4p^3

4p^4

4p^5

4p^6

4d^1

4d^2

4d^3

4d^4

4d^5

4d^6

4d^7

4d^8

4d^9

4d^10

Here,

1s, 2s, 2p, 3s and 3p get filled completely. There is one electron left which gets placed in 4s as 4s has a lower energy level than 3d.

This can be represented by the figure below.

Problems:

1. Write the electronic configurations of the following ions:

(a) (b) (c)

Solution:

(a)  ion

The electronic configuration of H atom is 1s¹.

A negative charge on the species indicates the gain of an electron by it.

∴ Electronic configuration of ion = 1s²

(b) ion

The electronic configuration of F atom is 1s² 2s² 2p⁵.

A negative charge on the species indicates the gain of an electron by it.

∴ Electron configuration of ion = 1s² 2s² 2p⁶

(c)  ion

The electronic configuration of O atom is 1s² 2s² 2p⁴.

A dinegative charge on the species indicates that two electrons are gained by it.

∴ Electronic configuration of  ion = 1s² 2s² 2p⁶

2. What are the atomic numbers of elements whose outermost electrons are represented by 

(a) 3s¹ (b) 2p³ (c) 3p⁵

Solution:

(a) 3s¹

Completing the electron configuration of the element as 1s² 2s² 2p⁶ 3s¹

∴ Number of electrons present in the atom of the element

= 2 + 2 + 6 + 1

= 11

∴ Atomic number of the element = 11

(b) 2p³

Completing the electron configuration of the element as 1s² 2s² 2p³

∴ Number of electrons present in the atom of the element

= 2 + 2 + 3

= 7

∴ Atomic number of the element = 7

(c) 3p⁵

Completing the electron configuration of the element as 1s² 2s² 2p⁵

∴ Number of electrons present in the atom of the element

= 2 + 2 + 5

= 9

∴ Atomic number of the element = 9

3. Which atoms are indicated by the following configurations?

(a) [He] 2s¹ (b) [Ne] 3s² 3p³ (c) [Ar] 4s² 3d¹

Solution:

(a) [He] 2s¹

The electronic configuration of the element is [He] 2s¹ = 1s² 2s¹.

∴ Atomic number of the element = 3

Hence, the element with the electronic configuration [He] 2s¹ is lithium (Li).

(b) [Ne] 3s² 3p³

The electronic configuration of the element is [Ne] 3s² 3p³ = 1s² 2s² 2p⁶ 3s² 3p³.

∴ Atomic number of the element = 15

Hence, the element with the electronic configuration [Ne] 3s² 3p³ is phosphorus (P).

(c) [Ar] 4s² 3d¹

The electronic configuration of the element is [Ar] 4s² 3d¹ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹.

∴ Atomic number of the element = 21

Hence, the element with the electronic configuration [Ar] 4s² 3d¹ is scandium (Sc).