Bond Angles (Simplified) - Video Tutorials & Practice Problems

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According to the VSEPR Model, bond angles result from surrounding elements and lone pairs around the central element positioning themselves at an optimal distance.

Ideal Bond Angles

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concept

Bond Angles (Simplified) Concept 1

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above angle is the angle formed by two adjacent neighbouring atoms in a molecule. And we're gonna say when the central element has zero lone pairs, it possesses what we call an ideal bond angle. Now this ideal bond angle is the optimal angle elements take in order to minimize repulsion between one another. And we're gonna say when the central element has one or more lone pairs, it's ideal, bond angle will be decreased. So for example, a particular here we said that the bond angle is the angle by two adjacent neighbouring atoms in a molecule. So you'll have a central element, It will be connected to two surrounding elements. And you're gonna say that the angle bond angle is this portion here, if we have three surrounding elements, the bond angle would be here. It would also be here as well as here. And if we had a lone pair, the bond angle would be in here. Now again remember once we have a lone pair on our central element, the bond angles going to decrease. This decrease will be represented as a blue image for a smaller bond angle. All right, so now that we know what a bond angle is and how lone pairs helped to reduce our ideal bond angle values, Let's pick on the next video and take a deeper look in terms of the exact values with these bond angles.

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Bond Angles (Simplified) Example 1

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here, we're told if the H C h angle within CH four molecule is one of 9.5, What is the H&H fund angle within NH 3? All right. So, if we were to draw out CH four carbon would go on the center. Remember bonds to hydrogen? Our only single bonds, This would represent CH four. We have four bonding groups on the carbon with zero lone pairs on the carbon. NH three. So in a three would have our bonds to hydrogen. Okay, But remember if we count up the total number of valence electrons, nitrogen has five and then we have three oxygen's each one for three hydrogen, each one with one electron. Because they're a group one A So maybe eight total valence electrons. We have 246 year and we have a long pair on the nitrogen two for remaining electrons Here, the Bongo is 195. Remember we said that the presence of a lone pair reduces or decreases our bond angle. So we'd expect an H. 3 to have a bond angle that is less than 109.5. If we take a look at our options, the only one that has an angle that's less than more than 95 is option C 107.3Â°.

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Bond Angles (Simplified) Concept 2

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here, we can say that bond angles can further differentiate molecules that possess the same number of electron groups. So when we have two electron groups, them is we have only one possibility are central element having to surrounding elements. In that case we have an ideal bond angle because our central element can't have a lone pair, So for electron group of to the ideal bond angle is 180Â°. When we have three electron groups around the central element, we have two possibilities. One were the central element just has three bonding groups And zero lone pairs. In this case because there are no lone pairs on the central element, we have an ideal bond angle of 120Â°. But remember, another possibility exists where we could have two bonding groups And one lone pair. The presence of the lone pair means that our bond angle will decrease from its ideal value. All you need to say at this point is if the ideal bond angle for three electron groups is 1 20 Then when it gets decreased it'll be less than 120. You don't have to give an exact number. You can just say less than 120. All right, So when we have four electron groups, we have three possibilities. We have four bonding groups, zero lone pairs, zero lone pairs means we have an ideal bond angle of 109. degrees, But we also have three bonding groups and one lone pair. So here we just say that our bond angle now is less than one of 9.5. And then we have our last possible option to bonding groups to lone pairs. Here We expect the bonding the bond angle to be again less than one of 9.5. And in fact, we'd say that it's even a little bit less than 195 than this one, because the presence of more lone pairs only helps to further reduce the bond angle. So just remember when we have no lone pairs on the central element, we have an ideal bond angle. The inclusion of any lone pairs after this means that our bond angle will be decreased from this ideal bond angle value.

Bond angles can further differentiate molecules that possess the same number of electron groups.

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example

Bond Angles (Simplified) Example 2

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here, we need to determine the H S N. H. Von and go for the following compound. So we have 10 connected to two hydrogen tim is in group four A. So it has four valence electrons Here we have two hydrogen. Each one has one valence electron since during group one a. So that's a total of six valence electrons involved here. We would place tin in the center. It is connected to the two hydrogen. Remember hydrogen can only make single bonds. They don't follow the octet rule, they only follow the too wet rule. So if you don't have to add any additional electrons to them. So at this point we've used four of our electrons because remember each co valent bond has two electrons within it. That means we have two valence electrons remaining, which we simply place on the 10. Now, how many electron groups does the tin half? It has one lone pair and to bonding groups. So to surrounding elements. So we have three electron groups. So remember For three electron groups, the ideal bond angle is 120Â°. But when we have the presence of a lone pair on the central element, the ideal bond angle decreases. So all you have to say here is that we have a bond angle that is less than 120Â° for the following compound.

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Problem

Problem

Determine the bond angle for the following compound:BeCl_{2}.

A

90Â°

B

180Â°

C

109.5Â°

D

120Â°

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Problem

Problem

Determine the bond angle for the thiocyanate ion, SCN^{â€“}.

A

180Â°

B

90Â°

C

120Â°

D

109.5Â°

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Problem

Problem

Determine the Clâ€“Oâ€“Cl bond angle for the OCl_{2} molecule.

A

>109.5Â°

B

<109.5Â°

C

>120Â°

D

<120Â°

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