Iridium orbital diagram

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In the iridium orbital diagram, the 1s subshell accommodates two electrons, the 2s subshell carries another pair, the 2p subshell encompasses six electrons, the 3s subshell contains two electrons, the 3p subshell carries six electrons, the 4s subshell holds two electrons, the 3d subshell accommodates ten electrons, the 4p subshell carries six electrons, the 5s subshell holds two electrons, the 4d subshell accommodates ten electrons, the 5p subshell holds six electrons, the 6s subshell holds two electrons, the 4f subshell carries fourteen electrons, and the 5d subshell has seven electrons, totaling seventy-seven electrons.

When depicting the iridium orbital diagram, start by determining the number of electrons from the periodic table. Use the electron configuration for reference and follow the three fundamental rules: the Aufbau principle, Pauli exclusion principle, and Hund’s rule. This systematic approach ensures an accurate representation of iridium’s orbital arrangement.

Steps

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 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁷.

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 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁷. 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 1s² indicates that the 1s subshell has 2 electrons. So draw two arrows in the 1s box showing two electrons as follows:

2s² indicates that the 2s subshell has 2 electrons. So draw two arrows in the 2s box showing two electrons as follows:

2p⁶ indicates that the 2p subshell has 6 electrons. So draw six arrows in the 2p box showing six electrons as follows:

3s² indicates that the 3s subshell has 2 electrons. So draw two arrows in the 3s box showing two electrons as follows:

3p⁶ indicates that the 3p subshell has 6 electrons. So draw six arrows in the 3p box showing six electrons as follows:

4s² indicates that the 4s subshell has 2 electrons. So draw two arrows in the 4s box showing two electrons as follows:

3d¹⁰ indicates that the 3d subshell has 10 electrons. So draw ten arrows in the 3d box showing ten electrons as follows:

4p⁶ indicates that the 4p subshell has 6 electrons. So draw six arrows in the 4p box showing six electrons as follows:

5s² indicates that the 5s subshell has 2 electrons. So draw two arrows in the 5s box showing two electrons as follows:

4d¹⁰ indicates that the 4d subshell has 10 electrons. So draw ten arrows in the 4d box showing ten electrons as follows:

5p⁶ indicates that the 5p subshell has 6 electrons. So draw six arrows in the 5p box showing six electrons as follows:

6s² indicates that the 6s subshell has 2 electrons. So draw two arrows in the 6s box showing two electrons as follows:

4f¹⁴ indicates that the 4f subshell has 14 electrons. So draw fourteen arrows in the 4f box showing fourteen electrons as follows:

5d⁷ 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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Related

• Periodic table

More topics

• Iridium
• Iridium Bohr model
• Iridium electron configuration