(For mobile users) 00:40 Reducible representation for 3N degrees of freedom 06:10 Reduction of reducible representation 14:37 Subtracting out rotations and translations 17:54 Effect of each symmetry operation on representative bond stretch 19:54 A1 stretch 22:04 E stretches 23:08 Accounting for degeneracy of E stretches 27:07 Effect of each symmetry operation on representative bond bend 29:24 A1 bend 30:50 E bends 31:52 Accounting for degeneracy of E bends 34:24 Visibility in IR and / or Raman
Great video. Later, can you solve this same problem, but using Kim's method? I'm trying it on my own, but I'm not able to choose with basis function to use, as the c3v character table presents all linear, quadratic and cubic functions for the irreducible representations. Take A1 for instance irrep linear quadratic cubic A1 z x^2 -+y^2, z^2 z^3, x(x^2 - 3y^2), z(x^2 + y^2) Depending on which function I chose, I get differenct orbital coefficient values
There's a very useful trick for A1 in any point group. Since it is totally symmetric, we can always use the basis function "1". Multiply each atomic orbital by "1", then add. So for molecular group orbitals the A1 is immediately (s1 + s2 +s3), unnornalized, where s1, s2, s3 are the H atomic 1s orbitals on hydrogens 1, 2, and 3, respectively. It is completely legitimate. Does this make sense?
@@lseinjr1 Thank you very much for your response, it made complete sense. Eventually I managed to solve this problem, but I had to redraw my coordinates system. I tried to make more or less a "3D" system, just like the methane video. Some time later, I redrawed it to a simple 2D case, with the NH3 lone pair ponting outside the notebook's plane. I'll try to solve it on a 3D coordinates system later on
My references are: 1. "Concepts and Models of Inorganic Chemistry" , 3rd edition, by Douglas, McDaniel, and Alexander 2. "inorganic Chemistry" by Miessler and Tarr 3. "Chemical Applications of Group Theory" F. Albert Cotton 4. "Group Theoretical Methods" Shoon K. Kim. I took a graduate course with Dr. Kim (ref 4) at Temple University. His book describes the theory of the projection operator method, but also demonstrates what I call the "Kim Method", which is much easier to apply to high symmetry molecules. I have videos demonstrating this method here: ua-cam.com/video/vPmeouMsbyo/v-deo.html, ua-cam.com/video/KjOUeGBYxdY/v-deo.html, ua-cam.com/video/QMU6SOLbvbs/v-deo.html, ua-cam.com/video/VdwlTLRV36s/v-deo.html, and ua-cam.com/video/Rbh-YVWk5z4/v-deo.html . For an introduction to group theory, see my videos here: ua-cam.com/video/GESob5sKeek/v-deo.html ua-cam.com/video/COjpFPk-m5Q/v-deo.html ua-cam.com/video/bhYEDa-RQfU/v-deo.html ua-cam.com/video/_9rNZ__aWSw/v-deo.html ua-cam.com/video/Y57erKDQObc/v-deo.html ua-cam.com/video/EDtRCLYTZC0/v-deo.html ua-cam.com/video/Fz4JLe7gcWw/v-deo.html ua-cam.com/video/zlUztdkBSh0/v-deo.html ua-cam.com/video/DgUmQdcs6Tk/v-deo.html ua-cam.com/video/Yme3FnWz7z4/v-deo.html
They are not (necessarily) "supposed" to be orthogonal, It is often very convenient for them to be so. If you know how to do Graham-Schmidt orthogonalization and normalization, that will give you the orthonormal set. Showing those in video would have made it too long. Does that make sense?
@@lseinjr1 Hello Dr. Sein! An identical procedure to that used to determine the form of the vibrational modes is used to deduce the LCAOs, and we all know that the resulting functions should be orthogonal. As the procedure used to obtain the second function for E is generally arbitrary, in my course I follow a slightly different approach. Specifically, I multiply by 2 the function resulting from C3 rotation of the three vectors and then add it to the initial one. This allows eliminating the b1 vector in the expression of the second function for E, which will be orthogonal with the other two. Brian Pfennig, in his book, attaches a Cartesian system to the central atom and looks at the projection of the three vectors on this system to define new vectors for the third function and obtain it orthogonal to the other two. Anyway, your videos are very nice and informative. Best regards!
