Why is CO2 not in the C∞v point group?

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It seems to me that CO2, since it can be linearly rotated around the z-axis without change of shape, ought to be in the C∞v point group. However, in all the character tables I can find, it's listed as a D∞h. What's the difference between these two, and why does CO2 fall in the D∞h and not the C∞v?






inorganic-chemistry symmetry







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  • 2




    In short, the two O atoms are equivalent, but $C_infty h$ fails to acknowledge this, so you add more symmetry elements that switch these two atoms, and so you end up with $D_infty h$ .
    – Ivan Neretin
    5 hours ago










  • Okay, I almost got that. What symmetry operations would switch the O's which wouldn't apply to a C∞v (assuming that's what you meant?) group?
    – anonymous2
    4 hours ago











  • Why, many. Think of all these perpendicular $C_2$ axes, to begin with.
    – Ivan Neretin
    4 hours ago










  • Oh, of course. Duh, my stupid. Sorry, and thanks! If you have time to post a complete answer, I'll accept.
    – anonymous2
    4 hours ago










  • Crazy how it is; now it's clicked I can't believe I didn't get it before. :)
    – anonymous2
    4 hours ago














up vote
2
down vote

favorite












It seems to me that CO2, since it can be linearly rotated around the z-axis without change of shape, ought to be in the C∞v point group. However, in all the character tables I can find, it's listed as a D∞h. What's the difference between these two, and why does CO2 fall in the D∞h and not the C∞v?






inorganic-chemistry symmetry







share|improve this question

















  • 2




    In short, the two O atoms are equivalent, but $C_infty h$ fails to acknowledge this, so you add more symmetry elements that switch these two atoms, and so you end up with $D_infty h$ .
    – Ivan Neretin
    5 hours ago










  • Okay, I almost got that. What symmetry operations would switch the O's which wouldn't apply to a C∞v (assuming that's what you meant?) group?
    – anonymous2
    4 hours ago











  • Why, many. Think of all these perpendicular $C_2$ axes, to begin with.
    – Ivan Neretin
    4 hours ago










  • Oh, of course. Duh, my stupid. Sorry, and thanks! If you have time to post a complete answer, I'll accept.
    – anonymous2
    4 hours ago










  • Crazy how it is; now it's clicked I can't believe I didn't get it before. :)
    – anonymous2
    4 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











It seems to me that CO2, since it can be linearly rotated around the z-axis without change of shape, ought to be in the C∞v point group. However, in all the character tables I can find, it's listed as a D∞h. What's the difference between these two, and why does CO2 fall in the D∞h and not the C∞v?






inorganic-chemistry symmetry







share|improve this question













It seems to me that CO2, since it can be linearly rotated around the z-axis without change of shape, ought to be in the C∞v point group. However, in all the character tables I can find, it's listed as a D∞h. What's the difference between these two, and why does CO2 fall in the D∞h and not the C∞v?






inorganic-chemistry symmetry




inorganic-chemistry symmetry






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 5 hours ago









anonymous2

1418




1418







  • 2




    In short, the two O atoms are equivalent, but $C_infty h$ fails to acknowledge this, so you add more symmetry elements that switch these two atoms, and so you end up with $D_infty h$ .
    – Ivan Neretin
    5 hours ago










  • Okay, I almost got that. What symmetry operations would switch the O's which wouldn't apply to a C∞v (assuming that's what you meant?) group?
    – anonymous2
    4 hours ago











  • Why, many. Think of all these perpendicular $C_2$ axes, to begin with.
    – Ivan Neretin
    4 hours ago










  • Oh, of course. Duh, my stupid. Sorry, and thanks! If you have time to post a complete answer, I'll accept.
    – anonymous2
    4 hours ago










  • Crazy how it is; now it's clicked I can't believe I didn't get it before. :)
    – anonymous2
    4 hours ago












  • 2




    In short, the two O atoms are equivalent, but $C_infty h$ fails to acknowledge this, so you add more symmetry elements that switch these two atoms, and so you end up with $D_infty h$ .
    – Ivan Neretin
    5 hours ago










