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# Worked example: Lewis diagram of the cyanide ion (CN⁻)

We can draw Lewis structures for polyatomic ions (ions containing multiple atoms) using the same stepwise procedure as for neutral molecules. In this video, we'll see how to construct the Lewis diagram of the cyanide ion (CN⁻). Created by Sal Khan.

## Want to join the conversation?

• Why is an extra electron on Carbon side, isn't Nitrogen more electronegative, so shouldn't an electron hover over instead? • I don’t understand how carbon has the last bond because carbon only has 4 valence electrons right, so how does it have 5 electrons on its side of the share. • Carbon begins with four valence electrons as a single atom, but as a second period element wants to follow the octet rule and reach eight valence electrons. The three bonds of the triple bond are a total of six electrons and the lone pair is another two electrons giving carbon a total of eight electrons which is what it would like to have.

Same thing with nitrogen, it starts with five electrons as a single atom but also wants to attain eight valence electrons following the octet rule. And so it arrives at eight electrons in the same way the carbon does.

Hope that helps.
• Why sometimes I see charges on atoms instead of a bracket? Which one is appropriate?
Is bracket for ions? • If the entire compound or atom has a charge, you'll see that represented as a '+' or '-' on the upper right of the chemical formula. For example, a fluoride ion is F^(-), a calcium ion is Ca^(2+), and a phosphate ion is PO4^(3-). If we have an ion as a Lewis structure, the ion should have a bracket around the entire structure to show the same thing as the chemical formula, the overall charge of the ion. So my listed examples would have their valance electrons drawn in with a bracket around the ion and the charge in the upper right. Sometimes in your Lewis structure you'll want to know exactly which atom in an polyatomic has the charge and so we call this formal charge. Formal charge tells you the charge of individual atoms in an ion (neutral molecules too). So in Sal's cyanide example the carbon would have a -1 formal charge and so we write that as a negative sign in a small circle next to the carbon. In my phosphate example, three of the four oxygens would have a -1 formal charge so we would write the same negative sign in a small circle next to those oxygens. Sometimes formal charge will be drawn in lieu of the overall charge in brackets in Lewis structures, and sometimes they will be drawn together. Hope this helps.
• If I started with the carbon atom and not the nitrogen atom when distributing valence electrons, would it make any difference?
(1 vote) • Are atoms just allowed to use two electrons from their end to make a bond? Like nitrogen in this example grants carbon access to electrons by using two of its own valence electrons, instead of sharing one in exchange for an electron of carbon. Is that a rule that applies in all chemical settings?
(1 vote) • So there's a difference between the electrons contributed by each atom for a bond and the electrons shared in the final bond. For cyanide the carbon contributes four valence electrons and the nitrogen contributes five (plus the additional electron for the negative charge). In the final ion though, the carbon and nitrogen are sharing collectively six electrons in a triple bond which ultimately came from both atoms.

Hope that helps.
• how does the the two other lines convay thats adding up more electrons to get the answer?
(1 vote) • Why does the bonding stop at a triple bond? Why not a quadruple bond, even a quintuple bond!
(1 vote) • There are higher order bonds beyond just triple bonds. Quadruple, quintuple, and even sextuple bonds are possible. You don't really view these higher bond orders for second and third period elements like carbon because the resulting molecules are often unstable. They accomplish greater stability by using smaller bond orders so there's no incentive for them to use higher ones.

Where these higher bond orders are more prevalent are the transition metal complexes, or molecules involving transition metals. Because atoms of these elements make use of more orbitals than second or third period elements they can share more electrons in bonds. Often this happens between two transition metal atoms within the same molecule.

As a side note, taking into consideration resonance, it also possible to get fractional bond orders instead of just whole number ones. It's possible to have things like 1.3 and 1.5 bonds as well.

Hope that helps.
(1 vote)
• Is helium also an exception to the octet rule?
(1 vote) ## Video transcript

- [Instructor] In this video we're gonna try to get more practice constructing Lewis diagrams. And we're gonna try to do that for a cyanide anion. So this is interesting. This is the first time we're constructing a Lewis diagram for an ion. So pause this video and see if you can have a go at that. All right, now let's do this together. So we've already seen in many videos, the first step is to essentially count the total valence electrons that we're dealing with. And the reason why we do that is to make sure that we're allocating all the valence electrons. To help us there we can look at a periodic table of elements. You might already know that carbon has one, two, three, four valence electrons in that second shell, it's in the second period, so you have four valence electrons from carbon. Nitrogen has one, two, three, four, five valence electrons in its second shell, it's in that second period. And so the valence electrons from a neutral carbon and a neutral nitrogen-free atom would be a total of nine valence electrons. But we are not done yet. Because this is not a neutral molecule. We have a negative charge here. It is an anion, it has a negative one charge. And so because of that negative one we can think about it having an extra valence electron. So let's add a valence electron here. Why do we do it? Because of this negative charge. So we're dealing with a total of 10 valence electrons. Now, the next step is to try to draw single bonds. Try single bonds, and identify a central atom. Now, we only have two atoms here, so really neither feels central, so let me just put a carbon and a nitrogen next to each other here. And then let me draw one single bond. So by drawing that one single bond I have now accounted for two valence electrons. So now I am left with eight valence electrons, and so that's the next step, allocate remaining valence electrons, allocate valence electrons. So let me start with the more electronegative. Let's try to get nitrogen to eight. It already has two. So let's give it three more lone pairs. So we have two, four, six, eight. So I have just used up six of these remaining valence electrons, six, so minus six, so I have two left to allocate. So let me give carbon two valence electrons, like that. And there I have used up all of my, all of my valence electrons. Now let's see how happy everyone is. Nitrogen has eight valence electrons hanging around, two, four, six, eight. But carbon only has four, two and four. So this is where we think about whether we would want to have some extra bonds, extra bonds, or higher-order bonds. So how can we give carbon more valence electrons? Well, what we could do is we could take some of these lone pairs around nitrogen and then use them, turn this single covalent bond into a higher-order bond. So let's see, if we were to take these two and turn it into another covalent bond, what is going to happen? Let me erase all of these, and then I'll just draw another covalent bond. So nitrogen still has eight electrons hanging around. Carbon now has six. So maybe we can do that again. So let me erase these two characters. Let me erase these two characters and make another covalent bond out of them. So let me make a covalent bond out of them. And so now what's going on? Carbon has two, four, six, eight valence electrons hanging around. Nitrogen has two, four, six, eight valence electrons hanging around. So this is looking pretty good. But are we done yet? The simple answer is no. We still haven't represented this negative charge in our Lewis diagram. The way that we would do that is say hey, this entire molecule, you put brackets around it, has a negative charge. And now we're done. We've allocated all of our valence electrons, we have our octet rule on all of our atoms that are not hydrogen, there's no hydrogen here, and we're showing that this indeed is an anion, and now we are done.