Understanding Lewis Structures and Octet Exceptions

In this video, we're going to go over Lewis structures that are exceptions to the octet rule. Now if you recall, the octet rule is that atoms should have 8 electrons around it. Octet means 8. The first example that violates the Altec rule is BH3. Go ahead and draw the Lewis structure of BH3. To draw it, we need to count the number of valence electrons. Boron has 3, and hydrogen has 1. 3 is 6. Now every hydrogen atom needs at least one bond, and every bond represents two electrons, 2, 4, 6. Now there's no way in which we can put eight electrons in boron because We don't have 8 valence electrons in this molecule, so therefore boron has no choice but to have an incomplete octet. And this is the correct Lewis structure of BH3. It has a trigonoplano-molecular geometry with a bond angle of 120. And it's just the way it is. So not every molecule will obey the octet rule. The octet rule is just a general rule. But this is one example of a substance. that has an incomplete octet. It has less than eight electrons. Now there are some other molecules that have more than eight electrons. On a periodic table, the second row elements that you'll see often is carbon, nitrogen, oxygen, and fluorine. These electrons, excuse me, not electrons, but these atoms cannot have more than eight electrons. So they cannot have an equivalent of two electrons. expanded octet. However, they can have an incomplete octet. It is possible. Now the elements below them, for example, like silicon, phosphorus, sulfur, chlorine, these elements, they can have what? what is known as the expanded octet. They can have more than eight electrons. The reason why nitrogen can have more than eight is because nitrogen is in the second row. And in the second row, you have the 2s sublevel and you have the 2p sublevel. Notice that in the second energy level, the maximum number of electrons is eight. So that's why second row elements cannot have more than eight electrons. It's just not possible. Phosphorus is in the third row. the third row so in the third energy level you have the 3s sub level the 3p sub level and the 3d sub level so therefore in the third energy level you can have up to 18 electrons and that's why elements like phosphorus sulfur and chlorine they can have expanded octets so if they're the center atom they can have more than eight electrons sometimes you might seem have 10 sometimes even 12 so just keep that in mind So anything that's in a third row or below can have an expanded octet. But the second row elements, like carbon, nitrogen, oxygen, and fluorine, they cannot have an expanded octet. They cannot have more than eight electrons. Now let's consider an example of an expanded octet, PCl5. go ahead and draw the Lewis structure of this substance phosphorus has five valence electrons and chlorine has 75 times seven is 35 plus five is 40 and to calculate the number of pairs it's equal to the valence electrons minus 8n divided by 2 so there's 40 valence electrons and n is the number of atoms that is not a center atom so n is 5 in this example so 8 times 5 is 40 so 40 minus 40 is 0 so there are no lone pairs on a center atom so to draw the Lewis structure All we need to do is put the 5-chlorine atoms around phosphorus because there's no lump here on phosphorus. And as you can see, phosphorus has more than 8 electrons around it. Because it has 5 bonds and each bond represents 2 electrons, it has a total of 10 electrons around it. So it has an expanded octet. So let me give you another example. Let's use ICl5. Go ahead and draw the Lewis structure for this molecule. So first, let's count the number of valence electrons. Iodine has 7. Chlorine has 7. 5 times 7 is 35, plus 7, that's 42. Now the number of lone pairs is going to be the valence electrons, which is 42, minus 8n. And n is the number of atoms. that is not the center atom which is 5 so 8 times 5 is 40 42 minus 40 is 2 so we have one lone pair on iodine the center atom and there's five other chlorine atoms attached to it And so this is the Lewis structure of iodine petachloride. It has a square pyramidal molecular geometry. And as you can see, it has an expanded octane. It has 2, 4, 6, 8, 10, 12 electrons around it. Now, there are some other examples that violate the octet rule. And this occurs when a molecule has an odd number of electrons. So let's consider nitrogen monoxide. Nitrogen has 5 valence electrons and oxygen has 6. So the total is 11. So how