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Current time:0:00Total duration:10:21

Voiceover: Now that we
understand hybridization states, let's do a couple of examples, and so we're going to
identify the hybridization states, and predict the geometetries for all the atoms in this molecule, except for hydrogen, and so, let's start with this carbon, right here. And so, the fast way of
identifying a hybridization state, is to say, "Okay, that carbon has "a double bond to it; therefore, it must "be SP two hybridized." And if it's SP two hybridized, we know the geometry around that
carbon must be trigonal, planar, with bond angles
approximately 120 degrees. This carbon over here,
also has a double-bond to it, so it's also SP two hybridized, with trigonal planar geometry. All right, let's move to
this carbon, right here, so that carbon has only
single bonds around it, and the fast way of
doing it, is if you see all single bonds, it must
be SP three hybridized, and if that carbon is SP three hybridized, we know the geometry is tetrahedral, so tetrahedral geometry
with ideal bond angles of 109 point five degrees
around that carbon. All right, let's move over to this carbon, right here, so this
carbon has a triple-bond on the right side of
it, and so the fast way of doing this, is if it has a triple-bond, it must be SP hybridized
here, so SP hybridized, and therefore, the
geometry would be linear, with a bond angle of 180 degrees. And, same with this
carbon; this carbon has a triple-bond to it, so it also must be SP hybridized with linear geometry, and so that's why I drew it
this way, so it's linear around those two carbons, here. Let's go ahead and count
up the total number of sigma and pi bonds for this, so that's also something we talked about in the previous videos here. So, first let's count up
the number of sigma bonds, so let's go back over to
start with this carbon, here. So here's a sigma bond to that carbon, here's a sigma bond to
that carbon; we know that our double-bond, one of
those bonds is a sigma bond, and one of those bonds is a pi bond, so let me go ahead, and also draw in our pi bonds, in red. So, already colored the
sigma bond blue, and so let's say this one is the pi bond. All right, let's continue
assigning all of our bonds here. So I know this single-bond
is a sigma bond, I know this single-bond is a sigma bond, so all of these single
bonds here are sigma. When I get to the triple
bond, I know one of those is a sigma bond, and two
of those are pi bonds. So, two of those are pi bonds, here. And then, finally, I have one
more bond; it's a single-bond, so I know that it is a sigma bond here, and if you count up all
of those sigma bonds, you should get 10, so let's
do that really quickly. So you get, let me go ahead
and change colors here, so you get one, two,
three, four, five, six, seven, eight, nine, and 10; so we have 10 sigma bonds total, and
in terms of pi bonds, we had three pi bonds, so three pi bonds for this molecule. You can also find hybridization states using a steric number, so let's go ahead and do that really quickly. So let's go back to this
carbon, and let's find the hybridization state of that carbon, using steric number. So let's use green for
this, so steric number is equal to the number of sigma bonds, plus lone pairs of electrons. So, one, two, three sigma
bonds around that carbon. So three plus zero gives me
a steric number of three, therefore I need three hybrid orbitals, and SP two hybridization gives
me three hybrid orbitals. All right, if I wanted
to do for this carbon I would have one, two, three
four; so the steric number would be equal to four sigma
bonds, and zero lone pairs of electrons, giving me a total of four for my steric numbers, so I
need four hybrid orbitals; I have four SP three hybridized
orbitals at that carbon. And then finally, let's
do it for this carbon, right here, so using steric number. Steric number is equal
to number of sigma bonds, plus numbers of lone pairs of electrons, so there are two sigma
bonds around that carbon, zero lone pairs of electrons,
steric number of two, means I need two hybridized orbitals, and an SP hybridization,
that's what you get: You get two SP hybridized
orbitals, like that. All right, let's move
onto another example; let's do a similar analysis. Before we do, notice I
excluded hydrogen here, and that's because hydrogen is only bonded to one other atom, so
there's no real geometry to talk about. All right, let's move on to this example. So this molecule is diethyl
ether, and let's start with this carbon, right here,
so the hybridization state. Well, the fast way of
doing it, is to notice that there are only
single-bonds around that carbon, only sigma bonds, and
