If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:9:47

AP.Chem:

SAP‑4 (EU)

, SAP‑4.C (LO)

, SAP‑4.C.3 (EK)

Voiceover: The concept of
steric number is very useful, because it tells us the
number of hybridized orbitals that we have. So to find the steric
number, you add up the number of stigma bonds, or
single-bonds, and to that, you add the number of
lone pairs of electrons. So, let's go ahead and do it for methane. So, if I wanted to find the steric number, the steric number is equal
to the number of sigma bonds, so I look around my carbon here, and I see one, two, three, and four
sigma, or single-bonds. So I have four sigma bonds;
I have zero lone pairs of electrons around that
carbon, so four plus zero gives me a steric number of four. In the last video, we saw
that SP three hybridized situation, we get four hybrid orbitals, and that's how many we
need, the steric number tells us we need four hybridized orbitals, so we took one S orbital,
and three P orbitals, and that gave us four,
SP three hybrid orbitals, so this carbon must be
SP three hybridized. So let's go ahead, and draw that in here. So this carbon is SP three hybridized, and in the last video, we
also drew everything out, so we drew in those four,
SP three hybrid orbitals, for that carbon, and we
had one valence electron in each of those four,
SP three hybrid orbitals, and then hydrogen had
one valence electron, in an un-hybridized S orbital,
so we drew in our hydrogens, and the one valence electron, like that. This head-on overlap; this
is, of course, a sigma bond, so we talked about this in the last video. And so now that we have this picture of the methane molecule,
we can think about these electron pairs,
so these electron pairs are going to repel each
other: like charges repel. And so, the idea of the
VSEPR theory, tell us these electron pairs are going to repel, and try to get as far away from each other as they possibly can, in space. And this means that the arrangement of those electron pairs,
ends up being tetrahedral. So let's go ahead and write that. So we have a tetrahedral
arrangement of electron pairs around our carbon, like that. When we think about
the molecular geometry, so that's like electron group geometry, you wanna think about the
geometry of the entire molecule. I could think about
drawing in those electrons, those bonding electrons, like that. So we have a wedge coming
out at us in space, a dash going away from us in space, and then, these lines mean,
"in the plane of the page." And so, we can go ahead
a draw in our hydrogens, and this is just one way to
represent the methane molecule, which attempts to show the geometry of the entire molecule. So the arrangement of the
atoms turns out to also be tetrahedral, so let's
go ahead and write that. So, tetrahedral. And, let's see if we can
see that four-sided figure, so a tetrahedron is a four-sided figure, so we can think about this being one face, and then let's go ahead
and draw a second face. And if I draw a line
back here, that gives us four faces to our tetrahedron. So our electron group
geometry is tetrahedral, the molecular geometry of
methane is tetrahedral, and then we also have a
bond angle, let me go ahead and draw that in, so a bond angle, this hydrogen-carbon-hydrogen
bond angle in here, is approximately 109 point five degrees. All right, let's go ahead
and do the same type of analysis for a
different molecule, here. So let's do it for ammonia, next. So we have NH three, if I want
to find the steric number, the steric number is equal
to the number of sigma bonds, so that's one, two, three;
so three sigma bonds. Plus number of lone pairs
of electrons, so I have one lone pair of electrons
here, so three plus one gives me a steric number of four. So I need four hybridized
orbitals, and once again, when I need four hybridized
orbitals, I know that this nitrogen must be SP three hybridized, because SP three hybridization gives us four hybrid orbitals,
and so let's go ahead and draw those four hybrid orbitals. So we would have nitrogen,
and let's go ahead and draw in all four of those. So, one, two, three, and four; those are the four hybrid orbitals. When you're drawing the
dot structure for nitrogen, you would have one
electron, another electron, another electron, and then you'd have two in this one, like that. And then you'd go ahead,
and put in your hydrogens, so, once again, each
hydrogen has one electron, in a hybridized S orbital, so we go ahead and draw in those
hydrogens, so our overlap of orbitals, so here's
a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds in ammonia, and then we have this lone pair up here. So the arrangement of
these electron pairs, is just what we talked about before: So we have this tetrahedral arrangement of electron pairs, or electron groups, so the VSEPR theory tells us that's how they're going to repel. However, that's not the
