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:12:29

AP.Chem:

SAP‑4 (EU)

, SAP‑4.C (LO)

, SAP‑4.C.1 (EK)

, SAP‑4.C.2 (EK)

in this video we're going to apply VSEPR theory to six 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 six valence electrons fluorine is in group seven so seven valence electrons but I have six of them so seven times six gives me 42 and 42 plus six 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 all right so I'll go ahead 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 two electrons we're going to give each fluorine six more so by giving each fluorine six more now each fluorine has an octet of electrons around it so if I'm adding six electrons to six atoms 6 times 6 is 36 and so therefore I have now represented 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 right so these bonding electrons here that's a region of electron density and I can just keep on going all the way around right 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 six 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 eight faces on it so let me see if I can sketch and and octahedron here so let's see if we can do it it's a little bit tricky to draw but let's see if we can do it right there so if we consider our sulfur to be at the center right here all right let's go ahead and put a point up here and then start connecting some lines all right so this is a sort of what it looks like all right so let's do that and then a point down here as well and so we connect those lines and once again this is a just a rough sketch of an octahedron right something like that so if you think about where your fluorines are right there here's a fluorine right here is a fluorine right here so at these corners right you can think about a fluorine being there like that so that's our octahedron all right so that's step three the geometry of the electron clouds around the central atom they occupy and 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 right 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 this top fluorine right and I go straight down like an axis to that other fluorine we would expect one of the ideal bond angles to be a hundred and eighty degrees for this octahedron here and the other ideal bond angles would be ninety degrees right so if I think about the angle that the axis I just drew makes with this one right here right so that's ninety degrees as well and again any anywhere you look you're also going to get ninety degrees let me go ahead and change colors here and we can look at another bond angle in here so this bond angle right that would also be ninety degrees alright so for an octahedral all six positions right we have 6 fluorines occupying the six positions are equivalent they are identical which means no axial or equatorial groups in an octahedral arrangement and that makes makes our life much easier because in the videos on five electron clouds we had to think about the axial and equatorial groups all right let's do let's do one for bromine pentafluoride here so be rf5 so valence electrons bromine has 7 it's in group 7 fluorine is also in group 7 and I have 5 fluorines 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 all right we have represented let's see 2 4 6 8 and 10 valence electrons so 4 so far 42 minus 10 is of course 32 valence electrons and we're going to put start putting those leftover electrons on our terminal atoms which are our fluorines so once again we're going to give each fluorine and octet all right so we're going to put six more valence electrons around each of our fluorine atoms and so we're putting six more electrons around five atoms so 6 times 5 is 30 so 30 so 32 minus 30 gives me two valence electrons left over and whenever you have a valence electrons left over after assign 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 all right 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 all right let's see how many electron clouds well we would have you would have these bonding electrons right in a region of electron density these bonding electrons these bonding electrons we keep on going around here so those are all electron clouds so that's 5 and then remember these nonbonding electrons is 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 like to want to point towards the corners of a regular octahedron so we're going to get an octahedral geometry for your electron clouds all right so let's think about let's think about this one though where would we put those loans that lone pair of electrons in an octahedron well sit since all six 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 this shape really fast so if I if I were to draw bromine right here I'm going to put a fluorine going in this direction another fluorine going back listen coming out a little bit and this is going away and then I'm going to put a fluorine going this way and then I would put the lone pair of electrons right down here and 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 all right 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 alright 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 we have a square base here and and if we connect up here to this top fluorine alright 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 right we know we know our ideal bond angles are going to be 90 degrees right so if we look at that let's use instances green here so it's just like we talked about before right so that bond angle is nine degrees right this bond angle in here is 90 degrees so our ideal bond angles are all 90 degrees for our square pyramidal geometry all right let's do one more example of six electron clouds and this is xenon tetrafluoride so we need to find our valence electrons right so xenon hat is in group eight eight valence electrons fluorine is in group seven so seven valence electrons times four gives me twenty eight twenty eight plus eight 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 four fluorines right so go ahead and put in our four fluorines surrounding our xenon and let's see we have represented two four six and eight valence electrons so 36 minus eight right so 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 six valence electrons on each one of our fluorine atoms so we are representing six more electrons on four atoms so 6 times 4 is 24 so 28-24 right 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 all right so after you draw your dot structure you count the number of electron clouds right and then you predict the geometry of those electron clouds and so let's count our electron clouds for this one all right so our regions of electron density so we can see that these bonding electrons our electron cloud right same with these bonding electrons and these over here as well and in this example we have we have two lone pairs of electrons in 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 Earth as they can so there they are going to be in an octahedral arrangement and so let's see if we can sketch out this molecule again alright so if the lone pairs of electrons want to get as far away from each other as they possibly can all right we're going to put those lone pairs 180 degrees from each other so here's one lone err and then here's our other lone pair all right it'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 right here would be another fluorine and then we would have two more back here so when we look at the shape of xenon tetrafluoride all right let's see if we can sketch in and what the shape would look like here so remember you 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 all right so we call this square planar so the geometry is square planar and in terms of ideal bond angles right that would be at 90 degrees so let me go ahead and show that real fast so in terms of bond angles right everything here would be 90 degrees for our square planer and so that's that's how to approach six electron clouds right now first example had zero lone pairs of electrons around the central atom our second example had one lone pair and our third example had two lone pairs and so even though the 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

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