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

##### 1 Answer

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*, *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"^(-)#

thesecond 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, **electrons** surrounding its nucleus. These electrons will be placed in *orbitals* in order of **increasing energy** in accordance to the *Aufbau Principle*.

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 *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 **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 **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