# Symmetry representations for CH3CN

Can anybody share the symmetry labels for acetonitrile ($\ce{CH3CN}$), please? I mean I to assign the symmetry labels to orbitals of acetonitrile

• Welcome to chemistry.se! This seems to be a homework question according to our policy, please share your thoughts and attempts towards the solution. For formatting help visit the help center center and for more information take the tour. – Geoff Hutchison Dec 11 '14 at 15:20

It depends a bit on how you handle the $\ce{CH3}$ group. Usually, you assume free rotation, so it's effectively a sphere. In that case, $\ce{CH3CN}$ is linear and asymmetric. Since the two ends are different, that places it in the $C_{\infty v}$ point group.

Typically, though, we don't assign molecular orbitals using $C_{\infty v}$ or $D_{\infty h}$ representations. It's hard to do the math when using an infinite rotation axis, so we usually use $C_{2v}$ and $D_{2h}$ instead.

That is, you'll want to use the irreducible representations from $C_{2v}$ for acetonitrile.

• "$\ce {CH3}$ is linear and asymmetric." How can it be linear ? – Babounet Dec 11 '14 at 17:13
• Oh ok you've edited it ! Sorry for my shambles, I was looking for quoting you and emphazise "linear" but it seems to be not working in comments :x – Babounet Dec 11 '14 at 17:17
• In acetonitrile if you treat the methyl as spherical, then you have three linear spheres in the molecule, but of course it's asymmetric because CH3 and N are different – Geoff Hutchison Dec 11 '14 at 17:22
• Yes I get it after your edit since it's now clear that's your are treating the geometry of the molecule considering $\ce {CH3CN}$ in a row. But when it was only about $\ce {CH3}$ it mades me a bit confused. Thank you for the precision ! – Babounet Dec 11 '14 at 17:30

I am afraid I disagree with Geoff's assignment of acetonitrile to the point group $C_{\infty v}$ and "approximation" with $C_{2v}$, so I will make another attempt to answer this question. The way I have structured this answer, the resources I used, and the level of detail included follow a similar question on this site.

# Molecule structure

First you need to know the geometrical structure of your compound. This can be found easily from Wikipedia. The image is reproduced here.

# Assigning the point group

Then you need to find the point group of acetonitrile. This can be done by following flow charts, one of which is available here. While it may be a little difficult at the beginning, with some practise you can look at a molecule and determine its point group accurately, so it is worth working through the flow chart for a range of molecules until you are confident with this. There are other arrangements of this flow chart, but all flow charts should give you the same result, so use the one provided in your course if possible.

With the flow chart I have linked to, we go through these steps (if you have a flow chart organised differently, you will encounter the same steps, but likely in a slightly different order - check this is the case!):

• molecule is not linear
• molecule contains only one $C_3$ axis
• molecule DOES contain a proper rotation axis which is $C_3$ (rotation over methyl group)
• the highest rotation axis is $C_3$ but there are no perpendicular $C_2$ axes
• the molecule does NOT contain a horizontal reflection plane ($\sigma_n$). This can sometimes be confusing -- in general, a horizontal plane is perpendicular to the highest rotation axis.
• the molecules DOES contain 3 vertical reflection planes ($\sigma_v$).
• therefore the point group is $C_{3v}$.

# "Orbitals"

In your question, you said that you need to "to assign the symmetry labels to orbitals of acetonitrile". This is a little unclear, because you can mean that you want to assign symmetry labels to atomic orbitals of acetonitrile, or molecular orbitals of acetonitrile. I am going to assume that you need the molecular orbitals here. But how do we find the molecular orbitals?

Molecular orbitals are constructed by making combinations of atomic orbitals. Which means that, a logical next step is to consider all the valence atomic orbitals (VAOs) on all the atoms of acetonitrile.

# Atomic orbitals and symmetry sets

To help us talk about VAOs, I have put numbers on all the atoms.

Carbon, atoms 1 and 2: $2s$, $2p_x$, $2p_y$, $2p_z$

Nitrogen, atom 3: $2s$, $2p_x$, $2p_y$, $2p_z$

Hydrogen, atoms 4, 5 and 6: $1s$

When we try to tackle these types of problems, it is common to construct symmetry sets of atomic orbitals that would transform in the same way under the operations of the point group of the molecule. You will also notice that you cannot use any of the operations of the group to transform certain atomic orbitals on the same atom into each other (cannot transform $2s$ to $2p$, cannot transform $2p_x$ to $2p_z$, etc), so these should be considered as separate sets. The $p_x$ and $p_y$ orbitals are mixed by the rotations and reflections of the group, so we should consider them as one symmetry set.

In acetonitrile, there is only one nitrogen, so we have to consider as separate sets:

• $2s$
• $2p_z$
• $2p_x$, $2p_y$

Carbon 1 and carbon 2, though both carbons, cannot be transformed from one to another, and your chemical intuition should also tell you they are different carbons because they have different bonds attached to them, so they are also in separate sets of their own. The three hydrogen atoms, on the other hand, are related by rotation (a $C_{3v}$ rotation on the principal axis, so they can form one set of 3 hydrogen 1s orbitals.

