(For mobile users) 01:02 Reducible representation for pi group orbitals 05:42 Reduction of reducible representation 12:23 Effect of each symmetry operation on representative pi orbital 15:52 A2u irreducible representation 19:00 B2u irreducible representation 19:52 Eg irreducible representation 21:45 Accounting for orbital degeneracy 22:17 Visualizing the group orbitals
Hello, thank you for teaching this topic. Please also explain the calculation of the share of each orbital in the cyclo-butadiene wave functions. Thank you very much.
Thanks for kind teaching! I have one question. How to determine the lowest energy state of the molecular orbital? (ex. how to know A2u case has lowest energy state?)
The simplest way to know is to look at the number of nodes in the orbital; the more nodes, the higher the energy. (The reason for this is that, the more nodes an orbital has, the more "curved" the wavefunction is, and the curvature of the wavefunction is related to the kinetic energy of the electrons in that orbital.) For the pi orbitals of cyclobutadiene, the lowest energy orbital (A₂ᵤ) has zero (0) nodes. The E𝓰 orbitals each have one (1) node, shown with the red or orange lines. The highest energy pi orbital is the B₂ᵤ, which has two (2) nodes. *Nodes* are places where the wavefunction goes through zero, so from positive to zero to negative, or from negative to zero to positive. Does that help?
Hello, thanks for this topic. I have a question: has cyclobutadiene D2h symmetry or D4h symmetry? I'm studying on Weller's Inorganic Chemistry and the book states that cyclobutadiene has rettangular symmetry (D2h).
Great question! The answer is somewhat convoluted. We know that cyclobutadiene should not be D4h, because of a Renner distortion (often lumped in with Jahn-Teller distortions). ON the other hand, because it has 4n pi electrons, it must be anti-aromatic, so unstable. In fact, it does not even really exist! It become aromatic if two (2) electrons are added, or two (2) are subtracted, since then will pass the test for aromaticity; the existence of these cat- or an- ions has been well-established. However, neutral cyclobutadiene itself can be stabilized by chelating it to a metal, and these transition metal complexes actually exist, with interesting properties. BTW, cyclobutadiene is not the only such non-existent molecule to be widely used for pedagogical examples. Borane (BH3), a hypothetical monomer of the fascinating diborane (B2H6) is used as a simple model of aD3h ,molecule. One can determine its Lewis dot structure, which leaves boron with only six (6) electrons, so "electron deficient". Such electron deficient molecules with boron or aluminum are very often used a strong Lewis acids. Does that help?
Hi, a question if anyone may help, does it matters if i define the C2 axis in a different way?, i got the same characters for reducible representation but in differents operations.
If you use a different set of axes for a point group, the orbitals derived will look the same, but you will get different Mulliken labels for them. For example, you might get an A2 instead of a B1 label.
can you explain why you seperate 2C4 operation into C4 and C4^-1 so that p1 become p4 ? in 2C2 you didn,t treat that like 2C4. I hope you understand my english :v
Great question! (I understand you perfectly.) Whenever we have a class of operations with C3's. C4's, C5.s, and so one, there is always both a C3 and C3-1, a C4 and a C4-1, a V5 and C5-1 in the class. Each member of the pair is a different operation, C3 being a 120 degree (2pi/3 radians) rotation counter-clockwise, (anti-clockwise) and C3-1 being a rotation clockwise. However, notice that C2 and C2-1 are the SAME operation. If we go clockwise 180 degrees (pi radians), or anti-clockwise 180 degrees, we end up at the same place. It can take some experience with the point group tables to learn equally which operations are involved in each class, because the tables do not tell us that directly. Did that help?
@@lseinjr1 I really understood what you say, but initially i think that the 2C4 is like C4 and C4 again (the p1 became p2, because the operation is C4 again, p2 became p3). Previously i always treat 2C4 or 2C3 be doubled the operation. Is this wrong?
