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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?

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well, make a question implies that you must be prepared for the answers... I must to say that to me is very difficult to select the best answer. all the answers have a great effort and each have very detailed information, thanks to all for the dedication on this question for all. –  moon.watcher May 29 '12 at 23:58
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Accept whichever one helped you most. Whichever one was at your level of understanding, and made you understand the most about the system. If you're still confused, accept the most-upvoted answer :) –  ManishEarth Jun 8 '12 at 12:40
    
2*1=2,2*(1+3)=8,2*(1+3+5)=18,2*(1+3+5+7)=32... –  user26143 Feb 8 at 14:08

6 Answers 6

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):

salt and oxygen in donation notation

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):

calcium sulfate and sulfite in donation notation

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:

sulfur hexafluoride in donation notation

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.

2014-04-04 Update

Finally, if you are tired of dots and arrows and yearn for something closer to standard valence bond notation, these two equivalences come in handy:

covalent and u-bond versions of donation notation

The upper straight-line equivalence is trivial, since the resulting line is identical in appearance and meaning to the standard covalent bond of organic chemistry.

The second u-bond notation is the novel one. I invented it out of frustration in high school back in the 1970s (yes I'm that old), but never did anything with it at the time.

The main advantage of u-bond notation is that it lets you prototype and assess non-standard bonding relationships while using only standard atomic valences. As with the straight-line covalent bond, the line that forms the u-bond represents a single pair of electrons. However, in a u-bond it is the atom at the bottom of the U that donates both electrons in the pair. That atom gets nothing out of the deal, so none of its bonding needs are changed or satisfied. This lack of bond completion is represented by the absence of any line ends on that side of the u-bond.

The beggar atom at the top of the U gets to use both of the electrons for free, which in turn means that two of its valence-bond needs are met. Notationally, this is reflected by the fact that both of the line ends of the U are next to that atom.

Taken as a whole, the atom at the bottom of a u-bond is saying "I don't like it, but if you are that desperate for a pair of electrons, and if you promise to stay very close by, I'll let you latch onto a pair of electrons from my already-completed octet."

Carbon monoxide with its baffling "why does carbon suddenly have a valence of two?" structure nicely demonstrates how u-bonds interpret such compounds in terms of more traditional bonding numbers:

carbon monoxide in u-bond notation

Notice that two of carbon's four bonds are resolved by standard covalent bonds with oxygen, while the remaining two carbon bonds are resolved by the formation of a u-bond that lets the beggar carbon "share" one of the electron pairs from oxygen's already-full octet. Carbon ends up with four line ends, representing its four bonds, and oxygen ends up with two. Both atoms thus have their standard bonding numbers satisfied.

Another more subtle insight from this figure is that since a u-bond represents a single pair of electrons, the combination of one u-bond and two traditional covalent bonds between the carbon and oxygen atoms involves a total of six electrons, and so should have similarities to the six-electron triple bond between two nitrogen atoms. This small prediction turns out to be correct: nitrogen and carbon monoxide molecules are in fact electron configuration homologues, one of the consequences of which is that they have nearly identical physical chemistry properties.

Below are a few more examples of how u-bond notation can make anions, noble gas compounds, and odd organic compounds seem a bit less mysterious:

calcium sulfate in u-bond notation

xenon tetroxide in u-bond notation

xenon trioxide in u-bond notation

fulminic acid in u-bond notation

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Terry, may I please buy you lunch? –  Chris K Jun 17 '12 at 22:21
    
Chris, sure, I appreciate the offer. I'll likely be in San Diego sometime in the next few months, visiting Navy folks, so I may actually be able to take you up on it! –  Terry Bollinger Jun 18 '12 at 1:00
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I hope a science cartoonist sees the bit about the flourine bullies –  DarenW Nov 11 '12 at 1:16
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I regret to have to make a critical comment about such a highly rated answer, but this is not an answer to the question, but rather a missive on an alternative graphical representation of resonance structures. –  Eric Brown May 31 at 23:31
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Eric my man, this is humiliating: You've only down voted 5 of my 11 chemistry answers! I mean, what must I do to get a grand slam of all eleven?? –  Terry Bollinger Jun 5 at 4:50

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:

enter image description here

(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.

