The **bromine**** electron configuration**, denoted as [Ar] 4s^{2} 3d^{10} 4p^{5} **or** 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{5}, showcases the precise placement of electrons within the atom. This configuration can be determined through various methods, including the aufbau principle, periodic table organization, Bohr model representation, or orbital diagram visualization.

## Methods

### Aufbau principle

- First, find electrons of bromine atom

The atomic number of bromine represents the total number of electrons of bromine. Since the atomic number of bromine is 35, the total electrons of bromine are 35.

- 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 35 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 35 – 2 = 33 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 33 – 2 = 31 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 31 – 6 = 25 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 25 – 2 = 23 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 23 – 6 = 17 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 17 – 2 = 15 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 15 – 10 = 5 electrons left.

After 3d subshell now comes 4p subshell. Each p-subshell can hold a maximum of 6 electrons, but here we have only 5 electrons left, so we will use that 5 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 ^{5}**. Where, 4p

^{5}indicates that the 4p subshell has 5 electrons.

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

Where, Ar is argon

### 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 bromine on periodic table

Bromine is the p block element located in group 17 and period 4. Hence, mark the location of bromine 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 bromine.

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

### 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, and 4 represents the 4^{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, and the 4^{th} electron shell contains 4s 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 bromine will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{5}.

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^{5} indicates that the 4p subshell has 5 electrons

Learn how to draw: Bromine Bohr model

### 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, and the 4p subshell has 5 electrons.

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

Learn how to draw: Bromine orbital diagram

**Next:** Tin electron configuration

## Related

## More topics

- Bromine
- Bromine Bohr model
- Bromine orbital diagram
- Bromine protons neutrons electrons
- Bromine valence electrons

## External links

- https://socratic.org/questions/electronic-configuration-of-bromine
- https://valenceelectrons.com/bromine-electron-configuration/
- https://materials.gelsonluz.com/2019/08/electron-configuration-of-bromine-br.html

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.