# What is the rotational symmetry number used to calculate the rotational entropy from an ORCA frequency calculation?

The quantum chemistry software ORCA gives the following warning on frequency calculations:

CAUTION: The rotational entropy is not quite correctly treated here
because it includes a symmetry number that is not yet correctly
implemented in ORCA!
For a nonlinear molecule the correct rotational entropy is:
S(rot) = R*(ln(qrot/sn)+1.5)
R    = 8.31441 J/mol/K = 1.987191683e-3 kcal/mol/K
qrot = 4922043.1563766
sn is the rotational symmetry number. We have assumed 3 here
if it is different for your molecule then you should correct
the printed rotational entropy by manually evaluating the equation
as given above

For convenience we print out the resulting values for sn=1 - 12:
sn= 1  qrot/sn= 4922043.1564 T*S(rot)=    10.02 kcal/mol T*S(tot)=    34.72 kcal/mol
sn= 2  qrot/sn= 2461021.5782 T*S(rot)=     9.61 kcal/mol T*S(tot)=    34.31 kcal/mol
sn= 3  qrot/sn= 1640681.0521 T*S(rot)=     9.37 kcal/mol T*S(tot)=    34.07 kcal/mol
sn= 4  qrot/sn= 1230510.7891 T*S(rot)=     9.20 kcal/mol T*S(tot)=    33.90 kcal/mol
sn= 5  qrot/sn=  984408.6313 T*S(rot)=     9.06 kcal/mol T*S(tot)=    33.77 kcal/mol
sn= 6  qrot/sn=  820340.5261 T*S(rot)=     8.96 kcal/mol T*S(tot)=    33.66 kcal/mol
sn= 7  qrot/sn=  703149.0223 T*S(rot)=     8.87 kcal/mol T*S(tot)=    33.57 kcal/mol
sn= 8  qrot/sn=  615255.3945 T*S(rot)=     8.79 kcal/mol T*S(tot)=    33.49 kcal/mol
sn= 9  qrot/sn=  546893.6840 T*S(rot)=     8.72 kcal/mol T*S(tot)=    33.42 kcal/mol
sn=10  qrot/sn=  492204.3156 T*S(rot)=     8.65 kcal/mol T*S(tot)=    33.36 kcal/mol
sn=11  qrot/sn=  447458.4688 T*S(rot)=     8.60 kcal/mol T*S(tot)=    33.30 kcal/mol
sn=12  qrot/sn=  410170.2630 T*S(rot)=     8.55 kcal/mol T*S(tot)=    33.25 kcal/mol


Now, I have got the following two questions:

1. What is the rotational symmetry number $s_n$, i.e. how is it defined?

2. What symmetry number $s_n$ would a molecule with no symmetry, i.e. $C_1$ point group have? The way I understand it, the rotational symmetry number reflects the rotational symmetry, e.g. $\ce{CHCl3}$ would have $s_n=3$ as there are three equivalent structures superimposed by the $C_3$ symmetry element. So that my guess is that $s_n = 1$ in that case. Is that correct?

• ad 1) The rotational symmetry number corrects for the fact that you have to take the Pauli principle into account when deriving the partition function. That is, it accounts for the spin statistical weights of the different rotational levels. – Paul Nov 16 '16 at 13:59
• @Paul "It accounts for the spin statistical weights of the different rotational levels" is not a really clear definition to me, sorry. In what way does it do this and what does it have to do with the Pauli principle? – logical x 2 Nov 16 '16 at 22:56
• On the link provided in the answer below the symmetry number is defined as: "The "rotational symmetry number or external symmetry number [...] is the number of unique orientations of the rigid molecule that only interchange identical atoms." I think that's a pretty clear-cut definition. – logical x 2 Nov 16 '16 at 22:59
• @Paul So given the definition from my comment above, I really don't see the connection to the Pauli principle, maybe you can enlighten me, I never really gotten warm with statistical thermodynamics : ) – logical x 2 Nov 16 '16 at 23:00
• Only for nonlinear molecules it is the case that the symmetry number is the order of the point group if only pure rotations (including E) are considered. $s=1$ for linear molecules of $C_{\infty\text{v}}$ and $s=2$ for $D_{\infty\text{h}}$ symmetry. – Paul Nov 18 '16 at 9:35

A key thing to remember is that the rotational symmetry number doesn't include reflections, so molecules in groups like $C_s$ still have $s_n = 1$ since there are no superimposable orientations achievable only with rotations.
• Thanks for the fast reply : ) The only thing I don't get then is why the ORCA people made $s_n=3$ the default setting, as obviously $s_n=1$ is by far the most common... – logical x 2 Nov 16 '16 at 13:55
• @ketbra Agreed, defaulting to $1$ would make more sense to me, too. Perhaps, though, the relative error associated with an incorrect $s_n$ is largest for small molecules, which have a much higher probability of $s_n > 1$, and thus they chose a compromise value. – hBy2Py Nov 16 '16 at 14:03