can we draw a Lewis structure in which we have 11 electrons? The first thing I like to take into account is the number of bonds that certain elements like to form. So let me just give you a quick review of how to determine that. Carbon has four valence electrons. Nitrogen has five. Oxygen has six. Fluorine has seven. Now, elements on the right side of the periodic table, they like... to acquire electrons. So carbon wants to have eight. It needs to get four more electrons in order to have eight. So carbon likes to form four bonds to get those four electrons that it needs. Nitrogen has five valence electrons. It needs three more to get to eight, so it likes to form three bonds to get the three electrons it needs. Oxygen likes to form two bonds. Fluorine likes to form one. Now, on the left side, these elements like to give away electrons instead of acquiring electrons. Boron has three valence electrons, so it forms three bonds to give away those three electrons. Beryllium has two valence electrons, so it likes to form two bonds to give away those two valence electrons. And so knowing that can help us to get a good idea of how to draw the Lewis structure for NO. So now, let's go ahead and draw NO. Now, nitrogen likes to form three bonds, but oxygen likes to form two. So, this molecule is either going to have two bonds or three bonds. So let's start with two. We need a total of 11 electrons. Right now we have four. Every bond represents two. So now we have six, and now we have eight. Now we have 9, now we have 10. So who's going to get the last electron? There's two ways in which we can draw it. We can give it to oxygen, or we can give the last electron to nitrogen. As you can see, either way you draw, one of these atoms will have an incomplete oxide. In the first example, oxygen has eight electrons, two, four, six, eight, so its oxide requirements are satisfied. but nitrogen has 7 2 4 6 7 in the second example nitrogen has 8 but oxygen has 7 2 4 6 7 so regardless of how you structure this molecule you're going to have an element that has an incomplete octet. So anytime you have an odd number of electrons, the octet rule will be violated. There's nothing you can do about it. It's just based on the numbers that you have. Now, which way is the best way to draw this structure? Should oxygen have 8 or should nitrogen have 8? Now, which element is more electronegative? Oxygen is more electronegative than nitrogen. Electronegativity increases as you go up and to the right towards fluorine. And because oxygen is closer to fluorine than nitrogen on a periodic table, oxygen is more electronegative. So oxygen is going to take away that one electron. And so this is going to be the most stable Lewis structure out of the two that's listed. Now let's try another example in which we have an odd number of electrons. Go ahead and draw the Lewis structure for NO2. So nitrogen has 5 electrons, oxygen has 6. So this is going to be 12 plus 5, so that's a total of 17. So chances are nitrogen is going to be the center atom. Since nitrogen can form more bonds than oxygen. Oxygen likes to have two, but nitrogen likes to have three. But when you have an odd number of electrons, something's going to be lacking. So each bond needs at least one single bond. Now, if we calculate the number of lone pairs, let's see what's going to happen. So we have 17 valence electrons minus 8n, and this 2. There's two atoms that are not center atoms, divided by 2. So 17 minus 8 times 2, that's 16. That's going to be 1. So what we have is half of a lone pair. Half of a lone pair is 1 electron. So now at this point, we have a total of 5 electrons. 2, 4, 5. Now we need to get as close to 8 as possible. So if we add another bond, nitrogen has 7. And if we add another one, it would have 9, which that cannot happen. Nitrogen cannot have 9 electrons. It can't have an expanded octet. So the... The best that we can do is 7. 7 is as close to 8 as possible, and nitrogen can't pass 8. So now we need to fill up the other atoms with lone pairs. So oxygen has a double bond, which means it needs 2 lone pairs to get to 8 electrons. 2, 4, 6, 8. The other oxygen atom on the right has a single bond, so it needs three lone pairs to have eight electrons. And so this is the Lewis structure of the nitrogen dioxide molecule. This is 8, 8, that's 16 plus 1, so we have a total of 17 valence electrons. And so anytime you have an odd number of electrons, typically it's going to violate the octet rule.

