# Bond angle

## Homework Statement

Explain why the H-N-H bond angle in ammonia is less than that of H-C-H in methane.

Though this is a worked example. I just want to understand the explanation given for the problem.

## The Attempt at a Solution

As I said, this is a worked example, am just battling to understand the explanation given for the problem.

Here is the explanation given for the problem:

The electron cloud formed by the unshared pair in NH3 spread out over a greater volume than that of the three pairs connected to hydrogen atoms. This would tend to force the bonding pairs closer to one another and thereby reduce the bond angle to less than the bond angle in the tetrahedral methane.

I don't want to memorise this explanation without understanding it. That is why I have decided to post the problem into the forum. Thank you!

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AGNuke
Gold Member
The electron pair on the top of N atom is more volumetrically dispersed than electrons participating in the bond formation. Thus, lone pairs offers more repulsion than bond electrons.

You can also see the structure of ammonia to confirm it. So, according to VSEPR theory, the bond angle decreases as the result of this extra repulsion.

What this VSEPR theory propose is that in a hybridised atom, the orbitals tends to occupy spaces which are possibly furthest. If electrons are more repulsive, bond angle decreases to compensate for that repulsion too.

Simon Bridge
Homework Helper
Oh - someone beat me to it.
Must ... resist ...

"He does chemistry with a Walther PPK?"
"Yep - it's his Bond angle."

<quietly sneaks off>

The electron pair on the top of N atom
Please what do you mean by electron pair on top of N atom?
is more volumetrically dispersed than electrons participating in the bond formation. Thus, lone pairs offers more repulsion than bond electrons.

You can also see the structure of ammonia to confirm it. So, according to VSEPR theory, the bond angle decreases as the result of this extra repulsion.

What this VSEPR theory propose is that in a hybridised atom, the orbitals tends to occupy spaces which are possibly furthest. If electrons are more repulsive, bond angle decreases to compensate for that repulsion too.

AGNuke
Gold Member
unshared pair in NH3
You said it.

You said it.
Those links shows the H-N-H bond angle in ammonia and than that of H-C-H in methane
http://www.meta-synthesis.com/webbook/45_vsepr/balloon3.jpg
http://www.chem.ucla.edu/harding/IGOC/S/sp3_07.jpg [Broken]
Respectively.
Can you use it to explain what they are trying to say here?:
The electron cloud formed by the unshared pair in NH3 spread out over a greater volume than that of the three pairs connected to hydrogen atoms. This would tend to force the bonding pairs closer to one another and thereby reduce the bond angle to less than the bond angle in the tetrahedral methane.
Please if you have more links that will help me in understanding this fast, please do well by sending them in, so that I can study them and ask question if necessary.

Last edited by a moderator:
AGNuke
Gold Member
Standard Data. H-C-H Bond = 109°28' (Better Remember it, useful data)

It is a perfect tetrahedral geometry. Look at the Carbon, it has sp3 hybridization, and there are 4 Hybrid orbitals, so that means each orbital is facing repulsions from 3 other orbitals, so they always have a tendency to move as far as possible in 3-Dimensional space. The result, a Tetrahedral Geometry.

Now, in CH4 see that there are 4 orbitals with equal electron density, as there all are involved in bonding. Due to their involvement in bonding, their electrons are not only limited to themselves, but to the bonded atom too, thus dispersing their electron density along the length of the bond.

Now, in NH3, Hybridization is the same but this time, a lone pair is present in one of the hybrid orbital. This lone pair cannot disperse its electron density along the length, so it just do so along the width. In doing so, the other orbitals, which were involved in bonding, faces more repulsion from lone pair.

This is because, bonded electrons are dispersed along the length, so they are not close to the N atom, but the lone pair is close to N atom. This is the reason bonded electrons can't constrain lone pair, instead, they get constrained, and bond angle reduces to 104°.

Standard Data. H-C-H Bond = 109°28' (Better Remember it, useful data)

It is a perfect tetrahedral geometry. Look at the Carbon, it has sp3 hybridization, and there are 4 Hybrid orbitals, so that means each orbital is facing repulsions from 3 other orbitals, so they always have a tendency to move as far as possible in 3-Dimensional space. The result, a Tetrahedral Geometry.

Now, in CH4 see that there are 4 orbitals with equal electron density, as there all are involved in bonding. Due to their involvement in bonding, their electrons are not only limited to themselves, but to the bonded atom too, thus dispersing their electron density along the length of the bond.

Now, in NH3, Hybridization is the same but this time, a lone pair is present in one of the hybrid orbital. This lone pair cannot disperse its electron density along the length, so it just do so along the width. In doing so, the other orbitals, which were involved in bonding, faces more repulsion from lone pair.

