The electron configuration of barium is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{2} **or** [Xe] 6s^{2}.

## Methods

We can write the electron configuration of barium using four different methods:

#1 Using aufbau principle

#2 Using periodic table

#3 From its bohr model

#4 From its orbital diagram

Let’s break down each method in detail.

### #1 Using Aufbau Principle

- First, find electrons of barium atom

The atomic number of barium represents the total number of electrons of barium. Since the atomic number of barium is 56, the total electrons of barium are 56.

- Second, make a table of subshell and its maximum electrons

Calculate the maximum number of electrons each subshell can hold using the formula: 4ℓ + 2

Where, ℓ = azimuthal quantum number of the subshell

For s subshell, ℓ = 0

For p subshell, ℓ = 1

For d subshell, ℓ = 2

For f subshell, ℓ = 3

subshell |
max. electrons |

s | 2 |

p | 6 |

d | 10 |

f | 14 |

This means that,

Each s subshell can hold maximum 2 electrons

Each p subshell can hold maximum 6 electrons

Each d subshell can hold maximum 10 electrons

Each f subshell can hold maximum 14 electrons

- Finally, use aufbau chart and start writing electron configuration

Remember that we have a total of 56 electrons.

According to the aufbau principle, 1s subshell is filled first and then 2s, 2p, 3s… and so on.

By looking at the chart, you can see that electrons are first filled in 1s subshell. Each s-subshell can hold a maximum of 2 electrons, so we will use 2 electrons for the 1s subshell.

So the electron configuration will be **1s ^{2}**. Where, 1s

^{2}indicates that the 1s subshell has 2 electrons.

Now we have used 2 electrons in the 1s subshell, so we have a total of 56 – 2 = 54 electrons left.

Looking at the chart, after 1s subshell now comes 2s subshell. Again, each s-subshell can hold a maximum of 2 electrons, so we will use 2 electrons for the 2s subshell.

So the electron configuration will be 1s^{2} **2s ^{2}**. Where, 2s

^{2}indicates that the 2s subshell has 2 electrons.

Again, we have used 2 electrons in the 2s subshell, so we have a total of 54 – 2 = 52 electrons left.

After 2s subshell now comes 2p subshell. Each p-subshell can hold a maximum of 6 electrons, so we will use 6 electrons for the 2p subshell.

So the electron configuration will be 1s^{2} 2s^{2} **2p ^{6}**. Where, 2p

^{6}indicates that the 2p subshell has 6 electrons.

Here, we have used 6 electrons in the 2p subshell, so we have a total of 52 – 6 = 46 electrons left.

After 2p subshell now comes 3s subshell. Each s-subshell can hold a maximum of 2 electrons, so we will use 2 electrons for the 3s subshell.

So the electron configuration will be 1s^{2} 2s^{2} 2p^{6} **3s ^{2}**. Where, 3s

^{2}indicates that the 3s subshell has 2 electrons.

Here, we have used 2 electrons in the 3s subshell, so we have a total of 46 – 2 = 44 electrons left.

After 3s subshell now comes 3p subshell. Each p-subshell can hold a maximum of 6 electrons, so we will use 6 electrons for the 3p subshell.

So the electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} **3p ^{6}**. Where, 3p

^{6}indicates that the 3p subshell has 6 electrons.

Here, we have used 6 electrons in the 3p subshell, so we have a total of 44 – 6 = 38 electrons left.

After 3p subshell now comes 4s subshell. Each s-subshell can hold a maximum of 2 electrons, so we will use 2 electrons for the 4s subshell.

So the electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} **4s ^{2}**. Where, 4s

^{2}indicates that the 4s subshell has 2 electrons.

Here, we have used 2 electrons in the 4s subshell, so we have a total of 38 – 2 = 36 electrons left.

After 4s subshell now comes 3d subshell. Each d-subshell can hold a maximum of 10 electrons, so we will use 10 electrons for the 3d subshell.

So the electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} **3d ^{10}**. Where, 3d

^{10}indicates that the 3d subshell has 10 electrons.

Here, we have used 10 electrons in the 3d subshell, so we have a total of 36 – 10 = 26 electrons left.

After 3d subshell now comes 4p subshell. Each p-subshell can hold a maximum of 6 electrons, so we will use 6 electrons for the 4p subshell.

So the electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} **4p ^{6}**. Where, 4p

^{6}indicates that the 4p subshell has 6 electrons.

Here, we have used 6 electrons in the 4p subshell, so we have a total of 26 – 6 = 20 electrons left.

After 4p subshell now comes 5s subshell. Each s-subshell can hold a maximum of 2 electrons, so we will use 2 electrons for the 5s subshell.

So the electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} **5s ^{2}**. Where, 5s

^{2}indicates that the 5s subshell has 2 electrons.

Here, we have used 2 electrons in the 5s subshell, so we have a total of 20 – 2 = 18 electrons left.

