Exceptions to the octet rule
While most atoms obey the duet and octet rules, there are some exceptions. For example, elements such as boron or beryllium often form compounds in which the central atom is surrounded by fewer than eight electrons (e.g., BF₃ or BeH₂). In contrast, many elements in the third-row and beyond have been observed to exceed the octet rule, forming compounds in which the central atom is surrounded by more than eight electrons (e.g. ICl₄⁻ or PF₅). Created by Sal Khan.
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- Why doesn't Aluminium form an ionic bond with the Hydrogens instead? It would satisfy the octet rule if it did this, and wouldn't it make it more stable?(6 votes)
- For an ionic bond to occur, one atom has to take the electrons away from another atom forming a cation/anion pair. The atom which takes the electrons in an ionic bond is the more electronegative atom since electronegativity measures how much an atom of an element desires electrons.
So for aluminum hydride (AlH3), the more electronegative atom is the hydrogen with a value of 2.20 on the Pauli scale compared to aluminum which has a value of 1.61 on the same scale. So if it were ionic, hydrogen would need to be the one taking the electrons from aluminum, but it is unable to do that because the electronegativity difference between the atom is not large enough. Hydrogen simply doesn't have enough of a pull over aluminum's electrons for the two atoms to engage in ionic bonding. Their electronegativity difference is 0.59 which would put their bonding in the polar covalent range.
If you're interested in predicting what type of bonding would occur between elements, you should look up the van Arkel-Ketelaar Triangle.
Hope that helps.(14 votes)
- How do we know if a bond is ionic or covalent? At2:37, Sal draws the ionic structure of AlH3 (The octet rule is not satisfied here). Why can't it be an ionic bond where aluminium (Al) gives its 3 electrons to the 3 hydrogens (H)?(8 votes)
- This is quite long and may look a bit intimidating but it is very simple!
Could be because the difference in electronegativity between H and Al is not very high. If their difference is large, we'd get an ionic bond, else we'd get a covalent bond (not a metallic bond as this ain't a clump of Al atoms),
(remember Al is sometimes considered both metal and non-metal ~ you know, a metalloid)
You can also think of it like this,
Al would have to lose 3 v.e- to achieve an octet(or Ne Electronic config.)=13 proton holding onto just 8 v.e-, and Al is quite a small atom, therefore it will have a moderate electronegativity (not too less like in metals),
this wouldn't be very stable, and why do atoms react?
To achieve stability, hence a covalent bond could be more probable.
(here, Al is a less extreme version of Boron-which gets its bizarre properties due to its small size = moderately high electronegativity but not high enough to gain 5 e-)
This is just my analogy,
if wrong, please do correct,
Hope this helps,
One more thing: ionic bonds can exhibit covalent properties (partial charges == polar) and ionic compounds can exhibit covalent characteristics too!
Here, the electronegativity of H = 2.20 and Al = 1.61, Δelectronegativity = 0.59 (all on the Pauli scale) as it is between o.5 and 1.9 Pauling units, AlH3 would be a polar covalent bond with Al gaining a partial +ve charge and the hydrogens a partial -ve charge which will sum up to zero as AlH3 as a whole is electrically neutral)
If this goes over your head, don't worry about it, thought it would be helpful to bring the point home :)
- Okay, this all makes sense, but how can we actually predict an exception to the octet in covalent bonds? I assume there would be some way to determine the likelyhood of an atom forming an unusual number of bonds without experimentation, right? Maybe using electronegativity?(6 votes)
- So probably the best way of predicting bond orders is through a method called molecular orbital theory. It's a different theory of bonding compared to valance bond theory which is what is seen here. Without getting too much into the details it allows us to calculate the number of bonds before testing it in experiments. So there is a way to predict exceptions to the octet rule, but it is fairly complicated.