im so anxious i see simple molecules like ammonia and water but when it comes to coordination complexes i cant see anything my 3d vision fades. i cant see the point group of iron pentacarbonyl
When there are five (5) ligands - and the ligands are all identical - there are usually only two (2) possibilities. One is trigonal bipyramid (TBP, which would have D₃ₕ symmetry) and the other is square pyramidal (SP, C₄ᵥ). I have videos with models of these structures, which might be helpful: ua-cam.com/video/1oao6HGYc0E/v-deo.html ua-cam.com/video/ogQ5vJqN8yM/v-deo.html ua-cam.com/video/7KwMEFrzHG8/v-deo.html ua-cam.com/video/Cuc7lifQD7k/v-deo.html ua-cam.com/video/vWLuGtnRcfc/v-deo.html Note that these videos discuss the cases where all ligands are identical AND cases where they are varied. Also note that the TBP and SP structures are not as different as they might seem at first, and many molecules rapidly interconvert between the two (2) structures, in a process called Berry pseudorotation.
@@lseinjr1 this might save my ass for the exam i have in a week, we only have to discuss the point group and its IR-Active vibrations compared to another coord. complex.
We note that the cause of degeneracy in a molecule is the presence of a Cₙ or Sₙ axis, where n greater than 2. We then perform a Cₙ (or Cₙ²) operation on one (1) member of the *E* pair. The result is (usually) not the other member of the pair, but it is a linear combination of the *E* members. We then subtract the result from the first member of the *E* pair, to get the other member of the pair. A key point that degeneracy is *always* caused by the presence of a Cₙ (or Sₙ) axis, with n greater than 2 - degeneracy is a direct consequence of *geometry*. Does that help?
@@lseinjr1 thanks Professor it really helps. So if am correct I perform a Cn operation of the highest order n that we have for example if molecule is C4v I should apply a C4 operation. Is that so
That is correct. You can see an example of this in two (2) videos (parts I and II): ua-cam.com/video/ufcVQVNCI7w/v-deo.html and ua-cam.com/video/_7P7CI-46WA/v-deo.html (This molecule is D4h rather than C4v, but the idea is the same.
Great question! There is actually just one (1) C3 axis, but there are two (2) C3 rotations along this axis. The first C3 is the 120 degree rotation counter-clockwise (anti-clockwise), and the second C3 rotation is the 120 degree rotation clockwise. The clockwise C3 (which can be written C3^-1) is equivalent to doing two (2) C3 rotations counter-clockwise.(C3^2). Does that make sense?
i am trying to find the vibrations of ethylene H2C-CH2, the modes that have changing bond lengths and changing angles i can derive, but there are 3 vibrations out-of-the-plane, where bonds and angles stay the same. how do I describe these motions to make the symmetry operations?
There is a subtle trick that you need to use. Draw the molecule on paper almost, but not completely, from the side. One carbon will be to the left, and the other to the right. ↑ ↑ ↑ ↑ H ↑ C =====C H ↑ H H Then draw a vector from each carbon or hydrogen perpendicular (normal) to the plane of the molecule. If you have drawn the molecule from the side, all, the vectors will point straight up on the paper. Label each vector. Now, take one of the H vectors, and do the projection operator step, seeing how this vector transforms under each symmetry operation, just as you had done earlier to find the bending vibrations. Note that you would have draw the vectors for those IN the plane of the molecule. Finally, do the same procedure for the carbons. If the carbon and hydrogen vectors have the same symmetry, combine them into a single vibrations that has both carbon and hydrogen displacements. Does that help?
Linear molecules have 3N - 5 vibrations, and non-linear molecules have 3N - 6, where N is the number of atoms. [Co(NH3)6]+3 is non-linear, and N = 25, so 75 - 6 = 69 vibrations.
I depends on what you are trying to do. If you are checking that you have found ALL of the 3N-6 (or #N-5 for a linear molecule) modes of vibration, then YES, you should include the coefficients. In the case 2A1 + 2E1, the two (2) A1's count as 2 vibrations, the 2 E1's count as 4 (because each E1 is double degenerate), giving the correct total of 6.
The "degrees of freedom" is the minimum number of variables required to fully specify the system. In a 3d space (like R³) there are three (3) degrees of freedom (x coordinate, y coordinate, z coordinate) for each particle. In rhis video, there are four (4) particles, each of which has its own x,y,z coordinates, so there are 4 x 3 = 12 degrees of freedom. Does that help?
