The orbital diagram of iridium 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.

## Steps

Here’s how you can draw the orbital diagram of iridium step by step.

#1 Find electrons of iridium

#2 Write electron configuration of iridium

#3 Draw orbital diagram of iridium

Let’s break down each step in detail.

### Find electrons

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.

### Write electron configuration

The 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}.

Learn how to find: Iridium electron configuration

Now in the next step, start drawing the orbital diagram for iridium.

### Draw orbital diagram

Before drawing the orbital diagram, you should know the three general rules.

- Aufbau principle – electrons are first filled in lowest energy orbital and then in higher energy orbital
- Pauli exclusion principle – two electrons with the same spin can not occupy the same orbital
- Hund’s rule – each orbital should be first filled with one electron before being paired with a second electron

Also, you should know the number of orbitals in each subshell.

We can calculate the number of orbitals in each subshell using the formula: 2ℓ + 1

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

So each s subshell has one orbital, each p subshell has three orbitals, each d subshell has five orbitals, and each f subshell has seven orbitals.

Now start to draw!

As mentioned above, the 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}. Hence, draw the blank orbital diagram of iridium up to 5d subshell as follows:

In the above orbital diagram, the box represents an orbital. Each orbital has a capacity of two electrons. And the arrows (↑↓) are drawn inside the box to represent electrons.

Now 1s^{2} indicates that the 1s subshell has 2 electrons. So draw two arrows in the 1s box showing two electrons as follows:

2s^{2} indicates that the 2s subshell has 2 electrons. So draw two arrows in the 2s box showing two electrons as follows:

2p^{6} indicates that the 2p subshell has 6 electrons. So draw six arrows in the 2p box showing six electrons as follows:

3s^{2} indicates that the 3s subshell has 2 electrons. So draw two arrows in the 3s box showing two electrons as follows:

3p^{6} indicates that the 3p subshell has 6 electrons. So draw six arrows in the 3p box showing six electrons as follows:

4s^{2} indicates that the 4s subshell has 2 electrons. So draw two arrows in the 4s box showing two electrons as follows:

3d^{10} indicates that the 3d subshell has 10 electrons. So draw ten arrows in the 3d box showing ten electrons as follows:

4p^{6} indicates that the 4p subshell has 6 electrons. So draw six arrows in the 4p box showing six electrons as follows:

5s^{2} indicates that the 5s subshell has 2 electrons. So draw two arrows in the 5s box showing two electrons as follows:

4d^{10} indicates that the 4d subshell has 10 electrons. So draw ten arrows in the 4d box showing ten electrons as follows:

5p^{6} indicates that the 5p subshell has 6 electrons. So draw six arrows in the 5p box showing six electrons as follows:

6s^{2} indicates that the 6s subshell has 2 electrons. So draw two arrows in the 6s box showing two electrons as follows:

4f^{14} indicates that the 4f subshell has 14 electrons. So draw fourteen arrows in the 4f box showing fourteen electrons as follows:

5d^{7} indicates that the 5d subshell has 7 electrons. So draw seven arrows in the 5d box showing seven electrons as follows:

That’s it! This is the final orbital diagram of iridium as we have used all 77 electrons.

**Next:** Platinum orbital diagram

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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.