Hope that helps.(4 votes)
- At3:50, Sal said Xenon Pentaflouride is a cation. How do we know the entire molecule has a +1 charge, instead of each flourine having a +1 charge?(1 vote)
- I'm a little confused about your question, but fluorine has a -1 charge as an anion, which is most often the fluorine ion you will encounter (a fluorine cation can be formed, but this would require a prohibitively high amount of energy to remove the electron).
Remember that Lewis diagrams are for molecules with covalent bonds, not ionic bonds (which is what I believe you meant by fluorine's charge). Essentially, covalent bonding is between two non-metal atoms (like Xe and F) and electrons are shared, while ionic bonding is between metal and non-metal ions, where the electron is taken by the (typically) more electronegative non-metal atom to form an anion (with the metal atom losing this electron and becoming a cation). See below for more info on the difference between the two.
Since these are covalent bonds between atoms, fluorine can't have any charge. As for why xenon pentafluoride is a cation here, that is simply because it has had an electron removed from the molecule, somewhere. There are no ions involved, just atoms, so the XeF5 molecule is a cation because an electron was removed. I hope this helps!
For more information on covalent vs. ionic bonding, this is a good place to start: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Covalent_Bonds_vs_Ionic_Bonds.
This is a bit higher-level but is more comprehensive: https://www.chemguide.co.uk/atoms/bonding/covalent.html
This link has some good visualizations: https://courses.lumenlearning.com/introchem/chapter/comparison-between-covalent-and-ionic-compounds/(8 votes)
- How do you know how many valence electrons are in the transition elements?(2 votes)
- It’s better to not worry about this. Transition metals do not behave like the main group elements, “valence electrons” do not determine how they react(5 votes)
- How would we be able to draw Lewis Structures if the number of electrons calculated was odd?(1 vote)
- You would simply have an unpaired electron. An atom/molecule/ion which has an unpaired electron like that is known as a radical. As the name suggests, it is highly reactive because it wants to find another electron to pair with the unpaired one. In a Lewis structure we represent that as a single dot in a similar location as a lone pair would go.
Hope that helps.(5 votes)
- Why did he put the 2 remaining electrons on Xe instead of on one of the Fluorine?(1 vote)
- Fluorine is a second period element which can only contain a maximum of 8 valence electrons. Xenon meanwhile can have an expanded octet and can hold more than 8 electrons. Basically any of the fluorine atoms wouldn't be able to hold onto an extra 2 electrons which only leaves the central xenon atom with the ability to.
Hope that helps.(4 votes)
- In the Molecule XeF5+, sal counts the VE and gets 43-1=42. Wouldn't the Plus account for all 5 of the florines to give you 43-5=38? Or how do you tell how many florines have the cation?(1 vote)
- The positive charge in the XeF^(+) formula applies to the entire ion, not a single atom in the ion. Sal shows this at6:18.
Hope that helps.(3 votes)
- Where does that plus one charge come from in xenon pentaflouride?(2 votes)
- If there were as many electrons as protons for all the atoms of the molecule, then it would have an overall 0 charge and be neutral. Since it has a +1 charge, then there is 1 less electron compared to if it was neutral. When Sal draws out the Lewis structure we can see that it is the Xenon which is missing an electron giving it a +1 charge if we work out the formal charge on it.
Hope that helps.(1 vote)
- How do we know when an atom (like Al, B, or on the other hand, stuff with d and f orbitals) will break the octet rule?
Also what makes it favorable for Al to form only 3 bonds? I tried looking at its electron configuration diagram, and also tried writing the hybridized electron configuration since it looks like Al has sp2 hybridization here, but it didn't help me in finding out that it is favorable to make 3 bonds.(1 vote)
- The octet rule is a general rule that only applies to main group elements (groups 1-2 & 13-18), but of course has plenty of exceptions within those groups. Most of chemistry is learning these simple rules and later discovering that actually it's a lot more complicated and full of subtle nuances.
