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### Course: Chemistry library > Unit 7

Lesson 4: Dot structures and molecular geometry- Drawing dot structures
- Drawing Lewis diagrams
- Worked example: Lewis diagram of formaldehyde (CH₂O)
- Worked example: Lewis diagram of the cyanide ion (CN⁻)
- Worked example: Lewis diagram of xenon difluoride (XeF₂)
- Exceptions to the octet rule
- Counting valence electrons
- Lewis diagrams
- Resonance
- Resonance and dot structures
- Formal charge
- Formal charge and dot structures
- Worked example: Using formal charges to evaluate nonequivalent resonance structures
- Resonance and formal charge
- VSEPR for 2 electron clouds
- VSEPR for 3 electron clouds
- More on the dot structure for sulfur dioxide
- VSEPR for 4 electron clouds
- VSEPR for 5 electron clouds (part 1)
- VSEPR for 5 electron clouds (part 2)
- VSEPR for 6 electron clouds
- Molecular polarity
- VSEPR
- 2015 AP Chemistry free response 2d and e

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# VSEPR for 6 electron clouds

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

## Want to join the conversation?

- At8:30, why doesn't the lone pair repel the fluorine atoms to create a bond angle less than 90 degrees?(13 votes)
- It does, the prediction made in the video was wrong. In BrF₅, the four F atoms that are predicted to be in the same plane with each other actually have bond angles of 89.5° with each other and have a bond angle of 84.8° with the remaining atom of F (the F that is linear with the lone pair and the Br).

So, you are correct. Unfortunately, some resources on VESPR don't take into account the fact that lone pairs take up more space than bond and "push" the bonds out of ideal geometries.(19 votes)

- Can I place one lone pair in the axial side and one at an equitorial side for XeF4 ?(7 votes)
- No. The 3D structure is unstable in that conformation. The lone pairs are more negatively charged than the bonds, so the lone pairs will try to be as far apart as possible. This would place them either both axial, or opposite equatorial. They cannot be both axial and equatorial.(13 votes)

- the lone pair must be on equitorial . but in BrF5 molecule it is on axial line?i(5 votes)
- Octahedral or square bipyramidal has a 90* bond angle between the axial and equitorial atoms both. Regarless of how you spin the molecule it looks the same from all sides. Just pick off an atom to get Square Pyramidal. (Axial and Equatorial don't really axis for Octahedral - play with a model kit)

Trigonal bipyramidal DOES have axial (90*) and equatorial (120*) bond angles. If you remove an equatorial, you still have 2 atoms forming tight 90* bonds to both the axials (Electron repulsion = yikes). If you remove an axial, you have the three equatorials with no 90* bonds formed to the axial thats removed (less electron repulsion). Try it on a model kit(6 votes)

- How come the lone pairs aren't assigned an equatorial position in this example? In the previous videos it was mentioned that lone pairs should be assigned an equatorial position? :/(5 votes)
- It may be because all the Fluorines are the same distance from each other here so it doesn't matter where the lone pairs are placed in relation to the Fluorines. When they have their free choice of where to go, they choose positions that are furtherest away from each other as they repel each other.(3 votes)

- With 6 electron clouds, can't there be cases with 3 lone pairs and 4 lone pairs as well? If so, what are the shapes of those?(3 votes)
- Can't there be 3 THREE lone pairs and 3 bond pairs ??

What will be the structure then??

Example -> XeF3(-1)(2 votes)- A molecule can be formed using 3 lone pairs and 3 bonds
**but**the central atom cannot have 3 lone pairs**and**3 bonds.

The reason being is because then the electron the central atom would have would be ((1/2) * bonds = 3) + (1 for each lone pair electron (2 * 3 = 6)) = 9.

So in short having 3 pairs of shared electrons and 3 pairs of lone electrons would mean it would have 9 valence electrons which is invalid.

Hope this helps,

- Convenient Colleague(2 votes)

- At2:00how do we figure out that the electron cloud geometry was a regular octahedron? Is it worked out experimentally or is it something that we could deduce just by the number of electron density regions?(2 votes)
- It was determined experimentally. This was made possible due to development of advanced technology such as spectroscopy.(2 votes)

- In drawing the structures, it is always mentioned "[...] is following the octet rule". How do I know which atom follows the octet rule? Don't they all?(1 vote)
- Only the atoms from C to Si
**must**follow the octet rule.

