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### Course: Class 11 Chemistry (India) > Unit 4

Lesson 5: VSEPR theory# VSEPR for 4 electron clouds

In this video, we apply VSEPR theory to molecules and ions with four groups or “clouds” of electrons around the central atom. To minimize repulsions, four electron clouds will always adopt a tetrahedral electron geometry. Depending on how many of the clouds are lone pairs, the molecular geometry will be tetrahedral (no lone pairs), trigonal pyramidal (one lone pair), or bent (two lone pairs). Created by Jay.

## Want to join the conversation?

- at6:07it is easy to understand how the unpaired electrons repel the hydrogens attached to nitrogen so that a trigonal pyramidal shape is achieved. But what forces drive the formation of a tetrahedral shape for methane? Why doesn't it assume a flat plane? Wouldn't that create the greatest distance among the hydrogens?(28 votes)
- If it assumed a flat plane, then the angles between the Hs would be 360°/4 = 90°.

A tetrahedral shape allows the angles between the Hs to be 109.5°, greater than the 90° that a flat plane would allow.(57 votes)

- At7:45, how do you know that the lone pairs on oxygen are adjacent to each other, instead of having one above and one below and the hydrogens forming a 180 degree angle with the oxygen?(30 votes)
- In compounds such as Methane, seen drawn at2:57, all of the 4 bonding electron clouds are a 109.5 degrees from each other. Since water also has 4 electron clouds, it will take the same shape, except with unbonding electron clouds taking the place of two of those bonding electron clouds. So the angle between the hydrogen atoms will always be ~109.5 degrees (104.45, to be presice, due to reasons he stated) and the structure will always be bent/angular, you'll just be looking at if from a different angle depending on where the lone pairs were placed.(11 votes)

- Why do lone pair of electrons occupy more space than bond pairs?(11 votes)
- The lone pair of electrons are situated more closer to the central atom than the bonding electron. Hence, they have more repulsion and thus, occupy more space than bond pairs.(15 votes)

- At9:22and two previous other times, how do you calculate the angles between the hydrogens?(14 votes)
- So if the center has electrons that aren't bond to any other atom is also counted as a electron cloud.(0 votes)

- how is trigonal pyramidal different from tetrahedral? A tetrahedron IS a triangular pyramid.(10 votes)
- A tetrahedron is a triangular pyramid but a triangular pyramid is not necessarily a tetrahedron(the regular polygon), so you're right but by using both terms it gives you just that little bit more precision in describing the shapes.(8 votes)

- How do you predict how much the lone pairs of electrons will push the hydrogens in water? Like, how do they know that the Lone pairs reduce the angle from 107º to 104.5º and not to 100º or 105.5º?(1 vote)
- They can measure the bond angles by techniques like X-ray spectroscopy.(8 votes)

- what is the basic concepT to understand the VSEPR THEORY....??(1 vote)
- The basic concept is that electrons repel each other and will try to get as far away from each other as possible.(8 votes)

- At3:50, how is it that 109.5 separates all the clouds better on a single plane compared to 90?(2 votes)
- But it isn't on a single plane, it's in 3D space. 109.5 degrees just happens to be the maximum angle when you have 4 equal groups.

If you need proof then Jay did do a mathy video that proves this on here, might be easier just accept it for what it is.(5 votes)

- Bond angles are given at 2;40,, at7:00and at9:11but there is no mention of how they are determined/calculated or their importance to a given molecule's structure. In fact, in over an hour of VSEPR there is no mention whatsoever of why a molecule's shape is important.... does anyone know?(2 votes)
- Molecular shape is used in determining the stability of the molecule and how it can bond with other molecules; this becomes more relevant in the section on hybrid orbitals and hybridization.(4 votes)

- How do we determine the bond angles of the different shapes?(1 vote)
- First consider the shape you're determining the bond angles of. An easy example is the angle of a trigonal planar shape. All angles are 120 degrees because they only occupy one plane and must add up to 360. Here are some common bond angles, a lot of it is simply memorization.
`1 Axis Shapes`

- These form a bond with one other atom and are considered to have "no shape" or bond angle. They are sometimes referred to as linear as well.`2 Axis Shapes`

- These are all linear (they only form bonds with two other atoms), therefore they all have bond angles of 180 degrees.*These are "sp" hybridized.*`3 Axis Shapes`