Anyhow within the main group elements, exceptions to this rule include the first period elements; hydrogen and helium. This is because they follow a duet rule by reaching stability filling their 1s subshell with two electrons. Group 13 elements like boron and aluminium also breaks the octet rule by forming compounds which give them only six valence electrons and I'll explain this later. Period 3 elements can also deviate by forming compounds which give them more than eight electrons such as with phosphorus pentachloride. This is because elements here fill the third electron shell which contains d orbitals and can exceed the eight electron limit second period elements are restricted to. Transition metal elements follow an 18-electron rule while the lanthanides and actinides follow the 32-electron rule. So we can see when viewing the entire periodic table that the octet rule accurately predicts a small selection of elements; most the second and third period elements.
Group 13 elements like boron and aluminium form compounds with three single bonds giving them six valence electrons mainly because of their electron configurations. Boron for example has an electron configuration of [He]2s^(2)2p^(1). When it would like to bond with something, it'll promote one the 2s electrons to the 2p subshell and hybridize the 2s orbital and two 2p orbitals which have electrons in them creating three sp2 orbitals. It doesn't have the electrons available like carbon does to form four bonds (through four sp3 hybrid orbitals) or lone pairs like nitrogen or oxygen to give it that octet. Instead it's limited to having an incomplete octet by the amount of electron it begins with. These sort of compounds leave the central atom electron deficient.
That said, boron and other group 13 elements can attain stability by other means once they've formed these trigonal planar compounds with three single bonds. Boron trifluoride or boron trichloride for example use something called backbonding to create stability. Backbonding is when an electron-rich atom in a bond donates some electron density to the electron-poor atom. In these boron compounds the central boron atom has an unfilled unhybridized p orbital perpendicular to the plane containing the atoms. The halogens have lone pairs and these lone pairs can backbond into the unhybridized p orbital of the boron making it more stable. This makes the boron 'feel' more like it has an octet.
The main way though these group 13 elements attain an octet is by acting like Lewis acids and accepting lone pairs from Lewis bases. A Lewis acid is a chemical which is likely to accept a lone pair, and a Lewis base is a chemical likely to donate a lone pair. A compound like the boron ones I listed above are good Lewis acids since they are electron deficient and have a strong desire to gain two more electrons (following the octet rule). For example boron trichloride (Lewis acid) could react with ammonia (Lewis base) where the boron accepts the lone pair from the nitrogen in ammonia thus gaining an octet for the boron. Another boron compound is borane (BH3) which is where a boron is bonded to three hydrogen atoms. This is an especially strong Lewis acid since it can't do the backbonding that boron trifluoride or boron trichloride does to stabilize it (since hydrogen has no lone pairs). It's such a reactive compound that borane molecules quickly react with each other to form diborane (which has these interesting bridging hydrogens) to fix their mutual electron deficiency problem.
So group 13 elements form these three bond compounds not necessarily because they are the most stable, but because they are limited to them due to their available electrons. Once they have formed they are often reactive trying to attain that full octet by gaining two additional electrons for the central group 13 atom.
