# Question #15a87

Dec 13, 2014

There are a few rules that we can apply to the ground state electron configuration of an element to find what ions it can form.

First it is important to know that the transition elements do not obey the octet rule. So predicting their ions is a little more difficult, but not much more.

Rules for obtaining the electron configuration of Cations:
1. The first electrons lost by an atom or ion are those from the shell with the largest value n ( the outer s and p electrons).
2. Inside a given shell, the electrons are removed from the highest energy subshell before the lower energy subshell is emptied (p emptied before s.)
3. Electrons in the d orbitals are emptied after both s and p are empty since they are the lowest in energy.
4. When the d orbital is present, the s and p orbitals lose all their electrons at once, the d orbitals lose their electrons one at a time, if at all .
5. For transition elements the first electrons lost are from the s. If additional electrons are lost, they come from the lower energy d subshell one at a time.

Example when d orbital is present:
Tin (Sn)
Ground state electron configuration:
$S n \left[K r\right] 5 {s}^{2} 4 {d}^{10} 5 {p}^{2}$

First rewrite in the order of how they will lose their electrons.
$S n \left[K r\right] 4 {d}^{10} 5 {s}^{2} 5 {p}^{2}$

Now we follow our rules to understand the different possible tin ions.

A) $S {n}^{2 +} \left[K r\right] 4 {d}^{10} 5 {s}^{2}$
Notice we followed rule 2 and 4. Since electrons are negatively charged, losing 2 negative charges leaves a 2+ positive charge (since we didn't remove positively charged protons-but that's another lesson entirely.)

B) $S {n}^{4 +} \left[K r\right] 4 {d}^{10}$
Since the full d orbital is pretty stable, it would take alot of energy to keep removing electrons from this point.

Example transition element:

notice: I wrote the electron configuration so that we can just begin removing electrons--skipping the first step shown in the last example

Iron (Fe)
Ground state electron configuration:
$F e \left[A r\right] 3 {d}^{6} 4 {s}^{2}$

A) $F {e}^{2 +} \left[A r\right] 3 {d}^{6}$
Iron loses it's s electrons fairly easily.

B) $F {e}^{3 +} \left[A r\right] 3 {d}^{5}$
Rule 5
Now that the 3d subshell is half filled it is in a decently stable state and would be hard pressed to lose more electrons.