- Forming bonds with three different atoms.*These are "sp2" hybridized.*The different 3 axis shapes are:**Trigonal Planar**(120 degrees) [3 single bonds]**Trigonal Planar**(120 degrees) [2 single bonds] [1 double bond]**Bent/V-Shaped**(104.5 degrees) OR (90<x<120 degrees) [1 lone pair] [1 double, 1 single bond]**No shape**[2 lone pairs] [1 double bond]`4 Axis Shapes`

*These are "sp3" hybridized.***Tetrahedron**(109.5 degrees) OR (90<x<120 degrees) [4 single bonds]**Trigonal Pyramidal**(107.5 degrees) OR (90<x<120 degrees) [1 lone pair] [3 single bonds]**Bent/V-Shaped**(104.5 degrees) OR (90<x<120 degrees) [2 lone pairs] [2 single bonds]**No shape**[3 lone pairs] [1 Single Bond](5 votes)

## Video transcript

Let's figure out the shape
of the methane molecule using VSEPR theory. So the first thing that you
do is draw a dot structure to show it the
valence electrons. So for methane, carbon
is in group four. So 4 valence electrons. Hydrogen is in group one, and I
have four of them, so 1 times 4 is 4, plus 4 is 8
valence electrons that we need to show
in our dot structure. Carbon goes in the
center and carbon is bonded to 4 hydrogens,
so I can go ahead and put my hydrogens in there like that. And this is a very
simple dot structure. We've already shown all 8
of our valence electrons. Let me go ahead and
highlight those here. 2, 4, six, and 8. So carbon has an
octet and we are done. The next thing we need
to do is count the number of electron clouds that
surround our central atom. So remember, electron clouds
are regions of electron density, all right? So we can think about
these bonding electrons here as being electron clouds. So that's one electron cloud. Here's another one down here. And then here's one. And then finally,
here's another one. So we have four electron clouds
surrounding our central atom. The next step is to predict
the geometry of your electron clouds around your central atom. And so VSEPR theory tells us
that those valence electrons are going to repel each other
since they are negatively charged. And therefore,
they're going to try to get as far away from each
other as they can in space. And when you have
four electron clouds, the electron clouds are
farthest away from each other if they point towards the
course of a tetrahedron, which is a four sided figure. So let me go ahead and
draw the molecule here, draw the methane molecule. I'm going to attempt to show
it in a tetrahedral geometry. And then I'll
actually show you what a tetrahedron looks like here. So here's a quick sketch of
what the molecule sort of looks like. And let me go ahead and
draw tetrahedron over here so you can get a little
bit better idea of the shape, all right? So four sided figure. And so there you go. Something like that. So you could think about the
corners of your tetrahedron as being approximately
where your hydrogens are and that just gives you a
little bit better visual picture of that tetrahedron, that
four sided figure here. And so we've created the
geometry of the electron clouds around our central atom. And in step four, we
ignore any lone pairs around our central atom,
which we have none this time. And so therefore, the
geometry the molecule is the same as the geometry
of our electron pairs. So we can say that methane is a
tetrahedral molecule like that. All right, in terms
of bond angles. So our goal now is to figure
out what the bond angles are in a tetrahedral molecule. Turns out to be 109.5
degrees in space. So that's having those
bonding electrons as far away from each
other as they possibly can using VSEPR theory. So 109.5 degrees turns
out to be the ideal bond angle for a
tetrahedral molecule. Let's go ahead and
do another one. Let's look at ammonia. So we have NH3. First thing we need to do
is draw the dot structure. So we start by finding our
valence electrons, nitrogen in group five. So 5 valence electrons. Hydrogen in group one,
and I have three of them. So 1 times 3 plus 5 is 8. So once again, we have 8 valence
electrons to worry about. We put nitrogen in
the center and we know nitrogen has
bonded to 3 hydrogens, so we go ahead and put our 3
hydrogens in there like that. Let's see how many valence
electrons we've used up so far. 2, 4, and 6. So 8 minus 6 is 2
valence electrons left. We can't put them on
our terminal atoms, because the
hydrogens are already surrounded by two electrons. So we go ahead and put
those two valence electrons on our central atom, which
is our nitrogen like that. And so now we've gone ahead and
represented those two valence electrons. So we have all eight
valence electrons shown for our dot structure. All right, we go back
up here to our steps to remind us what to do after
we've drawn our dot structure. And we can see that
now we're going to think about the
electron clouds that surrounded the central atom. So regions of electron density. And let's go ahead
and find those. So I can see that I have
these bonding electrons. That's a region of
electron density, so that's an electron cloud. Here's another one. So that's two. Here's another one. So that's three. And this lone pair of
electrons, this non-bonding pair of electrons is also going to
be counted as an electron cloud. It's a region of
electron density too. And so once again we have four
