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

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?

• 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 $D_{\infty h}$ . Nov 6 '18 at 13:48
• 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? Nov 6 '18 at 14:02
• Why, many. Think of all these perpendicular $C_2$ axes, to begin with. Nov 6 '18 at 14:06
• Oh, of course. Duh, my stupid. Sorry, and thanks! If you have time to post a complete answer, I'll accept. Nov 6 '18 at 14:08
• Crazy how it is; now it's clicked I can't believe I didn't get it before. :) Nov 6 '18 at 14:11

Both are linear with a $$C_{\infty}$$ axis, but $$D_{\infty h}$$ has a center of inversion and $$C_{\infty v}$$ does not.
• And that all-important $sigma_h$ mirror plane perpendicular to the rotation axis, giving the name of the point group. Nov 6 '18 at 18:41
• @GeoffHutchison - Yes but for a linear molecule you don't need to identify a center of inversion, $i$, and a mirror plane perpendicular to the rotation axis ($\sigma_h$). For a linear molecule, if the center of inversion exists so must the perpendicular mirror plane, and if the perpendicular mirror plane exists so must the center of inversion.