Relationship between the symmetry number of a molecule as used in rotational spectroscopy and point group












12












$begingroup$


While doing a problem set I noticed that the symmetry number of a molecule turns out (usually) to be half the number of symmetry elements that the point group the molecule belongs to.



When I say symmetry number I refer to the symmetry number used in rotational spectroscopy — i.e. the number of indistinguishable orientations of the molecule. This accounts for suitable symmetrisation of the rotational wavefunction and the nuclear wavefunction of the molecule.



For example benzene belongs to point group $D_mathrm{6h}$, which has 24 symmetry elements. It also has symmetry number 12, i.e. 1/2 the number of symmetry elements in it's point group.



Another few examples:




  • Water. Point group $C_mathrm{2v}$, with 4 symmetry elements. Has 2 as symmetry number.

  • Ammonia. Point group $C_mathrm{3v}$ with 6 symmetry elements. 3 as symmetry number.

  • Ethene. Point group $D_mathrm{2h}$ with 8 symmetry elements. 4 as symmetry number.


This seems to be a general trend. It sort of makes sense intuitively that the 2 should have a relationship, but I can't formalise the reason for this.



Obviously for diatomics/linear molecules this doesn't apply. This is because rotational motion around the axis of symmetry is not physical — moment of inertia around this axis is zero, as all mass along axis. Thus there are no rotational levels associated with this axis. But ignoring diatomics/linear molecules, where we have neglected/ignored an axis of rotation, the above relationship seems to hold.



Why is this the case? I'd appreciate both a mathematical derivation and a more intuitive argument.










share|improve this question











$endgroup$












  • $begingroup$
    Table II on this NIST webpage gives the number of symmetry elements for most of the common point groups and a brief explanation of how they can be obtained. It seems to align with the trend you are describing. @SwedishArchitect
    $endgroup$
    – Tyberius
    Mar 6 at 16:59


















12












$begingroup$


While doing a problem set I noticed that the symmetry number of a molecule turns out (usually) to be half the number of symmetry elements that the point group the molecule belongs to.



When I say symmetry number I refer to the symmetry number used in rotational spectroscopy — i.e. the number of indistinguishable orientations of the molecule. This accounts for suitable symmetrisation of the rotational wavefunction and the nuclear wavefunction of the molecule.



For example benzene belongs to point group $D_mathrm{6h}$, which has 24 symmetry elements. It also has symmetry number 12, i.e. 1/2 the number of symmetry elements in it's point group.



Another few examples:




  • Water. Point group $C_mathrm{2v}$, with 4 symmetry elements. Has 2 as symmetry number.

  • Ammonia. Point group $C_mathrm{3v}$ with 6 symmetry elements. 3 as symmetry number.

  • Ethene. Point group $D_mathrm{2h}$ with 8 symmetry elements. 4 as symmetry number.


This seems to be a general trend. It sort of makes sense intuitively that the 2 should have a relationship, but I can't formalise the reason for this.



Obviously for diatomics/linear molecules this doesn't apply. This is because rotational motion around the axis of symmetry is not physical — moment of inertia around this axis is zero, as all mass along axis. Thus there are no rotational levels associated with this axis. But ignoring diatomics/linear molecules, where we have neglected/ignored an axis of rotation, the above relationship seems to hold.



Why is this the case? I'd appreciate both a mathematical derivation and a more intuitive argument.










share|improve this question











$endgroup$












  • $begingroup$
    Table II on this NIST webpage gives the number of symmetry elements for most of the common point groups and a brief explanation of how they can be obtained. It seems to align with the trend you are describing. @SwedishArchitect
    $endgroup$
    – Tyberius
    Mar 6 at 16:59
















12












12








12





$begingroup$


While doing a problem set I noticed that the symmetry number of a molecule turns out (usually) to be half the number of symmetry elements that the point group the molecule belongs to.



When I say symmetry number I refer to the symmetry number used in rotational spectroscopy — i.e. the number of indistinguishable orientations of the molecule. This accounts for suitable symmetrisation of the rotational wavefunction and the nuclear wavefunction of the molecule.



