# Using symmetry and group theory arguments to explain iron(II) in a tetrahedral crystal field

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?