I refer you to 14:20 of the video. There are two (2) columns for C₄ (C₄ and C₄ ⁻¹) and then there is a column for C₂. All three (3) of these operations rotate around the x axis. The C₄ is a 90 degree rotation anti-clockwise: that take p1 to p2. The C₄ ⁻¹ is a 90 degree rotation clockwise: that takes p1 to p₄. The C₂ is a 180 degree rotation: that takes p1 to p3. It is definitely true that we could think of this C₂ as a C₄ followed by another C₄; however, that is not what "2C₄" means in the character table for D₄h. The "2" (two) means that there are two (2) DIFFERENT C₄'s (C₄ and C₄ ⁻¹), that are each a symmetry operation of the group. Usually, when we have "2Cₙ", it means that both "Cₙ" and "Cₙ ⁻¹" are in the same class, so long as n is greater than 2.
This will be true for all groups. Recall, that in a group we MUST have an inverse for EVERY element. If we have C₃, we also must have C₃ ⁻¹. If we have C₄, we must have C₄ ⁻¹, too. If we have C₂, we must have C₂ ⁻¹ - but C₂ ⁻¹ is exactly the same as C₂. (It is possible for an element to be its own inverse.) The character tables contain much information, but it is often not stated directly. Does this make sense?
Cyclobutadiene does not belong to D4h point group. Due to zero delocalization energy it is rectangular in shape and hence cyclobutadiene belongs to D2h point group......!!! Make sure please.
Actually, cyclobutadiene does not exact at all! Note that it has 4n pi electrons, which makes it "anti-aromatic", so unstable. You are correct that IF it existed as written, it would relax from D4h to D2h. The reason for this relaxation is a version of the Jahn-Teller theorem. However, if we form the cyclobutadiene dianion, that ion WOULD be D4h. We can stabilize cyclobutadiene as D4h by binding it to a metal, where the metal is formally a cation and the cyclobutadiene is a dianion. Such compounds are very well known.
omg thank you so much!!! you literally saved my by the bell.. hw's up and all other resources are off cause it's 2 am.. you the GOAT!
Glad I could help!
(For mobile users)
01:02 Reducible representation for pi group orbitals
05:42 Reduction of reducible representation
12:23 Effect of each symmetry operation on representative pi orbital
15:52 A2u irreducible representation
19:00 B2u irreducible representation
19:52 Eg irreducible representation
21:45 Accounting for orbital degeneracy
22:17 Visualizing the group orbitals
You're a legend
Hello, thank you for teaching this topic. Please also explain the calculation of the share of each orbital in the cyclo-butadiene wave functions. Thank you very much.
Thank you❤
Thanks for kind teaching! I have one question. How to determine the lowest energy state of the molecular orbital? (ex. how to know A2u case has lowest energy state?)
The simplest way to know is to look at the number of nodes in the orbital; the more nodes, the higher the energy. (The reason for this is that, the more nodes an orbital has, the more "curved" the wavefunction is, and the curvature of the wavefunction is related to the kinetic energy of the electrons in that orbital.)
For the pi orbitals of cyclobutadiene, the lowest energy orbital (A₂ᵤ) has zero (0) nodes. The E𝓰 orbitals each have one (1) node, shown with the red or orange lines. The highest energy pi orbital is the B₂ᵤ, which has two (2) nodes.
*Nodes* are places where the wavefunction goes through zero, so from positive to zero to negative, or from negative to zero to positive.
Does that help?
how we can determine the projection operator: pi molecular orbital of dibromomethane?
Could you please help me?
Dibromomethane has no pi molecular bonding.
Hello, thanks for this topic. I have a question: has cyclobutadiene D2h symmetry or D4h symmetry? I'm studying on Weller's Inorganic Chemistry and the book states that cyclobutadiene has rettangular symmetry (D2h).
Great question!
The answer is somewhat convoluted. We know that cyclobutadiene should not be D4h, because of a Renner distortion (often lumped in with Jahn-Teller distortions).
ON the other hand, because it has 4n pi electrons, it must be anti-aromatic, so unstable. In fact, it does not even really exist! It become aromatic if two (2) electrons are added, or two (2) are subtracted, since then will pass the test for aromaticity; the existence of these cat- or an- ions has been well-established.
However, neutral cyclobutadiene itself can be stabilized by chelating it to a metal, and these transition metal complexes actually exist, with interesting properties.