For example benzene belongs to point group $D_mathrm{6h}$, which has 24 symmetry elements. It also has symmetry number 12, i.e. 1/2 the number of symmetry elements in it's point group.



Another few examples:




  • Water. Point group $C_mathrm{2v}$, with 4 symmetry elements. Has 2 as symmetry number.

  • Ammonia. Point group $C_mathrm{3v}$ with 6 symmetry elements. 3 as symmetry number.

  • Ethene. Point group $D_mathrm{2h}$ with 8 symmetry elements. 4 as symmetry number.


This seems to be a general trend. It sort of makes sense intuitively that the 2 should have a relationship, but I can't formalise the reason for this.



Obviously for diatomics/linear molecules this doesn't apply. This is because rotational motion around the axis of symmetry is not physical — moment of inertia around this axis is zero, as all mass along axis. Thus there are no rotational levels associated with this axis. But ignoring diatomics/linear molecules, where we have neglected/ignored an axis of rotation, the above relationship seems to hold.



Why is this the case? I'd appreciate both a mathematical derivation and a more intuitive argument.










share|improve this question











$endgroup$




While doing a problem set I noticed that the symmetry number of a molecule turns out (usually) to be half the number of symmetry elements that the point group the molecule belongs to.



When I say symmetry number I refer to the symmetry number used in rotational spectroscopy — i.e. the number of indistinguishable orientations of the molecule. This accounts for suitable symmetrisation of the rotational wavefunction and the nuclear wavefunction of the molecule.



For example benzene belongs to point group $D_mathrm{6h}$, which has 24 symmetry elements. It also has symmetry number 12, i.e. 1/2 the number of symmetry elements in it's point group.



Another few examples:




  • Water. Point group $C_mathrm{2v}$, with 4 symmetry elements. Has 2 as symmetry number.

  • Ammonia. Point group $C_mathrm{3v}$ with 6 symmetry elements. 3 as symmetry number.

  • Ethene. Point group $D_mathrm{2h}$ with 8 symmetry elements. 4 as symmetry number.


This seems to be a general trend. It sort of makes sense intuitively that the 2 should have a relationship, but I can't formalise the reason for this.



Obviously for diatomics/linear molecules this doesn't apply. This is because rotational motion around the axis of symmetry is not physical — moment of inertia around this axis is zero, as all mass along axis. Thus there are no rotational levels associated with this axis. But ignoring diatomics/linear molecules, where we have neglected/ignored an axis of rotation, the above relationship seems to hold.



Why is this the case? I'd appreciate both a mathematical derivation and a more intuitive argument.







quantum-chemistry spectroscopy symmetry






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 7 at 5:24









andselisk

17.7k656117




17.7k656117










asked Mar 6 at 16:23









Swedish ArchitectSwedish Architect

1,05821017




1,05821017












  • $begingroup$
    Table II on this NIST webpage gives the number of symmetry elements for most of the common point groups and a brief explanation of how they can be obtained. It seems to align with the trend you are describing. @SwedishArchitect
    $endgroup$
    – Tyberius
    Mar 6 at 16:59




















  • $begingroup$
    Table II on this NIST webpage gives the number of symmetry elements for most of the common point groups and a brief explanation of how they can be obtained. It seems to align with the trend you are describing. @SwedishArchitect
    $endgroup$
    – Tyberius
    Mar 6 at 16:59


















$begingroup$
Table II on this NIST webpage gives the number of symmetry elements for most of the common point groups and a brief explanation of how they can be obtained. It seems to align with the trend you are describing. @SwedishArchitect
$endgroup$
– Tyberius
Mar 6 at 16:59






$begingroup$
Table II on this NIST webpage gives the number of symmetry elements for most of the common point groups and a brief explanation of how they can be obtained. It seems to align with the trend you are describing. @SwedishArchitect
$endgroup$
– Tyberius
Mar 6 at 16:59












1 Answer
1






active

oldest

votes


















14












$begingroup$

This is not in general true



Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide



Hydrogen Peroxide



The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms. Both the symmetry number and order of the group are 2.



