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Unformatted text preview: CHAPTER 4 GEOME TRY In this chapter. we will see how to predict the 3D shape ol‘ molecules. This is im
portant because it limits much of the reactivity that you will see in the second half
of this course. For molecules to react with each other, the pans of the molecules that
can react with each other must be able to get close in space. If the geometry of the
molecules prevents them front getting close. then there cannot be a reaction. This
concept is called sterics. Let’s use an analogy to help us see the importance of geometry. Imagine that
you are stufﬁng a turkey for Thanksgiving dinner and your hand gets stuck inside
the turkey. Just at that moment, someone wants to shake your hand. You can’t shake
the person’s hand because your hand is unavailable at the moment. It‘s kind of the
same way with molecules. When two molecules react with each other, there are spe—
ciﬁc sites on the molecules that are reacting with each other. If those sites cannot get
close to each other, the reaction won’t happen. There will be many times in the second half of this course when you will be
trying to determine which way a reaction will proceed from two possible outcomes.
Many times, you will choose one outcome. because the other outcome has steric
problems to overcome (the geometry of the molecules does not permit the reactive
sites to get close together). In fact, you will learn to make decisions like this as soon
as you learn your ﬁrst reactions: 8N2 versus SNl reactions. Now that we know why
geometry is so important, we need to brush up on some basic concepts. To determine the geometry of an entire molecule, we need to be able to deter
mine the geometry of each atom based on how it is connected to the atoms around
it. After all. that is what determines the geometry—how the atoms are connected in
3D space. Since atoms are connected to each other with bonds. it makes sense that
we need to take a close look at bonds. In particular, we need to know the exact lo—
cations and angles of every bond to every atom. This might sound difﬁcult. but it is
actually straightforward. and with a little bit of practice, you can get to the point
where you know the geometry of a molecule as soon as you look at it (without even
needing to think about it]. That is the point that we need to get to, and that is what
this Chapter is all about. 4.1 ORBITALS AND HYBRIDIZATION STATES To determine the geometry of a molecule, we need to know how atoms bond with
each other three dimensionally. so it makes sense for our discussion to start with or—
bitals. After all, bonds come from overlapping orbitals. 76 4.1 ORBITALS AND HYBRIDIZATION STATES A bond is formed when an electron of one atom overlaps with an electron of
another atom. The two electrons are shared between both atoms. and we call that a
bond. Since electrons exist in regions of space called orbitals. then what we really
need to know is; what are the locations and angles of the orbitals around every atom?
It is not so complicated, because the number of possible arrangements of orbitals is
very small. You need to learn the possibilities. and how to identify them when you
see them. So, we need to talk about orbitals. There are two simple orbitals: s and p orbitals (we don’t really deal with d and
f orbitals in organic chemistry). 5 orbitals are spherical and p orbitals have two lobes
(one front lobe and one back lobe): s orhital p orbital Atoms in the second row (such as C. N, O. and F) have one s orbital and three p
orbitals in the valence shell. These orbitals are usually mixed together to give us hy
bridized orbitals (sp3, sp2. and sp). We get these orbitals by mixing the properties of
s and p orbitals. What do we mean by mixing? Imagine one swimming pool shaped like a triangle and another shaped like a
pentagon; now we put them next to each other. We wave a magic wand and they
magically turn into two rectangular pools. That would be a neat trick. That’s what sp
orbitals are: we take one s orbital and one p orbital, then wave a magic wand, and
poof—we now have two equivalent orbitals that look the same. The two new orbital
have a different shape from the original two orbitals. This new shape is somewhat of
an average of the two original shapes. If we mix two p orbitals and one s orbital, then we get three equivalent sp2 or—
bitals. Let‘s go back to the pool analogy. Imagine two pools shaped like octagons and
one shaped like a triangle. We wave our magic wand and get three pools shaped like
hexagons. We started with three pools and we ended with three pools. But the three
pools in the end are all the same and their shape is an average of the shapes of the
original three pools. The same thing is true here with orbitals. We start with three or~
bitals (two p orbitals and one s orbital). Then we mix them together and end up with
three orbitals that all look the same. All three new orbitals have the “average” prop
