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## Polarity of bonds

Current time:0:00Total duration:9:22

# Molecular polarity

AP.Chem:

SAP‑4 (EU)

, SAP‑4.C (LO)

, SAP‑4.C.1 (EK)

, SAP‑4.C.2 (EK)

## Video transcript

Now that we understand
how to draw dot structures and we know how to predict
the shapes of molecules, let's use those
skills to analyze the polarity of
molecules, using what's called the dipole moment. So to explain what
a dipole moment is, let's look at this situation
over here on the right, where we have a positively
charged proton some distance away from a
negatively charged electron. And let's say they're separated
by a distance of d here. We know that a proton
and an electron have the same
magnitude of charge, so both have a magnitude of
charge Q equal to 1.6 times 10 to the negative 19. So of course, a proton would
have positively charged Q, so let's go ahead and make
this a positively charged Q. And an electron would have
a negatively charged Q, like that. If we were to calculate the
dipole moment, the definition of a dipole moment, symbolized
by the Greek letter mu, dipole moment is
equal to the magnitude of that charge, Q,
times the distance between those charges, d. So mu is equal to Q times d. And we're not really going to
get into math in this video, but if you were to go ahead
and do that calculation, you would end up with
the units of Debyes. So you would get a
number, and that number would be in Debyes here. So we're more concerned with
analyzing a dipole moment in terms of the molecular
structure, so let's go ahead and look at the dot
structure for HCl. So if I look at this covalent
bond between the hydrogen and the chlorine, I know
that that covalent bond consists of two electrons. And chlorine is more
electronegative than hydrogen, which means that those
two electrons are going to be pulled closer
to the chlorine. So I'm going to go ahead and
show that here with this arrow. The arrows is pointing in
the direction of movement of electrons, so those
electrons in yellow are going to move closer
towards the chlorine. So chlorine is going to get
a little bit more electron density around it,
and so we represent that with a partial
negative charge. So we do a lowercase
Greek delta here, and it's partially
negative since it has increase in
electron density, one way of thinking about it. And since hydrogen is losing a
little bit of electron density, it's losing a little
bit of negative charge, and so it is partially positive. So we go ahead and draw a
partial positive sign here. And so we're setting
up a situation where we are polarizing
the molecule. So this part of the molecule
over here on the right is increasing electron
density, and so that is our partial negative side. That's one pole. And then this other side here
is losing some electron density, and so it's partially positive,
so we have it like that. So that's where the
positive sign comes in. You can think about
on this arrow here, this little positive sign giving
you the distribution of charge in this molecule. And so you have these two
poles, a positive pole and a negative pole. And if you think
about those two poles as having a center
of mass, you could have a distance
between them, and you could calculate the dipole
moment for this molecule. And so when you calculate
the dipole moment for HCl, mu turns out to be equal to
approximately 1.11 Debyes. And so we have a
polarized bond, and we have a polarized molecule. And so therefore we can say
that HCl is relatively polar. It has a dipole moment. So that's kind of how
to think about analyzing these molecules. Let's do another one here. Let's do carbon dioxide. So I know that the CO2
molecule is linear, so after you draw the
dot structure you're going to get a linear shape,
which is going to be important when we're trying to
predict the dipole moment. If I analyze the electrons
in this carbon-oxygen bond-- so we have a double bond between
carbon and oxygen-- oxygen is more electronegative
than carbon. So oxygen's going to try to
pull those electrons closer to itself. And so we go ahead and draw
our arrow or vector pointing towards the right here. And so we have a bond
dipole situation here. On the left, we have the
exact same situation. Oxygen is more
electronegative than carbon, and so these electrons are
going to be pulled closer to this oxygen. So we draw another arrow or
another vector in this case. So even though we have these
individual bond dipoles, if you think about this
molecule as being linear-- and you can see we
have these two vectors that are equal in magnitude,
but opposite in direction-- those two vectors are
going to cancel out. And therefore we
would not expect to have a dipole moment
