ฝัง
- เผยแพร่เมื่อ 8 เม.ย. 2017
- 01:52 Reducible representation for group orbitals
03:03 Reduction of reducible representation
08:41 Effect of each symmetry operation on representative orbital
10:26 A1 irreducible representation
11:45 The E irreducible representation
13:06 Accounting for orbital degeneracy
17:16 Visualizing the group orbitals
20:56 Sorting molecular orbitals by energy
Derivation of the sigma molecular orbitals of NH₃ (ammonia), using the projection operator method.
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My professor and TAs at UCLA struggle to even make their explanations even half as clear and concise as this video. Thankyou so much!
Thank you so much for your kind words.
Hello Fellow UCLA 172 Student here, I think we're talking about the same Professor and YES it's still the same haha... This channel is a LIFESAVER
(For mobile users)
01:52 Reducible representation for group orbitals
03:03 Reduction of reducible representation
08:41 Effect of each symmetry operation on representative orbital
10:26 A1 irreducible representation
11:45 The E irreducible representation
13:06 Accounting for orbital degeneracy
17:16 Visualizing the group orbitals
20:56 Sorting molecular orbitals by energy
Hopefully you're getting some compensation for this! Cause you are a life saver!
Inorganic chemistry had it really hard for me since the beginning of my semester and I really had no hope for myself in it until now--
I really get everything now and everything seems so crystal clear!
Thanks! I really appreciate your videos!
Great job!
you saved my life with this video, thank you very much. Greetings for Roma
This was extremely helpful! Thank you very much for this
Well aren't you a lifesaver? Really grateful to u
Very helpful. Thank you!
Brilliant explanation.. thank you so much!!!
Love you man thanks for the demonstration
This video was exactly what i needed.❤❤❤ THANK YOU❤
Truly amazing video 🙂
hello l am a student who is studying chemistry in Korea. However, I did not have any information on how to the ammonia projection operator in Korea, so I watched your TH-cam. It was so kindly explained that it was easy to understand. Thank you !
I want you to consider translating the translator because I am not good at English😂
ni hao shi shi o wai ma
Thanks for the video! I understood the method how the molecular orbitals of ammonia can be written.
But seems that we have to learn group theory, such as ir/reducible representation, to understand from the bottom. I still wonder why this method works well.
I have a test tmrrw. ur saving my life and my grade. thank you
You can do it!
Thank you for grant teaching
really thanks a lot!
Glad it helped!
Similar comment as on your BF3 video: you derived your E symmetry SALCs incorrectly and arrived at the
incorrect form of one of them; s1-s2 is not orthogonal to 2s1-s2-s3, and therefore cannot be a valid solution. Instead you needed to project on a linear combination of two s-orbitals. In this case s2+s3 will yield the equivalent of E(a) = 2s1-s2-s3 while projection of s2-s3 will yield E(b) = s2-s3.
You bring up several excellent points. However,
1. The video is already very long. Demonstrating a Gram-Schmidt orthogonalization at the end would double the length of the video.
2. There are many excellent videos online which demonstrate the Gram-Schmidt orthogonalization process. I like the online lectures at MIT.
3. There is a method (the "Kim" method), that generates the orthonormal set _en passant_. One
can see the method demonstrated here: th-cam.com/video/vPmeouMsbyo/w-d-xo.html, th-cam.com/video/KjOUeGBYxdY/w-d-xo.html,
th-cam.com/video/QMU6SOLbvbs/w-d-xo.html, and
th-cam.com/video/Qllcn-2aJZ8/w-d-xo.html
.
4. The set of normal modes of vibration, or set of molecular orbitals, is a set of bases for a vector space, mathematically. It is *convenient* for the bases to be orthogonal, but certainly not *required*.
5. The set you propose is only orthogonal *IF* you assume all overlap integrals (for orbitals on different atoms) are identically 0.
