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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 the valence electrons so for methane carbon is in Group four so four valence electrons hydrogen is in Group one and I have four of them so one times four is four plus four is eight valence electrons that we need to show in our dot structure carbon goes in the center and carbon is bonded to four hydrogen's so I can go ahead and put my hydrogen's in there like that and this is a very simple dot structure we've already shown all eight of our valence electrons so let me go ahead and highlight those here two four six and eight 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 an electron cloud so that's one electron cloud here's another one down here and then here's one and then finally here's another one's 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 corners of a tetrahedron which is a four-sided figure so let me go ahead and draw the molecule here draw the methane molecule in to attempt to show it in a tetrahedral geometry and then we'll actually show you what a tetrahedron looks like here so here here's a quick sketch of what the molecule sort of looks like and let me go ahead and draw a tetrahedron over here so you get a little bit better idea of the shape right so four-sided figure and so there you go something like that so you could think about the corners of your tetrahedron right as being approximately where your hydrogen's are right and and and that just gives you a little bit better visual picture of that tetrahedron that four-sided figure here and so we've predicted 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 of 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 right so our goal now is to figure out what the bond angles are in a tetrahedral molecule turns out to be 109 point five 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've we start by finding our valence electrons nitrogen in group five so five valence electrons hydrogen in Group one and I have three of them so one times three plus five is eight so once again we have eight valence electrons to worry about we put nitrogen in the center and we know nitrogen is bonded to three hydrogen's so we go ahead and put our three hydrogen's in there like that let's see how many valence electrons we've used up so far two four and six so 8 minus 6 is 2 valence electrons left we can't put them on our terminal atoms because the hydrogen's 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 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 surround 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 right that's a region of electron density so that's a that's an electron cloud here's another one so that's two here's another ones that's three and then this lone pair of electrons right this this nonbonding pair of electrons is also going to be Cao 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 the geometry of those electron clouds there are those four electron clouds or want to going to once again try to point towards the corners of a tetrahedron so we can kind of sketch out the ammonia molecule all right we can kind of draw the base the same way we did before with our with our three hydrogen's 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 the the the electron clouds in a tetrahedral geometry let's go back up here and look at our steps again so in step three right in step three we predicted the geometry of electron clouds are going to attempt to being a tetrahedron at 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 this on the bottom part for the shape here and so when we do that we get we get something looks like it like a little squat pyramid here so if I'm I mean norm that lone pair of electrons up there at the top all right it's going to look something like that for the shape and we call this trigonal pyramidal all right so this is a this is a trigonal trigonal pyramidal shape so even though the 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 in terms of a bond angle this lone pair of electrons on the nitrogen actually occupies more space right this nonbonding these nonbonding electrons occupy a little more space than bonding electrons and because of that those nonbonding electrons are going to repel these bonding electrons and we go ahead and put them in 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 bond angle we saw before if nine point five for a tetrahedral arrangement of electron clouds and so it turns out that this bond angle between the atoms the hydrogen nitrogen the hydrogen nitrogen hydrogen bond angle gets a little bit smaller than hundred nine point five so it actually actually gets smaller to approximately 107 degrees here for a trigonal pyramidal situation all right let's do one more let's go ahead 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 six plus two is once again eight valence electrons to represent for our dot structure and we put oxygen in the center oxygen is bonded to two hydrogen's so we go ahead and draw those in there and let's see how many valence electrons have we represented so far that's two that's four so 8 minus 4 is 4 valence electrons left we first think about putting them on our terminal atoms but those are our hydrogen's so they're already happy with two electrons so we go ahead and put those 4 valence electrons on our central atom which is our oxygen and a four valence electrons means two lone pairs of electrons now and so now we 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 is being an electron cloud all right these nonbonding electrons is lone pairs an electron cloud and same thing for these nonbonding 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 however however in this case let me go and redraw the molecule here so let's go ahead and we have our water molecule all right 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 nonbonding electrons take up a little bit more space than bonding electrons and therefore are going to repeal they're going to repeal 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 it's even going it's going to be even smaller than 107 degrees and so this bond angle right here you'll see it listed as you know approximately 104 point five 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 all right so the the geometry of the electron clouds are attempting to be once again in 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 all right if you look at the shape of this I'll go ahead and draw the shape over here right 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 point five degree bond angle so those are those are a couple examples of for electron clouds and how to figure out the geometry while also thinking about the bond angles
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