This is because, bonded electrons are dispersed along the length, so they are not close to the N atom, but the lone pair is close to N atom. This is the reason bonded electrons can't constrain lone pair, instead, they get constrained, and bond angle reduces to 104°.
I quite agree with you that CH4 is tetrahedral in nature because I have heared and seen it many times in different chemistry textbooks and sites. My question now is what are obitals? What are hybrid obitals?

AGNuke
Gold Member
Well, Hybridization IS the concept which was introduced as a consequence of study of bond length and bond angles of methane.

Orbitals - Those regions in the space around an atom where the probability of finding an electron is maximum (>90%). Like s, p, d, f. Wikipedia it if unsure.

Carbon has configuration 1s2 2s2 2p2. In order to form 4 bonds, carbon atom gets excited (Yep! He's excited to make bonds) to 1s2 2s1 2p3. Now since 2p subshell contains 3 orbitals, each orbital has one electron (Pauli). Now, Carbon is capable of forming 4 covalent bonds, owing to 4 unpaired electron.

Since Carbon is to form 4 σ bonds, all 4 orbitals combines to form sp3 hybrid orbital. Now these orbitals possess tetrahedral shape. Previously, s orbital was spherical and p orbitals (px,py,pz) were all 3 orthogonal (perpendicular) dumb-bells.

Borek
Mentor
Orbitals - Those regions in the space around an atom where the probability of finding an electron is maximum (>90%)
That's a pretty lousy wording (at best), misinformation (at worst).

AGNuke
Gold Member
Lousy wordings.

Lousy wordings.
Am the one confusing my self all this while. I have seen the bond formation in ammonia and am begining to understand the concept better now!

Can I as well put my own (the way I understand it) explanation this way:
the lone or unshared pair in H-N-H bond of ammonia that is not participating in bond formation form electron cloud which spreads out a greater volume, thereby reducing the bond angle of the remaining three shared pairs that is involved in bond formation, therefore the bond angle of H-N-H in ammonia is less than that of H-C-H in methane??

I will have more question to ask on this but let me wait till you reply my last post.

AGNuke
Gold Member
Yes. It is true. But if you want to understand the structure, geometry, bond angles better, you may want to learn hybridization.

As Borek stated, I very loosely defined the term orbital, as it is a mathematical function of ψ (Blah Blah, don't bother with it) but since you don't know orbital, I thought it might be a fair definition to begin with, as I began with same definition. Try to rote learn some knowledge from some other sources.

Hybridization is simply the mixing of orbitals of similar(not much difference) energy leveled orbitals. This results in the formation of degenerate hybrid orbitals, which, according to VSEPR theory, decide their structure and angle between them.

The electron pair on the top of N atom is more volumetrically dispersed than electrons participating in the bond formation. Thus, lone pairs offers more repulsion than bond electrons.

You can also see the structure of ammonia to confirm it. So, according to VSEPR theory, the bond angle decreases as the result of this extra repulsion.

What this VSEPR theory propose is that in a hybridised atom, the orbitals tends to occupy spaces which are possibly furthest. If electrons are more repulsive, bond angle decreases to compensate for that repulsion too.
Thus, lone pairs offers more repulsion than bond electrons.
How does the lone pair offer repulsion that the effect of the repulsion tend to reduce the bond angle of bonded electrons?

AGNuke
Gold Member
Bonded electrons tends lies between the atoms participating in the bond, and they belong to both the atoms, so they keep making a trip from one atom to other, thus giving them a certain direction to go.

Now, lone pair are exclusively on the central atom, and since the electrons repel themselves, and they are in the same orbital, they tend to occupy the maximum volume. In doing so, they are also pushing the bonded electrons. Repulsions increases between the bonded electron and lone pair due to decrease in distance between them (Remember, LP expanded) so electrostatic forces of repulsion between the electrons also increased. Thus, to balance the electrostatic forces on each bond, they adjust themselves. Now you know how they adjust yourself.

Bonded electrons tends lies between the atoms participating in the bond, and they belong to both the atoms, so they keep making a trip from one atom to other, thus giving them a certain direction to go.

Now, lone pair are exclusively on the central atom, and since the electrons repel themselves, and they are in the same orbital, they tend to occupy the maximum volume. In doing so, they are also pushing the bonded electrons. Repulsions increases between the bonded electron and lone pair due to decrease in distance between them (Remember, LP expanded) so electrostatic forces of repulsion between the electrons also increased. Thus, to balance the electrostatic forces on each bond, they adjust themselves. Now you know how they adjust yourself.
Thanks for that explanation! I think the concept is clearer now.