Projection operator method: sigma orbitals of boron trifluoride

[@kishanky1352]

You are a great teacher. You deserve to be in some big university as a Professor. Thank you so much

[@TheFEDFEDFEDFED]

excellent job. I'm taking Inorganic chemistry at UC Davis right now, and your clear and concise explanation has made this method so much clearer in my head, especially the relation from the irreducible representations to their corresponding molecular orbitals. Thank you.

[@lseinjr1]

(For mobile users) 02:00 Reducible representation for sigma group orbitals 07:12 Reduction of reducible representation 20:08 Effect of each symmetry operation on representative sigma orbital 24:02 A1' irreducible representation 26:40 E' irreducible representation 29:54 Accounting for orbital degeneracy 33:12 Visualizing the group orbitals

[@AlexLi123ABC]

I just wanna say I'm taking inorganic right now and you're saving me from going insane thank you so much!

[@carrieyu7985]

This was amazing and educational. Thanks!

[@rafiqulislam-ej1qt]

Asalamuliakum sir, your video are very helpful for me,i wish to get more more videos,may Allah bless you💕💕💕

[@lanle2060]

Thank you. It helps me to understand about Molecular Orbitals.

[@perrinebahnana763]

Merci beaucoup pour ces explications cela m'a beaucoup aidé ! Thank you

[@JH-if5rv]

Outstanding save

[@kendialeminde966]

thank you so much, i understand this subject thanks to you sir!!

[@lseinjr1]

Glad to hear that!

[@muthiahpillaipalanichamy3939]

good. Thank you

[@jafarjhandeer9677]

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[@shivanibisht1803]

Wonderful 🌟

[@lseinjr1]

Thank you! Cheers!

[@dulce9613]

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[@lseinjr1]

Happy to help!

[@malikah14otrew]

Bless you ♥️

[@kuvenmalig-on8685]

I learned a lot from this. Thank you! If the molecule would be ML5, trigonal bipyramidal, can I use this the same approach?

[@lseinjr1]

Great question! Answer: Yes and No. Yes, in that the same basic method is used. The tricky part i s that the equatorial and axial positions of trigonal bipyramid are NOT symmetry equivalent. That means that we need a table with two lines in it, one for the equatorial positions s1, s2, and s3, and then a SECOND line for s4 and s5. Then when we form the orbitals, we get two (2) sets, one for s1, s2, and s3, and a second set for s4 and s5. We can combine the sets for (s1, s2, and s3) with (s4, s5), when they have the SAME symmetry label (say A1, for example). Then we combine them two (2) separate ways, one with positive phase (add them all together) and one with negative phase *subtract the (s4, s5) combination from the (s1, s2, and s3) combination. Does this help?

[@kuvenmalig-on8685]

@@lseinjr1 Thank you very much. I figured it out! It made a lot of sense doing it step by step and treating the axial and equatorial positions separately.

[@she_lby02]

How would you suggest doing this method for the p-orbitals for sulfur trioxide (SO3)?

[@lseinjr1]

It would work much like this: kzbin.info/www/bejne/hpC4aqiDqrV4l5Y

[@kuanrsoren9681]

Sir Have you any MO diagram videos of BF3 , Benzene etc.. Please send the link.

[@josefjungmann4109]

thanks

[@brahimourhzal8393]

Thank you for the video sir 💙. I'm trying to do the same thing on methane but i found a problem in Px, Py and Pz of the carbon; if i try to apply C2 on Px, i do not know wether C2 should be following x or y or z. And the same for the other P orbitals. What shall i do?

[@lseinjr1]

It depends on which C₂ you are referring to. It is often easiest to imagine a tetrahedron are being placed inside a cube. You can see this in detail here: kzbin.info/www/bejne/sJeWeamXoauklZY The part that deals with the C₂'s begins at 24:25.

[@brahimourhzal8393]

I may ask a stupid question, Why did you choose the S orbital of Fluorine instead of the Pz?