for the molecule. There's no molecular
dipole here. So mu turns out
to be equal to 0. A simplistic way of
thinking about this would be like a tug of war. You have these really
strong atoms, these oxygens, but they're equally strong. And if they're pulling
with equal force in opposite directions,
it's going to cancel out. So the individual bond
dipoles cancel out, so there's no overall dipole
moment for this molecule. And carbon dioxide is
considered to be nonpolar. Let's go ahead and analyze
a water molecule over here on the right. So the electrons in
this covalent bond between the hydrogen
and oxygen, oxygen is more electronegative
than hydrogen, so those electrons are going to
be pulled closer to the oxygen. Same thing for this
bond over here. And we also have lone pairs of
electrons on our central atom to think about. And that's of course going
to increase the electron density going in this
direction for that lone pair and in this direction
for that one pair. And so even though we know the
geometry of the water molecule is bent, and it's
hard to represent that on this two-dimensional
surface here. If you use a molymod
set, you will kind of see that your net dipole
moment would be directed upward in this case. And so the individual
bond dipoles are going to add to give you a
molecular dipole, in this case pointed up, and so
therefore you're going to have a dipole moment
associated with your water molecule. So mu turns out to be
approximately 1.85, and we could consider water
to be a polar molecule. Let's do two more examples. So on the left is CCl4,
or carbon tetrachloride. And so you can see that
we have a carbon bonded to chlorine here, and since
this is a straight line, this means in the
plane of the page. And so we know the
geometry is tetrahedral around out this carbon,
so let's go ahead and analyze that as well. So I have a wedge
drawn here, which means this chlorine is
coming out at you in space. And then I have a dash back here
meaning this chlorine back here is going away from you in space. So that's how to think about
it, but it's really much easier to go ahead and make
this using a molymod set. And you can see that however
you rotate this molecule, it's going to look the
same in all directions. So a tetrahedral
arrangements of four of the same atoms
around a central atom, you can turn the molecule over. It's always going to look
the same in three dimensions. And that's really
important when you're analyzing the dipole
moment for this molecule. So let's go ahead and do that. We'll start with our
electronegativity differences. So if I look at this top
carbon-chlorine bond-- these two electrons in this top
carbon-chlorine bond-- chlorine is more electronegative
than carbon. And so we could think
about those electrons being pulled closer to the chlorines. Let me go ahead and
use green for that. So those two electrons are
going in this direction. And it's the same thing
for all of these chlorines. Chlorine is more
electronegative than carbon, so we can draw these
individual bond dipoles. We can draw four of them here. And in this case we
have four dipoles, but they're going to cancel
out in three dimensions. So again, this is a
tough one to visualize on a two-dimensional surface. But if you have the
molecule in front of you, it's a little bit easier to
see that if you keep rotating the molecule, it looks the same. And so these individual
bond dipoles cancel, there's no dipole moment
for this molecule, and so mu is equal to 0. And we would expect the
carbon tetrachloride molecule to be nonpolar. Let's look at the
example on the right, where we have substituted
in a hydrogen for one of the chlorines. And so now we have
CHCl3, or chloroform. So now if we analyze
the molecule-- so let's think about
this bond in here-- carbon is actually a little
bit more electronegative than hydrogen, so we
can show the electrons in that bond in red moving
towards the carbon this time. And once again, carbon
versus chlorine, chlorine is more electronegative,
so we're going to have a bond dipole
in that direction, which we can do for all
our chlorines here. And so hopefully it's a little
bit easier to see in this case. In this case, the
individual bond dipoles are going to combine to give
you a net dipole located in the downward direction
for this molecule. So I'm attempting to draw the
molecular dipole, the dipole for the entire molecule,
going a little bit down in terms of how I've
drawn this molecule. And so since we have
a hydrogen here, there's no upward
pull in this case to balance out
the downward pull. And so we would
expect this molecule to have a dipole moment. And so mu turns out to
be approximately 1.01 for chloroform,
so it is certainly more polar than our carbon
tetrachloride example.