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Chemistry library
Course: Chemistry library > Unit 9
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 5 electron clouds (part 1)
In this video, we apply VSEPR theory to molecules and ions with five groups or “clouds” of electrons around the central atom. To minimize repulsions, five electron clouds will always adopt a trigonal bipyramidal electron geometry. Depending on how many of the clouds are lone pairs, the molecular geometry will be trigonal bipyramidal (no lone pairs), seesaw (one lone pair), T-shaped (two lone pairs), or linear (three lone pairs). This video focuses on the first two molecular geometries, trigonal bipyramidal and seesaw. Created by Jay.
Want to join the conversation?
- Why is the bond angle between the equatorial lone pair and the axial fluorines still 90 degrees? Shouldn't the lone pair repel the axial fluorines so that the angles between axial fluorines and the equatorial lone pair be more than 90 degrees, therefore causing the angles between the equatorial fluorines and the axial fluorines to be less than 90 degrees?(30 votes)
- Sharp thinking! Those are the theoretical bond angles.
The lone pair repels all the bond pairs and does just as you predicted.
The F-S-F bond angle between the equatorial fluorines is reduced from 120 ° to 102 °.
The F-S-F bond angle between the axial fluorines is reduced from 180 ° to 173 °.
The F-S-F bond angle between the axial fluorines and the lone pair is increased from 90 ° to 93.5 °.
The F-S-F bond angle between the equatorial and axial fluorines is reduced from 90 ° to 86.5 °.(56 votes)
- Is the trigonal bipyramidal geometry just observed behaviour or is there like a mathematical proof, that this configuration minimises electron-electron repulsion?(26 votes)
- It's both. In these exercises we are trying to make our best predictions following some basic rules. Scientific experiments will test whether our predictions were correct or not. If they are (in most cases they are) than all right! If they aren't, we want to know why is that. There are many ways to find out spatial distribution of atoms in a molecule.(22 votes)
- why are the lone pairs of electrons around chlorine not counted as electron clouds?(7 votes)
- I think it is because he is talking from the point of view of the CENTRAL atom. It is the electron clouds immediately around phosphorus (the central atom) that gives PCl5's shape as a whole, according to VSEPR model. In other words, it will be useless to analyze PCl5 shape from the point of view of each individual chlorine atoms.(13 votes)
- At... 1:18
Why is it okay for Phosphorus to have an expanded octet?
What is the relation between period 3 and the expanded octet of Phosphorus?
If possible please give me a link of the Khan Academy video for my question's answer....
Thank you.!(3 votes)- You will have to refer to a concept called VBT (Valence Bond Theory)(5 votes)
- In the last example why is the angle between the lone pair of electrons and the axial bonding electrons still 90 degrees? I thought the lone pair would repel the axial ones more... Thank you!(2 votes)
- In a geometry sense, we still consider the bond angles to be the same as the previous molecule; 90° between the axial and equatorial groups, and 120° between all the equatorial groups. In a chemistry sense these bond angles are different because, like you said, the lone pairs provide some repulsion which distorts the bond angles. Geometrically it is viewed as an ideal shape, but chemistry views it how it appears realistically taking into consideration nuanced electron interactions.
The actual distortion from the ideal bond angles differs for each molecule. In case you’re interested, for SF4 the axial-equatorial bond angles are 86.55°, and the equatorial bond angles are 101.6°.
Hope that helps.(8 votes)
- AtWhy is it said that the structure on the left is the right one? ALSO why is it that we have only two resonance structures for the molecule at 12:23? 12:15(3 votes)
- These are isomers, not resonance structures.
The five orbitals have a trigonal bipyramidal geometry.
There are only two places where the lone pair can go — the axial ot the equatorial location.
The see-saw geometry is more stable because the lone pair in the axial location has less total repulsion from the other electrons.(5 votes)
- Why wouldn't the bond angles on SF4 be less than ideal if the lone pair repels the bonding pairs?(3 votes)
- Yes, that's true: the bond angles are "less than ideal" but, to be honest, that is the ideal arrangement for a molecule with a structure like that.
The structure of SF4 is actually a "seesaw" which is a variation of "trigonal bipyramidal" because there is this extra pair of electrons. The bond angles are also smaller(4 votes)
- In the molecular structure, does the Sulfur, the central atom of SF4, has four electrons that is bonded to F?
Is that why we can put two more valence electrons when there are left overs?
