Sunday, October 09, 2005

Lecture 003

Instructor: Jean-Claude Bradley, Drexel University
View Lecture


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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