IR active vibrations and point groups

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ReidMerrill
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Homework Statement


In lab we synthesised cis and trans copper glycine and we have to use IR to differentiate the two so we have to figure out the number of IR active vibrations for each complex. It's been a year since I did anything with point groups so I'm not sure if I did it right.

Homework Equations


the point group for cis-copper glycine is C2V

The Attempt at a Solution


http://[URL=http://s350.photobucket.com/user/remerril/media/hs%20blackfff_zpsexbt6f6c.png.html][PLAIN]http://i350.photobucket.com/albums/q422/remerril/hs%20blackfff_zpsexbt6f6c.png [PLAIN]http://i350.photobucket.com/albums/q422/remerril/hs%20blackfff_zpsexbt6f6c.png[/URL]
http://s350.photobucket.com/user/remerril/media/hs blackfff_zpsexbt6f6c.png.html?filters[user]=146245371&filters[recent]=1&sort=1&o=0

Here's my attempt at the solution. I'm not confident I did it right at all because I've never seen numbers that high. Water has the same symmetry but only 3 active vibrations. Then again I've never done this on a molecule this large.
 
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You have made mistakes in your "unshifted" and "contribution" rows.
How many atoms are there? (Have you forgotten some that are not written?)
How many are unshifted under σv' (reflection in the plane of the paper)?
What is the "contribution" under E (per atom)?
 
mjc123 said:
You have made mistakes in your "unshifted" and "contribution" rows.
How many atoms are there? (Have you forgotten some that are not written?)
How many are unshifted under σv' (reflection in the plane of the paper)?
What is the "contribution" under E (per atom)?
I had σv as the plane in the page and σv' as the one perpendicular to the page. I don't think that should make a difference thought. I'm not sure what the contributions thing is. I was following along with a video of someone doing the same process but with water.

And I totally forgot tho count some of the hydrogens... there are 19 atoms.
 
ReidMerrill said:
I had σv as the plane in the page and σv' as the one perpendicular to the page.
That is not consistent with what you have written.
ReidMerrill said:
I'm not sure what the contributions thing is.
You are considering the possible motions of the atoms. Each atom has 3 degrees of freedom - to move in the x, y or z directions. So you have 3N motions to consider. Now the number in the representation Γ is the trace (sum of diagonal elements) of the matrix, i.e. the sum of all contributions of the type "x on one atom → x on the same atom". So first, you only consider those atoms that stay in the same position under a symmetry operation; then, what happens to x, y and z under that symmetry operation. For example, under C2, only the Cu atom is unshifted. If we call the rotation axis the z axis, then a 180° rotation converts x to -x, y to -y and z to z, so the "contribution" is -1 + -1 + 1 = -1, and the number for C2 in Γ is 1 (atoms unshifted) * -1 (contribution per atom) = -1.
 
mjc123 said:
That is not consistent with what you have written.

You are considering the possible motions of the atoms. Each atom has 3 degrees of freedom - to move in the x, y or z directions. So you have 3N motions to consider. Now the number in the representation Γ is the trace (sum of diagonal elements) of the matrix, i.e. the sum of all contributions of the type "x on one atom → x on the same atom". So first, you only consider those atoms that stay in the same position under a symmetry operation; then, what happens to x, y and z under that symmetry operation. For example, under C2, only the Cu atom is unshifted. If we call the rotation axis the z axis, then a 180° rotation converts x to -x, y to -y and z to z, so the "contribution" is -1 + -1 + 1 = -1, and the number for C2 in Γ is 1 (atoms unshifted) * -1 (contribution per atom) = -1.
Oh yeah that rings a bell. For the signma in the place of the paper what would it be since all of the atoms will go from x to -x and y to -y while Z stays the same. Would it be -19+-19+19 (if you consider all of the atoms?) or -1+-1+1
 
The hydrogens on the two carbons and on the nitrogen are not in the plane.