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Methyl chloride belongs to the $C_{3\mathrm{v}}$ point group. This is because it has a $C_3$ axis down the $\ce{Cl-C}$ bond.

However what I do not understand is how it has three vertical mirror planes? It has three planes down each of the $\ce{C-H}$ bonds, but don't these bisect a bond making them dihedral planes? Although I'm wrong I can't fathom why it isn't a $C_{3\mathrm{d}}$ point group. I have used a molecular viewer to visualise it but I still get the vertical mirror planes bisecting a bond...

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  • $\begingroup$ Confusingly, C3d groups don't exist in the nomenclature I learned. The "v" subscript is used for C groups and "d" is used for D groups, but the difference between the notations (or for that matter, the exact distinction between a vertical and dihedral reflection) has never been made clear to me. $\endgroup$ – chipbuster Nov 10 '14 at 20:45
  • $\begingroup$ In principal, the "vertical" mirror planes go along bond axes, and the "dihedral" mirror planes go between bond axes. (Consider, for example an octahedral complex.) In practice, $C_n$ point groups don't have dihedral "d" mirror planes, and plenty of exceptions exist (e.g., ferrocene, where the mirror planes are both "d" and "v" at the same time). $\endgroup$ – Geoff Hutchison Nov 11 '14 at 3:24
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You've correctly identified the $C_3$ axis that contains the $\ce{C-Cl}$ bond. Methyl chloride also has 3 planes of symmetry that contain the $C_3$ axis and each of the $\ce{C-H}$ bonds. Since these planes contain the $C_3$ axis they are referred to as $\sigma_{v}$ planes of symmetry. There are no other symmetry elements in methyl chloride so it belongs to point group $C_{3v}$ (n $\sigma v$ planes plus a $C_n$ axis).

There is no $C_{nd}$ point group. There is a $D_{nd}$ point group which contains a $C_n$ axis, n $\sigma_{v}$ planes (so far methyl chloride fits the description), but also n $C_2$ axis. Methyl chloride does not contain any $C_2$ axes, so it does not belong to the $D_{nd}$ point group.

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  • $\begingroup$ Thank you very much for your answer. I think I understand now! The $n$ $C_2$ axes would have to be perpendicular to the principal axes right? $\endgroup$ – RedPen Nov 11 '14 at 13:09
  • $\begingroup$ Yes, that correct. If you want to apply it to some molecules, allene is $D_{2d}$ and the chair form of cyclohexane is $D_{3d}$. $\endgroup$ – ron Nov 11 '14 at 14:06
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If I remember my symmetry correctly then a dihedral plane by definition is a vertical plane that bisects a perpendicular $C_2$ axis. Since we have a perpendicular $C_2$ axis then I think that would mean you label it $D_n$.

In this case there are no perpendicular $C_2$ axes so I think for that reason they are just vertical planes until such time that you they bisect a perpendicular $C_2$.

I don't know what would be the case if you had $C_n$ on the principal axis and then $m$ perpendicular $C_2$ axes, where $m\neq n$, maybe someone could shed some light? My guess would be that it wouldn't be $D$ but would qualify for the $\sigma_d$ plane.

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  • $\begingroup$ You can't have $m$ perpendicular $C_2$ axes where $m \neq n$ unless we're talking about a highly symmetric point group, like $I_h$ or possibly $T_h$.. I'd have to think about those. $\endgroup$ – Geoff Hutchison Nov 11 '14 at 3:26
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The picture shows the vertical symmetry planes along the principal $\ce{C3}$ axis in the $\ce{C_{3{\mathrm{v}}}}$ molecule $\ce{CH3CN}$ which has a $\ce{CN}$ group instead of Cl.

acetonitrile

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