so, therefore we know that carbon is SP three hybridized, with tetrahedral geometry,
so SP three hybridized, tetrahedral geometry. All right, let's look at
this carbon right here; it's the exact same situation, right, only sigma, or single bonds around it, so this carbon is also
SP three hybridized, and so, therefore tetrahedral geometry. Let's next look at the
oxygen here, so if I wanted to figure out the
hybridization and the geometry of this oxygen, steric
number is useful here, so let's go ahead and calculate the steric number of this oxygen. So that's number of sigma bonds, so here's a single-bond, so that's a sigma bond, and then here's another one; so I have two sigma bonds, so two plus
number of lone pairs of electrons around the
atom, so here's a lone pair of electrons, and here's
a lone pair of electrons. So, I have two lone pairs of electrons, so two plus two gives me
a steric number of four, so I need four hybridized
orbitals for this oxygen, and we know that occurs when you have SP three hybridization, so therefore, this oxygen is SP three hybridized: There are four SP three hybrid
orbitals around that oxygen. All right, let's do
geometry of this oxygen. So, the electron groups,
there are four electron groups around that oxygen, so each electron group is in an SP three hydbridized orbital. The geometry of those electron groups might be tetrahedral, but not the geometry around the oxygen here, so the
geometry around the oxygen, if you ignore the lone pairs of electrons, you can see that it is
bent, so even though that oxygen is SP three
hybridized, it's geometry is not tetrahedral; the geometry of that oxygen there is bent or angual. All right, and because
of symmetry, this carbon right here is the same as
this carbon, so it's also SP three hybridized, and
then this carbon over here is the same as this carbon, so it's also SP three hybridized, so symmetry made our
lives easy on this one. All right, let's do one more example. So, once again, our goal is
to find the hybridization states, and the geometries
for all the atoms, except for hydrogen, and so, once again, let's start with carbon; let's start with this carbon, right here. All right, so once again,
our goal is to find the hybridization state, so
the fast way of doing it, is to notice there's one
double-bond to that carbon, so it must be SP two
hybridized, and therefore the geometry is trigonal planar, so trigonal planar geometry. Let's do the steric
number way, so if I were to calculate the steric number: Steric number is equal to
the number of sigma bonds. So here's a sigma bond,
here's a sigma bond; I have a double-bond between
the carbon and the oxygen, so one of those is a sigma bond, and one of those is a pi bond,
which I'll draw in red here. So I have three sigma
bonds around that carbon, so three plus zero lone
pairs of electrons, gives me a steric number
of three, so I need three hybridized orbitals,
and so once again, SP two hybridization. All right, let's do the next carbon, so let's move on to this one. So, I see only single-bonds
around that carbon, therefore, it must be SP three hybridized, with tetrahedral geometry,
so SP three hybridized, tetrahedral geometry. Same thing for this carbon,
only single-bonds around it, only sigma bonds, so
it's SP three hybridized, with tetrahedral geometry. Let's finally look at this nitrogen here. So if I want to find the
hybridization state of this nitrogen, I could use steric number. So the steric number is equal
to number of sigma bonds. So around this nitrogen, here's a sigma bond; it's a single bond. Here's another one,
and here's another one, so I have three sigma bonds. I have one lone pair of electrons, so three plus one gives me
four, a steric number of four, means I need four hybridized orbitals, and that's our situation
with SP three hybridization. And so, this nitrogen
is SP three hybridized, but it's geometry is
not tetrahedral, so the geometry for that
nitrogen, as we discussed in an earlier video, so it has these three sigma bonds like this, and a lone pair of electrons, and that
lone pair of electrons is in an SP three hybridized orbital. And if we look at that
geometry, and ignore the lone pair of electrons,
'cause you always ignore the lone pairs of
electrons, when you're looking at geometry, we can see, we have this sort of shape here, so the nitrogen's bonded to three atoms:
the carbon, hydrogen, and hydrogen, and then we have this sort of a shape, like that,
so in the back there, and you can see, we call
this trigonal-pyramidal, so the geometry around that
nitrogen is trigonal pyramidal. All right, so that does
it for three examples of organic hybridization,
so practice a lot for this.