Atoms after Si often follow the octet rule, but they may "expand their octets" and disobey the octet rule.(4 votes)

- At6:52Mr Jay says it doesn't really matter where we put the lone pair of electrons.... But surely the axial position is preferable because, although it is 90° from the equatorial regions, it gives 180° from the other axial electron density region - whereas the equatorial position is 90° from everything...?(2 votes)
- See, just look at the figure of the octahedron very carefully. If you rotate the octahedron(or just tilt your head and see), the positions which seemed to be equatorial will become axial and the seemingly axial positions will become equatorial. The point Jay is making is that every position is exactly same-90 degree angle with 4 neighboring positions and 180 degrees angle with 1 position. It might feel a bit counter-intuitive but try drawing the octahedral structure of the first compound, choose any one equatorial position and look carefully whether it can also act as an axial position. The equatorial position is in the same plane as three other positions;i.e.-forms a rectangle;and two positions project upward and downward. Hence, every position is same, unlike the situation with 5 electron cloud species. That's why the position of the lone pair does not matter in the second example.(1 vote)

- what is equatorial and axial?(1 vote)
- Think of the atom like the earth. Equatorial refers to the electrons that are on the earth's equator. Axial refers to the electrons that are at the north and south poles - on the earth's axis.(3 votes)

## Video transcript

In this video, we're going
to apply VSEPR theory to 6 electron clouds. So if our goal is to find
the shape of the sulfur hexafluoride
molecule, once again we start with our dot structure. So sulfur is in group 6
on the periodic table, so 6 valence electrons. Fluorine is in group 7,
so 7 valence electrons, but I have 6 of them. So 7 times 6 gives
me 42, and 42 plus 6 gives me 48 valence
electrons that we need to show in
our dot structure. Sulfur goes in the center, so we
go ahead and put sulfur there. And we surround sulfur
with 6 fluorines. So let me go ahead and put in
those 6 fluorines surrounding our sulfur. Our next step is to
see how many valence electrons that
we've shown so far. So I go and highlight those--
2, 4, 6, 8, 10, and 12. So 48 minus 12 gives me 36
valence electrons left over, which we put on our terminal
atoms, which are our fluorines. So fluorine is going to
follow the octet rule. And since each
fluorine is already surrounded by 2
electrons, we're going to give each fluorine 6 more. So by giving each
fluorine 6 more, now each fluorine has an octet
of electrons around it. So if I'm adding 6 electrons
to 6 atoms, 6 times 6 is 36. And so therefore, I
have now represented of all my valence electrons. And we're done with
our dot structure. We can move on to
step two and count the number of electron clouds
surrounding our central atom, so regions of electron density. So these bonding electrons here,
that's a region of electron density. And I can just keep on
going all the way around. So all of these bonding
electrons surrounding our sulfur are regions
of electron density. Therefore, we consider
them to be electron clouds. VSEPR theory says that these
valence electrons are all negatively charged,
and therefore, they're all going to repel
each other and try to get as far away from
each other in space as they possibly can. And so when you have
6 electron clouds, they're going to point
towards the corners of a regular octahedron
to try to get as far away from each
other as they can. So an octahedron
with 8 faces on it. So let me see if I can
sketch in an octahedron here. So let's see if we can do it. It's a little bit
tricky to draw. So if we consider our sulfur to
be at the center right here-- let's go ahead and
put a point up here and then start
connecting some lines. So this is sort of
what it looks like. So let's do that and then
a point down here as well. And so we connect those lines. And once again, this is just a
rough sketch of an octahedron. Something like that. So if you think about where your
fluorines are-- right there. Here's a fluorine right here. There's a fluorine right here. So at these corners,
you could think about a fluorine
being there like that. So that's our octahedron. So that's step three. The geometry of the electron
clouds around the central atom, they occupy an
octahedral geometry. Step four, ignore any lone
pairs in your central atom and predict the geometry
of the molecule. Well, since we have no lone
pairs on our central sulfur, the geometry of the
molecule is the same as the geometry of
the electron clouds. And so therefore, we can
say that sulfur hexafluoride is an octahedral molecule. So let's go ahead and
write octahedral here. In terms of bond angles,
let's analyze our drawing a little bit more here. So if I look at
this top fluorine and I go straight down like an
axis to that other fluorine, we would expect one of
the ideal bond angles to be 180 degrees for
this octahedron here. And the other ideal bond
angles would be 90 degrees. So if I think about the angle
that the axis I just drew makes with this one right here,
so that's 90 degrees as well. And again, anywhere
you look, you're also going to get 90 degrees. Let me go ahead and
change colors here, and we can look at another
bond angle in here. So this bond angle, that
would also be 90 degrees. So for an octahedral,
all 6 positions-- we have 6 fluorines
occupying the 6 positions-- are equivalents. They are identical, which means
no axial or equatorial groups in an octahedral arrangement. And that makes our
life much easier, because in the videos
on 5 electron clouds, we had to think about the
axial and equatorial groups. Let's do one for bromine
pentafluoride here, so BrF5. So valence electrons,
bromine has 7. It's in group 7. Fluorine is also in group
7, and I have 5 fluorine. So 7 times 5 gives me 35. 35 plus 7 gives me
42 valence electrons. Bromine goes in the
center, and bromine is bonded to 5 fluorines. So I can go ahead and
put those 5 fluorines around our central atom. We have represented, let's
see, 2, 4, 6, 8, and 10 valence electrons so far. 42 minus 10 is, of course,
32 valence electrons. And we're going to start
putting those leftover electrons on our terminal atoms,
