Can an atom have more than 8 valence electrons? If not, why is 8 the limit?

Question

According to some chemistry textbooks, the maximum number of valence electrons for an atom is 8, but the reason for this is not explained.

So, can an atom have more than 8 valence electrons?

If this is not possible, why can't an atom have more than 8 valence electrons?

Answer 1

This answer is intended to supplement Manishearth's earlier answer, rather than compete with it. My objective is to show how octet rules can be helpful even for molecules that contain more than the usual complement of eight electrons in their valence shell.

I call it donation notation, and it dates back to my high school days when none of the chemistry of the texts in my small-town library bothered to explain how those oxygen bonds worked in anions such as carbonate, chlorate, sulfate, nitrate, and phosphate.

The idea behind this notation is simple. You begin with the electron dot notation, then add arrows that show whether and how other atoms are "borrowing" each electron. A dot with an arrow means that that the electron "belongs" mainly to the atom at the base of the arrow, but is being used by another atom to help complete that atom's octet. A simple arrow without any dot indicates that the electron has effectively left the original atom. In that case the electron is no longer attached to the arrow at all, but is instead shown as an increase in the number of valence electrons in the atoms at the end of the arrow.

Here are examples using table salt (ionic) and oxygen (covalent):

Notice that the ionic bond of $\ce{NaCl}$ shows up simply as an arrow, indication that it has "donated" its outermost electron and fallen back to its inner octet of electrons to satisfy its own completion priorities. (Such inner octets are never shown.)

Covalent bonds happen when each atom contributes one electron to a bond. Donation notation shows both electrons, so doubly bonded oxygen winds up with four arrows between the atoms.

Donation notation is not really needed for simple covalent bonds, however. It's intended more for showing how bonding works in anions. Two closely related examples are calcium sulfate ($\ce{CaSO4}$, better known as gypsum) and calcium sulfite ($\ce{CaSO3}$, a common food preservative):

In these examples the calcium donates via a mostly ionic bond, so its contribution becomes a pair of arrows that donate two electrons to the core of the anion, completing the octet of the sulfur atom. The oxygen atoms then attach to the sulfur and "borrow" entire electrons pairs, without really contributing anything in return. This borrowing model is a major factor in why there can be more than one anion for elements such as sulfur (sulphates and sulfites) and nitrogen (nitrates and nitrites). Since the oxygen atoms are not needed for the central atom to establish a full octet, it is possible for some of the pairs in the central octet to remain unattached. This results in less oxidized anions such as sulfites and nitrites.

Finally, a more ambiguous example is sulfur hexafluoride:

The figure shows two options. Should $\ce{SF6}$ be modeled as if the sulfur is a metal that gives up all of its electrons to the hyper-aggressive fluorine atoms (option a), or as a case where the octet rule gives way to a weaker but still workable 12-electron rule (option b)? There is some controversy even today about how such cases should be handled. The donation notation shows how an octet perspective can still be applied to such cases, though it is never a good idea to rely on first-order approximation models for such extreme cases.

Answer 2

In chemistry, and in science in general, there are many ways of explaining the same empirical rule. Here, I am giving an overview that is very light on quantum chemistry: it should be fairly readable at a novice level, but will not explain in its deepest way the reasons for the existence of electronic shells.

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The “rule” you are citing is known as the octet rule, and one of its formulations is that

atoms of low (Z < 20) atomic number tend to combine in such a way that they each have eight electrons in their valence shells

You'll notice that it's not specifically about a maximal valence (i.e. number of electron in the valence shell), but a preferred valence in molecules. It is commonly used to determine the Lewis structure of molecules.

However, the octet rule is not the end of the story. If you look at hydrogen (H) and helium (He), you will see that do not prefer an eight-electron valence, but a two-electron valence: H forms e.g. H₂, HF, H₂O, He (which already has two electrons and doesn't form molecules). This is called the duet rule. Moreover, heavier elements including all transition metals follow the aptly-named 18-electron rule when they form metal complexes. This is because of the quantum nature of the atoms, where electrons are organized in shells: the first (named the K shell) has 2 electrons, the second (L shell) has 8, the third (M shell) has 18. Atoms combine into molecules by trying in most cases to have valence electrons entirely filling a shell.

