@@Chemistryuniversity do you get the same results if you switch x and y? im not sure about my own answers but when i switch i get B2 instead of b1 symmetry
So the SALCs are based off the point group and the combinations that make up orbitals. I'm not sure if this answers your question. The relative energies are based off the atomic orbital energies and then symmetry and orbital mixing. For example the A1 orbitals mix so the A1 from the s orbital from oxygen pushes the bonding MO resulting from oxygen's px and the SALC from both hydrogens S orbitals. Then you have to consider how many nodes. The higher the number of nodes, generally the higher the energy. This is helpful for all orbitals, but shines in the anti bonding ones.
Quantum-mechanical analysis of the MO method and VB method from the position of PQS. vixra.org/pdf/1704.0068v1.pdf The MO method and the VB method are analyzed using the principle of quantum superposition (PQS) and the method of describing a quantum system consisting of several parts. It is shown that the main assumption of the molecular orbitals method (namely, that the molecular orbital can be represented like a linear combination of overlapping atomic orbitals) enters into an insurmountable contradiction with the principle of quantum superposition. It is also shown that the description of a quantum system consisting of several parts (adopted in quantum mechanics) actually prohibits ascribe in VB method to members of equation corresponding canonical structures. Using the quantum superposition principle, the MO method and the VB method were analyzed and it is shown that they are in contradiction with quantum mechanics. Also, using the quantum-mechanical description of a system consisting of several parts, it is shown that the attribution of canonical structures to the members of the equation is incorrect. Therefore, both the MO method and the VB method did not describe molecules with chemical bonds but actually, a lot of atoms (of which the described molecules consisted). That is, in the quantum chemical calculations, the chemical bond was "lost". Therefore, in order to "introduce" a chemical bond into calculations and avoid conflict with quantum mechanics, it is suggested to postulate the existence of MO as a new fundamental quality that describes a specific chemical bond and is not derived from simpler structural elements. Quantum-mechanical analysis of the MO method and VB method from the position of PQS: vixra.org/pdf/1704.0068v1.pdf 1. Structure of the benzene molecule on the basis of the three-electron bond. vixra.org/pdf/1606.0152v1.pdf 2. Experimental confirmation of the existence of the three-electron bond and theoretical basis ot its existence. vixra.org/pdf/1606.0151v2.pdf 3. A short analysis of chemical bonds. vixra.org/pdf/1606.0149v2.pdf 4. Supplement to the theoretical justification of existence of the three-electron bond. vixra.org/pdf/1606.0150v2.pdf 5. Theory of three-electrone bond in the four works with brief comments. vixra.org/pdf/1607.0022v2.pdf 6. REVIEW. Benzene on the basis of the three-electron bond (full version, 93 p.). vixra.org/pdf/1612.0018v5.pdf 7. Quantum-mechanical aspects of the L. Pauling's resonance theory. vixra.org/pdf/1702.0333v2.pdf 8. Quantum-mechanical analysis of the MO method and VB method from the position of PQS. vixra.org/pdf/1704.0068v1.pdf Bezverkhniy Volodymyr (viXra): vixra.org/author/bezverkhniy_volodymyr_dmytrovych Review. Benzene on the basis of the three-electron bond (full version). vixra.org/pdf/1612.0018v5.pdf Bezverkhniy Volodymyr (Scribd): www.scribd.com/user/289277020/Bezverkhniy-Volodymyr# Bezverkhniy Volodymyr (Amazon): www.amazon.com/Volodymyr-Bezverkhniy/e/B01I41EHHS/ref=dp_byline_cont_ebooks_1 This screenshots (foto) (most with explanation) see by this link. Bezverkhniy Volodymyr (Archive.org): archive.org/details/@threeelectronbond
If anyone could explain, why is the A1 MO from the oxygen's 2s orbital considered non-bonding? I know it's because it's different in energy levels from the A1 of the 2H orbitals, but wouldn't there still be at least some very little overlap?