erties of the original three orbitals. The three new orbitals (since they came from one
s orbital and two p orbitals) are called sp2 orbitals. Similarly, when you combine
three p orbitals and one s orbital, you get four equivalent sp3 orbitals. To truly understand the geometry of bonds. we need to understand the geom
etry of these three different hybridization states. The hybridization state of an atom
describes the type of hybridized atomic orbitals (sp3. spz, or sp) that contain the va—
lence electrons. Each hybridized orbital can be used either to form a bond with an—
other atom or to hold a lone pair. 78 CHAPTER 4 GEOMETRY It is not difﬁcult to determine hybridization states. If you can add, then you
should have no trouble determining the hybridization state of an atom. Just count
how many other atoms are bonded to your atom. and count how many lone pairs
your atom has. Add these numbers. Now you have the total number of hybridized or—
bitals that contain the valence electrons. This number is all you need to determine
the hybridization state of the atom. That probably sounded complicated, so let’s look
at an example to clear it up. Consider the molecule below: 0 A H H Let’s try to determine the hybridization state of the carbon in the center. We begin
by counting the number of atoms connected to this carbon atom. There are 3 atoms
(0, H. and H). The oxygen atom only cormzs as (me. Next we count the number of lone pairs on the carbon atom. There are no lone
pairs on the carbon atom. (If you are not sure how to tell that there are no lone pairs
there, go back to Chapter I and review the section on counting lone pairs.) Now
we take the sum of the attached atoms and the number of lone pairs—in this case,
3 + 0 = 3. Therefore, three hybridized orbitals are being used here. That means that
we have mixed two p orbitals and one s orbital (a total of three orbitals) to get three
equivalent sp2 orbitals. Thus, the hybridization is spz. Let’s take a closer look at how
this works. Recall that the second row elements have three p orbitals and one s orbital that
can be hybridized in one of three ways: sp3, sp3, or sp. If we are using three hy~
bn'dized orbitals, then we must have mixed two 1) orbitals with one s orbital: 4 original orbitals 4 orbitals after hybridization Used to form the second bond / I to the oxygen by overlapping
with a p orbital of the oxygen. e Three hybridized orbitals
used to form three bonds.
No hybridized orbitals are
needed for lone pairs. So here‘s the rule: Just add the number of bonded atoms to the number of lone pairs.
The number you get tells you how many hybridized orbitals you need according to
the following: 4.1 ORBITALS AND HYBRIDIZATION STATES 79 if the sum is 4. then you have 4 sp3 orbitals If the sum is 3, then you have 3 5p2 orbitals and one p orbital (as in our example) If the sum is 2, then you have 2 sp orbitals and two p orbitals There is one exception. which you will see in the chapter on aromaticity in your text
book. For now, let’s not worr)l about it. Once you get used to looking at drawings of molecules, you should not have
to count anymore. There are certain arrangements that are always sp3 hybridized.
and the same is true for sp3 and sp. Here are some common examples: 3 .. , .
Sp —‘T* //N\ /0\ If you can determine the hybridization state of any atom. you will be able to easily
determine the geometry of that atom. Let’s do another example. EXERCISE 4.1 Identify the hybridization state for the nitrogen atom in ammonia
(NH3). Answer First we need to ask how many atoms are connected to this nitrogen
atom. There are three hydrogen atoms. Next we need to ask how many lone pairs the
nitrogen atom has. It has 1 lone pair. Now. we take the sum. 3 + l = 4. [f we need
to have four hybridized orbitals. then the hybridization state must be so}. PROBLEMS For each compound below. identify the hybridization state for the
central carbon atom. 0 Cl
JL 4.2 HO OH 4.3 4.5 )95 4.6 4.4 CI ClCI 02020
G) 4.7 H—CECCH3 80 CHAPTER4 GEOMETRY 4.8 For each carbon atom in the Following molecule. identify the hybridization
state. Do not forget to count the hydrogen atoms (they are not shown). Use the fol«
lowing simple method: A carbon with 4 single bonds is sp3 hybridized. A carbon with
a double bond is sp2 hybridized, and a carbon with a triple bond is sp hybridized. 00 Once you get used to it, you do not need to count anymore—just look at the
number of bonds. If carbon has only single bonds, then it is sp3 hybridized. If the
carbon atom has a double bond. then it is sp2 hybridized. If the carbon atom has a
triple bond, then it is sp hybridized. Consult the chart of common examples on the
previous page. 4.2 GEOMETRY Now that we know how to determine hybridization states, we need to know the
geometry of each of the three hybridization states. One simple theory explains it all.
This theory is called the valence shell electron pair repulsion theory (VSEPR).
