The **iridium**** electron configuration**, represented as [Xe] 6s^{2} 4f^{14} 5d^{7} **or** 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{2} 4f^{14} 5d^{7}, showcases the specific 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 iridium atom

The atomic number of iridium represents the total number of electrons of iridium. Since the atomic number of iridium is 77, the total electrons of iridium are 77.

- 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 77 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 77 – 2 = 75 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 75 – 2 = 73 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 73 – 6 = 67 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 67 – 2 = 65 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 65 – 6 = 59 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 59 – 2 = 57 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 57 – 10 = 47 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 47 – 6 = 41 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 41 – 2 = 39 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 39 – 10 = 29 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 29 – 6 = 23 electrons left.

After 5p subshell now comes 6s subshell. Each s-subshell can hold a maximum of 2 electrons, so we will use 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.

Here, we have used 2 electrons in the 6s subshell, so we have a total of 23 – 2 = 21 electrons left.

After 6s subshell now comes 4f subshell. Each f-subshell can hold a maximum of 14 electrons, so we will use 14 electrons for the 4f 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} **4f ^{14}**. Where, 4f

^{14}indicates that the 4f subshell has 14 electrons.

Here, we have used 14 electrons in the 4f subshell, so we have a total of 21 – 14 = 7 electrons left.

After 4f subshell now comes 5d subshell. Each d-subshell can hold a maximum of 10 electrons, but here we have only 7 electrons left, so we will use that 7 electrons for the 5d 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} 4f^{14} **5d ^{7}**. Where, 5d

^{7}indicates that the 5d subshell has 7 electrons.

Therefore, the final electron configuration of iridium 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} 4f^{14} 5d^{7}. And the condensed/abbreviated electron configuration of iridium is [Xe] 6s^{2} 4f^{14} 5d^{7}.

Where, Xe is xenon

### 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 iridium on periodic table

Iridium is the d block element located in group 9 and period 6. Hence, mark the location of iridium 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 iridium.

So the electron configuration of iridium 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} 4f^{14} 5d^{7}.

### 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, 4d, and 4f subshells, the 5^{th} electron shell contains 5s, 5p, and 5d 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 iridium 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} 4f^{14} 5d^{7}.

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

4f^{14} indicates that the 4f subshell has 14 electrons

5d^{7} indicates that the 5d subshell has 7 electrons

Learn how to draw: Iridium 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, the 4p subshell has 6 electrons, the 5s subshell has 2 electrons, the 4d subshell has 10 electrons, the 5p subshell has 6 electrons, the 6s subshell has 2 electrons, the 4f subshell has 14 electrons, and the 5d subshell has 7 electrons.

So the electron configuration of tantalum 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} 4f^{14} 5d^{7}.

Learn how to draw: Iridium orbital diagram

**Next:** Mercury electron configuration

## Related

## More topics

## External links

- https://materials.gelsonluz.com/2019/08/electron-configuration-of-iridium-ir.html
- https://valenceelectrons.com/iridium-electron-configuration/

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.