So, we can put up to four more valence electrons in a central atom?(4 votes)- Sulfur is in the 3rd period, meaning it has a d-orbital to store up to 10 extra electrons. In general, the octet rule is less important for the 3rd period down because of these sorts of exceptions(1 vote)
- How do you know when the octet rule applies? is it just when the formal charge isn't 0?(1 vote)
- The octet rule applies for atoms in the second row of the periodic table (most notably: B, C, N, O, F, Ne). The first row (H and He) only has 1s orbitals available, so can only hold a maximum of 2 electrons. The beginning of the second row (Li, Be) often loses electrons to look more like the first row (i.e. 2 electrons). From the third row onwards, atoms can sometimes have more than an octet ("expanded octet"), since they now have d orbitals available, too.
Formal charges do not impact whether an atom follows the 'octet rule' or not.(6 votes)
- At, Why do non bonding electrons take up more space than bonding electrons? 9:30(1 vote)
- Bonding electrons have to spend most of their tine between two nuclei.
Nonbonding electrons are attracted to a nucleus only from one side, so they are free to wander further away.(5 votes)
Video transcript
Let's use VSEPR
theory to predict the structure of this molecule--
so phosphorus pentachloride. So the first thing we need
to do is draw a dot structure to show our valence electrons. We find phosphorus in Group 5. So 5 valence electrons. Chlorine in Group 7. So 7 valence electrons,
and I have 5 of them. So 7 times 5 is 35. Plus 5 gives us a total
of 40 valence electrons that we need to show
in our dot structure. So phosphorus goes in the
center because it is not as electronegative as chlorine. And we have five chlorines. So we go ahead and
put our five chlorines around our central
phosphorus atoms like that. If we see how many valence
electrons we've drawn so far, this would to be
2, 4, 6, 8, and 10. So 40 minus 10 gives us 30
valence electrons left over. And remember, you start putting
those leftover electrons on your terminal atom. So we're going to put
those on the chlorines. Each chlorine's going to
follow the octet rule. So that means each chlorine
needs 6 more electrons. Now, each chlorine is surrounded
by 8 valence electrons like that. So if I'm adding 6 more
electrons to 5 atoms, 6 times 5 is 30. So I have now represented
all of my valence electrons on my dot structure. Notice that phosphorus is
exceeding the octet rule here. There are 10 valence
electrons around phosphorus. And it's OK for phosphorus to
do that because it's in Period 3 on the periodic table. I like to think
about formal charge. And so if you assign a
formal charge to phosphorus, you'll see it has a
formal charge of 0. And that helps to explain-- for
me, anyway-- the resulting dot structure. Now, step two. We're going to count the
number of electron clouds that surround our central atom. Remember, an electron
cloud is just a region of electron density. So I could think about these
bonding electrons in here as a region of electron
density around my central atom. I could think about these
bonding electrons, too. So here's another
electron cloud. And you can see we have
a total of five electron clouds around our central atom. The next step is to predict
the geometry of the electron clouds. Those valence shell electrons
are going to repel each other. All right. So that's a VSEPR theory--
Valence Shell Electron Pair Repulsion. Since they're all
negatively charged, they're going to repel
and try to get as far away from each other as
they possibly can in space. When you have five
electron pairs, it turns out the furthest they
can get away from each other in space is a shape called a
trigonal bipyramidal shape. So let me see if I can draw
our molecule in that shape. We're going to have our
phosphorus in the center, and we're going to have three
chlorines on the same plane. So let me attempt to show three
chlorines on the same plane here. These are called the
equatorial positions because they're kind of along
the equator, if you will. So three chlorines in the
same plane, one chlorine above the plane, and one
chlorine below the plane. Those are called
axial positions. All right. So there's a quick sketch. Let me see if I can draw
a slightly better shape of a trigonal
bipyramidal shape here. So let me see if I can
draw one over here so you can see what it looks
like a little bit better. So we could have one pyramid
looking something like that. And then, down here,
let's see if we can draw another pyramid
in here like that. So that's a rough
drawing, but we're trying to go for a trigonal
bipyramidal shape here. So let's focus in on
those chlorines that are on the same plane first. If I'm looking at
these three chlorines and I go over here to my
trigonal bipyramidal shape, you could think about
those three chlorines as being at these corners here. So it's a little
bit easier to see. They're in the same plane. So those are the
equatorial chlorines. When I think about the
bond angle for those-- so those chlorines
being in the same plane, you have these three
bond angles here. And so when we did
trigonal planar, we talked about 360
degrees divided by 3-- giving us a bond
angle of 120 degrees. So you could think about that