Go ahead and draw the Lewis structure of BH3. To draw it, we need to count the number of valence electrons. Boron has 3, and hydrogen has 1. 3 is 6. Now every hydrogen atom needs at least one bond, and every bond represents two electrons, 2, 4, 6. Now there's no way in which we can put eight electrons in boron because We don't have 8 valence electrons in this molecule, so therefore boron has no choice but to have an incomplete octet.

And this is the correct Lewis structure of BH3. It has a trigonoplano-molecular geometry with a bond angle of 120. And it's just the way it is. So not every molecule will obey the octet rule. The octet rule is just a general rule.

But this is one example of a substance. that has an incomplete octet. It has less than eight electrons. Now there are some other molecules that have more than eight electrons. On a periodic table, the second row elements that you'll see often is carbon, nitrogen, oxygen, and fluorine.

These electrons, excuse me, not electrons, but these atoms cannot have more than eight electrons. So they cannot have an equivalent of two electrons. expanded octet.

However, they can have an incomplete octet. It is possible. Now the elements below them, for example, like silicon, phosphorus, sulfur, chlorine, these elements, they can have what? what is known as the expanded octet. They can have more than eight electrons.

The reason why nitrogen can have more than eight is because nitrogen is in the second row. And in the second row, you have the 2s sublevel and you have the 2p sublevel. Notice that in the second energy level, the maximum number of electrons is eight.

So that's why second row elements cannot have more than eight electrons. It's just not possible. Phosphorus is in the third row.

the third row so in the third energy level you have the 3s sub level the 3p sub level and the 3d sub level so therefore in the third energy level you can have up to 18 electrons and that's why elements like phosphorus sulfur and chlorine they can have expanded octets so if they're the center atom they can have more than eight electrons sometimes you might seem have 10 sometimes even 12 so just keep that in mind So anything that's in a third row or below can have an expanded octet. But the second row elements, like carbon, nitrogen, oxygen, and fluorine, they cannot have an expanded octet. They cannot have more than eight electrons.

Now let's consider an example of an expanded octet, PCl5. go ahead and draw the Lewis structure of this substance phosphorus has five valence electrons and chlorine has 75 times seven is 35 plus five is 40 and to calculate the number of pairs it's equal to the valence electrons minus 8n divided by 2 so there's 40 valence electrons and n is the number of atoms that is not a center atom so n is 5 in this example so 8 times 5 is 40 so 40 minus 40 is 0 so there are no lone pairs on a center atom so to draw the Lewis structure All we need to do is put the 5-chlorine atoms around phosphorus because there's no lump here on phosphorus. And as you can see, phosphorus has more than 8 electrons around it. Because it has 5 bonds and each bond represents 2 electrons, it has a total of 10 electrons around it. So it has an expanded octet.

So let me give you another example. Let's use ICl5. Go ahead and draw the Lewis structure for this molecule.

So first, let's count the number of valence electrons. Iodine has 7. Chlorine has 7. 5 times 7 is 35, plus 7, that's 42. Now the number of lone pairs is going to be the valence electrons, which is 42, minus 8n. And n is the number of atoms. that is not the center atom which is 5 so 8 times 5 is 40 42 minus 40 is 2 so we have one lone pair on iodine the center atom and there's five other chlorine atoms attached to it And so this is the Lewis structure of iodine petachloride. It has a square pyramidal molecular geometry.

And as you can see, it has an expanded octane. It has 2, 4, 6, 8, 10, 12 electrons around it. Now, there are some other examples that violate the octet rule.

And this occurs when a molecule has an odd number of electrons. So let's consider nitrogen monoxide. Nitrogen has 5 valence electrons and oxygen has 6. So the total is 11. So how can we draw a Lewis structure in which we have 11 electrons?

The first thing I like to take into account is the number of bonds that certain elements like to form. So let me just give you a quick review of how to determine that. Carbon has four valence electrons.