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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.


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, PO43–, SO42–, … Other elements include I (e.g. in IF7), Si (Ph3SiCl), Xe (XeF4) and Cl (ClF5).

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This may be a definition issue if the asker is in high school or a recent graduate. The first three current editions of high school text books I pulled from the shelf (AP and beginning chemistry) use the definition for valence electrons as "electrons in the highest occupied principal energy level". –  Janice DelMar May 19 '12 at 0:54
    
Note that the 18electron/EAN rule isn't always followed.. Paramagnetic, octahedral complexes never follow it. They can't. Neither can tetrahedral/square planar complexes. These are usually still superoctet, though. –  ManishEarth May 19 '12 at 1:03
    
@ManishEarth I'm very worried about some of the answers given on SE that cover electronic structure concepts. I am wondering if it would be useful to start a meta discussion on how to answer "why" questions from 1900's chemical bonding theory -- should the answer be in terms of the old chemical rules or in terms of quantum mechanics? –  Eric Brown May 31 at 23:36
    
There are some molecules with helium. For example Helium Hydride. –  caters Aug 13 at 14:21

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.

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(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."

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Something worth adding to this discussion that I'm surprised hasn't been mentioned about such "hypervalent" molecules like SF6

One of my professors at university informed me that the common explanation (that the empty d-orbitals are empty and are thus accessible) is actually most likely incorrect. This is an old-model explanation that is out of date, but is for some reason continuously taught in schools. (a quote from the wiki article: "In 1990, Magnusson published a seminal work definitively excluding the role of d-orbital hybridization in bonding in hypervalent compounds of second-row elements.")

When you actually look at the numbers (which I unfortunately do not have available to me at the moment, and am too lazy to look up), the energy associated with those orbitals is significantly higher than the bonding energy found experimentally within molecules like SF6, meaning that it is highly unlikely that the d-orbitals are involved at all in this type of molecular structure.

This leaves us stuck, in fact, with the octet rule. Since S cannot reach into it's d-orbitals, it cannot have more than 8 electrons in its valence (see other discussions on this page for definitions of valence etc, but by the most basic definition, yes, only 8). The common explanation is the idea of a 3-centered 4-electron bond, which is (most) basically the idea that Sulfur and two F at 180 degrees apart share only 4 e- between their Molecular Orbitals. One way of comprehending this is to consider a pair of resonance structures where sulfur is bonded covalently to one F and ionically to the other something like:

F:- +S --- F <-----> F --- S+ -:F

When you average these two structures out, you will notice that Sulfur maintains a positive charge, and that each fluoride has a sort of "half" charge. Also, note that Sulfur only has two electrons associated with it in both structures, meaning that it has successfully bonded to two fluorines while only accumulating 2 electrons. The reason they have to be at 180 degrees is due to the geometry of the molecular orbitals, which I don't really feel like getting into at the moment (a little off topic).

So, just to review, we've bonded to two Fluorines to the Sulfur accumulating two electrons and 1 positive charge on Sulfur. If we bonded the remaining 4 F from SF6 in the normal covalent way, we'd end up with still 10 e-, so by utilizing another 3 center 4 electron bond pair, we achieve 8 electrons (filling both the valence s and p orbitals) as well as a +2 charge on the sulfur and a -2 charge distributed around the four fluorines involved in the 3c4e- bonding. (of course, all of the fluorines have to be equivalent, so that charge will actually be distributed around all of the fluorines if you consider all of the resonance structures).

There actually is a lot of evidence to support this style of bonding, the simplest of which is observed by looking at bond lengths in molecules such as ClF3 (T-Shape geometry), where the the two F 180 degrees apart (again, MO stuff I don't want to talk about) have a slightly longer bond length than the other F, indicating a weakened order of covalence in those two F bonds (a result of averaging out a covalent and ionic bond).

TL;DR Hypervalency doesn't really exist, and having more than 8 e- in non-transition metals is much harder than you would think.

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