Stated simply, all orbitals containing electrons in the outermost shell (the valence
shell) want to get as far apart from each other as possible. This one simple idea is all
you need to predict the geometry around an atom. First. let’s apply the theory to the
three types of hybridized orbitals. 1. Four equivalent sp3—hybridized orbitals achieve maximum distance from
one another when they atTange in a tetrahedral structure: Think of this as a tripod with an additional leg sticking straight up in the air. In this
arrangement, each of the four orbitals is exactly 109.53 from each of the other three
orbitals. 4.2 GEOMETRY 81 2. Three equivalent spZ—hybridizeti orbitals achieve maximum distance from
one another when they arrange in a trigonal planar structure: All three orbitals are in the same plane, and each one is 120° from each of the other
orbitals. The remaining p orbital is orthogonal to (perpendicular to the plane of) the
three hybridized orbitals. 3. Two equivalent sphybridized orbitals achieve maximum distance from one
another when they arrange in a linear structure: Both orbitals are 1803 from each other. The remaining two p orbitals are 90° from
each other and from each of the hybridized orbitals.
So far its very simple: 1. Sp3 = tetrahedral
2. sp2 = trigonal planar 3. sp = linear But here’s where students usually get confused. What happens when a hybridized or
bital holds alone pair? What does that do to the geometry? The answer is that the geom
etry of the orbitals does not Change, but the geometry of the molecule is affected. Why? Let's look at an example. In ammonia (NIl3), the nitrogen atom is sp3 hy
bridized. so (II/four orbitals arrange in a tetrahedral structure, just as we would ex
pect. But only three of the orbitals in this arrangement are responsible for bonds. So,
if we look just at the atoms that are connected. we do not see a tetrahedron. Rather.
we see a trigonal pyramidal arrangement: 82 CHAPTER4 GEOMETRY Trigona], because there are three bonds pointing away from the central nitrogen
atom, and pyramidal because it‘s shaped like a pyramid. Similarly, in H20. the oxygen is sp3 hybridized. So all four orbitals are in a
tetrahedral arrangement. just as we would expect for an sp3 hybridized atom. But
only two of the orbitals are being used for bonds. So if we look just at the atoms that
are connected, we do not see a tetrahedron. Rather, we see a bent arrangement: Let’s now put all of this information together: sp3 with O lone pairs = tetrahedral sp‘ with l lone pair = trigonal pyramidal
sp3 with 2 lone pairs = bent sp2 with 0 lone pairs = trigonal planar
sp2 with l lone pair = bent sp with O lone pairs = linear That’s it. There are only six different types of geometry that we need to know. First
we determine the hybridization state. Then, using the number of lone pairs. we can
ﬁgure out which of the six different types of geometry we are dealing with. Let’s try
it out on a problem. EXERCISE 4.9 Identify the geometiy of the carbon atom below:
0 JL H H Answer First, we need to determine the hybridization state. We did this for this
molecule earlier in this chapter and found that the hybridization state is sp2 (there are
3 atoms connected and no lone pairs, so we need three hybridized orbitals; therefore,
it is spz). Next we remind ourselves how many lone pairs there are; in this case, there
are none. So the geometry must be trigonal planar. Once you can determine the geometry around an atom, you should have no
problem determining the geometry, or shape, of a molecule. Simply repeat your
analysis for each and every atom in the molecule. This may seem like a large task at 4.2 GEOMETHY 83 ﬁrst, but once you get the hang of it, you will be able to tell the geometry of an atom
immediately upon seeing it. For the next set of problems, you should get to the point where you can do
these problems very quickly. The ﬁrst few will take you longer than the last ones. If
the last problem is still taking you a long time. then you have not mastered the pro
cess and you will need more practice. If this is the case, open to any page in the sec
ond half of your textbook. You will probably see drawings of structures. Point to any
atom in a structure and try to determine what the geometry is. Use the list above to
help you. Go from one drawing to the next until you can do it without the list. That
is the important part—doing it without needing the list. PROBLEMS For each compound below, identify the hybridization state and
geometry for every atom in the compound. 4.10 : 4.11 >_< o O
\
.0 Q ' NH
N
4.12 4.13 l
O O /
/
f *0
N l
4.14 4.15 4.16 4.17 ...
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This note was uploaded on 07/31/2010 for the course CHEM 241 taught by Professor Burnett during the Fall '07 term at UVA.
 Fall '07
 Burnett

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