# Assigning symmetry labels to each set of valence atomic orbitals

The usual trick at this point is to simply read off the orbitals that have corresponding symmetry labels, e.g. the $p_x$ orbital would transform as $x$ on the character table. This definitely works for any atom that lies on the origin point of the point group, but it is not completely obvious where you would take the origin point for putting a (right handed) axis system on acetonitrile. The "middle" of this molecule would appear to be somewhere between the two carbons, and that may seem a little awkward.

Remember that in all cases, the choice of origin and axis system is actually arbitrary, however you should pick the most convenient one. Now, for acetonitrile you can pick any point on the principal rotation axis ($C_{3}$) and it would work, because that is an inherent feature of the $C_{3v}$ point group. If this is not obvious, I suggest looking at the answer I referenced previously, and also working through any other examples that you can find.

Luckily for us, $C_{3v}$ has quite a small character table (i.e. it does not have many symmetry elements), so we can work through it and check. The character table of the point group $C_{3v}$ is shown below and can also be found here. The character tables for many common point groups can be found here.

## $s$ orbitals that lie on the principal axis

Now let's consider the $2s$ orbital or carbon atom 1. It should be fairly obvious that it transforms back to itself under all the operations of the point group (unchanged under identity $E$, any of the $C_3$ rotations, any of the $\sigma_v$ reflections), and so it transforms as $A_1$. It should also become very obvious that this is also true for the $2s$ orbitals of carbon atom 2 and nitrogen atom 3, even though these atoms are not symmetry-related, their atomic orbitals can have the same symmetry under the point group. This will be important when we come to construct molecular orbitals.

## $p$ orbitals that lie on the principal axis

For p orbitals, we need to impose an axis system. Conventionally, the principal axis is taken as the $z$ direction. You have to follow these conventions when doing symmetry analysis.

Let's consider the $2p_z$ orbital or carbon atom 1, which lies along the principal axis. This again transforms as $A_1$.

The $2p_x$ and $2p_y$ orbitals are a little tricky. They remain identical under the identity operator $E$, giving a character of 2. For the reflection operator $\sigma_v$, one of the three $\sigma_v$ planes would keep one identical while inverting the other one (giving zero), and the other two $\sigma_v$ planes give characters of zero, so the overall character for $3 \sigma_v$ is zero. The $C_{3v}$ is the difficult one, where you have to put vectors on the axis, rotate them, work out the geometry with trigonometry, form the $2 \times 2$ matrix and work out the trace (i.e. the character) is -1. Or you can use the fact that the character tables show (x,y) transform as the irreducible representation $E$.

Again, the same symmetry labels apply for all the atomic orbitals for all atoms that lie on the principal axis.

## Set of hydrogen $1s$ orbitals that do not lie on the principal axis

For these hydrogen $1s$ orbitals we examine explicitly how they behave under all operations of the point group.

• $E$: 3 (all three $1s$ remain unchanged from themselves)
• $C_v$: 0 (all three change and do not remain at the same position)
• $\sigma_v$: 1 (for any one operation, only 1 of the orbitals remain the same, the rest change)

This is a reducible representation. You need to reduce it to find the symmetry labels for the hydrogen $1s$ orbitals.

## Symmetry labels for all valence atomic orbitals

For the whole molecule of acetonitrile, the valence atomic orbitals give the irreducible representation $4 A_1 \oplus 4 E$. From the discussion above you should be able to deduce the exact symmetry label for any atomic orbital on any atom of the molecule.

# Molecular orbitals

You know that you need to generate as many molecular orbitals as atomic orbitals, and only the atomic orbitals of the same symmetry can mix. Molecular orbitals use lower case letters, so you expect two types of symmetry labels based on the irreducible representations of the VAOs. You should be able to say how many molecular orbitals of each type you expect, remembering that the irreducible representation $E$ is two-dimensional and indicates a degenerate pair of VAOs.

Symmetry analysis cannot give you the energy of these molecular orbitals, however, so you cannot know how they are ordered relative to one another.

• I have no problem with working acetonitrile as $C_{3v}$. Clearly if you treat the methyl as fixed, that would be correct. However, in most cases, when bonds have free rotation (e.g., terminal $\ce{-CH3}$ or $\ce{-OH}$ you really should approximate as a spheroid) IMHO. – Geoff Hutchison Dec 15 '14 at 4:42
• Reality, of course, is somewhere in between. Depending on the spectroscopic technique, you'll either observe behavior approximating $C_{3v}$ or $C_{\infty v}$. – Geoff Hutchison Dec 15 '14 at 4:43
• I'm afraid I disagree. I believe the normal modes you find using a $C_{3v}$ and $C_{\infty v}$ would have different symmetry inherently, as $C_{3v}$ has a $E$ irreducible representation while $C_{\infty v}$ has $E_1, E_2 ... E_n$, so a simple approximation would not give you normal modes with the same IR. I haven't gone through the whole normal mode analysis, so I can be wrong here. But that is my gut feeling! Of course, what you actually observe as normal modes may have accidental degeneracy as well, but that is a possibility with any symmetry analysis. – selkie222 Dec 15 '14 at 8:44
• For vibrational spectroscopy, if you used $C_{\infty v}$ instead, you will miss out on all the asymmetric modes. webbook.nist.gov/cgi/… – selkie222 Dec 18 '14 at 16:35