### Lecture 003

Instructor: Jean-Claude Bradley, Drexel University

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1

Chem. 241 - Lectures 3

1 DISCLAIMER: This text is being provided in a

rough-draft fashion. Communication Access

2 Realtime Translation (CART) is provided in order

to facilitate communication accessibility and may

3 not be a totally verbatim record of the

proceedings.

8 LECTURE 3

9 Okay. So what we're going to do

10 now is look at especially covalent bonds and

11 how we can figure out how atoms are

12 connected together to form more complicated

13 molecules. So we're going to learn how to

14 draw Lewis structures. You probably already

15 know how to do this, but I'll show you a way

16 of doing it so that you can find the

17 solution to the Lewis structure very quickly

18 for more complicated cases.

19 Okay. So I'm going to show you

20 that method and that method is on the first

21 blog. Right here, Lewis dot approach to

22 molecular structure. So we're going to

23 apply this today and you should be able to

24 do any Lewis structure after that.

25 So I'm going to start with the

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11

Chem. 241 - Lectures 2 and

1 same example, hydrogen, since we already

2 know the answer. We're talking about Lewis

3 structures. So if we follow this little

4 algorithm that I gave you, it has six steps.

5 The first step is to compound the total

6 number of electrons available, and that's

7 where you pull out that valence periodic

8 table, all right?

9 So carbon is four, nitrogen is

10 five, oxygen is six, fluorine is seven, neon

11 is eight, and hydrogen is one. Okay? So

12 that tells us how many electrons we have

13 available. We have 1 plus 1. So I'm only

14 talking about valence electrons.

15 Now the next step is to figure out

16 this number 2: Find out how many electrons

17 are needed. When I refer to "needed," I'm

18 talking about the octet rule. And the octet

19 rule states that you want eight electrons

20 for each carbon, nitrogen, oxygen, and

21 fluorine and related elements. So sulfur,

22 phosphorous, iodine, all those little other

23 elements that we drew in that little

24 periodic table, the same thing. They all

25 need eight electrons because of the octet

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12

Chem. 241 - Lectures 2 and

1 rule. Hydrogen only needs two.

2 So what we're going to try to do

3 is figure out how we can have a molecule so

4 that every atom satisfies the octet rule.

5 Okay?

6 So we need -- well, I have two

7 hydrogens, and each hydrogen needs two

8 electrons. So that's 2 plus 2 is 4. Okay?

9 So that's the octet rule part.

10 So the next step is, we figure out

11 how many electrons are shared. And then we

12 divide by 2 to get the number of bonds in

13 the molecule. So this should make some

14 sense.

15 I have two electrons available;

16 however, I need four. So I'm going to have

17 to share a certain number of electrons. How

18 many? Exactly 4 minus 2. I need to share 2

19 electrons. By definition, a bond, a single

20 bond, is two electrons.

21 So I divide 2 by 2, and I come up

22 with 1, okay? So for this example, it

23 actually takes a lot more time than the way

24 we did it, you know, earlier, where we just

25 drew the electrons.

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13

Chem. 241 - Lectures 2 and

1 You see, when you do molecules

2 that are far more complicated than knowing

3 the number of bonds in the molecules, it is

4 critical to being able to solve the problem

5 quickly. So I'm going to be doing examples

6 that are increasingly more complex until we

7 cover examples that you should be able to do

8 anything.

9 So knowing that we have one bond,

10 the next instead of is you write out the

11 atoms. In this case, we have two. There's

12 only one way to link two atoms together with

13 a single bond, and that's like that.

14 Sometimes we will have a choice

15 and we will talk about how we decide how to

16 put the bonds. When you only have two

17 atoms, you don't have a choice and you have

18 to put it like that.

19 So it looks like we're done, but

20 there's actually a couple more things to

21 worry about in order to make sure that we

22 have a correct answer.

23 The next thing we're going to do

24 is number 4 here. After placing the bonds,

25 complete the octets. What I'm doing here

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14

Chem. 241 - Lectures 2 and

1 is, I'm looking at each atom and counting

2 how many electrons are around it. If I look

3 at the left hydrogen, I count all of the

4 electrons until the bond. So I have two

5 electrons that are surrounding that hydrogen

6 for the octet rule. Well, hydrogen needs

7 two electrons so, I don't have to add any

8 electrons if I had some things that were

9 missing. We're going to do that in the next

10 example that we do.

11 So, you know, by symmetry, the

12 other hydrogen also has two electrons. So

13 we have satisfied the octet rule.

14 The last thing you have to worry

15 about is counting electrons around the atoms

16 and placing the charges. So when I'm

17 counting for charges, I only count one

18 electron per bond because charges have

19 nothing to do with the octet rule; charges

20 have to do with on average how many

21 electrons are around each hydrogen.

