Iridium Bohr model

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Iridium Bohr model
Iridium Bohr model

The Bohr model of iridium contains a nucleus having 77 protons and 115 neutrons in the center, and around this nucleus, there are six electron shells containing 77 electrons.

Steps

Here’s how you can draw the Bohr model of iridium step by step.

#1 Find protons, neutrons, and electrons of iridium atom
#2 Draw nucleus of iridium atom
#3 Draw 1st electron shell
#4 Draw 2nd electron shell
#5 Draw 3rd electron shell
#6 Draw 4th electron shell
#7 Draw 5th electron shell
#8 Draw 6th electron shell

Let’s break down each step in detail.

#1 Find protons, neutrons, and electrons of iridium atom

Iridium has 77 protons, 115 neutrons, and 77 electrons.

Iridium protons

  • Protons = atomic number

From the periodic table, find the atomic number of iridium.

Iridium location on periodic table

The atomic number of iridium is 77. Hence, iridium has a total of 77 protons.

Iridium neutrons

  • Neutrons = atomic mass – atomic number
Iridium neutrons
Iridium neutrons

The atomic mass of iridium is 192.217, so we’ll take the roundup value as 192. And the atomic number of iridium is 77.

Subtract the atomic number (77) from the atomic mass (192). Hence, iridium has a total of 192 – 77 = 115 neutrons.

Iridium electrons

  • Electrons = atomic number
Iridium electrons
Iridium electrons

The atomic number of iridium is 77. Hence, iridium has a total of 77 electrons.

#2 Draw nucleus of iridium atom

The nucleus of an iridium atom contains 77 protons and 115 neutrons. So draw the nucleus of iridium atom as follows:

Iridium nucleus
Iridium nucleus

Now in the next step, draw the 1st electron shell and start marking electrons.

#3 Draw 1st electron shell

Remember that we have a total of 77 electrons.

The 1st electron shell (containing s subshell) can hold up to a maximum of 2 electrons. So draw the 1st electron shell as follows:

Iridium shell 1
Iridium 1st electron shell drawn

In the above image, 1 represents the 1st electron shell that contains 1s subshell. And the green color represents the number of electrons in that subshell. This means that the 1st electron shell has a total of 2 electrons.

Since we have already used 2 electrons in the 1st electron shell, now we have 77 – 2 = 75 electrons left. So in the next step, we have to draw the 2nd electron shell.

#4 Draw 2nd electron shell

The 2nd electron shell (containing s subshell and p subshell) can hold up to a maximum of 8 electrons. So draw the 2nd electron shell as follows:

Iridium shell 2
Iridium 2nd electron shell drawn

In the above image, 2 represents the 2nd electron shell that contains 2s and 2p subshells. And the green and orange color represents the number of electrons in that subshell. This means that the 2nd electron shell has a total of 8 electrons.

Now we have already used 10 electrons in 1st and 2nd electron shells, so we have 77 – 10 = 67 electrons left. So in the next step, we have to draw the 3rd electron shell.

#5 Draw 3rd electron shell

The 3rd electron shell (containing s subshell, p subshell, and d subshell) can hold up to a maximum of 18 electrons. So draw the 3rd electron shell as follows:

Iridium shell 3
Iridium 3rd electron shell drawn

In the above image, 3 represents the 3rd electron shell that contains 3s, 3p, and 3d subshells. And the green, orange, and pink color represents the number of electrons in that subshell. This means that the 3rd electron shell has a total of 18 electrons.

Now we have already used 28 electrons in 1st, 2nd, and 3rd electron shells, so we have 77 – 28 = 49 electrons left. So in the next step, we have to draw the 4th electron shell.

#6 Draw 4th electron shell

The 4th electron shell (containing s subshell, p subshell, d subshell, and f subshell) can hold up to a maximum of 32 electrons. So draw the 4th electron shell as follows:

Iridium shell 4
Iridium 4th electron shell drawn

In the above image, 4 represents the 4th electron shell that contains 4s, 4p, 4d, and 4f subshells. And the green, orange, pink, and blue color represents the number of electrons in that subshell. This means that the 4th electron shell has a total of 32 electrons.

Now we have already used 60 electrons in 1st, 2nd, 3rd, and 4th electron shells, so we have 77 – 60 = 17 electrons left. So in the next step, we have to draw the 5th electron shell.

#7 Draw 5th electron shell

The 5th electron shell can hold up to a maximum of 50 electrons. So draw the 5th electron shell as follows:

Iridium shell 5
Iridium 5th electron shell drawn

In the above image, 5 represents the 5th electron shell that contains 5s, 5p, and 5d subshells. And the green, orange, and pink color represents the number of electrons in that subshell. This means that the 5th electron shell has a total of 15 electrons.

The 5th electron shell contains only 5s, 5p, and 5d subshells, and not a 5f subshell. This is because according to the aufbau principle, the 6s subshell is filled first and then 4f, 5d, 6p… and so on.

Now we have already used 75 electrons in 1st, 2nd, 3rd, 4th, and 5th electron shells, so we have 77 – 75 = 2 electrons left. So in the next step, we have to draw the 6th electron shell.

#8 Draw 6th electron shell

The 6th electron shell can hold up to a maximum of 72 electrons. So draw the 6th electron shell as follows:

Iridium shell 6
Iridium 6th electron shell drawn

In the above image, 6 represents the 6th electron shell that contains 6s subshell. And the green color represents the number of electrons in that subshell. This means that the 6th electron shell has a total of 2 electrons.

That’s it! This is the final Bohr model of iridium atom as we have used all 77 electrons: 2 electrons in the 1st electron shell, 8 electrons in the 2nd electron shell, 18 electrons in the 3rd electron shell, 32 electrons in the 4th electron shell, 15 electrons in the 5th electron shell, and 2 electrons in the 6th electron shell.

Next: Platinum Bohr model

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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.

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