Finally, there are elements which, in some chemical compounds, break the duet/octet/18-electron rules. The main exception is the family of hypervalent molecules, in which a main group element has more than 8 electrons in its valence shell. Phosphorus and sulfur are most commonly prone to form hypervalent molecules, including PCl₅, SF₆, PO₄^3–, SO₄^2–, … Other elements include I (e.g. in IF₇), Si (Ph₃SiCl), Xe (XeF₄) and Cl (ClF₅).

Answer 3

Yes, it can. We have molecules which contain "superoctet atoms". Examples:

$\ce{PBr5, XeF6, SF6, HClO4, Cl2O7, I3- , K4[Fe(CN)6], O=PPh3 }$

Almost all coordination compounds have a superoctet central atom.

Nonmetals from Period 3 onwards are prone to this as well. The halogens, Sulphur, and Phosphorous are repeat offenders, while all noble gas compounds are superoctet. Thus sulphur can show +6 valency, Phosphorous shows +5, and the halogens show +1, +3, +5, and +7 valency. Note that these are still covalent compounds--valency applies to covalent bonds as well.

The reason why this isn't usually seen is as follows. We basically deduce it from the properties of atomic orbitals.

By the Aufbau principle, electrons fill up in these orbitals for period $n$:

$ns, (n-2)f,(n-1)d,np$

(theoretically, you'd have $(n-3)g$ before the $f$, and so on. But we don't have atoms with those orbitals, yet)

Now, the outermost shell is $n$. In each period, there are only eight slots to fill in this shell by Aufbau Principle- 2 in $ns$, and 6 in $np$. Since our periodic table pretty much follows this principle, we don't see any superoctet atoms usually.

But, the $d,f$ orbitals for that shell still exist (as empty orbitals) and can be filled if the need arises. By "exist", I mean that they are low enough in energy to be easily filled. The examples above consist of a central atom, that has taken these empty orbitals into its hybridization, giving rize to a superoctet species(since the covalent bonds add an electron each)

I cooked up a periodic table with the shells marked. I've used the shell letters instead of numbers to avoid confusion. $K,L,M,N$ refer to shell 1,2,3,4 etc. When a slice of the table is marked "M9-M18", this means that the first element of that block "fills" the ninth electron in the M(third) shell, and the last element fills the eighteenth.

Click to enlarge:

_{(Derivative of this image)}

Note that there are a few irregularities, wuth $\ce{Cu}$, $\ce{Cr}$, $\ce{Ag}$, and a whole bunch of others which I've not specially marked in the table.

Answer 4

This question may be difficult to answer because there are a couple of definitions of valence electrons. Some books and dictionaries define valence electrons as "outer shell electrons that participate in chemical bonding" and by this definition, elements can have more than 8 valence electrons as explained by F'x.

Some books and dictionaries define valence electrons as "electrons in the highest principal energy level". By this definition an element would have only 8 valence electrons because the $n-1$ $d$ orbitals fill after the $n$ $s$ orbitals, and then the $n$ $p$ orbitals fill. So, the highest principal energy level, $n$, contains the valence electrons. By this definition, the transition metals all have either 1 or 2 valence electrons (depending on how many electrons are in the $s$ vs. $d$ orbitals).

Examples:

• Ca with two $4s$ electrons would have two valence electrons (electrons in the 4th principal energy level).
• Sc with two $4s$ electrons and one $3d$ electron will have two valence electrons.
• Cr with one $4s$ electron and five $3d$ electrons will have one valence electron.
• Ga with two $4s$ electrons, ten $3d$ electrons, and one $4p$ electron would have three valence electrons.

By the other definition, they could have more since they have more "outer shell" electrons (until the $d$ shell is filled).

Using the "highest principal energy level" definition for valence electrons allows you to correctly predict the paramagnetic behavior of transition metals ions because valence electrons (the $d$ electrons) are lost first when a transition metal forms an ion.

Answer 5

(Please accept other answers if you are looking for homework help. My answer is for people who have mastered freshman chemistry rules, and think it's a load of garbage.)

There is a big difference between a "rule" and a law of nature. The "Octet Rule" is a turn-of-the-last-century concept that somehow managed to get into introductory chemistry books and never got kicked out with the advent of modern quantum mechanics. (Circumstantial Proof: it is impossible to identify individual electrons to label them "valence" or "not valence")

Therefore, you won't find any answer based on physical evidence as to why/why not a rule that is not based on physical evidence will hold.

Atoms take their spatial configuration because it happens to be an electrostatically-favorable circumstance, not because electrons avail themselves like "slots."