# ligand pi system irreducible representation for trigonal bipyramidal transition metal complex

I am considering a D$_{3h}$ trigonal bipyramidal transition metal complex MX$_5$. I have found that the irreducible representation for the ligand s and $\sigma$ systems is 2A$_1'$+A$_2''$+E$'$. I have \begin{array}{|c|c|c|c|c|c|} \hline &E& 2C_3 & 3C_2 & \sigma_h & 2S_3&3\sigma_v\\ \hline \Gamma_s&5& 2& 1&3&0&3\\ \hline \Gamma_\sigma&5& 2& 1&3&0&3\\ \hline \Gamma_{\pi+\pi'}&10&0&-2&0&0&0\\ \hline \end{array} However, I think that my $\Gamma_{\pi+\pi'}$ is incorrect since I get fractional values for coefficients for the irreducible representations when I try to reduce it to irreducible representations. But from sketching the trigonal pyramidal complex, I cannot see how my $\chi(R)$ are wrong for the pi system representation, $\Gamma_{\pi+\pi'}$. Any help would be appreciated.

## 1 Answer

I believe you have the wrong character for the $\hat{C_3}$ rotation. None of the p orbitals in the equatorial positions remain in place so they all contribute zero. The axial p orbitals are rotated by $120^°$ which means they each contribute $\cos(120^°)=-\frac{1}{2}$. Since there are two of them at each of the two axial position, $\chi(\hat{C_3})=-2$. Using this new value for the character (assuming all the others are right, which they seemed to be), I obtained: $a_2',a_2'',2e',2e''$

• Thank you! I did not notice that the rotated orbitals work like that (although I should have, since they are part of a basis)... for some reason, I thought each orbital must count as 1, 0, or -1. – HBHSU Mar 10 '18 at 5:34
• please can you explain '..two of them at each ..' do you mean rotation by $+120$ and $-120$ ? – porphyrin Mar 10 '18 at 10:05
• Two perpendicular coplanar p orbitals that are part of the pi system at each axial ligand. – HBHSU Mar 10 '18 at 14:56
• @porphyrin there is a px and py orbital each axial end (assuming the pz the one that forms a sigma bond with the central atom). When we perform the C3 rotation, they are both shifted by 120° from their original position. – Tyberius Mar 10 '18 at 15:41
• @HBHSU it is usually glossed over at least when I have seen it taught. I think it would be much clearer if they said you get a character of $0$ if it moves away and a character of $\cos(\theta)$ if it rotates in place. – Tyberius Mar 10 '18 at 15:46