Bond Angles - Online Tutor, Practice Problems & Exam Prep

Bond angles are critical in determining the shape and structure of molecules. They are defined as the angle between two bonds originating from the same central atom. When a central atom has no lone pairs, it exhibits an ideal bond angle, which is the optimal angle to minimize electron repulsion. However, the presence of lone pairs reduces the ideal bond angle as they exert additional repulsion, pushing bonded atoms closer together. For example, a linear shape with two electron groups has a bond angle of:

$180°$

while a trigonal planar shape with three electron groups has:

$120°$

As the number of electron groups increases, the complexity of possible shapes and bond angles also increases, such as in molecules with five electron groups that have equatorial and axial positions leading to different bond angles. It's important to understand that the presence of lone pairs can significantly alter these angles, and while exact bond angles don't need to be memorized, recognizing the trends and how lone pairs affect geometry is essential.

Bond Angles Concept 1

A bond angle represents the angle between two bonds that begins from the same element with a molecule. So if we take a look at this, we have here our 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 bond angle 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 bond angle will be here to here. Now we're going to say when the central element has 0 lone pairs, it possesses an ideal bond angle. Like this example and this example.

Here we're going to say that the ideal bond angle 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 center element has one or more lone pairs, its 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 center element has no loanpayers, we're going to have a perfect ideal bond angle. But once it starts gaining loanpayers, that ideal bond angle no longer exists, and the angle itself starts to get smaller and smaller in value.

Bond Angles Example 1

If the HCH angle within the methane molecule is 109.5�, what is the H&H bond angle within ammonia? All right, so here we're to draw chapter 4. Carbon will go in the center. Carbon has four valence electrons. Hydrogens are in Group 1A, so they only have one valence electron. OK, so then here's our structure and they're just saying here that our bond angle between hydrogen, carbon, hydrogen is 109.5�.

Now if we're to draw ammonia, nitrogen's in Group 5A, so it has five valence electrons. Three of them would be used to connect to the three hydrogens that we have. Now here we have a lone pair on the nitrogen. And 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 O 9.5. It'd be less than one O 9.5. So if we take a look here, the only degree that we see as an option that's less than 109.5 is option C. The bond angle in the ammonia molecule between H&H is approximately one O 7.3�.

Bond Angles Concept 2

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 lone pairs. And then we start looking at what happens when we have 1, 2, or three lone pairs involved.

Now here, if we take a look at electron groups, there's only one possible shape that exists which is linear, its bonding angle would be 180°. There's no possibilities of 1, 2, 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 angle 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 it would be impossible to memorize every single one of those bond angles.

Now, if you have four electron groups, you can have an ideal bond angle if there are no lone pairs. So that would be 109.5°. Once you start adding lone pairs, or substituting in lone pairs, your bond angle decreases. So this will be less than 109.5°. And this one has two lone pairs. So this one is even more less than 109.5°. Now it's not going to be a huge difference. Less than 109.5° here could be like 107 degrees. And here this might be 105 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-electron 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 and an axial bond, then that's 90°. Once we start adding lone pairs, these are just going to decrease. So this will be less than 120° and this will be less than 90°. 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 others. 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 electron groups, we're going to have three possible shapes here. Each one of these gaps would be 90°. But once we have one lone pair, it becomes less than 90° for each one. And then if you have two lone 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.

Bond Angles Example 2

Here we need to determine the FIF bond angle for the following ion if 4 minus. So iodine is in Group 7A, so it has seven valence electrons. It will use four of them to connect to the fluorines. Fluorines are also in Group 7A, so they have 7 valence electrons, and they also only make single bonds.

We've used four out of iodine 7 valence electrons and we're going to say here it has three more that are not being used. But remember -1 charge means we'd gain an outside electron, which would give to the iodine. Ince it has a charge, you put it in brackets with the charge on the outside.

Now here we would have what we have, two loan payers and four bonding groups or surrounding elements connected to iodine. Here we'd say that the bond angles here and here from this particular shape would be approximately 90�. So that would be our bond angle for IF₄^-.

Determine the bond angle for the thiocyanate ion, SCN^–.

• A. 180°
• B. 90°
• C. 120°
• D. 109.5°

 In the PCl­₃F₂ molecule the chlorine atoms exist in the equatorial positions and the fluorine atoms exist in the axial positions. Based on this information, predict the Cl–P­–Cl bond angle.

• A. 90°
• B. 180°
• C. 109.5°
• D. 120°

Determine the O–N–O bond angle for N_­2O₄, which exists as O₂N–N­O_­2.

• A. 120°
• B. 109.5°
• C. 180°
• D. 90°

Here’s what students ask on this topic:

What is a bond angle?

A bond angle is the geometric angle formed between two adjacent bonds originating from the same atom in a molecule. To visualize this, consider a molecule with a central atom and at least two other atoms bonded to it. The angle between the lines representing the bonds to these atoms is the bond angle. This angle is a crucial aspect of molecular geometry and helps determine the shape of the molecule.

For example, in a water molecule (H₂O), the bond angle is the angle between the two hydrogen atoms with the oxygen atom at the vertex. The bond angles are influenced by the electronic repulsions between electron pairs located in the valence shell of the central atom, which can be described by the Valence Shell Electron Pair Repulsion (VSEPR) theory. In water, the bond angle is approximately 104.5 degrees, which is less than the ideal tetrahedral angle of 109.5 degrees due to the repulsion between the lone pairs of electrons on the oxygen atom.

How do you determine bond angle?

To determine bond angles in a molecule, you first need to understand the molecule's geometry, which is dictated by the number of electron pairs surrounding the central atom, both bonding pairs and lone pairs. This is described by the VSEPR (Valence Shell Electron Pair Repulsion) theory, which states that electron pairs will arrange themselves as far apart as possible to minimize repulsion.

Here's a step-by-step guide to determine bond angles:

1. Draw the Lewis structure of the molecule to identify the number of bonding pairs and lone pairs on the central atom.
2. Use VSEPR theory to determine the molecular geometry based on the number of electron pairs. Common geometries include linear (180°), trigonal planar (120°), tetrahedral (109.5°), trigonal bipyramidal (90°, 120°, and 180°), octahedral (90° and 180°), and bent or angular (less than 120° or 109.5° depending on the number of lone pairs).
3. Once the molecular geometry is known, you can infer the bond angles. For example, in a tetrahedral molecule, the bond angles are approximately 109.5°.
4. If there are lone pairs, they will repel more strongly than bonding pairs, slightly reducing the bond angle between the bonding pairs.

In which cases do you expect deviations from the idealized bond angle?

Deviations from idealized bond angles occur due to several factors. One primary reason is the presence of lone pairs of electrons, which occupy more space than bonding pairs and therefore push the bonding pairs closer together, reducing the bond angle. For example, in water (H₂O), the ideal bond angle for a tetrahedral arrangement is 109.5 degrees, but due to the two lone pairs on the oxygen, the H-O-H bond angle is reduced to about 104.5 degrees.

Another factor is differences in the size and electronegativity of the atoms involved in the bonds. Larger and more electronegative atoms can affect the spatial distribution of other bonds in the molecule, leading to deviations from the expected angles.

Additionally, the presence of double or triple bonds can also alter bond angles, as these bonds occupy more space than single bonds. In molecules with multiple types of atoms and bonds, the various repulsions between electrons can lead to adjustments in geometry that deviate from the idealized bond angles predicted by simple models like the Valence Shell Electron Pair Repulsion (VSEPR) theory.