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Current time:0:00Total duration:10:48

VSEPR for 5 electron clouds (part 2)

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Video transcript

the previous video we used VSEPR theory 4/5 electron clouds in this video we're just going to do a few more examples of five electron clouds so let's say we want to find the shape of chlorine trifluoride we start by drawing a dot structure all right which means we have to figure out the valence electrons so for chlorine in in group 7 therefore seven valence electrons fluorine is in group seven and we have three fluorines so seven times three is twenty-one plus seven means we have a total of 28 valence electrons that we need to show in our dot structure chlorine is going to go in the center since it is not as electronegative as fluorine and we know that the chlorine is bonded to three fluorines so we can go ahead and put in our three fluorines in here like that we have represented two four and six valence electrons so far so 28 minus six means we have 22 valence electrons left and we know we start by putting those those leftover valence electrons on our terminal atoms which are our fluorines fluorine is going to follow the octet rule and each fluorine already has two electrons around it and so therefore each fluorine needs six more so when we put six more valence electrons around each fluorine now each fluorine has an octet we just represented six times three more valence electrons so 6 times 3 is 18 and 22 minus 18 is 4 so we're left with 4 valence electrons and when you have leftover valence electrons after assign them to your terminal atoms you're going to give them to your central atom so we're going to assign 4 valence electrons to our central atom and it makes sense to put those in in lone pairs so there's one lone pair and there's another lone pair on the chlorine and so now that takes care of all of our valence electrons in terms of looking at the dot structure and thinking about chlorine there it actually is exceeding the octet rule and that's ok for chlorine to exceed the octet rule because of its position on the 3rd period on the periodic table I like to think about formal charge so if you look at if you actually assign a formal charge that chlorine you'll see it has a formal charge of 0 which to me just helps me under and these dot structures a little bit more and so we've drawn a dot structure that makes sense and so we can move on now to the second step which is where we count the number of electron clouds that surround our central atom so remember electron clouds are regions of electron density which means that bonding and non-bonding electrons have fit into that category so here we have a bonding pair of electrons right occupying an electron cloud here's another bonding pair of electrons in an electron cloud here's another one and then we have our lone pairs of electrons our nonbonding electrons are still regions of electron density and so there is an electron cloud and then our last lone pair there so a total of five electron clouds surrounding our central atom in step three predict the geometry of those electron clouds around your central atom and in the previous video we used VSEPR theory to to talk about why five electron clouds are going to are going to form a trigonal bi-pyramidal geometry so they're going to repel each other as much as they possibly can turns out that's a trigonal bi-pyramidal geometry here so our electron clouds are going to be in the same geometry here so the only tricky part for this dot structure is where do you put your lone pairs of electrons lone pairs of electrons take up more space alright these non-bonding lone pairs of electrons take up more space and bonding electrons and therefore they repel more so where you put them is extremely important for the overall structure of the molecule the last video we talked in a lot of detail about the fact that you're going to put nonbonding electrons or lone pairs of electrons in the equatorial position to minimize electron pair repulsion and so we're going to do the same thing here we're going to put our lone pairs of electrons equatorial so let me go ahead and put my put in my central chlorine atom I'm going to put a lone pair of electrons in equatorial here and then the other lone pair of electrons also equatorial right here and that means that I have one more one more spot and put one of the fluorines in the last equatorial position and so that leaves two more fluorines which we're going to put axial so one fluorine axial here and one fluorine axial here again putting your lone pairs equatorial minimizes electron pair repulsion and watch the previous video for more details about that this is at a more complicated example than what I talked about in the previous video but the same ideas are apply here as well I could probably spend a whole video talking about just this one molecule but we don't really have time for that so when you're talking about five electron clouds just think about putting your lone pairs in the equatorial position and so now we have now we have the general structure here let's go ahead and talk about the final shape right so when you're talking about the final shape you ignore any lone pairs and you predict the geometry of the molecule here so if we ignore lone pairs let me go ahead and redraw that so we're going to go ahead and put in our