I didn't get your point. What i understood is if I have 12 variables, the degrees of freedom must be less than 12, except if they can not be reduced in this example. I will be thankful if you can give me an example where the degrees of freedom is less than the total variables number.
The degrees of freedom are the same as the number of variables. There are three variables for each particle. Particles x 3 = total degrees of freedom. We need this because we know that there are 3N-6 *vibrational* degrees of freedom for a non-linear molecule, and 3N-5 for a linear one. "3N" is the total number of degrees of freedom, "3" is the number of *translational* degrees of freedom for the molecule as a WHOLE (x, y, z coordinates of the center of mass), and "2" *rotational* degrees of freedom for rotation for a linear molecule, and "3" *rotational* degrees of freedom for a non-linear molecule. The number of translational, rotational, and vibrational degrees of freedom for the *molecule* has to equal the total number of degrees of freedom for the *atoms* that make up the molecule. Does that make more sense now?
This video was so great! Thank you for being so thorough in your explanations!!!!
Highly inspirational and uniform teaching! Thanks a lot sir!
just let you know your vid is still helping us
Thank you for saving my assignment. Very clear and concise
Great to hear!
(For mobile users)
00:40 Reducible representation for 3N degrees of freedom
06:10 Reduction of reducible representation
14:37 Subtracting out rotations and translations
17:54 Effect of each symmetry operation on representative bond stretch
19:54 A1 stretch
22:04 E stretches
23:08 Accounting for degeneracy of E stretches
27:07 Effect of each symmetry operation on representative bond bend
29:24 A1 bend
30:50 E bends
31:52 Accounting for degeneracy of E bends
34:24 Visibility in IR and / or Raman
you sound like the intelligent version of chris griffin and I love it!
Thank you so much! This video was so helpful and helped me with my assignment.
Thank you so much. You are a star!
so relevant as always!
Very nice representation thank you
Thank you, your explanation was brilliant and really helpful!
De nada. :)
I think by looking at the character table for C3V, we can find Gamma-Trace by adding A1 to E.
thank you for these videos!
so helpful thankyouu!
Great video. Later, can you solve this same problem, but using Kim's method?
I'm trying it on my own, but I'm not able to choose with basis function to use, as the c3v character table presents all linear, quadratic and cubic functions for the irreducible representations. Take A1 for instance
irrep linear quadratic cubic
A1 z x^2 -+y^2, z^2 z^3, x(x^2 - 3y^2), z(x^2 + y^2)
Depending on which function I chose, I get differenct orbital coefficient values
There's a very useful trick for A1 in any point group. Since it is totally symmetric, we can always use the basis function "1". Multiply each atomic orbital by "1", then add. So for molecular group orbitals the A1 is immediately (s1 + s2 +s3), unnornalized, where s1, s2, s3 are the H atomic 1s orbitals on hydrogens 1, 2, and 3, respectively.
It is completely legitimate. Does this make sense?
@@lseinjr1 Thank you very much for your response, it made complete sense. Eventually I managed to solve this problem, but I had to redraw my coordinates system. I tried to make more or less a "3D" system, just like the methane video. Some time later, I redrawed it to a simple 2D case, with the NH3 lone pair ponting outside the notebook's plane. I'll try to solve it on a 3D coordinates system later on
i really like the reference to good vibrations (beachboys song)
You are amazing. Thank you so much. Can you please tell me the references(books) you have utilized for group theory? I need to dive into group theory.
My references are:
1. "Concepts and Models of Inorganic Chemistry" , 3rd edition, by Douglas, McDaniel, and Alexander
2. "inorganic Chemistry" by Miessler and Tarr
3. "Chemical Applications of Group Theory" F. Albert Cotton
4. "Group Theoretical Methods" Shoon K. Kim.
I took a graduate course with Dr. Kim (ref 4) at Temple University. His book describes the theory of the projection operator method, but also demonstrates what I call the "Kim Method", which is much easier to apply to high symmetry molecules.
I have videos demonstrating this method here: ua-cam.com/video/vPmeouMsbyo/v-deo.html,
ua-cam.com/video/KjOUeGBYxdY/v-deo.html,
ua-cam.com/video/QMU6SOLbvbs/v-deo.html,
ua-cam.com/video/VdwlTLRV36s/v-deo.html, and
ua-cam.com/video/Rbh-YVWk5z4/v-deo.html
.
