I am trying to figure out how to explain $1s \rightarrow 3d$ spectroscopic transitions for $\ce{Fe^{2+}}$ in $T_\mathrm{d}$ symmetry. These transitions make up the pre-edge region in K edge X-ray absorption spectroscopy (XAS). My goal is to rely on group theory and symmetry arguments as much as possible. Still, I am interested in the problem from a chemist perspective. I will first discuss what I already know and subsequently formulate my question.

$\ce{Fe^{2+}}$ is a $d^6$ ion with the ground state Russell–Saunders term $^5\!D$. In the event of a $1s \rightarrow 3d$ transition we find the final state configuration to be $1s^13d^7$. To find the corresponding terms we couple the core hole $s^1$ and the $d^7$ terms $$^2S \otimes \{{}^4\!F, {}^4\!P, {}^2\!H, {}^2G, {}^2\!F, {}^2\!P, {}^2\!D(2)\} = {{}^{5,3}F, {}^{5,3}\!P, {}^{1,3}\!H, {}^{1,3}G, {}^{1,3}\!F, {}^{1,3}\!P, {}^{1,3}\!D(2)}$$ From the $^5D$ ground state only the quintet $F$ and quintet $P$ terms can be reached following the quadrupole selection rule $\Delta S = 0$. For the isolated ion (without spin orbit coupling) we thus have two transitions: $^5\!D \rightarrow {}^5\!F$ and $^5\!D \rightarrow {}^5\!P$. So far so good. Now we can include the $T_\mathrm{d}$ crystal field.

In a $T_\mathrm{d}$ crystal field the $5D$ terms splits into $^5\!E_u$ and $^5T_{2u}$. Of these, the $^5\!E_u$ represents the lowest energy $(e_2)^3(t_2)^3$ configuration. Our initial state irreducible representations $\Gamma_i$ thus is $\Gamma_i = {}^5\!E_u$.

$$\Gamma_i = {}^5\!E_u$$

The final state irreducible representations $\Gamma_f$ are obtained by branching the $^5\!P$ and $^5\!F$ terms from $O_3$ to $O_h$: $$\begin{align} ^5\!F &\rightarrow {}^5\!A_{2u} \oplus {}^5T_{2u} \oplus {}^5T_{1u} \\ ^5\!P &\rightarrow {}^5T_{1u} \\ \Gamma_f &= {}^5\!A_{2u} \oplus {}^5T_{2u}\oplus {}^5T_{1u}(F) {}\oplus {}^5T_{1u}(P) \end{align}$$

The quadrupole transition operator in $T_\mathrm{d}$ symmetry is given by $\Gamma_{\hat{T}} = T_{2u} \oplus E_u$. Now I understand that a transition is possible if the matrix element $\langle f|\hat{T}|i \rangle$ is non zero. To see which final states irreducible representations $\Gamma_f$ are accessible through the transition operator $\hat{T}$ we can take the direct product $\Gamma_i \otimes \Gamma_{\hat{T}}$. This will give: $${}^5\!E_u \otimes (T_{2u} \oplus E_u) = {}^5\!E_u \otimes T_{2u} \oplus {}^5\!E_u \otimes E_u = {}^5\!A_{1u} \oplus {}^5\!A_{2u} \oplus{} ^5\!E_u \oplus {}^5T_{1u} \oplus {}^5T_{2u}$$

Please correct me if I already made some mistakes. From here I am not so sure how to continue. All the final state irreps $\Gamma_f$ are contained in the direct product $\Gamma_i \otimes \Gamma_f$.

  1. Can i conclude that from the $^5\!E_g$ ground state all of the final state irreps $\Gamma_f$ can and will be reached?
  2. If the only possible final states are $(e_2)^4(t_2)^3$ and $(e_2)^3(t_2)^4$ which $\Gamma_f$ irreps correspond to these configurations?
  3. It is well known that the K pre edge gains intensity for non-centrosymmetric ions with respect to centrosymmetric ions. This is because the $4p$ orbitals can mix with the $d_{xy}$, $d_{xz}$ and $d_{yz}$ orbitals. How does this fit the above storyline?
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    $\begingroup$ Since 24+ months have gone by with no answers or comments, I think it's fair to let you know about Matter Modeling Stack Exchange which also answers group theory questions, but make sure to delete the question from here if you're cross-posting there (or to at least put a note at the top of your question, saying that it's cross-posted). $\endgroup$ Commented Aug 8, 2023 at 2:31


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