Matrix representations



All rotations in 3D space can be represented by a orthogonal 3x3 matrix with determinant 1, and the composition of rotations about the origin can be represented by matrix multiplication of the transformation.



More formally: the group SO$(3)$ (3D rotations) is isomorphic to the group of orthogonal real 3x3 matrices with determinant 1 under multiplication.



The number of symmetry elements in a point group formed from rotations (and which hence preserve chirality) will be the same as the number which can be represented by such matrices and the same as the number of equivalent orientations of the molecule (the symmetry number) as they are all representation of the same thing - 3D rotations.



Inversions



The group of all transformations that keep the origin fixed, including reflections as well as rotations is O$(3)$ - or the product of SO$(3)$ with the set $lbrace I,-Irbrace$ or the inversion operation. It maps rotations onto improper rotations (which include reflections - improper rotation by 0 degrees) and hence all symmetries in 3D space.



All point groups with inversion symmetry, such as the 3 listed in the question, contains this $-I$ element. $-I$ will multiple all the elements of the rotation matrix by $-1$ and create an orthogonal 3x3 matrix with determinant -1. By composition with each of the rotations elements of the point group it creates another element - an inversion, improper rotation or reflection - doubling the size of the group without increasing the symmetry number.



Conclusion: chirality matters



This rule is true for non-chiral groups. You can invert the molecule to get another symmetric copy for every rotationally equivalent copy, so the order of the group is twice the symmetry number.



For chiral groups that will create the enantiomer which isn't symmetrically equivalent so the order of the group and symmetry number are the same.






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Just to clarify: the molecules in the question do not have inversion symmetry, but they have an improper symmetry operation (a mirror plane, inversion center, or improper rotation). The more direct way of saying this is the symmetry number corresponds to the order of the largest (purely) rotational subgroup of the point group. For $C_{2v}$, that's $C_2$. For $C_{3v}$, that is $C_3$. For $D_{nh}$, it's $D_n$.
    $endgroup$
    – levineds
    Mar 12 at 0:03











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1 Answer
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1 Answer
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active

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14












$begingroup$

This is not in general true



Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide



Hydrogen Peroxide



The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms. Both the symmetry number and order of the group are 2.



Matrix representations



All rotations in 3D space can be represented by a orthogonal 3x3 matrix with determinant 1, and the composition of rotations about the origin can be represented by matrix multiplication of the transformation.



More formally: the group SO$(3)$ (3D rotations) is isomorphic to the group of orthogonal real 3x3 matrices with determinant 1 under multiplication.



The number of symmetry elements in a point group formed from rotations (and which hence preserve chirality) will be the same as the number which can be represented by such matrices and the same as the number of equivalent orientations of the molecule (the symmetry number) as they are all representation of the same thing - 3D rotations.



Inversions



The group of all transformations that keep the origin fixed, including reflections as well as rotations is O$(3)$ - or the product of SO$(3)$ with the set $lbrace I,-Irbrace$ or the inversion operation. It maps rotations onto improper rotations (which include reflections - improper rotation by 0 degrees) and hence all symmetries in 3D space.



All point groups with inversion symmetry, such as the 3 listed in the question, contains this $-I$ element. $-I$ will multiple all the elements of the rotation matrix by $-1$ and create an orthogonal 3x3 matrix with determinant -1. By composition with each of the rotations elements of the point group it creates another element - an inversion, improper rotation or reflection - doubling the size of the group without increasing the symmetry number.



Conclusion: chirality matters



This rule is true for non-chiral groups. You can invert the molecule to get another symmetric copy for every rotationally equivalent copy, so the order of the group is twice the symmetry number.



For chiral groups that will create the enantiomer which isn't symmetrically equivalent so the order of the group and symmetry number are the same.