For an introduction to group theory, see my videos here:
ua-cam.com/video/GESob5sKeek/v-deo.html
ua-cam.com/video/COjpFPk-m5Q/v-deo.html
ua-cam.com/video/bhYEDa-RQfU/v-deo.html
ua-cam.com/video/_9rNZ__aWSw/v-deo.html
ua-cam.com/video/Y57erKDQObc/v-deo.html
ua-cam.com/video/EDtRCLYTZC0/v-deo.html
ua-cam.com/video/Fz4JLe7gcWw/v-deo.html
ua-cam.com/video/zlUztdkBSh0/v-deo.html
ua-cam.com/video/DgUmQdcs6Tk/v-deo.html
ua-cam.com/video/Yme3FnWz7z4/v-deo.html
@@lseinjr1 Thank you so much for replying so fast. Sure I'll look into your videos and more.
Thank you so much sir
Thank you ! This was very helpful
The two functions 2b1-b2-b3 and b1-b2 are not orthogonal but they are supposed to be. I will greatly appreciate it if you can explain this. Thank you!
They are not (necessarily) "supposed" to be orthogonal, It is often very convenient for them to be so. If you know how to do Graham-Schmidt orthogonalization and normalization, that will give you the orthonormal set. Showing those in video would have made it too long. Does that make sense?
@@lseinjr1 Hello Dr. Sein! An identical procedure to that used to determine the form of the vibrational modes is used to deduce the LCAOs, and we all know that the resulting functions should be orthogonal. As the procedure used to obtain the second function for E is generally arbitrary, in my course I follow a slightly different approach. Specifically, I multiply by 2 the function resulting from C3 rotation of the three vectors and then add it to the initial one. This allows eliminating the b1 vector in the expression of the second function for E, which will be orthogonal with the other two. Brian Pfennig, in his book, attaches a Cartesian system to the central atom and looks at the projection of the three vectors on this system to define new vectors for the third function and obtain it orthogonal to the other two. Anyway, your videos are very nice and informative. Best regards!
im so anxious i see simple molecules like ammonia and water but when it comes to coordination complexes i cant see anything my 3d vision fades. i cant see the point group of iron pentacarbonyl
When there are five (5) ligands - and the ligands are all identical - there are usually only two (2) possibilities. One is trigonal bipyramid (TBP, which would have D₃ₕ symmetry) and the other is square pyramidal (SP, C₄ᵥ).
I have videos with models of these structures, which might be helpful:
ua-cam.com/video/1oao6HGYc0E/v-deo.html
ua-cam.com/video/ogQ5vJqN8yM/v-deo.html
ua-cam.com/video/7KwMEFrzHG8/v-deo.html
ua-cam.com/video/Cuc7lifQD7k/v-deo.html
ua-cam.com/video/vWLuGtnRcfc/v-deo.html
Note that these videos discuss the cases where all ligands are identical AND cases where they are varied. Also note that the TBP and SP structures are not as different as they might seem at first, and many molecules rapidly interconvert between the two (2) structures, in a process called Berry pseudorotation.
@@lseinjr1 this might save my ass for the exam i have in a week, we only have to discuss the point group and its IR-Active vibrations compared to another coord. complex.
Sorry I’m little confuse about 25:00, why after c3 rotation that will keep -r2 to -r3, but change -r3 to +r1? Thanks for answering!
Good catch - I caught it myself on the video at 25:39.
Thanks a lot professor , so how do i explain using a Cn axis due to degeneracy to find the 2nd projection of E in my exam next month .
We note that the cause of degeneracy in a molecule is the presence of a Cₙ or Sₙ axis, where n greater than 2. We then perform a Cₙ (or Cₙ²) operation on one (1) member of the *E* pair. The result is (usually) not the other member of the pair, but it is a linear combination of the *E* members. We then subtract the result from the first member of the *E* pair,
to get the other member of the pair.
A key point that degeneracy is *always* caused by the presence of a Cₙ (or Sₙ) axis, with n greater than 2 - degeneracy is a direct consequence of *geometry*.
Does that help?
