Worked example: Lewis diagram of the cyanide ion (CN⁻) | AP Chemistry | Khan Academy

In this video, we're going to try to get more practice constructing Lewis diagrams, and we're going to 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. The reason why we do that is to make sure that we're allocating all the valence electrons. To help us, 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. 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, has a negative one charge. 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 did 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.

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. I have just used up six of these remaining valence electrons, so minus six means 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 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 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 take some of these lone pairs around nitrogen and then use them to 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, and 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 to 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.