22 So if I count one electron per

23 bond, I have one electron on each hydrogen

24 for charges. Hydrogen normally has one

25 electron; so, therefore, it is not charged.

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15

Chem. 241 - Lectures 2 and

1 So it is really important to do

2 that last step because if you get everything

3 else right but you don't get the charge

4 right, you will have a wrong answer, and

5 it's going to lead you to wrong solutions

6 for some of the problems we're going to do.

7 So when we come up with a charge,

8 you will see how that works. But in this

9 one, it doesn't have a charge so we're

10 pretty much done.

11 If you had an issue with the with

12 the charges, you can always recheck your

13 molecules to make sure that the octet rule

14 is satisfied everything. So that number 6

15 is just check up on it.

16 So any questions on that? We're

17 going to do a bunch of example so is if it's

18 not all clear now, it will be all clear.

19 Okay. Next example is fluorine,

20 F2. So we're going to have available, need,

21 share, and then bonds. Always the same

22 thing.

23 Okay. So how many do I have

24 available in this molecule? Each fluorine

25 is what? Seven. So how do I know that?

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16

Chem. 241 - Lectures 2 and

1 This is where you go back to the valence

2 periodic table. You look up fluorine --

3 seven, and that's it.

4 So we have seven, but we have two

5 fluorines. So that's 7 plus 7=14. How many

6 do we need? 16. How do I know that?

7 Again, I look here, and it tells me that for

8 each carbon -- nitrogen, oxygen, fluorine --

9 I need eight. So 8 plus 8 is 16.

10 I'm going to share 16 minus 14=2

11 electrons. And divide that by 2 and find

12 out that I have one bond.

13 So I have two atoms. I have to

14 connect them together and make a molecule,

15 so I have to at least put the bond between

16 the two atoms, right?

17 So now if we stop here, this is

18 incorrect, because a Lewis structure a

19 complete Lewis structure, satisfies the

20 octet rule, and all the valence electrons

21 are shown and all the charges are shown, so

22 we're not done here.

23 We're going to look at the next

24 step. So after placing the bond which we

25 just finished, we complete the octets.

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17

Chem. 241 - Lectures 2 and

1 So the fluorine has, for octets,

2 two electrons around it, so it's actually

3 missing six to be able to come up to eight.

4 So I'm going to add six electrons around the

5 fluorine, but I'm going to do it in pairs.

6 So two on the top, two on the left, two on

7 the bottom. Okay? So I'm completing the

8 octet rule at this point. And by symmetry,

9 the fluorine on the right, I have to add six

10 electrons.

11 Okay. So the next step is, we are

12 going to count and place the charges. So

13 we're going to count the electrons per

14 charges. So when I count the charges, I

15 only count one electron per bond, but I

16 count all the other electrons.

17 So one for the bonds, two, three

18 four, five, six, seven. So for charges, I

19 have seven electrons around the fluorine.

20 Fluorine normally has seven electrons;

21 therefore, it's not charged. So I don't

22 have to put a plus or a minus anywhere

23 Okay? So it's not very

24 complicated, but it's one step above

25 hydrogen. There's something new in that

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18

Chem. 241 - Lectures 2 and

1 example.

2 Oxygen is the next one, O2. So

3 available, need, share, number of bonds to

4 be filled out.

5 So oxygen is in the column of six

6 valence electrons. So that's 6 plus 6=12.

7 How many do we need? So we have, you know,

8 any atom that's listed here, oxygen is

9 listed. It needs eight. So that would be

10 8+8=16.

11 Share is 16. So we're going to be

12 sharing 16-12=34. 4 divided by 2=2.

13 Okay? So it's useful to know that

14 there are exactly two bonds in oxygen. So

15 let's see how we are going to put them.

16 Well, if you only have two atoms,

17 it's not really very difficult. There's

18 only one place to put them, and that's

19 between the two atoms. So the first bond

20 links the two atoms together. But I have to

21 put two bonds total, so the only place I can

22 put it is here.

23 Okay. So now I'm going to go a

24 little bit more quickly with each example,

25 okay?

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19

Chem. 241 - Lectures 2 and

1 For each oxygen, now we're going

2 to complete the octets. How many electrons

3 do I have to put on? Four. I have four

4 electrons per octet already there. I only

5 need four more.

6 So, again, putting them as pairs,

7 I put the four electrons like that. And by

8 symmetry, I will do the same thing with the

9 oxygen on the right. Okay, now we have the

10 eight electrons per octet, so it's

11 satisfied.

12 Next thing we're going to count

13 the charges. One two tree four, five six.

14 Again, it's only one electron per bond, so I

15 have six valence electrons -- I have six

16 electrons per charge. Oxygen normally has

17 six electrons, so it's uncharged.