@@lseinjr1 thanks Professor it really helps. So if am correct I perform a Cn operation of the highest order n that we have for example if molecule is C4v I should apply a C4 operation. Is that so
That is correct. You can see an example of this in two (2) videos (parts I and II): ua-cam.com/video/ufcVQVNCI7w/v-deo.html
and ua-cam.com/video/_7P7CI-46WA/v-deo.html
(This molecule is D4h rather than C4v, but the idea is the same.
Why are there two C3 axes? (also, thank you for this video!)
Great question! There is actually just one (1) C3 axis, but there are two (2) C3 rotations along this axis.
The first C3 is the 120 degree rotation counter-clockwise (anti-clockwise), and the second C3 rotation is the 120 degree rotation clockwise. The clockwise C3 (which can be written C3^-1) is equivalent to doing two (2) C3 rotations counter-clockwise.(C3^2).
Does that make sense?
i am trying to find the vibrations of ethylene H2C-CH2, the modes that have changing bond lengths and changing angles i can derive, but there are 3 vibrations out-of-the-plane, where bonds and angles stay the same. how do I describe these motions to make the symmetry operations?
There is a subtle trick that you need to use. Draw the molecule on paper almost, but not completely, from the side. One carbon will be to the left, and the other to the right.
↑ ↑ ↑ ↑
H ↑ C =====C H ↑
H H
Then draw a vector from each carbon or hydrogen perpendicular (normal) to the plane of the molecule. If you have drawn the molecule from the side, all, the vectors will point straight up on the paper. Label each vector.
Now, take one of the H vectors, and do the projection operator step, seeing how this vector transforms under each symmetry operation, just as you had done earlier to find the bending vibrations. Note that you would have draw the vectors for those IN the plane of the molecule.
Finally, do the same procedure for the carbons. If the carbon and hydrogen vectors have the same symmetry, combine them into a single vibrations that has both carbon and hydrogen displacements.
Does that help?
@@lseinjr1 ohh yeah yeah, thank you
please any example for point group C4v
يا ودي خرجتلنا كيما نتا بصح راهم كاين فالحقيقة 6 أنماط اهتزاز!! الله يحغظك ڤوللنا وين راحو هذوك الزوج
How many vibrations does it have [Co(NH3)6]+3?
Linear molecules have 3N - 5 vibrations, and non-linear molecules have 3N - 6, where N is the number of atoms. [Co(NH3)6]+3 is non-linear, and N = 25, so 75 - 6 = 69 vibrations.
Sir should we include coefficient from vibrational modes while writing Raman active modes and IR active modes??
i.e, Raman active mode = 2A1 + 2E??
I depends on what you are trying to do. If you are checking that you have found ALL of the 3N-6 (or #N-5 for a linear molecule) modes of vibration, then YES, you should include the coefficients.
In the case 2A1 + 2E1, the two (2) A1's count as 2 vibrations, the 2 E1's count as 4 (because each E1 is double degenerate), giving the correct total of 6.
What does the degree of freedom mean?
The "degrees of freedom" is the minimum number of variables required to fully specify the system. In a 3d space (like R³) there are three (3) degrees of freedom (x coordinate, y coordinate, z coordinate) for each particle. In rhis video, there are four (4) particles, each of which has its own x,y,z coordinates, so there are 4 x 3 = 12 degrees of freedom.
Does that help?
I didn't get your point. What i understood is if I have 12 variables, the degrees of freedom must be less than 12, except if they can not be reduced in this example. I will be thankful if you can give me an example where the degrees of freedom is less than the total variables number.
The degrees of freedom are the same as the number of variables. There are three variables for each particle. Particles x 3 = total degrees of freedom.
We need this because we know that there are 3N-6 *vibrational* degrees of freedom for a non-linear molecule, and 3N-5 for a linear one. "3N" is the total number of degrees of freedom, "3" is the number of *translational* degrees of freedom for the molecule as a WHOLE (x, y, z coordinates of the center of mass), and "2" *rotational* degrees of freedom for rotation for a linear molecule, and "3" *rotational* degrees of freedom for a non-linear molecule.
The number of translational, rotational, and vibrational degrees of freedom for the *molecule* has to equal the total number of degrees of freedom for the *atoms* that make up the molecule.
Does that make more sense now?
Yeap much better. Thank you very much sir.
Projection operator from 19 minutes