Nitrogen has five. Oxygen has six. Fluorine has seven.

Now, elements on the right side of the periodic table, they like... to acquire electrons. So carbon wants to have eight. It needs to get four more electrons in order to have eight.

So carbon likes to form four bonds to get those four electrons that it needs. Nitrogen has five valence electrons. It needs three more to get to eight, so it likes to form three bonds to get the three electrons it needs. Oxygen likes to form two bonds.

Fluorine likes to form one. Now, on the left side, these elements like to give away electrons instead of acquiring electrons. Boron has three valence electrons, so it forms three bonds to give away those three electrons. Beryllium has two valence electrons, so it likes to form two bonds to give away those two valence electrons.

And so knowing that can help us to get a good idea of how to draw the Lewis structure for NO. So now, let's go ahead and draw NO. Now, nitrogen likes to form three bonds, but oxygen likes to form two. So, this molecule is either going to have two bonds or three bonds.

So let's start with two. We need a total of 11 electrons. Right now we have four.

Every bond represents two. So now we have six, and now we have eight. Now we have 9, now we have 10. So who's going to get the last electron?

There's two ways in which we can draw it. We can give it to oxygen, or we can give the last electron to nitrogen. As you can see, either way you draw, one of these atoms will have an incomplete oxide.

In the first example, oxygen has eight electrons, two, four, six, eight, so its oxide requirements are satisfied. but nitrogen has 7 2 4 6 7 in the second example nitrogen has 8 but oxygen has 7 2 4 6 7 so regardless of how you structure this molecule you're going to have an element that has an incomplete octet. So anytime you have an odd number of electrons, the octet rule will be violated.

There's nothing you can do about it. It's just based on the numbers that you have. Now, which way is the best way to draw this structure?

Should oxygen have 8 or should nitrogen have 8? Now, which element is more electronegative? Oxygen is more electronegative than nitrogen.

Electronegativity increases as you go up and to the right towards fluorine. And because oxygen is closer to fluorine than nitrogen on a periodic table, oxygen is more electronegative. So oxygen is going to take away that one electron. And so this is going to be the most stable Lewis structure out of the two that's listed.

Now let's try another example in which we have an odd number of electrons. Go ahead and draw the Lewis structure for NO2. So nitrogen has 5 electrons, oxygen has 6. So this is going to be 12 plus 5, so that's a total of 17. So chances are nitrogen is going to be the center atom. Since nitrogen can form more bonds than oxygen. Oxygen likes to have two, but nitrogen likes to have three.

But when you have an odd number of electrons, something's going to be lacking. So each bond needs at least one single bond. Now, if we calculate the number of lone pairs, let's see what's going to happen. So we have 17 valence electrons minus 8n, and this 2. There's two atoms that are not center atoms, divided by 2. So 17 minus 8 times 2, that's 16. That's going to be 1. So what we have is half of a lone pair.

Half of a lone pair is 1 electron. So now at this point, we have a total of 5 electrons. 2, 4, 5. Now we need to get as close to 8 as possible.

So if we add another bond, nitrogen has 7. And if we add another one, it would have 9, which that cannot happen. Nitrogen cannot have 9 electrons. It can't have an expanded octet. So the... The best that we can do is 7. 7 is as close to 8 as possible, and nitrogen can't pass 8. So now we need to fill up the other atoms with lone pairs.

So oxygen has a double bond, which means it needs 2 lone pairs to get to 8 electrons. 2, 4, 6, 8. The other oxygen atom on the right has a single bond, so it needs three lone pairs to have eight electrons. And so this is the Lewis structure of the nitrogen dioxide molecule.

This is 8, 8, that's 16 plus 1, so we have a total of 17 valence electrons. And so anytime you have an odd number of electrons, typically it's going to violate the octet rule.

Transcript for:Understanding Lewis Structures and Octet Exceptions

Transcript for:
Understanding Lewis Structures and Octet Exceptions