After 5s subshell now comes 4d subshell. Each d-subshell can hold a maximum of 10 electrons, so we will use 10 electrons for the 4d subshell.

So the electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} **4d ^{10}**. Where, 4d

^{10}indicates that the 4d subshell has 10 electrons.

Here, we have used 10 electrons in the 4d subshell, so we have a total of 18 – 10 = 8 electrons left.

After 4d subshell now comes 5p subshell. Each p-subshell can hold a maximum of 6 electrons, so we will use 6 electrons for the 5p subshell.

So the electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} **5p ^{6}**. Where, 5p

^{6}indicates that the 5p subshell has 6 electrons.

Here, we have used 6 electrons in the 5p subshell, so we have a total of 8 – 6 = 2 electrons left.

After 5p subshell now comes 6s subshell. Each s-subshell can hold a maximum of 2 electrons, and we also have 2 electrons left, so we will use that 2 electrons for the 6s subshell.

So the electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} **6s ^{2}**. Where, 6s

^{2}indicates that the 6s subshell has 2 electrons.

Therefore, the final electron configuration of barium is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{2}. And the condensed/abbreviated electron configuration of barium is [Xe] 6s^{2}.

Where, Xe is xenon

### #2 Using Periodic Table

- First, get periodic table chart with spdf notation

The above image shows periodic table blocks.

The ‘s’ in s block represents that all s block elements have their valence electrons in s subshell. Similarly, the ‘p’ in p block represents that all p block elements have their valence electrons in p subshell. And so on for d block and f block.

- Second, mark location of barium on periodic table

Barium is the s block element located in group 2 and period 6. Hence, mark the location of barium on the periodic table as follows:

- Finally, start writing electron configuration

Remember that: each s subshell can hold maximum 2 electrons, each p subshell can hold maximum 6 electrons, each d subshell can hold maximum 10 electrons, and each f subshell can hold maximum 14 electrons.

Start writing electron configuration from the very first element (i.e., hydrogen) all the way up to barium.

So the electron configuration of barium will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{2}.

### #3 From its Bohr Model

In the above image, 1 represents the 1^{st} electron shell. Similarly, 2 represents the 2^{nd} electron shell, 3 represents the 3^{rd} electron shell, 4 represents the 4^{th} electron shell, 5 represents the 5^{th} electron shell, and 6 represents the 6^{th} electron shell.

The 1^{st} electron shell contains 1s subshell, the 2^{nd} electron shell contains 2s and 2p subshells, the 3^{rd} electron shell contains 3s, 3p, and 3d subshells, the 4^{th} electron shell contains 4s, 4p, and 4d subshells, the 5^{th} electron shell contains 5s and 5p subshells, and the 6^{th} electron shell contains 6s subshell.

We know that each s subshell can hold maximum 2 electrons, each p subshell can hold maximum 6 electrons, each d subshell can hold maximum 10 electrons, and each f subshell can hold maximum 14 electrons.

Also, we have to make sure that the electron configuration will match the order of aufbau principle (i.e., the 1s subshell is filled first and then 2s, 2p, 3s… and so on).

So the electron configuration of barium will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{2}.

Where,

1s^{2} indicates that the 1s subshell has 2 electrons

2s^{2} indicates that the 2s subshell has 2 electrons

2p^{6} indicates that the 2p subshell has 6 electrons

3s^{2} indicates that the 3s subshell has 2 electrons

3p^{6} indicates that the 3p subshell has 6 electrons

4s^{2} indicates that the 4s subshell has 2 electrons

3d^{10} indicates that the 3d subshell has 10 electrons

4p^{6} indicates that the 4p subshell has 6 electrons

5s^{2} indicates that the 5s subshell has 2 electrons

4d^{10} indicates that the 4d subshell has 10 electrons

5p^{6} indicates that the 5p subshell has 6 electrons

6s^{2} indicates that the 6s subshell has 2 electrons

Learn how to draw: Barium Bohr Model

### #4 From its Orbital Diagram

The above orbital diagram shows that the 1s subshell has 2 electrons, the 2s subshell has 2 electrons, the 2p subshell has 6 electrons, the 3s subshell has 2 electrons, the 3p subshell has 6 electrons, the 4s subshell has 2 electrons, the 3d subshell has 10 electrons, the 4p subshell has 6 electrons, the 5s subshell has 2 electrons, the 4d subshell has 10 electrons, the 5p subshell has 6 electrons, and the 6s subshell has 2 electrons.

So the electron configuration of barium will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{2}.

Learn how to draw: Barium Orbital Diagram

**Next:** Radium Electron Configuration

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**Related:**

**Also Read:**

- Barium Element
- Barium Bohr Model
- Barium Orbital Diagram
- Barium Protons Neutrons and Electrons
- Barium Valence Electrons

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