# Question b1dad

Oct 27, 2015

"Fe"^(2+): ["Ar"] 3d^6

#### Explanation:

I'll show you how to write the ground state electron configuration for the iron(II) cation, and leave the second ion to you as practice.

The first thing to do when dealing with the electron configuration of ions is to start by writing the electron configuration of the neutral atom.

In the case of a cation, which is a positively charged ion, doing this will help you determine out of which orbitals the removed electrons will come from.

So, iron is located in period 4, group 8 of the periodic table, and has an atomic nubmer equal to $26$.

This means that a neutral iron atom will have $26$ electrons surrounding its nucleus.

Iron is a transition metal, which means that its occupied 4s-orbital will actually be higher in energy than the 3d-orbitals. Keep this in mind when writing the electron configuration

$\text{Fe: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{6} 4 {s}^{2}$

Using noble gas shorthand notation, you have

"Fe: " ["Ar"] color(white)(x) 3d^6 4s^2

Now, in order for the iron(II) cation to be formed, an iron atom must lose $2$ electrons.

These electrons will come from the orbitals that are highest in energy. In the case of the ${\text{Fe}}^{2 +}$ cation, both electrons will come from the 4s-orbital

"Fe"^(2+):" "["Ar"] 3d^6

Another cool thing about transition metals is that can use the electrons that are outside the nobile gas core as valence electrons. This means that, for example, you can have a ${\text{Fe}}^{3 +}$ cation that will have the electron configuration

"Fe"^(3+): ["Ar"] 3d^5#

Now use the same technique to find the electron configuration of ${\text{S}}^{2 +}$.