shape of the molecule, so if I go ahead and
draw in another picture over here, to talk about
the molecular geometry, and go ahead and draw in
the bonding electrons, like that, and then I'll
put in my non-bonding electrons, up here: this
lone pair right here, housed in an SP three hybridized orbital. So, the arrangement of the atoms turns out not to be tetrahedral, and that has to do with this lone pair of
electrons up here, at the top. So, this lone pair of
electrons is going to repel these bonding
electrons more strongly than in our previous
example, and because it's going to repel those
electrons a little bit more strongly, you're not
gonna get a bond angle of 109 point five; it's going
to decrease the bond angle. So let me go ahead, and use the same color we used before, so this bond
angle is not 109 point five; it goes down a bit, because
of the extra repulsion, so it turns out to be
approximately 107 degrees. And in terms of the shape of the molecule, we don't say "tetrahedral";
we say "trigonal-pyramidal." So let me go ahead, and write that here, so the geometry of the ammonia molecule is trigonal-pyramidal, and
let's analyze that a little bit. So, "trigonal" refers to the fact that nitrogen is bonded
to three atoms here, so nitrogen is bonded to three hydrogens, so that takes care of the trigonal part. The "pyramidal" part comes
in, because when you're doing molecular geometry, you ignore lone pairs of electrons. So if you ignore that
lone pair of electrons, and just look at this nitrogen, at the top of this pyramid right here, so that's where the
"pyramidal" term comes in. So bonded to three other atoms, like this, this, and
this; for our pyramid. So trigonal-pyramidal is the geometry of the ammonia molecule, but the nitrogen is SP three hybridized. All right, let's do one more
example; let's do water. So, first we calculate the steric number, so the steric number is equal
to the number of sigma bonds, so that's one, two; so, two sigma bonds. Plus numbers of lone pairs of electrons, so here's a lone pair, here's a lone pair; so we have two plus two,
which is equal to four, so we need four hybridized orbitals. As we've seen in the
previous two examples, when you need four hybridized orbitals, that's an SP three
hybridization situation; you have four SP three
hybridized orbitals. So this oxygen is SP
three hybridized, so I'll go ahead and write that in here, so oxygen is SP three hybridized. So we can draw that out, showing oxygen with its four SP three hybrid orbitals; so there's four of them. So I'm gonna go ahead
and draw in all four. In terms of electrons,
this orbital gets one, this orbital gets one, and these orbitals are going to get two, like that; so that takes care of oxygen's
six valence electrons. When you're drawing in your hyrdogens, so let's go ahead and
put in the hydrogen here, so, once again, each
hydrogen with one electron, in a un-hybridized S orbital, like that. So in terms of overlap
of bonds, here's one sigma bond, and here's another sigma bond; so that's our two sigma bonds for water. Once again, the arrangement
of these electron pairs is tetrahedral, so VSEPR theory
says the electrons repel, and so the electron group
geometry, you could say, is tetrahedral, but
that's not the geometry of the entire molecule, 'cause I was just thinking about electron groups,
and these hybrid orbitals. The geometry of the molecule is different, so we'll go ahead and draw that over here. So we have our water
molecule, and draw in our bonding electrons, and
now let's put in our non-bonding electrons,
like that, so we have a different situation than with ammonia. With ammonia, we had one
lone pair of electrons repelling these bonding electrons up here; for water, we have two
lone pairs of electrons repelling these bonding electrons, and so that's going to
change the bond angle; it's going to short it even more than in the previous example. So the bond angle decreases,
so this bond angle in here decreases to approximately 105 degrees, rounded up a little bit. So, thinking about the molecular geometry, or the shape of the water molecule, so we actually call this
"bent," or "angular," so this is, "bent geometry,"
because you ignore the lone pairs of
electrons, and that would just give you this oxygen
here, and then this angle; so you could also call this, "angular." So we have this bent
molecular geometry, like that, or angular, and once again,
for molecular geometry, ignore your lone pairs of electrons. So these are examples of three molecules, and the central atom in all
three of these molecules is SP three hybridized, and so, this is one way to figure out your overall molecular geometry, and to think about bond angles, and to think about how
those hybrid orbitals affect the structure of these molecules.

AP® is a registered trademark of the College Board, which has not reviewed this resource.