chlorine here alright so there's our chlorine we ignore lone pairs so we have our two axial fluorines and then we have our equatorial fluorine right ninety degrees to our axial fluorines and so that's of course an ideal bond angle so let's talk about let's talk about the bond angles here so if you look at the shape right so it looks kind of like like it's like these are linear here and then this is ninety degrees to that so you can see kind of a T shape here and so we actually call this molecule the shape t-shaped all right because we're ignoring the lone pair of electrons and so we see a t shape right here so this is a t-shaped geometry in terms of the ideal bond angle since it is t-shaped it makes it simple alright so you can think about you think about for your bond angles right this one you would expect to find a ninety degree bond angle for the fluorine chlorine fluorine bond angle here and then 180 degrees on this side but once again those are just ideal bond angles that's what you would predict thinking about a simple T shape the actual bond angles you'd have to get experimentally so so t-shaped geometry here all right let's do them let's do one more example of five electron cloud so this is the triiodide ion so i3 with a negative charge here so we need to draw the dot structure so we need two valence electrons right iodine is in group 7 so 7 times 3 gives us 21 valence electrons but this is an ion so it has a negative charge it has an extra electrons we have to add one to so 21 plus 1 gives us a total of 22 valence electrons to show in our dot structure so we have 3 iodine's so we can go ahead and show our 3 iodine's bonded together like that and we've already represented 4 valence electrons so 2 here and 2 here so 22 minus 4 means we have 18 valence electrons left and we start by putting those leftover valence electrons on our terminal atoms which are iodine's so we're going to think about our terminal atoms is following the octet rule here and so we're going to put 6 more electrons on each iodine to give each iodine and octet and so I just represented 12 more valence electrons so 18 minus 12 right gives us 6 valence electrons left over and those leftover electrons are of course going to be assigned to our central atom and so with 6 valence electrons you would expect three lone pairs of electrons alright so let's go ahead and put those last 6 valence electrons around our central iodine right in the form of 3 lone pairs and since this is an ion we should put this in brackets and put a negative 1 charge outside like that so this is our dot structure and if we look at our central iodine right it's once again exceeding our octet so it's once again okay for iodine to expand its valence shell because of its position on the periodic table and if you assign a formal charge that iodine you'll see it's a negative 1 formal charge which once again I always like to do just to help me understand what's going on a little bit better here so this is a dot structure that makes sense that follows all of our rules and so let's go back up here to to our next step so number one we've already drawn a dot structure to show our valence electrons in step two we count the number of electron clouds that surround our central atom so let's go ahead and look at our central atom and figure out how many electron clouds surround it right so these bonding electrons are an electron cloud these bonding electrons are an electron cloud and our nonbonding electrons right our lone pairs of electrons are still regions of electron density city and so you can see we have a total of five electron clouds for this dot structure - so for five electron clouds right that's going to be a trigonal by pyramidal arrangement of our electron clouds around our central atom so go ahead and put in our central atom here we think about our electron clouds being in a trigonal by pyramidal arrangement and we have seen that we put lone pairs of electrons into the equatorial positions to minimize electron pair repulsion so we have three lone pairs of electrons that surround our central iodine so we're going to put those three lone pairs equatorial all right so we have our three lone pairs of electrons equatorial like that alright and then we still have two iodine's and so one iodine would have to go axial up here alright the same thing for the other iodine down here and so I'll just leave off the lone pairs of electrons and so this is this is what our ion looks like let's go back up here to our steps alright and let's look at on right let's see we've already done step three we predicted the geometry of the electron clouds to be trigonal bi-pyramidal but when you're trying to predict the geometry of in this case the ion you ignore any lone pairs of electrons so let's go back down and look at it again alright so we're going to ignore the lone pairs of electrons around the central atom and when we do that we look for the shape and we can see the shape is just a straight line alright so we say it's a linear shape so this is a linear shape and since it's linear we would predict the bond angle to be approximately 180 degrees since it's a straight line and so that's the way to approach that's way to approach drawing this dot structure and so we've done four examples of five electron clouds right and so we've and each example has been slightly different in terms of the number of lone pairs around the central atom