Question

The atom cobalt has 27 electrons. How many energy levels will its electrons use?

Answer 1

Four energy levels.

Explanation:

The number of electrons each energy level can hold increases as you add more and more energy levels to an atom.

The relationship that exists between the energy level, `n`, and the number of electrons it can hold can be written like this

`color(blue)(|bar(ul(color(white)(a/a)"no. of e"^(-) = 2n^2color(white)(a/a)|)))`

You can use this equation to find the maximum number of electrons that can be added to each energy level. You will have

• the first energy level, `n=1`

`"no. of e"^(-) = 2 * 1^2 = "2 e"^(-)`

• the second energy level `n=2`

`"no. of e"^(-) = 2 * 2^2 = "8 e"^(-)`

• the third energy level, `n=3`

`"no. of e"^(-) = 2 * 3^2 = "18 e"^(-)`

• the fourth energy level, `n=4`

`"no. of e"^(-1) = 2 * 4^2 = "32 e"^(-)`

and so on.

In your case, cobalt, `"Co"`, is said to have a total of `27` electrons surrounding its nucleus. These electrons will be placed in orbitals in order of increasing energy in accordance to the Aufbau Principle.

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Now, it's very important to remember that when you're adding electrons to an atom, the 3d-orbitals, which are located on the third energy level, are higher in energy than the 4s-orbital.

This means that you must fill the 4s-orbital first, then distribute the rest of the electrons to the 3d-orbitals.

So, a neutral cobalt atom will have

`n=1 -> "2 e"^(-)` in the `1s` subshell

`n=2 -> "8 e"^(-)` in the `2s` and `2p` subshells

Now, these two energy levels will hold

`"2 e"^(-) + "8 e"^(-) = "10 e"^(-)`

Now comes the tricky part. The third energy level can hold `"18 e"^(-)`, so in theory it can hold the remaining

`"27 e"^(-) - "10 e"^(-) = "17 e"^(-)`

that the neutral cobalt atom has. You could thus say that

`color(red)(cancel(color(black)(n=3 -> "17 e"^(-))))` in the `3s`, `3p`, and `3d` subshells

and conclude that the electrons that surround the nucleus of a cobalt atom are spread out on `3` energy levels. You would be wrong.

Taking it one subshell at a time, you will have

`"2 e"^(-) ->` in the `3s` subshell

`"6 e"^(-) ->` in the `3p` subshell

You now have

`"17 e"^(-) - ("2 e"^(-) + "6 e"^(-)) = "9 e"^(-)`

to distribute. Because the 4s orbital is filled before the 3d-orbitals, the next two electrons are going to be distributed on the fourth energy level

`n=4 -> "2 e"^(-)` in the `4s` subshell

The remaining `"7 e"^(-)` will now be distributed in the 3d-subshell.

Therefore, a neutral cobalt atom will have

`n=1 -> "2 e"^(-)` in the `1s` subshell

`n=2 -> "8 e"^(-)` in the `2s` and `2p` subshells

`n= 3 -> "15 e"^(-)` in the `3s`, `3p`, and `3d` subshells

`n=4 -> "2 e"^(-)` in the `4s` subshell

![http://www.slideshare.net/crumpjason/electron-configuration-5788473]()