# Calculating the electrons an atom wants to gain/lose to reach a noble gas

I'm trying to understand ions.

From what I understand, an ion is when the atom gains or loses electrons. More electrons means it is negatively charged (anion). Less electrons means it is positively charged (cation).

My problem is with the following premise:

Many atoms gain/lose electrons with the hope of having the same number of electrons as the closest noble gas in the periodic table.

The book uses Potassium as an example. Looking at my table, the closest noble gas to Potassium is Argon, which has $18$ electrons. Therefore Potassium wants to lose one electron...

$$\ce{K^+}$$

Nice, I get that. But then I noticed something odd about all the examples in my book: all the elements they picked are elements conveniently one or two steps away from a noble gas. They never picked an element in the middle of the table like, say, iron.

Iron has $26$ electrons. Its closest noble gas is Argon, which has $18$ electrons. So iron wants to lose $8$ electrons (becoming a cation):

$$\ce{Fe^{8+}}$$

Is this correct? My book literally avoids picking elements in the middle of the table, so I feel there's something weird about this.

Also, I noticed that in every example, the book mentions the Group of the element (Potassium is group 1A, etc...). How is that even relevant? All you have to do is count the number of electrons, no?

• The explanation is simple: the story is that nice and smooth, when you have to loose one or two electrons. Things in the middle don't just lose 8 electrons or get 5 extra. Extra charge means extra energy: an ion with a large charge is actually getting unstable because of the coulomb energy. So whenever ionization happens with the elements in the middle, 1) they loose less electron / smaller formal charge,l 2) they are not purely ionic substances, but stabilized by other mechanism (e.g complexion with water). – Greg Jun 15 '15 at 9:24
• Check out this answer, which goes into some detail about why the octet rule exists, and also why it is more of a "loose guideline" than a rule. – thomij Jun 15 '15 at 17:44

The book's explanation about a noble gas configuration is somewhat accurate, but fairly incomplete.

The elements on the right and on the left of the periodic table (the alkali (earth) metals, the halogens, the chalcogens (the group that starts with Oxygen) and the pnictogens (Nitrogen group)) have electron configurations that make it somewhat easier to lose/gain electrons to "look" like noble gasses (this is why alkali metals and halogens are so reactive: the destabilization from the charge is very minimal compared to the stability of the electron configuration).

However, as you have observed, the book goes to some effort to avoid talking about transition metals. There is a reason for that.

Noble gas configurations are a subset of the stable configurations of electrons. In reality, what's actually being aimed for is an element with no incomplete electron shells.

What is an electron shell? The actual quantum mechanical definition may be a bit more complicated than you need, but for your purposes, it suffices to say that it's how electrons will tend to be distributed around a nucleus. Note that this is extremely distinct from saying that the electron "orbits" the nucleus.

If you look at a periodic table, there are four general regions to look at:

The labels "s", "p", "d", and "f" all refer to the various electron orbitals, and the blocks all indicate what the outermost (or "valence") shell is.

So while elements in the s and p blocks can lose a small number of electrons get complete shells (which will resemble the noble gases), transition metals are a bit more complicated. And by a bit more complicated, I mean that there are very rough rules in place that have almost more exceptions than they have actual examples of following the rule.

But in very general terms, a transition metal will lose electrons. If they are in the first half of the transition metals in their row, they can lose a number of electrons less than or equal to the number of d electrons plus the number of s electrons. This maximum number of electrons lost will make them resemble the noble gasses in configuration, but it's only appreciably stable when the transition metal is in the first half of transition metals in its group ($\ce{Mn^{7+}}$ is well documented in ionic compounds, but $\ce{Fe^{8+}}$ doesn't exist a lot because it has too many electrons to try to lose to comfortably fall into a noble gas configuration). But transition metals can lose electrons up to this point. The issue is, most transition metals can adopt multiple charge configurations for various reasons, so generally, when writing an ionic compound with a transition metal in words, you need to specify the charge of the transition metal (actually, its oxidation state, but in an ionic compound that's mostly splitting hairs). So you can have "Iron (II) Oxide", which is a black powder and "Iron (III) Oxide" which is rust. These compounds will have different chemistries because of the charge on the transition metal.

Suffice to say, your textbook is avoiding transition metals because transition metals are weird.

• Also, I avoided talking about the second half of the transition metal elements: this is because the way the electrons are lost in there are a bit wonkier (as some transition metals in there will do some sincerely weird things to their electron configuration.) The explanation requires some quantum mechanics. – Breaking Bioinformatics Jun 15 '15 at 16:42
• $\ce{Mn^{7+}}$ doesn't exist stably as a bare cation so saying that it is well documented as an example of a noble gas configuration is highly misleading. – bon Jun 15 '15 at 16:49
• I thought a counterion to it would have been implied. Let me edit that in. – Breaking Bioinformatics Jun 15 '15 at 17:03
• I'm not talking about counterions, I'm talking about the fact that manganese in the +7 oxidation state is always strongly bonded to something else. Take $\ce{MnO4-}$ as an example. The manganese 'ion' is strongly bonded to four oxygens and does not exist as a bare $\ce{Mn^{7+}}$ cation. – bon Jun 15 '15 at 17:19
• I think we're meaning the same thing, but talking past each other. When I "counterion" I generally mean "ion of the opposite charge that the first ion is bonded ionically to" – Breaking Bioinformatics Jun 15 '15 at 17:24