6. There are cases where we specifically do *NOT* want an orthonormal set of bases. One might be familiar with the set of five (5) d atomic orbitals. Clearly, these span a five (5) dimensional space. In several computational chemistry "basis sets", these five (5) orbitals are represented by a basis of SIX (6) d orbitals. It is clear that there is no way to construct a mutually orthogonal set of SIX (6) vectors that span a FIVE (5) dimensional space.
This technique is not even unique to d atomic orbitals. There are f atomic orbitals, and it is not uncommon in
computational chemistry to span this seven (7) dimensional space with a set of TEN (10) vectors.
thanks you so much!
You're welcome!
Thanks a lot professor this was really helpful . I have a question though. In your previous video where you calculated reducible representation for all atoms (lambda of 3N), you multiplied lamda (not moving due to operation) by lamda(trace of symmetry matrix, but you didn’t do that here. If you were to do that then your final reducible representation would have been 9E+sigmaV instead of 3E+ sigmaV. Could you please explain the reasoning behind this?
Great question!
There are two (2) different procedures: one for vibrations, and one for orbitals. When we are finding vibrations, we need to multiply by the trace. When we are finding orbitals, we do NOT need (or want) the trace.
In vibrational motion, we have ultimately need 3N -6 (or 3N-5 for linear) modes. With orbitals, we need N molecular orbitals from N atomic orbitals. As a result, deriving orbitals is much less complicated than finding vibrations.
Does this help?
@@lseinjr1 Thanks a lot professor. This helps a lot. Your explanation is very clear
what about the pz orbital??? does it hve A1 symmetry.. so it can bond with 1s hydorgen??? im confused :/
See previous comments below.
In short, the pz orbital is a lone pair on nitrogen for reasons of energy, even though it has the correct symmetry for A1.
Hi I'm a student studying this part in south Korea. Thank you for the nice explanation and exact pronunciation!
I'm really appreciated. But, I have a question. I'm confused on finding the second SALC of 'E'. Can you generalize the method once more? I mean the way I can apply for another kind of molecule other than ammonia.
When a molecule has degeneracy (Eg or Tg or whatever), the cause is a rotation operation Cn or Sn, where n is greater than 2. In ammonia, this is a C3; in a square planar molecule like XeF4, it is a C4.
To find the second member of Eg ( the "other" Eg), do a C3 or C4 (whichever is found in the point group) on the first Eg. Take this result, and subtract it from the first Eg. The result is the second Eg.
To see this idea for other molecules, see:
th-cam.com/video/zf3HvalFq_4/w-d-xo.html (methane)
th-cam.com/video/sTuv-9WoMn4/w-d-xo.html (BF3)
th-cam.com/video/KVLeo2EN_fc/w-d-xo.html (cyclobutadiene).
Does this help?
@@lseinjr1
Thank you. I've just watched the video about the CH4! I understood my question above.
However, I got curious of one of the step finding the projection operator of CH4. How do I know if the thing I found is the linear combination of MO or the MO we wanted? You did several more times of substration to find SALC of T2.
I thought that the first expression we've got (3sigma1-sigma2-sigma3-sigma4) was the wavefunction(T2_1) and the second one 2sigma-2sigma3 was the second SALC of T2.
Sorry if my english is bad.
Great question! If were do the C3 or C4 operation on the first orbital, and the result is just a re-labeling of the first orbital, then we know that it is a linear combination.
For example, we might have s1 - s2 as the first orbital. If the "new" orbital is s2 - s1, that is just a switching of the labels of atom 1 and atom 2 - so a linear combination. On the other hand, if the new version is s3 -s4, these are completely different atoms than those that were involved in the first orbital, so this second orbitals is the MO we wanted.
In general, we tend to get linear combinations after C3 (say NH3 or BF3), but get different MO after C4 (XeF4, CH4, or SF6). This is not a rule, but it is usually true.
Does that help?