[@lseinjr1]

@@brahimourhzal8393 Great question. For sigma bonding, either the 2s or 2p orbital of fluorine has the correct symmetry. We know from other reasoning (the fact that molecular orbitals form most strongly between atomic orbitals with similar energies. Because of the energy requirement, the sigma bonds will involve 2s and 2p orbitals with the 2p orbital on fluorine that points toward the boron atom. Since the 2s orbital CAN make a molecular orbital of the proper (sigma) symmetry , and drawing p orbitals on fluorine would make the diagram more cluttered, it is common to sketch s1, s2, and s3 as if they were "s" orbitals.

[@brahimourhzal8393]

@@lseinjr1 But sir, when i tried to study the BF3 using the Pz, I had A''2 + E" as irreducible presentation. I applied the projection operator but all i got is 0 for both of A''2 and E" .

[@lseinjr1]

@@brahimourhzal8393 The plane of the BF3 molecule is defined to be the xy-plane (note the basis functions for the E' irreducible representation). A pz orbital on fluorine is pointing *perpendicular* (*normal*) to the plane. It is incapable of making a sigma bond with any orbitals on C. (It is capable of making a Pi bond with the pz of carbon, but that is another story). This is another reason why it is useful to think of the orbitals on fluorine as being s orbitals, rather than p ones. It makes it easier to figure out the symmetry adapted linear combinations on the outer atoms. Only px or py orbitals on fluorine have the correct symmetry to make a sigma bond with orbitals on carbon. It makes sense that you got O's - since there are no sigma orbitals of those symmetries. If you take a look at one of the several videos I have made on deriving pi orbitals, you will see what you have done.

[@nurarifahmohdkamal3774]

When do we need to use the method where we calculate the contribution per atom

[@lseinjr1]

I do not know what you mean by "contribution per atom." We know how many electrons each atom has from the Periodic table, and we know which atomic orbitals are the valence ones: 2s and 2p for boron, 2p for fluorine.

[@morninghorizon5752]

NIce video sir. But why didn't we need to use the trace property when obtaining the irreducible representation?

[@lseinjr1]

Great question! We do use the trace property here. I assume you mean the property that the trace of a matrix is invariant (does not change) during a similarity transformation. In the point group D3h, the C2 operations all belong to the same class. We can transform one C2 into another by means of similarity operation. Matrices in the same class are similar matrices, so they have the exact same character for each irreducible representation. For example, in the E' representation, all three (3) c2's have the same character of 0. There is a strong relationship between the trace of a matrix and the character of an irreducible representation. Does that help?

[@heyjoshualogronio5879]

Hi sir! Just for clarification, may I ask why Boron is not included in the count for E, identity operation?

[@lseinjr1]

Great question! It is possible to solve the problem two (2) different ways - (1) to include the "central atom" (which here is boron), and (2) to NOT include the central atom. It is usually the easiest to do this the second way. Once we have found the linear combinations of the "outer" atoms (here, fluorine) for each irreducible representation, then figure out which atomic orbitals on the "central" atom combine with them. This can be done by inspection, since boron has a 2s and three 2p orbitals. Suppose that we DID include boron in the identity operation. We would also have to include boron in every OTHER symmetry operation as well. Notice that, in the reducible representation gs that several of the operation classes have a character of "0". If we include boron, there would be NO 0 characters. The effect would be that we would still get the same (correct) answer as before, but the computations in reducing the reducible representation would be slightly longer. Therefore, once we reduced the reducible representation, we would have boron being a member of each molecular orbital, just as we got the other way. We would know that boron was involved, but we would still have to figure out WHICH atomic orbitals on boron we involved in each molecular orbital. (we already know that, for fluorine, the atomic orbitals involved will be 2p's). In summary, we COULD do that, but it would make the work required a little longer, and still require us to find, by inspection, which atomic orbitals on boron we involved in each case. Does that make sense?

[@heyjoshualogronio5879]

@@lseinjr1 yes sir, thank you so much for the response! I learned a lot from your video

[@carllundell2930]

Great video! However I have a question that is also related to this video and your MO diagram of ammonia. How did you know that there were two E' irreducible orbitals and not 2 A1'? As in the case of ammonia you mention that E is double degenerate - but is to due to their position on the molecule? I understand that there needs to be three, but I can't figure out how you intuitively knew it was 2 E'. Thanks.

