Carbon Dioxide has a degree of freedom of 6, yet from what I understand, it is a linear molecular shape. Therefore, the number of axis of linear movement is 3 (x,y,z) but there are only two axis of rotational movement since the center of mass essentially does not rotate on one of the axis due to its linear shape. Yet I'm pretty sure my textbook is saying that carbon dioxide has a degree of freedom of 6 when that should be wrong. Can someone explain?

  • $\begingroup$ There are 3N degrees of freedom (trans+rot+vib) for all molecules, regardless of their geometry, so maybe you want to specify what types of degrees of freedom you are looking at. For carbon dioxide, there are 3 translational, 2 rotational, and 4 vibrational degrees of freedom. chemwiki.ucdavis.edu/Core/Physical_Chemistry/… $\endgroup$ Commented Apr 16, 2016 at 21:07
  • $\begingroup$ Thank you, but why did my textbook say it only had 5? I just looked it up on wikipedia and the formulas and degree assignments dont match up. And some places say the same thing as my textbook and some say different. I am so confused. $\endgroup$
    – phi2k
    Commented Apr 16, 2016 at 21:11
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    $\begingroup$ Look, I can't speak for the author of whatever textbook you're using. At least give some context.. Which degrees of freedom are you trying to consider? Exactly what is the problem? Are you trying to calculate the constant-volume heat capacity using equipartition or something? If you are, then the vibrational modes don't contribute to the heat capacity (they aren't excited at normal temperatures), so you only have 3 trans + 2 rot = 5 degrees of freedom for a heat capacity $C_{v,m} = 5R/2$. $\endgroup$ Commented Apr 16, 2016 at 23:28
  • $\begingroup$ Thanks, that last part was actually what I needed. Thanks for clearing it up! $\endgroup$
    – phi2k
    Commented Apr 17, 2016 at 21:28
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    $\begingroup$ No problem. It's always a good idea to be clear exactly what you're talking about - if the term "degree of freedom" is not qualified any further, there are always $3N$ of them, including all trans/rot/vib ones. Depending on the specific application, some of them may not be relevant, so just make sure you know what's going on. $\endgroup$ Commented Apr 17, 2016 at 21:30

1 Answer 1


There are always 3N total independent degrees of freedom for a molecule, where N is the number of atoms. These come about because when each atom moves, it has three independent degrees of freedom: its position in each of the x, y, and z directions.

Now, having independent degrees of freedom for each atom isn't all that useful. Instead, we can make combinations of different degrees of freedom. The important thing when doing so is that the number of independent degrees of freedom are preserved: it's just that what a particular degree of freedom does to the atoms changes.

The standard breakdown of degrees of freedom subtracts out global movement in each of the three directions. So you have 3N total degrees of freedom, but you can set aside 3 of them as translation of the whole molecule in each of the x, y and z directions, leaving (3N-3) degrees of freedom.

Likewise, it's standard to subtract out the whole molecule rotation. For most larger molecules, there's three different degrees of rotational freedom: rotation around each of the x, y, and z directions. But for linear molecules like $\ce{CO2}$, one of those rotations (around the axis of the molecule) doesn't actually change the position of the atoms. Therefore it's not a "degree of freedom" which counts against the 3N total. So while for non-linear molecules there are (3N-3-3) = (3N-6) degrees of freedom which are independent from the global rotational and translational ones, for linear molecules there are (3N-3-2) = (3N-5) degrees of freedom which are independent from the global rotational and translational ones. -- A quick clarification. The reason why we ignore this rotation is not because the center of mass doesn't move. The center of mass doesn't move for any of the global rotations: in the typical assignment of degrees of freedom the axis of rotation goes through the center of mass. Instead, the reason the rotation is ignored is that none of the atoms move due to the "rotation".

So since $\ce{CO2}$ has three atoms and is linear, it has 3*3 - 5 = 4 degrees of freedom which are independent of the global rotation and translation. We call these the vibrational modes. There's different ways you can decompose them, but the most useful one is 1) the symmetric stretch (both oxygens going out and in at the same rate while the carbon sits still), 2) the asymmetric stretch (the carbon going back and forth while the oxygens sit more-or-less still), and 3&4) two different out-of plane bending modes. (If the molecule is aligned along the x-axis, one each where the carbon moves in the y and z directions.)

That's it for $\ce{CO2}$: any movement you make - any position you put the atoms in - can be thought of as a combination of those nine movements (three global translations, two global rotations, and four internal vibrations/bending).

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    $\begingroup$ By the way. if that still isn't making sense with respect to your textbook, it would help to edit your question to include an exact quote of the sentence/paragraph you think is in conflict. $\endgroup$
    – R.M.
    Commented Apr 17, 2016 at 19:32

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