which are our fluorines. So once again, we're going to
give each fluorine an octet. So we're going to put 6 more
valence electrons around each of our fluorine atoms. And so we're putting 6 more
electrons around 5 atoms. So 6 times 5 is 30. So 32 minus 30 gives me 2
valence electrons left over. And whenever you have
valence electrons left over after assigning them
to your terminal atoms, you put them on
your central atom. And so there's going to be
a lone pair of electrons on our central
bromine like that. So we've drawn
our dot structure. Let's go back up here and
look at our steps again. So after drawing
our dot structure, we next count the number
of electron clouds that surround our central
atom and then predict the geometry of those
electron clouds. And so if we look at our
central bromine here, let's see how many
electron clouds. Well, we would have
these bonding electrons, a region of electron density,
these bonding electrons, these bonding electrons. And we keep on
going around here. So those are all
electron clouds. So that's 5. And then remember, these
non-bonding electrons, this lone pair of
electrons, is also a region of electron density. And so we have 6
electron clouds. And so we just saw in the
previous example, when you have 6 electron
clouds, the electron clouds are going to want to
point towards the corners of a regular octahedron. So you're going to get
an octahedral geometry for your electron clouds. Let's think about
this one, though. Where will we put
those that lone pair of electrons in an octahedron. Well, since all 6
positions are identical, it doesn't really
matter which one you put that lone
pair of electrons in. And so let me see if I can
just go ahead and sketch out the shape really fast. So if I were to draw
a bromine right here, I'm going to put
a fluorine going in this direction, another
fluorine going back, this one coming out a little
bit, and this one going away. And then I'm going to put
a fluorine going this way, and then I'm going to put
the lone pair of electrons right down here. Again, it didn't
really matter which one I chose since they're
all identical. I just chose it this way
because it's a little bit easier to see the geometry. Because when you're looking at
the geometry of the molecule, you ignore any lone pairs of
electrons on your central atom. And so if we ignore that
lone pair of electrons now, and we look at the
shape-- so let's see if we can connect
these dots here. So we're just going to connect
this to look at a shape. So we have a square base here. And if we connect up here
to this top fluorine, well, that's kind of a pyramid. So we have a pyramid
with a square base. And so we call this
square pyramidal. So let's go ahead
and write that. This shape is referred to
as a square pyramidal shape. And in terms of
bond angles, we know our ideal bond angles are
going to be 90 degrees. So if we look at that,
let's use this green here. So it's just like we
talked about before. So that bond angle
is 90 degrees. This bond angle in
here is 90 degrees. So our ideal bond angles
are all 90 degrees for our square
pyramidal geometry. Let's do one more example
of 6 electron clouds. And this is xenon tetrafluoride. So we need to find
our valence electrons. So xenon is in group
8, 8 valence electrons. Fluorine is in group 7. So 7 valence electrons
times 4 gives me 28. 28 plus 8 gives me
36 valence electrons. Xenon goes in the center. So we go ahead and put
xenon in the center here. And xenon is bonded
to 4 fluorines. So we go ahead and
put in our 4 fluorines surrounding our xenon. And let's see, we have
represented 2, 4, 6, and 8 valence electrons. So 36 minus 8, that would
give me 28 valence electrons left over, which we will put
on our terminal fluorines here. So each fluorine is
going to get an octet. And so we need to put
6 valence electrons on each one of our
fluorine atoms. So we are representing 6
more electrons on 4 atoms. So 6 times 4 is 24. So 28 minus 24 gives us 4
valence electrons left over. And we're going to put those
on our central atom here, so we're going to put
those on the xenon. So we go ahead and add
in those 4 electrons in the form of two lone
pairs to our central atom. All right. Let's go back up and
refresh our memory about what we do after we
draw our dot structure. So after you draw
your dot structure, you count the number
of electron clouds. And then you predict the
geometry of those electron clouds, and so let's count our
electron clouds for this one. So our regions of
electron density, so we can see that
these bonding electrons are an electron cloud, same
with these bonding electrons, and these over here as well. And in this example, we have
2 lone pairs of electrons, and each one of those is a
region of electron density. And so we have a
total of 6 electron clouds for this example. So once again, 6
electron clouds, they are going to want to get
as far away from each other as they can. So they are going to be in
an octahedral arrangement, and so let's see if we can
sketch out this molecule again. So if the lone
pairs of electrons want to get as far
away from each other as they possibly
can, we're going to put those lone pairs 180
degrees from each other. So here's one lone
pair, and then here's our other lone pair. That's as far away from
each other as they can get. And then we're going to
put our fluorines in here. So here's one fluorine. Here would be another fluorine. And then we would have
two more back here. So when we look at the shape
of xenon tetrafluoride, let's see if we
can sketch in what the shape would look like here. So remember, when
you're predicting the geometry of
the molecule, you ignore the lone
pairs of electrons. And so that makes it
much easier to see that we have a square
that is planar. So we call this square planar. So the geometry
is square planar. And in terms of
ideal bond angles, that would be 90 degrees. So let me go ahead and
show that real fast. So in terms of bond
angles, everything here would be 90 degrees
for our square planar. And so that's how to
approach 6 electron clouds. And our first example
had 0 lone pairs of electrons around
the central atom. Our second example
had 1 lone pair. And our third example
had 2 lone pairs. And so even though the electron
clouds have the same geometry, the actual molecule is said
to have a different shape, because you ignore the
lone pairs of electrons on your central atom.