# How to get the reducible representations of the B-B and the B-Cl bond of B2Cl4

I just can't seem to get the correct answer to this symmetry question in my book. "Introduction to molecular symmetry J.S.Ogden."

I am trying to find the reducible representations of both the $\ce{B-Cl}$ bond and the $\ce{B-B}$ bonds in $\ce{B_2Cl_4}$ in the point group of $D_\mathrm{2d}$.
The correct answer for both is $\Gamma_{\ce{B-Cl}}=A_1+B_1+E$ and $\Gamma_{\ce{B-B}}=A_1$.
My reducible representations are:

• $\Gamma_{\ce{B-Cl}}=4,0,0,0,2$ for the $\ce{B-Cl}$ bond in the order $E, 2S_4, C_2, 2C', 2\sigma_\mathrm{d}$.
• $\Gamma_{\ce{B-B}}= 1,0,1,0,1$.

Clearly either of these are not correct but honestly I can't see why! Your help is much appreciated! (I worked them out based on the number of bonds that remain unshifted during the respective symmetry operation. I have also made a little model for myself in Gaussian but it didn't really help.

• Why do you think that your reducible representations are not correct? Have you tried reducing them? If not, you should try it and add the result and how you got to it to your question. Nov 16, 2014 at 2:30
• yes I have tried and they do not give the correct answer, So the $B-Cl$ bond representation gives me $A_1+B_2+E$ while the $B-B$ bond gives me fractional answers so I get a half $A_1$ which I know must be wrong. Both of these answers are wrong. I think my error lies in the determination of $\Gamma$ but two days later I still can't see where I'm wrong. Nov 16, 2014 at 10:25

There seem to be two problems: the first is that your reducible representation for the $\ce{B-B}$ bond is wrong but your reducible representation for the $\ce{B-Cl}$ bonds seems to be correct and the second is that your book is apparently completely wrong. If I didn't make a mistake and my second assumption is correct then maybe you should think about switching to another book as it seems to raise more questions than it answers. But I'm not working regularly with group theory, so my analysis of the problem might be faulty, but maybe someone else can support or refute my answer.

Ok, here is my analysis: First, make yourself familiar with the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group. Have a look at this video for example. Now, I looked at the effect each symmetry operation has on the bonds I'm looking at. For that I will use the following depiction of $\ce{B2Cl4}$: For the $\ce{B-B}$ bond I used an arrow to indicate the direction of the bond for better emphasis as I will need that soon. In the following I will assume that you are familiar with the way characters are related to your reducible representation and how you get those characters from analyzing the changes of your basis vectors (in this case the bonds) under the symmetry operations of the group since you seem to have already used it to determine the reducible representations in your question. Furthermore, I use the following notation: The reducible representation of the $\ce{B-Cl}$ bonds will be denoted as $\Gamma_{\ce{B-Cl}}$ and the character of the symmetry operation $C_{2}$ for this reducible representation will be denoted as $\Gamma_{\ce{B-Cl}}(C_{2})$.

Ok, let's see what the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group do to this structure, so that we can get the reducible representations: The $E$ operation does nothing to the structure, i.e. all four $\ce{B-Cl}$ bonds and the one $\ce{B-B}$ bond stay where they are. Thus you get a character of $4$ for $\Gamma_{\ce{B-Cl}}(E)$ and a character of $1$ for $\Gamma_{\ce{B-B}}(E)$.

Now, we come to the improper axis $S_{4}$. It rotates along the long the $\ce{B-B}$ bond and mirrors perpendicular to it: As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(S_{4}) = 0$. Since the mirror plane is perpedicular to the $\ce{B-B}$ bond the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(S_{4}) = -1$.

Next, we look at the principle $C_{2}$ rotation along the $\ce{B-B}$ bond. It converts your molecule to the following: As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}) = 0$. But as the rotation is conducted along the $\ce{B-B}$ bond axis the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}) = 1$.

Now, for the two $C_{2}^{\prime}$ axes which are perpendicular to $C_{2}$. One disects the dihedral angle between the dark red $\ce{Cl}$ and the cyan $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other disects the dihedral angle between the red $\ce{Cl}$ and the blue $\ce{Cl}$ - its effect on the structure is shown in the right picture: $\qquad \qquad \qquad$ Again the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}^{\prime}) = 0$. But now the rotation is conducted along an axis perpedicular to the $\ce{B-B}$ bond and so the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}^{\prime}) = -1$.

Finally there are the two $\sigma_{\mathrm{d}}$ planes which lie along the $C_{2}$. The one plane contains the dark red and the red $\ce{Cl}$ while disecting the angle betweeen the cyan and the blue $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other contains the cyan and the blue $\ce{Cl}$ while disecting the angle betweeen the dark red and the red $\ce{Cl}$ - its effect on the structure is shown in the right picture: $\qquad \qquad \qquad$ In this case either the red and dark red $\ce{Cl}$ atoms or the blue and cyan ones remain unchanged while the other $\ce{Cl}$ atoms change their places. So, two $\ce{B-Cl}$ bonds keep their location and direction which gives $\Gamma_{\ce{B-Cl}}(\sigma_{\mathrm{d}}) = 2$. The $\ce{B-B}$ bond axis lies within the mirror plane and so the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(\sigma_{\mathrm{d}}) = 1$.

So, in conclusion you end up with the reducible representations (I kept the order of the symmetry elements you chose, i.e. $E$, $2 \, S_4$, $C_2$, $2 \, C_{2}^{'}$, $2 \, \sigma_{d}$):

\begin{align} \Gamma_{\ce{B-Cl}} &= 4, 0, 0, 0, 2 \\ \Gamma_{\ce{B-B}} &= 1, -1, 1, -1, 1 \\ \end{align}

and they reduce to

\begin{align} \Gamma_{\ce{B-Cl}} &= A_1 + B_2 + E \\ \Gamma_{\ce{B-B}} &= A_2 \\ \end{align}

As you can see both results differ from the ones your book provides.

• This is so helpful thank you very much! I completely agree with your working and I think the only difference between my own answer and yours is that instead of putting -1 I erroneously put a zero instead. It must be a typo in the answers! Thank you so much for your effort :) Nov 16, 2014 at 13:52
• @RedPen Glad I could help. May I ask for what future purpose you are learning group theory? Is it only to understand the basics so that the meaning of symmetry labels, e.g. in transition metal complexes become clear, or are you going to really work with group theory, e.g. in quantum mechanics or spectroscopy? Nov 16, 2014 at 14:02
• At the moment I am a masters student in physical chemistry so I am just becoming familiar with the matrix formulation of symmetry, but I am hoping to study for a PHd in theoretical chemistry such as the development of new computational algorithms and similar simulation theory. My main interests are in understanding processes in photochemistry/intersections of excited states and the time evolution of systems. So I think I'd better learn quick haha. Thanks again! Nov 16, 2014 at 15:36
• I must say though I don't know too much about actual computational chemistry, I don't have a lot of experience with the different programmes in use, I'm much more familiar with the equations and theory behind it! :) Nov 16, 2014 at 15:38
• @RedPen Then I can recommend this book by Inui. It takes a more formal approach to group theory (more mathematically oriented) but it gives you a very solid foundation. Nov 16, 2014 at 15:39