Chapter 9 Covalent Bonding
In Chapter 6 we used the wave function to understand the orbitals of an individual
atom, and we talked about the different shapes of the s, p, and d orbitals In the last
chapter we started looking at the structure of fairly complicated molecules and came up
some interesting geometries for central atoms that bore no little or no resemblance to
the atomic geometries we had previously. In this chapter we will bridge the gap
between atoms and molecules and show where the orbitals that will be used to make
covalent bonds come from.
9-1 Molecular orbitals
In chapter 6 we put a single + proton in the center of an atom, and then used the
Schrödiner equation to solve the wave function that told us where the electron
could be. We then moved on to He. We put two protons in the middle and
solved the Schrödinger equation for two electrons.
Now all we have to do is to not put the two proton in the center, but put a little bit
of distance between then and now solve the Schrödinger equation for this
situation.
See Figure 9.1
As you can see. When the two protons are widely separated, we get two
individual 1s orbitals. But as the two nuclei get closer together the electrons start
to overlap, and we get a large electron density between the two atoms that
represents the electrons holding the atoms together in a covalent bond.
Key Concept:
A molecular orbital is a wave function ( or orbital) that extends over two (or more)
atoms.
The calculation we get by doing this with a computer is in excellent agreement
with reality. If give the distance between the two atoms in the H2 molecule as 74
pm exactly what we find in experiments
9-2 Molecular orbitals of H2+
OK, I jumped the gun a little. To really to this right I should start with a simpler
system: 2 separate protons sharing a single electron, This corresponds to the H2+
ion, the hydrogen molecular ion. This is actually good because the H2+ ion is
stable. It occurs in those hydrogen lamps when you run current through
hydrogen gas. As a stable ion it can actually be studied, and we can use the
experimental data to confirm or deny the theory that I am about to build for you.
2
When you solve the Schrödinger equation for this system you get the various
orbital shown in Figure 9.2 In this figure the orbitals with the lowest energy are at
the bottom, and the ones with the highest energy are at the top.
Let me explain the labels. Look at the bottom 4 orbitals. Notice how the
electrons are along the line connecting the atoms? (Called the internuclear axis).
We call these orbital ó orbitals
Key Concept:
ó molecular orbitals are orbital in which the electrons are located along the
internuclear axis, the line connecting the atoms
Notice how in the lowest ó orbital the orbital has an electron density in the space
between the nuclei, but in the next highest orbital there is a node, so there are
no electrons between the atoms?
In the lowest orbital the electron density between the atoms would serve to hold
the positive nuclei of the two atoms together. In the other orbital, with no
electron density between the nuclei, there would be nothing to hole the atoms
together so the molecule would not be stable. To differentiate between these
two orbital we call the lower orbital the bonding (or probonding) orbital,
designated ó, and we call the upper orbital an anti-bonding orbital, designated ó*
Key Concept:
Probonding orbital have electron density between atoms to hold molecules
together. Antibonding orbitals (designated with a *) do not have an electron
density between nuclei and serve to destabilize a molecule.
There is another way to get these orbitals. Rather than placing two nuclei in
space and solving the Schrödinger equation for the system, we can simply start
with the s orbitals for the two different atoms and simply add them together or
subtract them from each other (Figure 9.3)
This is where the subscripts like 1s and 2s come from on the diagram. The
lowest energy ó bond that comes from adding the 1s orbitals of the two atoms
together is designated ó1s, while the antibonding orbital that comes from
subtracting one atom’s 1s orbital from the other atoms is designated ó*1s
As can be seen in Figure 9.4 the energy of the ó1s orbital does have a negative
energy, so it would hold the atoms together. Not only that, but it has a minimum
at about 100 pm that corresponds nicely with the interatomic distance in the H2+
ion Figure 9.4. On this same figure you can see that the energy for the ó* bond
is positive at all distances, so it cannot hold a molecule together.
3
The next two orbitals are the ó2s and the ó*2s. Same story, second verse. The
only comment I would have is that you know the 1s orbitals are smaller than the
2s orbitals, so there really should be a difference in size that is not shown in this
figure.
