I have a question, why we only checked 6 irreducible representations of D6h group in order to decide which of the irreducible representations of D6h group making linear combination, as you said d6h group consists of 12 irreducible representations sorry for my English, and tnx a lot
We do a procedure where we reduce the reducible representation to a linear combination of irreducible representations. Once we have found six (6), we know that we have found all of them (starting with six atomic orbitals, we get six molecular orbitals) that are not equal to 0. Does that make sense?
There are six (6) orbitals in benzene (labelled in the sketch at the top of the whiteboard. To generate the "reducible representation", we see how many of those pi orbitals DON'T MOVE when we perform the symmetry operation. For example, the "E" operation is the identity - every orbitals stays where it was. Therefore, all of the orbitals (i.e., "6") don't move, so we write down a "6". For the C6 operation, we rotate the ring counter-clockwise by 1/6 of a turn. p1 goes to p2, p2 goes to p3, and so on. Since ALL the p's move, NONE of them don't move, so we write down a "0". Does that help?
Brilliant! All the websites I checked were useless. You're a proper legend mate!
Sir the second part of this vedio is private...
11:53 why is it assumed that B2g is included in th pi conj system?
Thank you sooooooooooooo much. This was very easy to understand
(For mobile users)
00:09 Reducible representation for pi group orbitals
08:01 Reduction of reducible representation
thanks a lot for your explanation:)
I have a question, why we only checked 6 irreducible representations of D6h group in order to decide which of the irreducible representations of D6h group making linear combination, as you said d6h group consists of 12 irreducible representations sorry for my English, and tnx a lot
We do a procedure where we reduce the reducible representation to a linear combination of irreducible representations. Once we have found six (6), we know that we have found all of them (starting with six atomic orbitals, we get six molecular orbitals) that are not equal to 0.
Does that make sense?
@@lseinjr1 yes 👍
You speeded the derivation as you said in the video
Sir, could you please explain better why there is a six under E, a zero under 2C6 and so on? Thanks a lot.
There are six (6) orbitals in benzene (labelled in the sketch at the top of the whiteboard. To generate the "reducible representation", we see how many of those pi orbitals DON'T MOVE when we perform the symmetry operation.
For example, the "E" operation is the identity - every orbitals stays where it was. Therefore, all of the orbitals (i.e., "6") don't move, so we write down a "6".
For the C6 operation, we rotate the ring counter-clockwise by 1/6 of a turn. p1 goes to p2, p2 goes to p3, and so on. Since ALL the p's move, NONE of them don't move, so we write down a "0".
Does that help?
@@lseinjr1 Yes!!!! Thanks a lot!! Great work and cheers from Brazil
Thank you very much! I was struggling because we did not deal with the pi cases very much in class. It was super clear :)))
Glad it helped!
thank u
You're welcome!
Sir please can you explain cyclopropenium molecule..
Yes!
Here are the pi orbitals derived:
ruclips.net/video/P_U9uMuPEa4/видео.html
Here are the Hückel energies:
ruclips.net/video/8f0dhomaVFo/видео.html