Bond Angles

a bond angle represents the angle between two bonds that begins from the same element within a molecule. So if we take a look at this, we have here are central element and it's connected to two surrounding elements. The bond angle would be the distance from here to here. Here are central element. So the bongo will be the distance from here to here. And then finally, if we have a lone pair involved, we are going to ignore the lone pair. We're looking at the bond between the central element and the two surrounding elements. So the bank will be here to here. Now we're going to say when the central element has zero lone pairs, it possesses an ideal bond angle like this example in this example here, we're going to say that the ideal bond angles is the optimal angle elements take in order to minimize repulsion between one another. So it's that perfect angle they take when the central element has no lone pairs on it. And what we need to realize here is that when a central element has one or more lone pairs it's ideal, bond angle will be decreased. So if we took a look here at this third one, because our central element possesses a lone pair, What this lone pair does is it pushes the bonds further away from itself, causing them to come closer and tighter together. So we'd expect the bond angle here to be different from the bond angle here, although both look kind of V shaped or bent, their bond angles would not be the same because again the lone pairs. they push for more repulsion. They push those bond angles slightly closer together, causing a decrease in our bond angle. So keep this in mind when our central element has no lone pairs, we're going to have a perfect ideal bond angle, but once it starts gaining lone pairs, that ideal bond angle no longer exists, and the angle itself starts to get smaller and smaller and value.

if the H ch angle within the methane molecule is one of 9. degrees, what is the H NH bond angle within ammonia? All right. So here we're gonna draw ch four carbon will go in the center. Carbon has four valence electrons, Hydrogen Zara Group one a. So they only have one valence electron. So then here's our structure. And they're just saying here that our bond angle between hydrogen carbon, hydrogen is one of 9.5°. Now, if we draw ammonia nitrogen in group five A sort has five valence electrons. Three of them would be used to connect to the three hydrogen. Is that we have. Now here we have a lone pair on the nitrogen. Remember when we have that lone pair on the nitrogen, it causes our bond angle to decrease to become smaller. So we would expect the bond angle to be a number smaller than one of 9.5 would be less than one or 9.5. So, if we take a look here, The only degree that we see as an option, that's less than 195 is option seat. The bond angle in the ammonia molecule between H&H is approximately 107.3°

Bond angles can further differentiate molecules that possess the same number of electron groups. So if we take a look here, we have electron groups going from 2-6. In the first column we have our ideal bond angles because those central elements have zero long pairs. And then we start looking at what happens when we have 12 or three lone pairs involved. Now here, if we take a look to electron groups, there's only one possible shape that exists which is linear, it's bonding, go would be 180°. There's no possibilities of 12 or three lone pairs. So we take those out. When you have three electron groups, you could have possibly zero lone pairs which gives you an ideal bond angles of 120°. Or you can have one lone pair Here. All you need to remember is that we have three electron groups for this second structure and one of them is a lone pair. That just means our bond angle would be expected to be less than 120°. You're not expected to memorize an exact number. That's because there's a ton of different molecules that fit this description where we have a central element connected to two surrounding elements and one lone pair and be impossible to memorize every single one of those bond angles. Now, if you have four electron groups, you can have ideal Bonnie go if there are no lone pairs. So that would be one of 9.5. Once you start adding loan pairs or substituting in lone pairs, your bond angle decreases. So this will be less than 1 9.5. And this one has two lone pairs. So this one is even more less than one of 9.5. Now it's not gonna be a huge difference. Uh less than 19.5 here could be like 107 degrees. And here this might be one oh five or 106 degrees. So it's not a huge difference in bond angles. Next, if you have five electron groups, you have four different shapes. This one's a little bit different because here we have two different types of bond angles involved. Now remember if we have a five elektron system and imagine it being a sphere. Remember we have surrounding elements that exist along the equator and then we have ones that exist in the axial positions. And because of this, that's why we have two different types of bond angles involved. Now here for looking along the equator. So this and this, the bond angle would be 120°. And then if we're looking at the difference between the equatorial position in an actual bond, then that's what, that's 90°. Once we start adding loan pairs, these are just gonna decrease. So this will be less than 120 and this will be less than For the next one. This one here would be exactly 90°. And then here this would be linear in shape. So this would be 180°. So when it comes to five electron systems, it's pretty complex compared to the other. So this one you do have to remember a little bit in terms of certain bond angles. And then finally here, if we have six electrons groups were going to have three possible shapes here. Each one of these Gaps would be 90°. But once we have one loan period becomes less than 90° for each one. And then if you have too long pairs, it goes back to being 90° again. Okay, so those are the different types of bond angles we can discuss when it comes to these different shapes.

here, we need to determine the F I F bond angle for the following. I am I F four minus. So iodine in group seven A sword has seven valence electrons. It will use four of them to connect to the for ins Florins are also in group seven a. So they have seven valence electrons. When and they also only make single bonds. We've used four out of Aydin, seven valence electrons. Mhm. And we're going to say here it has three more that are not being used. But remember -1. Charge means would gain an outside electron which would give to the iodine since it has a charge. You put it in brackets with the charge on the outside. Now here we would have what we have to long pairs and four bonding groups or surrounding elements connected to. I don here we'd say that the bond angles here and here from this particular shape would be approximately 90 degrees. So that would be our bond angle for I F four minus