JR0002 - Triple bond rotation – Conversation in Particle Accelerator

Guru Vishnu

Jan 31, 2020 14:33

@JohnRennie: Hi sir. Kindly reply when you find time. Thank you.

Few days ago, we discussed about the rotational barrier of a triple bond. I liked the following explanation of yours:

in Problem Solving Strategies, Jan 14 at 11:14, by John Rennie

Although I'd expect the barrier to be lower than a double bond because with a triple bond it only takes a quarter turn to recreate the bonds rather than a full turn.

But, now I realised, that "quarter turn" isn't possible due to different in phase of the orbitals. So will the final outcome change or will it remain the same?

Earlier, I understood this based on the following image (without the indication of different phases)

Now, it seems the "cylindrical symmetry" will not hold true. I think due to difference in phases, there will be two half hollow cylinders with opposite signs.

Guru Vishnu

Feb 1, 2020 07:18

@JohnRennie: Hi sir. Good morning :-)

John Rennie

@GuruVishnu hi :-)

I'm working at the moment. Sorry :-(

Guru Vishnu

Fine sir. No problem :-)

Could you ping me once you're free?

Or may I post the question and you'd look at if afterwards?

John Rennie

Feb 1, 2020 08:27

@GuruVishnu hi, I've finished work now.

Guru Vishnu

Feb 1, 2020 08:42

@JohnRennie: Hi sir. Did I made you wait so long? I'm sorry, my eyes were on my book and not on the screen and so I missed your notification.

John Rennie

@GuruVishnu no problem, I have plenty to do checking Facebook and my e-mail :-)

Guru Vishnu

:-)
Sir, do you remember, few days ago in the PSS chatroom, we had a discussion on the rotatability of a triple bond?

John Rennie

Yes

Guru Vishnu

Did you see this and the following messages? :

18 hours ago, by Guru Vishnu

@JohnRennie: Hi sir. Kindly reply when you find time. Thank you.

Just a little above this conversation.

John Rennie

Yes, and I see what you mean. The two p orbitals when added together must have a nodal plane. So there must be a plane in which the electron density is zero.

Guru Vishnu

Feb 1, 2020 08:46

@JohnRennie Yes and the two lobes are of different phases.

Now, I don't see how the system could be cylindrically symmetric.

John Rennie

@GuruVishnu And yet Googling suggests that the pi electron density in acetylene is cylindrically symmetric.

Guru Vishnu

@JohnRennie Yes sir. So, I think, the symmetry they're talking is "structural" rather than on the molecular orbital basis.

In other words, we can't really determine whether a molecule bonded in a triple bond has rotated or not, on the basis of structures alone.

John Rennie

Ah, OK, I get it.

Guru Vishnu

:-)

John Rennie

Suppose we take the x axis to be along the triple bond, then look in the yz plane through one of the nuclei i.e. we're looking at a plane normal to the bond.

The electron density is proportional to $|\psi|^2$. Yes?

Guru Vishnu

Feb 1, 2020 08:59

Yes sir.

John Rennie

But we don't calulate $|(p_y + p_z)|^2$, we calculate $|p_y|^2 + |p_z|^2$ to get the electron density.

Guru Vishnu

@JohnRennie I don't think I understand this statement properly, sir.

I haven't learnt something like this.

Could you explain a little bit, if possible?

John Rennie

I'll have to think how best to explain this. I'm not sure it will be easy to explain unless you've done more QM.

But the function $|(p_y + p_z)|^2$ has a nodal plane because of the sign changes in the p orbitals. Yes? (I can draw a diagram if it will help)

Guru Vishnu

@JohnRennie I can understand between different lobes of opposite phases, there will be a nodal plane. Is this what you were referring to, sir?

John Rennie

Yes

Guru Vishnu

Feb 1, 2020 09:05

Is it possible to continue with this diagram itself, sir? I think it could save some effort.

Unfortunately, it doesn't show different phases.