Next we move up to the p orbitals. Remember that the p orbital is the peanut
shape, and there are three of them, one along each major axis Figure 9.6
Not only that, but here is another thing they didn’t tell you in kindergarten (or
chapter 5 for that matter). When we solved the Schrödinger equation, that
wavefunction we came up with actually has a sign associated with different parts
of it. In this case the sign on one lobe is + while the sign on the other lobe is - .
Now let’s start combining orbitals, by starting with the p orbital that lie along the
line connecting the atoms. Figure 9.7.
Again there is one orbital that has lots of density between the atoms to hold them
together. This orbital is called a ó2pz orbital because it has electron density along
the line connecting the atoms so it is ó, but it came from the overlap of 2p
orbitals so we give it a 2p subscript. Since we assume that the z axis is the one
connecting the two atoms we add the z subscript to say this came specifically
from the pz atomic orbitals
There is also a second orbital along the line connecting the atoms, but it doesn’t
have enough electron density between the atoms to hold them together so this is
a ó*2pz
What about the other two p orbitals, the ones to either side of the line connecting
the atoms along the X and Y planes? Figure 9.8
For each pair of p orbitals we get two molecular orbital. These orbital are not
along the line connecting the atoms so we call them ð orbitals
Key Concept:
ð molecular orbital have electron density on either side of the line connecting the
atoms
One of the ð orbitals has a lot of density between the atoms, but not right
between them, and this is enough to stabilize the molecule so this becomes a
probonding ð2p orbital, and since we have one in each plane we have two
orbitals ð2px and ð2py, while the other orbitals has nothing near the middle, so
these become two antibonding ð*2px and ð*2py orbitals.
4
One final note before we leave this section. We did not try to combine orbitals
with n=1 with orbitals with n=2, also we did not try to combine s orbital with p
orbitals. While you can do this with the computer, you rapidly find out that the
only orbitals that combine to make good combinations are the one with the same
energy
9-3 Bond Order
Now that we know that we can calculate and look at the orbitals to a 2 atoms
system, we can start adding more charges on the nuclei and more electrons and
move on to diatomic molecules more complicated than H2+.
The way we will do this is simple, just like filling in the atomic orbitals. You
arrange the orbitals from lowest to highest energy, and then start putting pairs of
electrons into the orbitals starting at the lowest energy orbital
For instance: H2+, H2, H2-, He2+ and He2
ó*1s
ó1s
8
H2+
89
H2
8
89
H2-
8
89
89
89
He2+ He2
Not only can we start filling in orbitals, but knowing that probonding orbitals
stabilize molecules, and antibonding orbitals destroy molecules, we can evaluate
if we have a stable molecule.
We do this with a simple calculation called the bond order
Key Equation/Concept:
Applying the to the above example
ó*1s
ó1s
BO
8
H2+
½
89
H2
1
8
89
H2½
8
89
He2+
½
89
89
He2
0
This says that H2 has a single bond, H2- and He2+ have a very weak bond
equivalent to about ½ of a normal bond, and He2 is not stable and does not exist
See table 9.1
5
A bond order of 2 indicates a double bond, and a bond order of 3 tell you that
you have a triple bond.
9-4 Molecular electron configurations
So what happens beyond H2 and He2
Let’s try Li2 (Captain Kirk, we need more dilithium crystals)
For Li2 we will have 6 electrons, two from each Li, and we just fill in or molecular
orbital ‘ladder’
Higher orbitals
ó*2s
ó 2s 89
ó*1s 89
ó1s
89
Or, in linear form (ó1s)2 (ó*1s)2 (ó2s)2
With a bond order of (4-2)/2 =1, saying we have a single bond in the molecule
Now for the tricky part. Remember how all the orbitals at a given n level were
degenerate for the H atom, but when we added more electrons the energy levels
were not degenerate, but the s was a lower energy than the p? And then ass we
added more and more charge to the nucleus things shifted more and more so
the 4s orbital has a lower energy than the 3d orbital?
The same kind of shifting goes on with molecular orbitals as you add more
charge to the atoms and more electrons to the system. Look at
Figure 9.11. In particular notice how the ó2pz orbital energy level is above the
ð2px,y energy levels from Li2, Be2, B2, C2, and N2, but is lower in O2, F2, and Ne2.
This change in order is important so you need to memorize it.