as being a bond angle of 120. All right. So same idea. Those bonding electrons are
going to repel each other. When we focus in on our axial
chlorines-- so this one up here and this one down here. You could think about those
as being here and here on your trigonal
bipyramidal shape like that. And if you draw the axis, if
you draw a line down this way connecting those,
it's easy to see those are 180 degrees
from each other. So you could think
about a bond angle of 180 degrees between
your chlorines like that. And then, finally, if we think
about the bond angle between, let's say, this axial
chlorine up here at the top and then one of these
green chlorines right here, I think it's a little bit easier
to see that's 90 degrees here. So this bond angle right
here would be 90 degrees. And so those are your
three ideal bond angles for a trigonal bipyramidal
situation here. It's important to understand
this trigonal bipyramidal shape because all of the five electron
cloud drawings that we're going to do are going to
have the electron clouds want to take this shape. So it's important to
understand those positions. For step four,
ignore any lone pairs and predict the geometry
of the molecule. Well, there are no lone pairs
on our central phosphorus. So the electron clouds take
a trigonal bipyramidal shape and so does the molecule. Let's go ahead and
do another example. Sulfur tetrafluoride here. So we're going to start by
drawing the dot structure, and we need to count our
valence electrons, of course. So sulfur's in Group 6,
so 6 valence electrons. Fluorine is in Group 7. So 7 valence electrons. I have 4 of them. 7 times 4 is 28. 28 plus 6 is 34
valence electrons. We know sulfur is going
to go in the center because fluorine is much
more electronegative. We put sulfur in
the center here. We know sulfur is
bonded to 4 fluorines. So we put our fluorines
around like that. And let's see how many valence
electrons we've shown so far-- 2, 4, 6, and 8. So 34 minus 8 gives us
26 valence electrons we still need to account
for on our dot structure. We're going to start by
putting those leftover electrons on our terminal
atoms, which are our fluorines. Fluorine's going to have an
octet of electrons around it. Therefore, each
fluorine needs six, since each fluorine
already has two around it. So we go ahead and put
6 valence electrons around each one of
our fluorine atoms. All right. So we are showing 6 more
valence electrons on 4 atoms. 6 times 4 is 24. So 26 minus 24 gives us 2
leftover valence electrons. And remember your rules
for drawing dot structures. When you get some
leftover electrons, you're going to go ahead and put
them on your central atom now. So we have a lone pair of
electrons on our sulfur. And by adding that lone pair
of electrons to our sulfur, the sulfur now exceeds
the octet rule. But once again,
it's OK for sulfur to have an expanded
valence shell. It's in Period 3 on
our periodic table. And once again, I like to
think about formal charge. And if you assign a formal
charge to that sulfur, it has a formal charge of 0. So that just helps me
understand these dot structures a little bit better. So we've drawn
our dot structure. Let's go back up and remind
ourselves of the next step here. So once you complete
step one, next is the electron cloud step. So how many electron
clouds do you have surrounding
your central atom? So we go back down, and we
look at our electron clouds that surround our central atom. Here's the regions
of electron density. So we know that these
bombing electrons here, that would be one
electron cloud. Same with these
bonding electrons. Same with these
bonding electrons. And same with these
bonding electrons. And then we have a lone pair
of electrons on our sulfur. Well, that's also a region of
electron density surrounding our central atom. So that lone pair you could
think of as being an electron cloud as well. And so we have five
electron clouds. So just like in the
previous example. And when you have
five electron clouds, those electron clouds
are going to try to adopt a trigonal bipyramidal
shape-- just like we saw, again, in the previous example. So let's go ahead and draw two
possible versions of the dot structure for this molecule. All right. So I'm going to
draw one right here. And for this first
version, I'm going to show the lone
pair of electrons on the sulfur in the
equatorial position here. So I'm going to put the lone
pair of electrons right here, equatorial here. And so that means there are
two fluorines also equatorial. And then that means that there's
one fluorine here, axial. Another fluorine here, axial. So that is one
possible dot structure. The other possibility
would be, of course, to put to the lone
pair of electrons in the axial position. So if we do that, we
would have a sulfur bonded to three fluorines. Those would be the
equatorial fluorines. And then, we would have
a lone pair of electrons. Let's just put it right
here in the axial position. And then, another fluorine
in the axial position. So here are our
two possibilities. So let's see if we can
analyze this structure. Now, when you have lone pairs
of electrons in your dot structure, lone pairs
take up more space. Or non-bonding electrons--
I should say-- take up more space than
bonding electrons. And so since they