BTW, cyclobutadiene is not the only such non-existent molecule to be widely used for pedagogical examples. Borane (BH3), a hypothetical monomer of the fascinating diborane (B2H6) is used as a simple model of aD3h ,molecule. One can determine its Lewis dot structure, which leaves boron with only six (6) electrons, so "electron deficient". Such electron deficient molecules with boron or aluminum are very often used a strong Lewis acids.
Does that help?
@@lseinjr1 a lot, thanks!
Hi, a question if anyone may help, does it matters if i define the C2 axis in a different way?, i got the same characters for reducible representation but in differents operations.
If you use a different set of axes for a point group, the orbitals derived will look the same, but you will get different Mulliken labels for them. For example, you might get an A2 instead of a B1 label.
@@lseinjr1 I see, thank you!
Awesome
can you explain why you seperate 2C4 operation into C4 and C4^-1 so that p1 become p4 ? in 2C2 you didn,t treat that like 2C4. I hope you understand my english :v
Great question! (I understand you perfectly.)
Whenever we have a class of operations with C3's. C4's, C5.s, and so one, there is always both a C3 and C3-1, a C4 and a C4-1, a V5 and C5-1 in the class. Each member of the pair is a different operation, C3 being a 120 degree (2pi/3 radians) rotation counter-clockwise, (anti-clockwise) and C3-1 being a rotation clockwise.
However, notice that C2 and C2-1 are the SAME operation. If we go clockwise 180 degrees (pi radians), or anti-clockwise 180 degrees, we end up at the same place.
It can take some experience with the point group tables to learn equally which operations are involved in each class, because the tables do not tell us that directly.
Did that help?
@@lseinjr1 I really understood what you say, but initially i think that the 2C4 is like C4 and C4 again (the p1 became p2, because the operation is C4 again, p2 became p3). Previously i always treat 2C4 or 2C3 be doubled the operation.
Is this wrong?
I refer you to 14:20 of the video. There are two (2) columns for C₄ (C₄ and C₄ ⁻¹) and then there is a column for C₂.
All three (3) of these operations rotate around the x axis. The C₄ is a 90 degree rotation anti-clockwise: that take p1 to p2. The C₄ ⁻¹ is a 90 degree rotation clockwise: that takes p1 to p₄.
The C₂ is a 180 degree rotation: that takes p1 to p3. It is definitely true that we could think of this C₂ as a C₄ followed by another C₄; however, that is not what "2C₄" means in the character table for D₄h. The "2" (two) means that there are two (2) DIFFERENT C₄'s (C₄ and C₄ ⁻¹), that are each a symmetry operation of the group. Usually, when we have "2Cₙ", it means that both "Cₙ" and "Cₙ ⁻¹" are in the same class, so long as n is greater than 2.
@@lseinjr1 Okay, but how we know that the 2C3, 2C4, etc is C3^-1 or C4^-1 in another character table? or it was for character table of "D..." only?
This will be true for all groups.
Recall, that in a group we MUST have an inverse for EVERY element. If we have C₃, we also must have C₃ ⁻¹. If we have C₄, we must have C₄ ⁻¹, too.
If we have C₂, we must have C₂ ⁻¹ - but C₂ ⁻¹ is exactly the same as C₂. (It is possible for an element to be its own inverse.)
The character tables contain much information, but it is often not stated directly.
Does this make sense?
Should not principle axis be C2 instead of C4 as we have di en..
We imagine that the molecule is stabilized by resonance, so square.
@@lseinjr1 thanks for clarify. 😄
Cyclobutadiene does not belong to D4h point group. Due to zero delocalization energy it is rectangular in shape and hence cyclobutadiene belongs to D2h point group......!!! Make sure please.
Actually, cyclobutadiene does not exact at all! Note that it has 4n pi electrons, which makes it "anti-aromatic", so unstable. You are correct that IF it existed as written, it would relax from D4h to D2h. The reason for this relaxation is a version of the Jahn-Teller theorem.
However, if we form the cyclobutadiene dianion, that ion WOULD be D4h. We can stabilize cyclobutadiene as D4h by binding it to a metal, where the metal is formally a cation and the cyclobutadiene is a dianion. Such compounds are very well known.
@@lseinjr1 Ah so beyond the Born-Oppenheimer.
Yes!