Problem Statement

An isotopic variant of the molecule SF$_6$ is S$^{19}$F$_4^{18}$F$_2$, with the two $^{18}$F nuclei oriented axially, so that the $^{18}$F-S-$^{18}$F angle is 180°and the four $^{19}$F nuclei form a square in a plane perpendicular to the $^{18}$F-S-$^{18}$F axis. All of the bond lengths are equal. In the figure, the regions surrounding the $^{18}$F nuclei are shown in blue, and the regions around the $^{19}$F nuclei are shown in green.Two of the moments of inertia of S$^{19}$F$_4^{18}$F$_2$ are identical, and the third is slightly larger: I$_A$= I$_B$ = $2.9942 ×10^{-45}$ kg m$^2$ and I$_C$= $3.0749×10^{-45}$ kg m$^2$. Wikipedia describes SF$_6$ as an “extremely potent greenhouse gas.”

What is the symmetry number of S$^{19}$F$_4^{18}$F$_2$?

I have considered it and think there is one 4-fold axis, and two 2 fold axes, but still unsure if that is right

  • $\begingroup$ Welcome to Chemistry Stack Exchange! Please edit your question to show your efforts at solving the problem $\endgroup$
    – jezzo
    Commented Mar 19, 2020 at 0:04
  • $\begingroup$ I made one edit, but haven't really been able to get anywhere as I am very confused! $\endgroup$
    – Lauren
    Commented Mar 19, 2020 at 0:23
  • 2
    $\begingroup$ One way to think about it is: Start from SF$_6$ without any isotope labeling, and see what point group you get. What symmetry operations do you lose when you change the two axial fluorines to $^{18}$F? $\endgroup$
    – jezzo
    Commented Mar 19, 2020 at 0:37
  • 3
    $\begingroup$ See chemtube3d.com/octahedral-sulfur-hexafluoride-sf6-symmetry for exploring the O$_h$ symmetry of the untagged (all $^{19}$F SF$_6$. Which operations do you lose when you do the replacement I mentioned above? $\endgroup$
    – jezzo
    Commented Mar 19, 2020 at 3:37
  • 1
    $\begingroup$ If you assume that your molecule is of the form $\ce{AB4C2}$ then one form is that shown in your figure and belongs to the $D_{4h}$ point group, the other has the two blue atoms at 90 deg to one another and belongs to $C_{2v}$ $\endgroup$
    – porphyrin
    Commented Mar 19, 2020 at 10:08


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