  • Okay, I almost got that. What symmetry operations would switch the O's which wouldn't apply to a C∞v (assuming that's what you meant?) group?
    – anonymous2
    4 hours ago











  • Why, many. Think of all these perpendicular $C_2$ axes, to begin with.
    – Ivan Neretin
    4 hours ago










  • Oh, of course. Duh, my stupid. Sorry, and thanks! If you have time to post a complete answer, I'll accept.
    – anonymous2
    4 hours ago










  • Crazy how it is; now it's clicked I can't believe I didn't get it before. :)
    – anonymous2
    4 hours ago







2




2




In short, the two O atoms are equivalent, but $C_infty h$ fails to acknowledge this, so you add more symmetry elements that switch these two atoms, and so you end up with $D_infty h$ .
– Ivan Neretin
5 hours ago




In short, the two O atoms are equivalent, but $C_infty h$ fails to acknowledge this, so you add more symmetry elements that switch these two atoms, and so you end up with $D_infty h$ .
– Ivan Neretin
5 hours ago












Okay, I almost got that. What symmetry operations would switch the O's which wouldn't apply to a C∞v (assuming that's what you meant?) group?
– anonymous2
4 hours ago





Okay, I almost got that. What symmetry operations would switch the O's which wouldn't apply to a C∞v (assuming that's what you meant?) group?
– anonymous2
4 hours ago













Why, many. Think of all these perpendicular $C_2$ axes, to begin with.
– Ivan Neretin
4 hours ago




Why, many. Think of all these perpendicular $C_2$ axes, to begin with.
– Ivan Neretin
4 hours ago












Oh, of course. Duh, my stupid. Sorry, and thanks! If you have time to post a complete answer, I'll accept.
– anonymous2
4 hours ago




Oh, of course. Duh, my stupid. Sorry, and thanks! If you have time to post a complete answer, I'll accept.
– anonymous2
4 hours ago












Crazy how it is; now it's clicked I can't believe I didn't get it before. :)
– anonymous2
4 hours ago




Crazy how it is; now it's clicked I can't believe I didn't get it before. :)
– anonymous2
4 hours ago










1 Answer
1






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3
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Both are linear with a $C_infty$ axis, but $D_infty h$ has a center of inversion and $C_infty v$ does not.






share|improve this answer






















  • And that all-important $sigma_h$ mirror plane perpendicular to the rotation axis, giving the name of the point group.
    – Geoff Hutchison
    15 mins ago










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1 Answer
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1 Answer
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active

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active

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up vote
3
down vote













Both are linear with a $C_infty$ axis, but $D_infty h$ has a center of inversion and $C_infty v$ does not.






share|improve this answer






















  • And that all-important $sigma_h$ mirror plane perpendicular to the rotation axis, giving the name of the point group.
    – Geoff Hutchison
    15 mins ago














up vote
3
down vote













Both are linear with a $C_infty$ axis, but $D_infty h$ has a center of inversion and $C_infty v$ does not.






share|improve this answer






















  • And that all-important $sigma_h$ mirror plane perpendicular to the rotation axis, giving the name of the point group.
    – Geoff Hutchison
    15 mins ago












up vote
3
down vote










up vote
3
down vote









Both are linear with a $C_infty$ axis, but $D_infty h$ has a center of inversion and $C_infty v$ does not.






share|improve this answer














Both are linear with a $C_infty$ axis, but $D_infty h$ has a center of inversion and $C_infty v$ does not.







share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 2 hours ago









MaxW

14.3k12156




14.3k12156











  • And that all-important $sigma_h$ mirror plane perpendicular to the rotation axis, giving the name of the point group.
    – Geoff Hutchison
    15 mins ago
















  • And that all-important $sigma_h$ mirror plane perpendicular to the rotation axis, giving the name of the point group.
    – Geoff Hutchison
    15 mins ago















And that all-important $sigma_h$ mirror plane perpendicular to the rotation axis, giving the name of the point group.
– Geoff Hutchison
15 mins ago




And that all-important $sigma_h$ mirror plane perpendicular to the rotation axis, giving the name of the point group.
– Geoff Hutchison
15 mins ago

















 

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