There should be some overlap -13.6 ev for H 1s and -32.4 ev for O 2s. There will be more electron density on the O, but probably some contribution to bonding. I was attempting to relate it to what we see in a lewis structure.
Every diagram I have seen has A1 above the B1. Yes there is more symmetry for A1, but there could be orbital mixing from the S orbital A1. This can push up the orbital (they have the same symmetry so they can have some interaction if they are close enough in energy).
@@roohanib99 The experimentally determined ionization energies are A1 is 14.7 and B1 18.5 eV. Since B1 requires more energy to ionize, it is lower than the A1. Here is a reference, but they exchanged the x and y axis so what is B1 in my video is B2 in the reference. chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/10%3A_Bonding_in_Polyatomic_Molecules/10.4%3A_Photoelectron_Spectroscopy
if an atom moves = 0, but if an atom stays the same = 1. that's why for the c2 operation for hydrogens when both hydrogens switch place after the flip, they are 0.
same thing for sigma yz-plane. imagine a plane on the yz-plane, coming out of the screen towards you. the hydrogens are mirrored across so they switch places and = 0.
Thank you for explaining every single step you did!
Thank you so much! You are way better than my professor.
Thank you SOOOOOO much, I was so lost by this concept, and you have been the only one to explain in so well!
Thank you for the explanation! probably the only one that is worth watching on RUclips.
P.S: You sound like Kevin Spacey sometimes
Thanks.
Finally a good explanation!
can you please explain why antibonding B1 is higher than antibonding A1?
According to the convention, y is axis points to outer atoms, but why did you use x as axis that points to outer atoms?
I was used to it from math classes.
@@Chemistryuniversity do you get the same results if you switch x and y? im not sure about my own answers but when i switch i get B2 instead of b1 symmetry
Thank you a lot! But I wonder how do we consider the situation where the O in NO2 has p orbitals?
Please explain how to write the SALC's and explain how you found or reasoned the relative engergies of the MO.
So the SALCs are based off the point group and the combinations that make up orbitals. I'm not sure if this answers your question. The relative energies are based off the atomic orbital energies and then symmetry and orbital mixing. For example the A1 orbitals mix so the A1 from the s orbital from oxygen pushes the bonding MO resulting from oxygen's px and the SALC from both hydrogens S orbitals. Then you have to consider how many nodes. The higher the number of nodes, generally the higher the energy. This is helpful for all orbitals, but shines in the anti bonding ones.
Quantum-mechanical analysis of the MO method and VB method from the position of PQS.
vixra.org/pdf/1704.0068v1.pdf
The MO method and the VB method are analyzed using the principle of quantum superposition (PQS) and the method of describing a quantum system consisting of several parts. It is shown that the main assumption of the molecular orbitals method (namely, that the molecular orbital can be represented like a linear combination of overlapping atomic orbitals) enters into an insurmountable contradiction with the principle of quantum superposition. It is also shown that the description of a quantum system consisting of several parts (adopted in quantum mechanics) actually prohibits ascribe in VB method to members of equation corresponding canonical structures.
Using the quantum superposition principle, the MO method and the VB method were analyzed and it is shown that they are in contradiction with quantum mechanics. Also, using the quantum-mechanical description of a system consisting of several parts, it is shown that the attribution of canonical structures to the members of the equation is incorrect. Therefore, both the MO method and the VB method did not describe molecules with chemical bonds but actually, a lot of atoms (of which the described molecules consisted). That is, in the quantum chemical calculations, the chemical bond was "lost". Therefore, in order to "introduce" a chemical bond into calculations and avoid conflict with quantum mechanics, it is suggested to postulate the existence of MO as a new fundamental quality that describes a specific chemical bond and is not derived from simpler structural elements.