Key concept:
Remember the two different ‘ladders’ of molecular orbital bond energy
Li, Be, B, C, N
ó*2p ___
ð*2px,y ___ ___
ó2pz ___
ð2px,y ___ ___
ó*2s ___
ó2s
___
ó*1s ___
ó1s
___
O, F, Ne
ó*2p ___
ð*2px,y ___
ð2px,y ___
ó2pz ___
ó*2s ___
ó2s
___
ó*1s ___
___
ó1s
___
___
6
One of the important outcomes of the molecular theory is that it correctly predicts
another property of these compounds, paramagnetism and diamagnetism.
First of all what are paramagnetism and diamagnetism.
Key concepts:
A paramagnetic substance is a substance that is weakly attracted to a magnetic
field. It occurs in compounds that have unpaired electrons.
A diamagnetic substance is one that is slightly repelled by a magnetic field. In
occurs in compounds where all the electrons are paired together.
Practice problems:
Construct the molecular orbital ‘ladder’ for N2, O2 and F2using that ladder predict determine the Bonds order, the type of bond the relative
length of the bond and the magnetic properties of each diatomic species
N2
Electrons:
7x2=14
ó*2p
ð*2px,y
ó2pz 89
ð2px,y 89 89
ó*2s 89
ó2s
89
ó*1s 89
ó1s
89
BO
(10-4)/2 =3
Bond
triple bond
Length
Shortest
Magnetic properties
diamagnetic
repelled
O2
8x2=16
ó*2p
ð*2px,y
ð2px,y 8 8
ó2pz 89
ó*2s 89
ó2s
89
ó*1s 89
89
ó1s
F2(9x2) +1=19
ó*2p 8
ð*2px,y 89 89
ð2px,y 89 89
ó2pz 89
ó*2s 89
ó2s
89
ó*1s 89
89
ó1s
(8-4)/2=2
(10-9)/2=1/2
double bond
weak single (½ a bond)
Medium
Longest
Paramagnetic
attacted
Paramagnetic
Attracted
Notice that in solving this problem with O2 we used Hund’s rule: spread
electrons between all degenerate orbitals before you pair them up.
You can use the same molecular orbital energy ‘ladders’ on heteronuclear (two
different atoms) diatomic molecules as well, with two cautions. First the two
atoms have to be close on the periodic table. Second, both atoms have to use
the same ‘ladder’
7
So you can do things like CN, CN- (The cyanide ion), or OF. You actually can do
NO, but you need to be told which ‘ladder’ to use (the Li-N ladder)
9-5 sp Orbitals
Well, we have taken care of diatomic molecules. What about all those
interesting molecules with 3,4, or more atoms? The polyatomic molecules?
The purist carries on just as we have done. We put in more nuclei and more
electrons and solve for the corresponding molecular orbitals, then fill the orbitals
from lowest to highest energy with our electrons. These molecular orbitals get
spread out across all the atoms in the molecule and can end up looking pretty
bizarre. Not only that, each molecule is a rule unto itself, so it makes
generalizing difficult. So let’s ditch that approach.
Instead we will work with a simpler model called the localize bond model. In this
model we will keep most of the electrons localized to the bond region between
the two atoms instead of letting them roam around the entire molecule.
This actually works quite well with our Lewis formulas, since each bond in this
model has a pair of electrons localized to the space between the atoms to form
the covalent bond between the atoms.
In order to describe localized covalent bonds we will introduce the concept of a
hybrid orbital.
Key Concept:
A hybrid orbital is a combination of atomic orbital on the same atom.
Sort of like we were doing earlier when we combined orbitals on different atoms
to make a molecular orbital, only now we use different orbitals on the same
atom.
Let’s start with BeH2. Your Lewis formula is H-Be-H (Be does not have a filled
octet, but that is OK for Be). VSEPR says that to get the two electron regions as
far apart as possible you need a linear structure with bond angles of 180o
As shown in figure 9.16 you can get two bonds, 180o apart if you hybridize the 2s
and the 2pz orbitals together. Notice how we have to remember that the p orbital
has + and a - lobe to do this.
