- [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.