I observe the UV–Vis spectrum of $\ce{[Co(H2O)6]^{2+}}$ and there is a little peak at $\sim\pu{640 nm}$. By observing the Tanabe–Sugano diagram I see that the spin allowed transitions are:

$$ \begin{align} \mathrm{^4T_{1g}(F)} &\to \mathrm{^4T_{1g}(P)} \\ \mathrm{^4T_{1g}(F)} &\to \mathrm{^4A_{2g}} \\ \mathrm{^4T_{1g}(F)} &\to \mathrm{^4T_{2g}} \end{align} $$

The answer about why there is this little peak is that it is explained by the $\mathrm{^4T_{1g}(F)} \to \mathrm{^4A_{2g}}$ transition and it has to do with a two-electron transition. I can't understand why it is explained by this transition. Also, how do I know that it has to do with two electrons?

  • $\begingroup$ The transition seems to go from $t_{2g}^3e_g^4 \to t_{2g}^5e_g^2$ thus the photon shifts two electrons but the overall transition is still spin allowed. You might expect that such a transition would be very weak as the matrix element probably involves some product of terms involving a transient configuration. $\endgroup$
    – porphyrin
    Apr 24, 2018 at 17:11
  • $\begingroup$ @chemistrylove all of these allowed transitions could only be two photon or vibronically allowed since for one photon, g<->g is not allowed $\endgroup$
    – Tyberius
    Apr 25, 2018 at 12:58


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