18 So, once again, with the simple

19 examples, you know they're not going to be

20 charged. Now it's complete. That's the

21 full Lewis structure.

22 Okay. Let's do nitrogen next.

23 N2. Available, need, share, and number of

24 bonds.

25 Okay. So going back to the table,

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20

Chem. 241 - Lectures 2 and

1 nitrogen is in the column with five valence

2 electrons. So I have five plus five

3 available. Okay? Available valence

4 electrons. Nitrogen is one of the atoms

5 that requires eight to satisfy the octet.

6 So 8+8=16. I have to share 16-10, which is

7 6, which tells me that I have 6 divided by

8 2=3 bonds. So once again, it's useful to

9 know that going in.

10 I only have two atoms. So there's

11 only one place I can put those three

12 bonds -- between the two nitrogens.

13 Now I'm going to satisfy the octet

14 rule. I have six electrons on each nitrogen

15 for the octet rule, so I'm only missing two.

16 Put them in a pair like that. Opposite for

17 the triple bond. By symmetry, the nitrogen

18 on the right would have two electrons.

19 Counting for charges each one of

20 the bonds contributes one electron so

21 there's one, two, three. And then we count

22 all the electrons. Four, five. So we have

23 five electrons. Nitrogen normally has five

24 electrons. So, again, uncharged.

25 Okay. Let's start to put more

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21

Chem. 241 - Lectures 2 and

1 atoms and see how this works out. So if I

2 try methane, CH 4, the same thing -- we are

3 going to look at available, how many we

4 need, how many we share, and then the number

5 of bonds.

6 So the carbon is four. And each

7 hydrogen is one. So to make it very

8 explicit here, 4+4x1, that totals up to 8.

9 So how many we need is going to be

10 8 for the carbon, and this is where people

11 often make a mistake. Hydrogen is 2, it

12 requires 2. So I have 4 of them, and that's

13 4x2=8+8, total of 16

14 So I'm going to have to share

15 16-8, or 8 electrons. Number of bonds is 8

16 divided by 2=4. Again, not a bit surprised

17 for this molecule because I'm starting

18 something simple, so we can build up to it.

19 We have carbon. I have 4 bonds

20 and I have 4 atoms to connect it to so

21 there's only one way to do that. Hydrogen

22 can't have more than one bond, so the only

23 thing I can do is put the four bonds around

24 the carbon. So how many electrons does

25 carbon need to complete the octet in this

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22

Chem. 241 - Lectures 2 and

1 case?

2 It's already full. It already has

3 eight electrons. So there's nothing to add.

4 For charges, I have one, two,

5 three, four. Carbon normally has four, so

6 it's uncharged so we're done here.

7 Carbon dioxide, CO2. Available,

8 need, share, and number of bonds.

9 So carbon is going to need four.

10 And each oxygen is 6. So 6x2=12+4=16. So

11 we're going to need the carbon. And the

12 oxygens each require 8. 8x3=24. So we have

13 to share 24-16, so we have to share 8

14 electrons. That tells us we have 8 divided

15 by 2, and that's 4 bonds. So we have four

16 bonds and we have three atoms.

17 So now we have more than one way

18 of solving this issue, right? We can join,

19 you know, the three atoms in a triangle. We

20 can do all kinds of things like this. So

21 this is where you have to use heuristics to

22 tell you what is likely to be the most

23 stable form of this.

24 Generally, when you have one atom

25 followed by, you know, a whole bunch of

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23

Chem. 241 - Lectures 2 and

1 another one of oxygen, you're going to put

2 that single atom in the middle and surround

3 it. So if I follow that, it's going to work

4 95 percent of the time.

5 We will see maybe a couple of

6 exceptions to that, but that's pretty good

7 heuristics to follow. So following that, I

8 would put the carbon and then I would put

9 the oxygens on either side.

10 Okay? So what's the minimum I

11 need to hold the structure together? One

12 bond on the left, one bond on the right.

13 But I have to do something with all the four

14 bonds that I know are in that molecule.

15 So, you know, again, here I could,

16 you know, draw a bond between the two

17 oxygens and maybe make a triangular

18 structure, but as we will find out a little

19 bit later, three-membered rings are unstable

20 usually. So that's not going to be your

21 first choice.

22 So if I can't make a triangle out

23 of it, the only thing I can do is to put

24 double bonds. So I'm going to add C double

25 bond 0 on the left, C double bond 0 on the

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24

Chem. 241 - Lectures 2 and

1 right.

2 Okay? So now I'm going to

3 complete the octets. So I add four

4 electrons on the oxygens, and the carbon

5 already is satisfying the octet rule.

6 Looking at charges -- one, two, three, four,

7 five, six electrons around the oxygen.

