I'm studying about molecular symmetry and its representations. Today I got a little confused about the planar $\ce{N2H2}$ molecule.

It looks like this

enter image description here

The basis in the example was chosen as $\Delta r_1, \Delta r_2$, i.e. lengths of N-H bonds. Then, it was stated, that the molecule belongs to $\mathrm C_{2h}$ point group, which I agree with.

The problem is, the example stated, that the reducible representation is

\begin{array}{|l|l|l|l|} \hline {} & \mathrm E & \mathrm C_2 & i & \sigma_h\\ \hline \Gamma & 1+1=2 & 0+0=0 & 0+0=0 & 1+1=2\\ \hline \end{array}

And here I don't understand, what's going on. I understand, that $\mathrm E$ and $\sigma_h$ are 2, as both $\Delta r_1$ and $\Delta r_2$ stay the same after those operations, i.e. bond length doesn't change.

But, bond lengths don't change after $\mathrm C_2$ and $i$ either, so why are there zeros in the table?

I'd understand it if we used vectors of Cartesian coordinates on every atom as a basis, but in this case, I'm quite lost.

Could you, please, explain it to me?

The example is taken from this presentation (slide 18).

  • $\begingroup$ $r_{1} \rightarrow r_{2}$ and $r_{2} \rightarrow r_{1}$, so there's no $\pm 1$. It's a different object. $\endgroup$
    – Zhe
    Sep 27, 2018 at 14:01
  • $\begingroup$ @Zhe Would you mind to look at the presentation and explain it in a more detail? It would help me tremendously. $\endgroup$
    – Eenoku
    Sep 27, 2018 at 14:02
  • $\begingroup$ I think you are confused because you're trying to just look at the bond length as a number, but the bond is an object. When rotate it, it goes somewhere else, whereas if you look at the identity or the reflect, it stays in the same place. $\endgroup$
    – Zhe
    Sep 27, 2018 at 14:07
  • $\begingroup$ @Zhe Ok, so they think about it as bond itself, not only its length, if I understand you correctly? $\endgroup$
    – Eenoku
    Sep 27, 2018 at 14:10
  • $\begingroup$ If that's the reducible representation you're showing me, then yes. $\endgroup$
    – Zhe
    Sep 27, 2018 at 14:14

1 Answer 1


Instead of using the $\Delta r$ as in your figure redraw it with a vector shown as an arrow parallel to each NH bond and label them 1 and 2. You know from the point group that the NH bonds are the same length and at the same angle so these vectors are similar but point in different directions.

Operate on the whole diagram including arrows (with labels) according to the operations in the point group. If, after an operation the figure is indistinguishable from your starting point, then count $1$ for each vector otherwise count $0$.

In your figure the point in the point group is at the centre of the NN bond. The $C_2$ axis points out of the page centred at this point. The $\sigma_h$ plane is the plane of the figure.

The identity, which is an 'I exist' or 'do nothing' operation count 2. The $C_2$ operation is rotation by 180 degrees and so interchanges the vectors 1 and 2 so is not indistinguishable and so each vector counts 0. Similarly for the inversion $i$ through the centre. Reflection in the mirror plane (the plane of the image) leaves the molecule indistinguishable so counts 2. Thus you have the reducible representation you give in your question.


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