0

How to determine the characters of px atomic orbital in $\text{C}_{\text{4v}}$ point group? I understand (px,py) is a set of basis function of E, and px and py are mixed by $\text{C}_4$ and $\sigma_{\text{d}}$. However, based on the figure below, how can I determine the characters of only px orbital? Are the characters the same as those of E?

px

meTchaikovsky
  • 273
  • 2
  • 14
  • Operate on the orbital with each of the symmetry operations, $\sigma_v$ etc and compare them with the irreps for each symmetry species in the point group, count 1 for unchanged , -1 if changed, see https://molecule-viewer.com/ for many examples. A few molecules in each point group also have orbitals included. – porphyrin Nov 01 '22 at 08:09
  • @porphyrin Yes, I understand this, but for the px orbital in the figure, $\sigma_\text{d}$ will transform it into py orbital, does this count as zero? Also for the two $\sigma_\text{v}$ operations, one character is 1 and the other is -1, so for the overall $2\sigma_\text{v}$ operation in the character table, does it count as 0? – meTchaikovsky Nov 01 '22 at 08:20
  • Yes zero, also look at the third column in a point group, see entries for x, y, z and $R_x $ etc so you can check your answers. – porphyrin Nov 01 '22 at 09:35
  • @porphyrin Thanks. However, for $C_{4v}$ point group $p_x$ is a basis function of the doubly degenerate irreducible representation, is it correct to consider how only one of the basis functions transform? Since the calculators to decompose reducible representatons just return error, if I consider $p_x$ as a reducible representation. – meTchaikovsky Nov 18 '22 at 13:24
  • see the last part of this answer for a worked example. https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613 , a fuller example is given by R. Carter, chapter 3 'Molecular Symmetry and group theory' publ Wiley. – porphyrin Nov 18 '22 at 15:44
  • Since the question has received 0 answers or votes here, I wonder if you considered asking it at MMSE? – user1271772 Aug 08 '23 at 02:28

0 Answers0