Quantum-mechanical analysis of the MO method and VB method from the position of PQS:
vixra.org/pdf/1704.0068v1.pdf
1. Structure of the benzene molecule on the basis of the three-electron bond.
vixra.org/pdf/1606.0152v1.pdf
2. Experimental confirmation of the existence of the three-electron bond and theoretical basis ot its existence.
vixra.org/pdf/1606.0151v2.pdf
3. A short analysis of chemical bonds.
vixra.org/pdf/1606.0149v2.pdf
4. Supplement to the theoretical justification of existence of the three-electron bond.
vixra.org/pdf/1606.0150v2.pdf
5. Theory of three-electrone bond in the four works with brief comments.
vixra.org/pdf/1607.0022v2.pdf
6. REVIEW. Benzene on the basis of the three-electron bond (full version, 93 p.).
vixra.org/pdf/1612.0018v5.pdf
7. Quantum-mechanical aspects of the L. Pauling's resonance theory.
vixra.org/pdf/1702.0333v2.pdf
8. Quantum-mechanical analysis of the MO method and VB method from the position of PQS.
vixra.org/pdf/1704.0068v1.pdf
Bezverkhniy Volodymyr (viXra): vixra.org/author/bezverkhniy_volodymyr_dmytrovych
Review. Benzene on the basis of the three-electron bond (full version).
vixra.org/pdf/1612.0018v5.pdf
Bezverkhniy Volodymyr (Scribd):
www.scribd.com/user/289277020/Bezverkhniy-Volodymyr#
Bezverkhniy Volodymyr (Amazon): www.amazon.com/Volodymyr-Bezverkhniy/e/B01I41EHHS/ref=dp_byline_cont_ebooks_1
This screenshots (foto) (most with explanation) see by this link.
Bezverkhniy Volodymyr (Archive.org):
archive.org/details/@threeelectronbond
If anyone could explain, why is the A1 MO from the oxygen's 2s orbital considered non-bonding? I know it's because it's different in energy levels from the A1 of the 2H orbitals, but wouldn't there still be at least some very little overlap?
There should be some overlap -13.6 ev for H 1s and -32.4 ev for O 2s. There will be more electron density on the O, but probably some contribution to bonding. I was attempting to relate it to what we see in a lewis structure.
You are my new religion.
Among the bonding orbitals, shouldn't the A1 go below B1? coz A1 is more symmetric, hence lesser energy than B1
Every diagram I have seen has A1 above the B1. Yes there is more symmetry for A1, but there could be orbital mixing from the S orbital A1. This can push up the orbital (they have the same symmetry so they can have some interaction if they are close enough in energy).
@@Chemistryuniversity In college, we did A1 below B1. Also in UCI lecture videos it's the same.
@@roohanib99 The experimentally determined ionization energies are A1 is 14.7 and B1 18.5 eV. Since B1 requires more energy to ionize, it is lower than the A1. Here is a reference, but they exchanged the x and y axis so what is B1 in my video is B2 in the reference.
chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/10%3A_Bonding_in_Polyatomic_Molecules/10.4%3A_Photoelectron_Spectroscopy
But sigma v contain 0 in other Book
Why you write 2 ... Becoz of we passing plane in the o Molecule and both H atom reflects each other so the tau oh Streching of unshifted atom is 0 naa
Thanks for the great explanation
I don’t understand why C2 & sigma is 0
if an atom moves = 0, but if an atom stays the same = 1. that's why for the c2 operation for hydrogens when both hydrogens switch place after the flip, they are 0.
same thing for sigma yz-plane. imagine a plane on the yz-plane, coming out of the screen towards you. the hydrogens are mirrored across so they switch places and = 0.
why not A2 and B2? They also sum up to 2 0 2 0.
A2 and B2 sum to 2 0 -2 0.
alright I got it, oof. sorry I miscalculated!! :(
Do you even enjoy explaining this topic to other people?
Well, I intend to make other videos and probably should redo the ones I have. I would like to make them more concise.
I just turn the playback speed to 1.25, so it doesn't seem like he's tired of what he's talking about
thanks!
do you realize that you just saved my ass?