But I'll try to understand it based on "top" lobe as positive and the "bottom" one as negative phase.

John Rennie

That's the orbitals seen in the yz plane. I've drawn the nodal plane as the dashed line.

Guru Vishnu

It looks great!

Yes sir. Understood the diagram.

John Rennie

Guru Vishnu

I understood the first and the second diagrams sir.

John Rennie

When we add $p_y$ and $p_z$ we get a function that still has the nodal plane.

That's the middle diagram.

The right diagram shows the result when we square $p_y + p_z$. Now this is positive everywhere because anything squared is positive everywhere, and there is still a nodal plane. OK so far?

Guru Vishnu

Feb 1, 2020 09:15

@JohnRennie Ok sir. So the probability of finding electrons in both the grey lobes is equal. Am I right?

John Rennie

The probability of finding an electron is given by $\psi^2$. Yes?

Guru Vishnu

@JohnRennie Yes sir.

John Rennie

In the middle diagram $p_y+p_z$ is antisymmetric about the nodal plane i.e. if we reflect in the nodal plane the magnitude is the same but we get a sign change.

OK so far?

Guru Vishnu

Yes sir.

John Rennie

And when we square the function the result is positive everywhere. So the square is symmetric about the nodal plane.

Guru Vishnu

Feb 1, 2020 09:20

Yes sir.

John Rennie

But this is not how you calculate the electron density, so the function (with a nodal plane) that we have calculated is not the electron density.

The electron density is proportional to $p_y^2 + p_z^2$ not $(p_y + p_z)^2$

Guru Vishnu

@JohnRennie I'll consider this to be true for the time being sir. Even though I didn't understand this, I could understand your immediately above statement. So according to this, the orbital (region of maximum probability of finding an electron) looks much different from that of the third grey lobes diagram. Am I right, sir?

John Rennie

Yes. If we square the $p_y$ and $p_z$ orbitals we get the middle diagram and this does not have a nodal plane because now there is no sign change.

Guru Vishnu

@JohnRennie Now, I see how intricate this is. I can see the triple bond is cylindrically symmetric even on the basis of electronic arrangement.

John Rennie

In fact when we add the squared orbitals we get the symmetric function on the right.

Guru Vishnu

Feb 1, 2020 09:28

@JohnRennie Ok sir, so electrons in one lobe of one phase can move happily into another lobe of opposite phase?

I thought they wouldn't.

John Rennie

I'm not sure it makes sense to talk about the electrons moving.

Rememner the electrons are not little balls whizzing around the orbital. The electrons are delocalised into a cloud in the shape of the orbital.

Guru Vishnu

Ok sir. So it's meaningless to call an electron labelled "A" can be present in either positive lobe or negative lobe at different points of time.

I understand that we couldn't fix a momentum or position of an electron accurately as per Heisenberg's uncertainty principle. Is this why it's meaning less, sir?

John Rennie

@GuruVishnu yes, the electron is simultaneously present everywhere in the orbital.

Guru Vishnu

@JohnRennie Ah yes! There is only one electron. I totally forgot this fact.

John Rennie

@GuruVishnu no. I think you'll have to take this on faith until you start doing QM at college.

Guru Vishnu

Feb 1, 2020 09:34

@JohnRennie Ok sir :-)

So, could you tell the moral of the story, sir?

Will it be easier to rotate a triple bonded group over double bonded group?

As per our previous discussion and the textbook.

John Rennie

The end result is that there is no barrier to rotation along the axis of a triple bond.

Guru Vishnu

@JohnRennie Easier like single bonds?

John Rennie

Same as single bonds.

Only the double bond has a barrier to rotation.

Guru Vishnu

That's great sir.

Thank you very much for explaining this in a way I could understand :-)

John Rennie

:-)

Guru Vishnu

Feb 1, 2020 09:38

I think we'd have more to discuss on this in future.

JR0002 - Triple bond rotation

♦ Particle Accelerator

Participants