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Can anyone help me with this question on vibrational modes of free base porphyrin (C20H14N4)?

Rule of mutual exclusion: is the converse true?Can anyone explain this condensation pattern?Should the coupling between vibrational modes increase or decrease with increasing temperature/vibrational energy level?How to systematically find the reducible representation of a molecule? (eg: XeF4)I need some help on this Chemistry question can any one help please!! I need to find the cost of a single atom of aluminumHow to distinguish two molecules with same formula but different geometry with vibrational spectroscopy

3

$begingroup$

4 a. The structure of free-base porphyrin $(ceH2P$, chemical formula: $ceC20H14N4)$ is shown below. Given that this molecule belongs to the $D_mathrm2h$ point group (character table given above) identify the symmetries (irreps) for all the vibrational modes of this planar polyatomic molecule;

(i) Identify the number of vibrational modes in $ceH2P$. [3 marks]

(ii) Work out the reducible representation for the vibrational modes. [6 marks]

(iii) Use the reduction formula to identify the (irreps) for all the vibrational modes. [6 marks]

(iv) Identify which of the vibrational modes are Raman-active and which are IR-active. [4 marks]

In particular I'm getting stuck with figuring out the unmoved atoms when applying symmetry operations. I know i need to calculate gamma vib in order to use the reduction formula but I cannot picture how to apply the symmetry operations to calculate unmoved atoms for each symmetry class (aside from E) because it is a more complicated structure than I'm used to dealing with (h20, benzene etc)

physical-chemistry spectroscopy molecular-structure ir-spectroscopy

share|improve this question

edited May 13 at 12:04

Gillian

asked May 13 at 9:02

GillianGillian

214

New contributor

Gillian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  One thing to think about in the question: is the free base porphyrin really D2h symmetric?
  $endgroup$
  – Martin - マーチン♦
  May 13 at 22:48
  

  
  
  
  

add a comment | 

3

$begingroup$

4 a. The structure of free-base porphyrin $(ceH2P$, chemical formula: $ceC20H14N4)$ is shown below. Given that this molecule belongs to the $D_mathrm2h$ point group (character table given above) identify the symmetries (irreps) for all the vibrational modes of this planar polyatomic molecule;

(i) Identify the number of vibrational modes in $ceH2P$. [3 marks]

(ii) Work out the reducible representation for the vibrational modes. [6 marks]

(iii) Use the reduction formula to identify the (irreps) for all the vibrational modes. [6 marks]

(iv) Identify which of the vibrational modes are Raman-active and which are IR-active. [4 marks]

In particular I'm getting stuck with figuring out the unmoved atoms when applying symmetry operations. I know i need to calculate gamma vib in order to use the reduction formula but I cannot picture how to apply the symmetry operations to calculate unmoved atoms for each symmetry class (aside from E) because it is a more complicated structure than I'm used to dealing with (h20, benzene etc)

physical-chemistry spectroscopy molecular-structure ir-spectroscopy

share|improve this question

edited May 13 at 12:04

Gillian

asked May 13 at 9:02

GillianGillian

214

New contributor

Gillian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  One thing to think about in the question: is the free base porphyrin really D2h symmetric?
  $endgroup$
  – Martin - マーチン♦
  May 13 at 22:48
  

  
  
  
  

add a comment | 

$begingroup$

4 a. The structure of free-base porphyrin $(ceH2P$, chemical formula: $ceC20H14N4)$ is shown below. Given that this molecule belongs to the $D_mathrm2h$ point group (character table given above) identify the symmetries (irreps) for all the vibrational modes of this planar polyatomic molecule;

(i) Identify the number of vibrational modes in $ceH2P$. [3 marks]

(ii) Work out the reducible representation for the vibrational modes. [6 marks]

(iii) Use the reduction formula to identify the (irreps) for all the vibrational modes. [6 marks]

(iv) Identify which of the vibrational modes are Raman-active and which are IR-active. [4 marks]

In particular I'm getting stuck with figuring out the unmoved atoms when applying symmetry operations. I know i need to calculate gamma vib in order to use the reduction formula but I cannot picture how to apply the symmetry operations to calculate unmoved atoms for each symmetry class (aside from E) because it is a more complicated structure than I'm used to dealing with (h20, benzene etc)

physical-chemistry spectroscopy molecular-structure ir-spectroscopy

share|improve this question

edited May 13 at 12:04

Gillian

asked May 13 at 9:02

GillianGillian

214

New contributor

Gillian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited May 13 at 12:04

Gillian

asked May 13 at 9:02

GillianGillian

214

New contributor

Gillian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited May 13 at 12:04

Gillian

asked May 13 at 9:02

GillianGillian

214

New contributor

Gillian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 13 at 9:02

GillianGillian

214

New contributor

Gillian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 13 at 9:02

GillianGillian

214

1 Answer
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4

$begingroup$

If you rotate the molecule by $45^circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$.

Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. You can ignore the position of the double bonds because it is an aromatic system.

Whenever an atoms lies on a mirror plane or a pure rotation axis, its position will be unaffected by the symmetry operation.

share|improve this answer

edited May 13 at 14:59

answered May 13 at 14:04

Karsten TheisKarsten Theis

6,005947

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4

$begingroup$

If you rotate the molecule by $45^circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$.

Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. You can ignore the position of the double bonds because it is an aromatic system.

Whenever an atoms lies on a mirror plane or a pure rotation axis, its position will be unaffected by the symmetry operation.

share|improve this answer

edited May 13 at 14:59

answered May 13 at 14:04

Karsten TheisKarsten Theis

6,005947

$endgroup$

add a comment | 

4

$begingroup$

If you rotate the molecule by $45^circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$.

Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. You can ignore the position of the double bonds because it is an aromatic system.

Whenever an atoms lies on a mirror plane or a pure rotation axis, its position will be unaffected by the symmetry operation.

share|improve this answer

edited May 13 at 14:59

answered May 13 at 14:04

Karsten TheisKarsten Theis

6,005947

$endgroup$

add a comment | 

$begingroup$

If you rotate the molecule by $45^circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$.

Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. You can ignore the position of the double bonds because it is an aromatic system.

Whenever an atoms lies on a mirror plane or a pure rotation axis, its position will be unaffected by the symmetry operation.

share|improve this answer

edited May 13 at 14:59

answered May 13 at 14:04

Karsten TheisKarsten Theis

6,005947

$endgroup$

share|improve this answer

edited May 13 at 14:59

answered May 13 at 14:04

Karsten TheisKarsten Theis

6,005947

answered May 13 at 14:04

Karsten TheisKarsten Theis

6,005947

answered May 13 at 14:04

Karsten TheisKarsten Theis

6,005947