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.
Answer
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.
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