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Just to clarify: the molecules in the question do not have inversion symmetry, but they have an improper symmetry operation (a mirror plane, inversion center, or improper rotation). The more direct way of saying this is the symmetry number corresponds to the order of the largest (purely) rotational subgroup of the point group. For $C_{2v}$, that's $C_2$. For $C_{3v}$, that is $C_3$. For $D_{nh}$, it's $D_n$.
    $endgroup$
    – levineds
    Mar 12 at 0:03
















14












$begingroup$

This is not in general true



Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide



Hydrogen Peroxide



The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms. Both the symmetry number and order of the group are 2.



Matrix representations



All rotations in 3D space can be represented by a orthogonal 3x3 matrix with determinant 1, and the composition of rotations about the origin can be represented by matrix multiplication of the transformation.



More formally: the group SO$(3)$ (3D rotations) is isomorphic to the group of orthogonal real 3x3 matrices with determinant 1 under multiplication.



The number of symmetry elements in a point group formed from rotations (and which hence preserve chirality) will be the same as the number which can be represented by such matrices and the same as the number of equivalent orientations of the molecule (the symmetry number) as they are all representation of the same thing - 3D rotations.



Inversions



The group of all transformations that keep the origin fixed, including reflections as well as rotations is O$(3)$ - or the product of SO$(3)$ with the set $lbrace I,-Irbrace$ or the inversion operation. It maps rotations onto improper rotations (which include reflections - improper rotation by 0 degrees) and hence all symmetries in 3D space.



All point groups with inversion symmetry, such as the 3 listed in the question, contains this $-I$ element. $-I$ will multiple all the elements of the rotation matrix by $-1$ and create an orthogonal 3x3 matrix with determinant -1. By composition with each of the rotations elements of the point group it creates another element - an inversion, improper rotation or reflection - doubling the size of the group without increasing the symmetry number.



Conclusion: chirality matters



This rule is true for non-chiral groups. You can invert the molecule to get another symmetric copy for every rotationally equivalent copy, so the order of the group is twice the symmetry number.



For chiral groups that will create the enantiomer which isn't symmetrically equivalent so the order of the group and symmetry number are the same.






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Just to clarify: the molecules in the question do not have inversion symmetry, but they have an improper symmetry operation (a mirror plane, inversion center, or improper rotation). The more direct way of saying this is the symmetry number corresponds to the order of the largest (purely) rotational subgroup of the point group. For $C_{2v}$, that's $C_2$. For $C_{3v}$, that is $C_3$. For $D_{nh}$, it's $D_n$.
    $endgroup$
    – levineds
    Mar 12 at 0:03














14












14








14





$begingroup$

This is not in general true



Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide



Hydrogen Peroxide



The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms. Both the symmetry number and order of the group are 2.



Matrix representations



All rotations in 3D space can be represented by a orthogonal 3x3 matrix with determinant 1, and the composition of rotations about the origin can be represented by matrix multiplication of the transformation.



More formally: the group SO$(3)$ (3D rotations) is isomorphic to the group of orthogonal real 3x3 matrices with determinant 1 under multiplication.



The number of symmetry elements in a point group formed from rotations (and which hence preserve chirality) will be the same as the number which can be represented by such matrices and the same as the number of equivalent orientations of the molecule (the symmetry number) as they are all representation of the same thing - 3D rotations.



Inversions



The group of all transformations that keep the origin fixed, including reflections as well as rotations is O$(3)$ - or the product of SO$(3)$ with the set $lbrace I,-Irbrace$ or the inversion operation. It maps rotations onto improper rotations (which include reflections - improper rotation by 0 degrees) and hence all symmetries in 3D space.



All point groups with inversion symmetry, such as the 3 listed in the question, contains this $-I$ element. $-I$ will multiple all the elements of the rotation matrix by $-1$ and create an orthogonal 3x3 matrix with determinant -1. By composition with each of the rotations elements of the point group it creates another element - an inversion, improper rotation or reflection - doubling the size of the group without increasing the symmetry number.