Hope that helps.(2 votes)
- [Instructor] In this video, we're gonna start talking about exceptions to the octet rule, which we've talked about in many other videos. The octet rule is this notion that atoms tend to react in ways that they're able to have a full outer shell. They're able to have eight valence electrons. Now, we've already talked about some exceptions, things like hydrogen. Its outer shell is that first shell which gets full with two electrons. So it's trying to get to that duet rule. But as we'll see, there are other exceptions. Boron and aluminum, for example, they can form stable molecules where the boron or the aluminum only have six valence electrons, not eight. And there are exceptions in the other direction. As you get to the third period and beyond, we'll actually see atoms that can maintain more than eight valence electrons. And we're actually going to see an example of that with xenon. So let's just go into a few examples. Given what I've told you, see if you can come up with the Lewis diagram for aluminum hydride. So aluminum hydride has one aluminum and three hydrogens. See if you can draw the Lewis diagram for that. All right, now let's do this together. So the first thing you wanna do is account for all of the valence electrons. Aluminum's outer shell is the third shell, so the third period here, and it has one, two, three valence electrons. And then we have three hydrogens, and each hydrogen has one valence electron. And so you add all of this up together, three plus three is equal to six valence electrons in aluminum hydride. Now, the next step after that is to try to draw the structure with some covalent bonds. We don't wanna make hydrogen our central atom. That would be very atypical. And so let's put aluminum in the center. And then we're gonna have three hydrogens. So one, two, and three. And then let's put some covalent bonds in here. And so let's see, how many valence electrons have we now accounted for? This is two in this covalent bond. Another two gets us to four. Another two gets us to six. So we have just accounted for all six valence electrons. So we have no more valence electrons to play with. Let's think about how the various atoms are doing. So the hydrogens are all meeting their duet rule. These two electrons in this bond are hanging around hydrogen and around the aluminum. But from hydrogen's point of view, it has a full duet, that hydrogen as well and that hydrogen as well. But notice the aluminum over here, it has two, four, six electrons, valence electrons around it, and so it's not a full octet. But aluminum hydride is actually something that has been observed. Let's think about another example. Let's think about xenon pentafluoride. Xenon pentafluoride cations, a positively charged ion here. Pause this video and see if you can draw the Lewis diagram for this. All right, now let's do this together. If any of this seems unfamiliar, I encourage you to watch the video on introduction to drawing Lewis diagrams. But what we'd wanna do is first think about our valence electrons. So xenon right over here, it's actually a noble gas. It already has a full octet in its outer shell, so it has eight valence electrons. So xenon has eight valence electrons. And then fluorine, we've seen this multiple times, has one, two, three, four, five, six, seven valence electrons, but there's five of them. So five times seven. I'm gonna be drawing a lot of electrons in this. So this gives us a total of eight plus 35, which is 43 valence electrons. But we have to be careful. This is a cation. It is a positively charged molecule. It has a positive one charge. So we have to take one electron away because of that. So let's take away one valence electron to get that cation. And so we are left with 42. 42 valence electrons. So the next step is to try to draw its structure with some basic single covalent bonds. And xenon would be our preferred central atom because fluorine is more electronegative. It's actually the most electronegative element. So let's put xenon in the middle, and then let's put some fluorines around it, five of them to be specific. So one, two, three, four. I'm having trouble writing an F. Four and then five fluorines. And now let me make five covalent bonds. One, two, three, four, five. So just like that, I have accounted for 10 valence electrons because you have two valence electrons in each of these covalent bonds, two, four, six, eight, 10. So let me subtract 10 valence electrons. And then we are left with 32 valence electrons. Now, the next step is to try to allocate some more of these valence electrons to the terminal atom so that they get to a full octet. So let me do that to the fluorines. Each of these fluorines already are participating in a covalent bond, so they already have two valence electrons hanging out with them, so let's give 'em each six more. So let's give that fluorine six, and that fluorine gets six, and that fluorine gets six valence electrons, and that fluorine gets six valence electrons, and then last but not least, this fluorine gets six valence electrons. So I have just given away six valence electrons to each of five fluorine atoms. So that is 30 valence electrons that I have just allocated. And then what does that leave me with? That leaves me with two valence electrons that have gone unallocated. And the only place to now put them is on the xenon. And as I said, things that are lower down in that periodic table of elements, especially as we get below the third period, these can defy the octet rule. Xenon already has 10 valence electrons, and I'm about to allocate it two more to it just like that. So you allocate those two more. And then we have allocated all of our valence electrons. And I wanna make sure I remind myself and everyone that this is a cation. So I have to put that plus charge just like this, but this is something that has been observed where you can actually have a central atom like this that goes beyond an octet number of valence electrons. In this case, it has two, four, six, eight, 10, 12 valence electrons. Now, an interesting question is how do these atoms that are in the third period or beyond handle more than eight valence electrons? And it is a matter of debate, but some chemists believe that it's possible because they're able to place their electrons in their empty valence d-orbitals. But once again, this is controversial in the chemistry community.