8 Oxygen normally has six, so that's

9 uncharged. Carbon -- two, three four,

10 carbon, it only has four, so it's also

11 uncharged. So that's complete, the CO2

12 Okay? So let's look at BF3, boron

13 trichloride. Available, need, share, number

14 of bonds. So available, let's go back to

15 the table here. Boron has three valence

16 electrons. And the chlorine has seven.

17 That's 24, right?

18 How many do we need? How many

19 does fluorine need to complete the octet?

20 Eight. How many does boron need to complete

21 the octet? Boron is not listed in the list

22 of atoms to complete the octet rule, so

23 that's one thing you have to watch out for.

24 Boron will not satisfy the octet

25 rule, so you simply cannot use this

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25

Chem. 241 - Lectures 2 and

1 procedure. You're going to end up with a

2 molecule. If it doesn't satisfy the octet

3 rule, that doesn't mean you can't form a

4 molecule. It just means that it will not

5 satisfy the octet rule, so you don't do

6 this.

7 I'm not going to give you problems

8 that are terribly complicated with boron, so

9 you can pretty much figure out what has to

10 happen. The boron has three valence

11 electrons and it's bonding with three atoms.

12 So there's really not much else for it to do

13 except take each one of those valence

14 electrons and use it to form a covalent bond

15 with each fluorine like this.

16 So notice that in the structure,

17 which, if I wanted to show the bonds, this

18 would still not be complete because I have

19 not shown the long pairs on the fluorines.

20 So you'll notice with this

21 example, the borine follows the octet rule,

22 but the boron only has six electrons, and

23 that's just the way it is, okay? So if it's

24 a boron, it just doesn't satisfy.

25 Let's look at ammonia, NE3. So

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26

Chem. 241 - Lectures 2 and

1 nitrogen is five. NE hydrogen is 1. So we

2 have 8 electrons available. We need 8 for

3 the nitrogen and 2 for each hydrogen.

4 3x2=6+8=14. We have to share 14-8, which is

5 6. So we have 6 divided by 2, so it's 3

6 bonds.

7 So each hydrogen cannot have more

8 than one bond to it, so the only way I can

9 hook this together with the nitrogen in the

10 middle is to use up my 3 bonds. And now if

11 I go to the next step, I have to complete

12 the octets. Each hydrogen is satisfied.

13 The nitrogen only has six

14 electrons. It's missing two, which I will

15 put. And for charges, one, two, three,

16 four, five. Nitrogen normally has five

17 electrons, so uncharged.

18 Okay. Let's look at carbon

19 monoxide, CO. Available, need, share, and

20 number of bonds. Available, we have four

21 from the carbon and 6 from the oxygen, and

22 that gives me 10 total. We need 8. Plus 8,

23 that's 16. So we have to share 16-10, which

24 is 6. That means we have 6 divided by 2, so

25 3 bonds.

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27

Chem. 241 - Lectures 2 and

1 As we start to get into more

2 complicated examples, it becomes more and

3 more useful to know how many bonds we have.

4 There's only two atoms. I have to

5 put the three bonds between the carbon and

6 the oxygen like that. So the next step is

7 to complete the octets. The carbon has 6

8 electrons around it already. It's only

9 missing two 2. Same thing for the oxygen.

10 Now, if we look at charges, for

11 the carbon, one, two, three from each bonds,

12 four, five. But carbon normally has four

13 electrons, so it has an extra electron.

14 Electrons are negativity charged; therefore,

15 I have a formal charge on the carbon of

16 minus 1. So I don't have to write 1. If I

17 just put a minus, it implies I have one.

18 The oxygen -- I have one, two,

19 three, four, five electrons. Oxygen

20 normally has six. It's missing an electron.

21 So if I'm missing a negative charge, I go to

22 positive. So it has a plus 1 charge. I

23 just put a plus on it.

24 Now, after you're finished placing

25 charges, the total charge in the molecule,

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28

Chem. 241 - Lectures 2 and

1 you draw has to be the same as the one in

2 the problem. So my total charge in CO is

3 zero. I have minus 1 and plus 1, so that

4 cancels out. That has to happen. So if it

5 doesn't happen, you made a mistake

6 somewhere, okay?

7 Okay. Now let's get some other

8 kinds of examples. SO, one of the oxides of

9 sulfur. Available, need, share, and bonds.

10 So sulfur is 6. Oxygen is also 6.

11 That's 12 electrons. We need 8+8=16. We

12 have to share 16-12 which is 4. That means

13 we have 4 divided by 2, or 2 bonds. So I

14 only have two atoms. That makes it pretty

15 straightforward. I have to put the two

16 bonds between the sulfur and the oxygen.

17 Completing the octets. Sulfur is missing 4.

18 Oxygen is missing 4. And if we count the

19 charges, we will find six electrons on the

20 sulfur. Okay?

21 * * *

22

23

24

25

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