We call these two new hybridized orbital sp orbital because the are derived from
one s and one p orbital
8
The next step, the, is to merge each sp orbital with the 1s orbitals on the two H
atoms to make our ó covalent bonds between the atoms. (Figure 9.18) Now we
will fill these orbital with electrons. 1 valence electron from each H and 2
valence eletrons from the Be, for a total of 4 electrons or two pairs, one for each
bond.
Notice this is a difference from our molecular orbital theory. In the molecular
orbital treatment we used all electrons, core and valence, to fill all our molecular
orbitals. When we use the localized electron model we will use only the valence
electrons and assume that the core electrons are localized to their respective
atoms, and so do not take part in the bonding orbitals
Also note: I will ignore the X and Y p orbitals because they don’t contain
electrons. I will also ignore the fact that there would actually be some ó* orbitals
as well, again because they don’t have any electrons.
9-6 sp2 Orbitals
Now try BF3
:::F-B-F:::
(Unfilled octet for B, but that is OK for B)
|
::F:
Trigonal planar, 120o bond angle
To get the right geometry here we will hybridise one s and 2 p orbitals
Figure 9.20
Again we have to invoke + and - lobes on the two p orbitals, and you have to use
your imagination to see how combining these two orbitals with the s gives you
the three equivalent sp hybrid orbitals in the x-z plane.
From this example and the last example you might have already formulated and
idea about hybrizing orbitals.
Key Concept:
The principle of conservation of orbitals - When you combine orbitals on the
same atom to form hybrid orbitals, then the number of resulting hybrid orbitals is
equal to the number of atomic orbitals that were combined.
9-7 sp3 Orbitals
I think you can do the VSEPR structure of CH4 in your head. To come up with 4
equivalent orbital to give us our tetrahedral structure, we will have to put together
4 atomic orbitals, one s and three p orbitals. Figure 9.22. And each of these
hybridized orbitals will then merge with the atomic 1s orbital of the hydrogen to
form the localized covalent bonding orbitals. Figure 9.23
9
We can use the sp3 construct to explain the structure of ethane, C2H6.
Here Figure 9.25 we string together sp3 hybridized orbitals on adjacent carbons
to get our final structure.
One thing you should remember, don’t put the cart before the horse. The
geometry of a hybridized orbital does not determine the geometry. It is the other
way around. The VSEPR theory tell you what the geometry has to be to achieve
the lowest energy state. And it is then that geometry that tells you which
hybridization you need.
9-8 Bonding in molecules with lone pairs
What do you do with lone pairs? How about something simple as water. I think
you can come up with the appropriate Lewis structure and VSEPR construct.
You should end up with 4 electrons regions, tetrahedral structure around the
central atom, and the says that you use sp3 hybridization.
Figure 9.26
In this structure you will have 6 + 2 =8 electrons or 4 pairs of electrons. There
will be two localized bonding orbitals each containing a pair of electrons: one
between an one sp3 hybridized orbital on the C and a 1s of one hydrogen and
another between a second sp3 orbital on the C and the 1s on the other H.
The two remaining pari of electrons will each be stuffed into the two remaining
sp3 orbitals. The sp3 orbital joined with the atomic 1s orbital will be a little more
constrained than the sp3 orbital that is not joined with an atomic orbital. That in
turn explains why the lone pairs electrons take up more room than the bonding
pari electrons, and why the bond angle in water is <109o
9-9 d hybrid orbitals
How do we get the 5 bonds in a trigonal bipyramid and the 6 bonds in an
octahera?
Well with our conservation of orbital theory you can probably guess we need 5
and 6 atomic orbitals so we will start mixing d orbitals in as well
To obtain a trigonal bipyramid with 5 bonds we hybridize an s, 3 p’s and a d to
make 5 sp3d orbitals
To make an octahedral with 6 bonds we hybridize an s, 3 p’s and a two d’s to
make 6 sp3d orbitals
Figures 9.30 & 9.31
10
Could you get these geometries in a second row element by hybridizing 2s,2p
and 3d orbitals together? No. Remember you can only hybridize orbital of
similar energies together so this won’t work. That is why we only see these
geometries in compounds make with a central atom in th third (or higher) row,
9-10 Double Bonds
How do we do deal with double bonds?
