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

1 Answer
Jun 26, 2016

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, #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.

http://www.chemguide.co.uk/atoms/properties/atomorbs.html

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