Conclusion: chirality matters



This rule is true for non-chiral groups. You can invert the molecule to get another symmetric copy for every rotationally equivalent copy, so the order of the group is twice the symmetry number.



For chiral groups that will create the enantiomer which isn't symmetrically equivalent so the order of the group and symmetry number are the same.






share|improve this answer











$endgroup$



This is not in general true



Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide



Hydrogen Peroxide



The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms. Both the symmetry number and order of the group are 2.



Matrix representations



All rotations in 3D space can be represented by a orthogonal 3x3 matrix with determinant 1, and the composition of rotations about the origin can be represented by matrix multiplication of the transformation.



More formally: the group SO$(3)$ (3D rotations) is isomorphic to the group of orthogonal real 3x3 matrices with determinant 1 under multiplication.



The number of symmetry elements in a point group formed from rotations (and which hence preserve chirality) will be the same as the number which can be represented by such matrices and the same as the number of equivalent orientations of the molecule (the symmetry number) as they are all representation of the same thing - 3D rotations.



Inversions



The group of all transformations that keep the origin fixed, including reflections as well as rotations is O$(3)$ - or the product of SO$(3)$ with the set $lbrace I,-Irbrace$ or the inversion operation. It maps rotations onto improper rotations (which include reflections - improper rotation by 0 degrees) and hence all symmetries in 3D space.



All point groups with inversion symmetry, such as the 3 listed in the question, contains this $-I$ element. $-I$ will multiple all the elements of the rotation matrix by $-1$ and create an orthogonal 3x3 matrix with determinant -1. By composition with each of the rotations elements of the point group it creates another element - an inversion, improper rotation or reflection - doubling the size of the group without increasing the symmetry number.



Conclusion: chirality matters



This rule is true for non-chiral groups. You can invert the molecule to get another symmetric copy for every rotationally equivalent copy, so the order of the group is twice the symmetry number.



For chiral groups that will create the enantiomer which isn't symmetrically equivalent so the order of the group and symmetry number are the same.







share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 11 at 23:56









levineds

2,677723




2,677723










answered Mar 6 at 17:41









user213305user213305

1,064516




1,064516








  • 1




    $begingroup$
    Just to clarify: the molecules in the question do not have inversion symmetry, but they have an improper symmetry operation (a mirror plane, inversion center, or improper rotation). The more direct way of saying this is the symmetry number corresponds to the order of the largest (purely) rotational subgroup of the point group. For $C_{2v}$, that's $C_2$. For $C_{3v}$, that is $C_3$. For $D_{nh}$, it's $D_n$.
    $endgroup$
    – levineds
    Mar 12 at 0:03














  • 1




    $begingroup$
    Just to clarify: the molecules in the question do not have inversion symmetry, but they have an improper symmetry operation (a mirror plane, inversion center, or improper rotation). The more direct way of saying this is the symmetry number corresponds to the order of the largest (purely) rotational subgroup of the point group. For $C_{2v}$, that's $C_2$. For $C_{3v}$, that is $C_3$. For $D_{nh}$, it's $D_n$.
    $endgroup$
    – levineds
    Mar 12 at 0:03








1




1




$begingroup$
Just to clarify: the molecules in the question do not have inversion symmetry, but they have an improper symmetry operation (a mirror plane, inversion center, or improper rotation). The more direct way of saying this is the symmetry number corresponds to the order of the largest (purely) rotational subgroup of the point group. For $C_{2v}$, that's $C_2$. For $C_{3v}$, that is $C_3$. For $D_{nh}$, it's $D_n$.
$endgroup$
– levineds
Mar 12 at 0:03




$begingroup$
Just to clarify: the molecules in the question do not have inversion symmetry, but they have an improper symmetry operation (a mirror plane, inversion center, or improper rotation). The more direct way of saying this is the symmetry number corresponds to the order of the largest (purely) rotational subgroup of the point group. For $C_{2v}$, that's $C_2$. For $C_{3v}$, that is $C_3$. For $D_{nh}$, it's $D_n$.
$endgroup$
– levineds
Mar 12 at 0:03


















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