regions of electron density. When you're thinking about
the geometry of those electron clouds, those four
electron clouds are going to, once
again, try to point towards the corners
of a tetrahedron. So we can kind of sketch
out the ammonia molecule. And we can draw the base
the same way we did before, with our three
hydrogens right here. And then we're going
to go ahead and put our lone pair of
electrons right up here. And so again, it's an attempt
to show the electron clouds in a tetrahedral geometry. Let's go back up here and
look at our steps again. So in step three, we predicted
the geometry electron clouds are going to attempt to
be in a tetrahedron shape around our central atom. But when we're actually talking
about the geometry or shape of the molecule, we're going
to ignore any lone pairs when we predict the geometry
of the molecule. So when we look at
the ammonia molecule, we're going to ignore that
lone pair of electrons on top of the nitrogen
and we're just going to focus in on the
bottom part for the shape here. And so when we do
that, we get something that looks like a little
squat pyramid here. So if I'm [? ignoring ?]
that lone pair of electrons up there at the top, it's
going to look something like that for the shape. And we call this
trigonal pyramidal. So this is a trigonal
pyramidal shape. So even though the
electron clouds are attempting to be in
a tetrahedron fashion, the shape is more
trigonal pyramidal because we ignore any
lone pairs of electrons. In terms of a bond
angle, this lone pair of electrons on the nitrogen
actually occupies more space. These non-bonding electrons
occupy a little more space than bonding electrons. And because of that, those
non-bonding electrons are going to repel
these bonding electrons. I'm going to go
ahead and put them in blue here just as an example. Repel these a little bit more
than in the previous example that we saw. And that's actually
going to make the bond angle a little bit smaller
than the ideal bonding angle we saw before for 109.5 for
a tetrahedral arrangement of electron clouds. And so it turns out that this
bond angle between the atoms, the hydrogen nitrogen
hydrogen bond angle gets a little bit
smaller than 109.5. So it actually gets smaller
to approximately 107 degrees here for a trigonal
pyramidal situation. All right, let's do one more. Let's go ahead and do
the water molecule. All right, so we have H2O. And to follow our steps, we know
that hydrogen's in group one. I have two of them. And we know that
oxygen is in group six. So 6 plus 2 is once
again 8 valence electrons to represent for
our dot structure. And we put oxygen in the center. Oxygen is bonded
to two hydrogens, so we go ahead and
draw those in there. And let's see, how many
valence electrons have we represented so far? That's 2, that's 4. So 8 minus 4 is 4
valence electrons left. We first think about putting
them on our terminal atoms, but those are our hydrogens,
so they're already happy with two electrons. So we go ahead and put
those four valence electrons on our central atom,
which is our oxygen. And four valence electrons means
two lone pairs of electrons now. And so now we've represented
all eight valence electrons for water. Our next step is, of course, to
count how many electron clouds we have around our central atom. So once again, we could think
about these bonding electrons as being an electron cloud. These bonding electrons
as being electron cloud, these non-bonding electrons
as lone pairs in an electron cloud, and same thing for
these non-bonding electrons, electron cloud
over there as well. And so once again we have
four electron clouds. And those four
electron clouds are going to attempt to be in
a tetrahedral arrangement around that central atom. Using VSEPR theory, they're
going to repel each other and get as far away from each
other as they possibly can. Let me go and redraw
the molecule here. So let's go ahead. And we have our
water molecule and we have our lone pairs of
electrons like that. And in this case we have
two lone pairs of electrons. Remember, lone pairs or
non-bonding electrons take up a little bit more space
than bonding electrons. And therefore, they're going
to repel these electrons right in here a little bit more. And that's going to make
our bond angle even smaller than before. So it's going to be even
smaller than 107 degrees. And so this bond
angle right here, you'll see it listed as
approximately 104.5 degrees, or some textbooks
will say 105 degrees. So that's approximately
what it is for this. In terms of the geometry
of the molecule. So the geometry of the electron
clouds are attempting to be, once again, a
tetrahedral fashion, but the geometry of the
molecule is different because you ignore lone
pairs of electrons. And so when you
look at the shape, if you look at the
shape of this-- I'll go ahead and draw
the shape over here. If you're ignoring lone pairs of
electrons, it looks like that. And we've seen
that shape before. That's bent or angular. So we say that the geometry
of the water molecule is bent or angular with an
approximately 104.5 degree bond angle. So those are a couple examples
of four electron clouds and how to figure out
the geometry while also thinking about the bond angles.