How about C2H4
H
\
H
/
C=C
/
\
H
H
The geometry says 3 electron regions, 120o so that says we have to use sp2
hybridization. We have 12 electrons so 6 pairs and if we make bonds with all our
hydrogen 1s orbitals and our C sp2 bonds we use up 5 of those pairs making ó
bonds between all the atoms.
So where does the last pair go and where is the double of the double bond?
In making the planar sp2 bonds we did not use the p orbital that goes above and
below the plane of the bonds. Up to now we have ignored these unused p
orbitals, but now they become important.
What we will to is to twist the molecule until the unused p orbitals in both atoms
line up. When this happens the p orbital between the atoms can join together to
make a ð bond on either side of the line connecting the atoms, and this is where
we stash the remaining pair of electrons
Figure 9.34
This is model is correct and consistent with some properties of double bonds that
we haven’t looked at before:
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Key Concepts:
1.) A double bond does NOT consist of two equal single bonds. Instead it
is made of a strong ó bond between to atoms and a second, weaker, ð
bond above and below the line connecting the atoms
2.) While the ð bond has two regions of electron density, it is only
occupied by one pair of electrons and is treated like a single bond
3.) The overlap between adjacent p orbitals that make a ð bond locks the
molecule into a planar configuration
One last thing before we leave this molecule. Let’s formally describe the bonds
in this molecule:
There are four C(2sp2) + H(1s) ó bonds
There is one C(2sp2) + C(2sp2) ó bond
There is one C(2p) + C(2p) ð bond
9-11 cis-trans isomers
I want to reiterate the last key concept I gave on double bonds. The presence fo
a double bond lock a molecule into a specific configuration and it takes a
significant amount of heat or light energy to rotate a molecule around that bond.
Why is that important? Look at the following two molecules:
1.
2.
H
H
H
Cl
\
/
\
/
C=C
C=C
/
\
/
\
Cl
Cl
Cl
H
Here are two molecules with same atom to atom connections but different spatial
arrangement are a special class of isomers called stereoisomers (Section 8-10)
Key concept:
Isomers that vary by the rotation around a double bond are called cis-trans
isomers
Cis-trans isomers can have very different physical and chemical properties. In
this case 1. Is polar and 2. Is nonpolar (Do you see why?)
#1. Is the cis isomer because we have the same substituents on the same side
#2 Is the trans isomer because the same substituents are on opposite sides
12
Cis-trans isomerization can be very important. Have you ever hear the phrase
‘low in trans fats’? Or how about the cis-trans isomerization that occurs in retinal
that lets you see. Figure 9.36
9-12 Triple bonds
You can probably figure this one out for yourself.
H-C/C-H
10 electrons, 5 pairs
Two electron regions around each C say that we use a linerar geometry, and this
tells us that we use sp hybridization.
This means we will have a C (2sp) + H (1s) ó bond between each C and H
and a C(2sp) + C(2sp) ó bond between the 2 carbons.
This leaves 2 pairs of electrons to be placed into two different C(2p) + C(2p) ð
bonds to complete our triple bond. (Figure 9.39)
9-13 Delocalized ð electrons
Last subject resonance structures or delocalized electrons
When we were doing Lewis structure we would occasionally come up with two or
more structures where the atoms don’t move, but there were different ways to
arrange the electrons. Benzene is a classic example:
Remember with resonance I said it is not one switching to the other, but rather
some kind of average between the two where the electrons in the bonds get
delocalized over more than two atoms. How does that fit in here?
Lets try it.
6C and 6H for 6(4) + 6 = 30 electrons or 15 pairs
Each C has 3 electron regions so will use sp2 hybridization.
So we will have:
6 pairs of electrons is 6 C(2sp2) + 1H ó bonds between the C and the H
6 pairs of electrons in 6 C(2sp2) + C(2sp2) ó bonds between the carbons
Figure 9.40
That leave unused p orbitals and 3 pair of electrons
We merge all 6 of the p orbitals together to make 6 molecular orbitals
(Remember conservation of orbitals?)
13
These potential molecular orbitals are shown in figure 9.41
And we put our remaining 3 pair of electrons into these orbitals as before.
You can see that these orbitals are indeed spread out over several atoms, so the
term ‘delocalized’ is truly appropriate