take up more space, they're going to repel
a little bit more. And so that means
that when you're trying to figure out valence
shell electron pair repulsion, the lone pairs of electrons
are more important to focus on in terms of where you're
going to put them. Let's focus in on those
lone pairs of electrons, and let's think
about how they're going to repel the other
electrons in these two dot structures. Let's look at the
left here where we had the lone
pair of electrons in the equatorial position here. And if you're thinking about
how they're interacting with, let's say, these bonding
electrons in the same plane here, this is about 120 degrees
between the bonding electrons and the non-bonding electrons. And it turns out that
120 degrees is not as important in terms of
repelling as, say, something like 9 degrees. You tend to ignore the 120
degree interactions when you're analyzing
these structures. However, a 90-degree angle
between a bonding pair and a non-bonding pair--
and we had that example for-- let me go ahead
and show you right here. So let's think about this lone
pair of electrons repelling these bonding electrons. So in the axial position. Well, these two are
only 90 degrees away. So remember, 9 degrees,
of course, being closer, you're going to get more
repulsion from this interaction than in the previous
interactions. So we're going to focus in
on the 90-degree interactions here. Those bonding electrons
and non-bonding electrons repel each other. And you have one possibility
with the axial fluorine. You also have
another possibility with this axial fluorine. Essentially, you have a lone
pair of electrons, 90 degrees, from two pairs of
bonding electrons from the example on the left. And of course, that's going
to destabilize it somewhat. But let's compare
this dot structure with the one on the right now. So we have our lone pair of
electrons in the axial position this time. And you can see that
we have three fluorines in the equatorial positions. So you have these
bonding electrons in the equatorial
position, which means that that lone
pair of electrons is 90 degrees to
all three of those. And so that, of course,
is going to cause some serious repulsion,
so 90 degrees to 3. In the example on the right, you
have these three interactions-- 90 degrees. An example on the left,
you have only two of these. The goal, of course, is
to minimize electron pair repulsion. So VSEPR theory
actually predicts that this dot structure on
the left is the correct one. You're going to see--
in the next video-- that non-bonding
electrons are placed in equatorial positions
in trigonal bipyramids to minimize electron
pair repulsion. So just think about putting
your lone pairs of electrons in the equatorial position. So the structure
on the left wins. Let's go ahead and
redraw that so we can analyze it a
little bit better. All right. So I have my sulfur
in the center here. Let's go ahead
and change colors. So I'm going to put the
sulfur in the center. I have my fluorine in a plane. Another fluorine in the plane. My lone pair of
electrons in a plane. All right. Those are my equatorial ones. I have a fluorine this way. And I have a fluorine that way. So when you're looking at
bond angles, of course, between this sulfur
fluorine bond angle-- the ideal bond angle anyway--
would be 120 degrees. So we can say and we would
expect it to be 120 degrees. If you're talking about
this axial fluorine and this equatorial
one, we would expect that to be 90 degrees. And then, finally,
between the two axial fluorines-- so this
bond angle back here would, of course,
be 180 degrees. OK. So we've done a lot of talking,
and we still haven't even talked about the final name
for the shape of this molecule. So let's go back up here and
look at our rules really fast. So we've done a lot
of work to predict the geometry of the electron
clouds around the central atom and draw it. And finally, we get to predict
the shape of the molecule. And we do that by
ignoring any lone pairs. So let's go ahead and do that. So we're going to ignore
the lone pair of electrons on the sulfur when we're
talking about the shape. So if we ignore the lone
pair and we actually turn this molecule on its side--
so let's go ahead and do that. We're going to put
our sulfur here. If we turn it on its
side, the axial fluorines would now be horizontal. Now, it's horizontal like that. So I'll go ahead and
put in my-- so these are the two fluorines that
used to be axial there. And my two fluorines
that were equatorial, they would look
something like this. And it helps if you
actually build this molecule with a MolyMod set. So that would be what the
molecule kind of looks like here, and we call
this a seesaw shape. So this is a seeshaw
shape or geometry. And let's think about why. So if you've ever been on a
playground and used a seesaw-- I'm going to draw
a little kid here on one side of our
seesaw like that. And so if the little
kid puts his weight on this side, of course,
this side of the seesaw salt would go down. And then, this side of
the seesaw would go up. So just a little
bit of intuition as to why you would call
this a seesaw shape. All right. So I think we'll
have to stop there. In the next video, we'll
do two more examples of molecules and ions that
have five electron clouds.