[@lseinjr1]

Great questions! If we have any orbitals of E symmetry, there have to be exactly two (2) such orbitals. Any orbitals with labeled of "A" or "B" are non-degenerate (so there is only one such orbital); any orbitals with an "E" label are doubly degenerate by definition (so there are two such orbitals); and if any orbitals are labeled "T" they are triply degenerate by definition (so there are three such orbitals). We can always predict when we must have degenerate orbitals. If the highest rotational axis has order three (3) or greater, there MUST be degenerate orbitals. Therefore, degeneracy is a direct consequence of the molecular geometry. To be specific, if the molecule has a C3 , C4, C5, or C6 axis, it must have degenerate orbitals / vibrations. If it only has C2's, it will have only NON-degenerate orbitals / vibrations. Does that help?

[@carllundell2930]

@@lseinjr1 Extremely helpful! Thanks!

[@Shaggyguy119]

Can I use the same method for NO3- ?

[@lseinjr1]

Yes, since nitrate (NO₃ -) is trigonal planar (D₃ₕ) just like BF₃.

[@raphaelberger1834]

I‘ve got a question: When focussing on on Fluorid, we actually focus on p-orbitals which is why the s-orbitals are filled. So why do we focus on s-Orbitals?

[@lseinjr1]

Excellent question! The relevant orbitals on fluorine happen to be the 2p's. But we cannot know this from symmetry alone. From symmetry, we know that we can get sigma orbitals from 1s or 2s orbitals on fluorine, or the 2px or 2py orbitals. We know that the 2pz on fluorine do not have the correct symmetry to form sigma molecular orbitals. To know exactly which atomic orbitals on fluorine are involved, we would need to know their energies, since formation of molecular orbitals is most effective when the contributing orbitals have similar energies. For convenience, I drew the group orbitals on fluorine as if they were s orbitals, but it would have been more accurate (for BF₃, anyway) to draw them as p orbitals. Drawn as in the video, it gives the correct combination for the hypothetical molecule borane (BH₃), which actually exists as a strange molecule diborane (B₂H₆) with bonding that only makes sense using M.O. theory.

[@danielaalejandraonate7094]

amazing Thanks

[@anjalivijayan8984]

Sir how many c3 axis bf3 have kindly reply

[@lseinjr1]

Two (2). See: symmetry.jacobs-university.de/cgi-bin/group.cgi?group=603&option=4

[@anjalivijayan8984]

@@lseinjr1 thank you sir

[@rimjhimshukla365]

Would the answer still be correct if I rotate it clockwise for C3 to find the second degenerate set for E (at 31:11)

[@lseinjr1]

Yes.

[@rimjhimshukla365]

@@lseinjr1 okay, but if I do it clockwise twice, I do not get the same result as that when I do anticlockwise once. Thank you for your quick reply

[@lseinjr1]

The orbitals that we derive are not "unique". If we add or subtract any two orbitals together, we get another valid solution. (This is a result of the Schroedinger equation being a linear differential equation. We often pick a specific set by using Gramm-Schmidt orthogonalization, by I have omitted that step here.)

[@rimjhimshukla365]

@@lseinjr1 Thank you so much for a great explanation.

[@sadman814]

why was the last one a 1 instead of a 3? I'm confused

[@lseinjr1]

Do you mean for 3s₁ - 3s₂ = s₁ - s₂ ? We are using the fact that if ψ is a solution of the Schoedinger equation, then cψ is a solution as well. (in effect, we reduced the coefficients to lowest terms). In physical chemistry courses, you will learn about the normalization constant, which would be 1/√2, so that the normalized solution would be 1/√2 (s₁ - s₂)

[@fluoridemindcontrol]

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.

[@lseinjr1]

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: kzbin.info/www/bejne/rIHQlqKrgtiVr9E, kzbin.info/www/bejne/gZuyhph9d76rmrs, kzbin.info/www/bejne/h364Z4aFgcepmNU, and kzbin.info/www/bejne/h53PlKFjZ8Z9kJo . 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.

[@신재용-h4e]

May i suggest the idea that the sum of the squares for normalization requirements are all 1? or the sum of squared SALC coefficients are all 1?

[@ivetterodriguez8982]

How do we calculate the h?

[@lseinjr1]

Do you mean the order (h) of the group?

[@PreciousMental]

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[@khadimndiaye3681]

Do you speak french please

[@lseinjr1]

Sorry, I do not.