Sorry I have a question about the pz orbitals. You said that pz orbitals doesn't have the right simmetry to combine with s orbital , but they have it... I've checked on character tables that they have A1 simmetry too. So my question is: Why nitrogen's pz orbitals does not interact with hydrogen's s orbital?
Great question! I should have explained this more carefully in the video. For ammonia, a C3v molecule, both the nitrogen 2s and the nitrogen 2pz have the right symmetry to combine with the A1 combination of hydrogen orbitals. So which one combines?
There are two (2) possibilities. Possibility #1 is that 2s combines with A1, and the 2pz orbital has the lone pair. Possibility #2 is that 2pz combines with A1, and 2s has the lone pair.
Neither possibility is obviously unreasonable. Water (H2O) has two lone pairs, one in an oxygen 2s orbital, and one in an oxygen 2p orbital.
We can do some reasoning about energy and basicity that the molecule favors having a lone pair in the higher energy atomic orbital (2p is higher energy than 2s for nitrogen). Ultimately, we can use techniques such as XPS (X-ray photoelectron spectroscopy) to prove that the lone pair of ammonia is a nitrogen 2pz, rather than a 2s.
Does that help?
@@lseinjr1 Thank you a lot for your help.
I've read that nitrogen's pz orbital have a quite weak interaction with hydrogen's s orbital, so instead of a non-bonding MO it going to form an ALMOST non-bonding MO. Is that right
That is correct. To simplify things, we often assume (in M.O. theory) that the 2pz and 2s orbitals do not overlap ("mix") at all. The fancy way to say this is that the two atomic orbitals are "orthogonal".
In Linus Pauling's theory of bonding, they DO interact - the 2s interacts with all three of the 2p's to make four equivalent sp3 hybrids. This idea is useful for organic chemists, but leads to some erroneous conclusions. It was "state of the art" in the 1930's.
In more advanced work, we consider that ALL the atomic orbitals of the correct symmetry interact. We call this idea "configuration interaction."
This was really helpful, thank you! Can you do the same with CH2Cl2 and HCCl3 molecules, please? :)
Great suggestion!
lseinjr1 Thank you very much!!
Sorry!I got three questions.
How to know that high order rotation leads to the degeneracy of E?If we take C3 two times,the result might be different.The result diagram for s1-s2 do not match Px or Py symmetry.
To show that a high order rotation axis greater than 2 leads to degeneracy, we look at the matrix representation of a Cn rotation in the xy plane. th-cam.com/video/Fz4JLe7gcWw/w-d-xo.html
In three dimensions, we have
cos (2π/n) -sin (2π/n) 0
sin (2π/n) cos (2π/n) 0
0 0 1
For C1 and C2, the "off diagonal" entries (from the sin terms) are zero (0). For any n greater than 2, the sin values will NOT be zero.
When the off diagonal terms are NOT zero, the x and y axes are effectively "mixed", and it is the "mixture" of axes is the cause of degeneracy.
It is a powerful insight of group theory that degeneracy is caused by GEOMETRY.
Does that help?
This is a fantastic video, but there is a problem with the second E group orbital. It isn't orthogonal to the first one. The node should be on s1 rather than s3. This clear when trying to draw the MO, px doesn't fit properly. As of right now, the contribution of s1 to the entire E set is too large and the contribution from s3 is too small.
Yes, that is true.
However, the molecular orbitals do not HAVE to be orthogonal to each other. In fact, the M.O combination you are suggesting is only orthogonal If we assume all overlap integrals are zero.
We can use a version of Gram-Schmidt orthogonalization to create an orthogonal set, but this procedure would double the length of the video.
Also, keep in mind that there is nothing special about orbitals s1, s2, and s3. Because we are in symmetry group C3v, we can rotate any molecular orbital by 120 degrees to yield an equivalent M.O.
Your 2nd E group orbital for the hydrogens is slighty wrong. It needs to be rotated 120°. Right now, the two E group orbitals are not orthogonal, which is impossible. The correct one is s2 - s3.