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exam question

This makes little sense to me, both of the electron configurations are d6, and would presumably be low spin - therefore both reactants should be inert, and yet my notes suggest that an electron transfer would occur to form the products Co(II) and Ru(III). Surely this is a less stable arrangement?

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    $\begingroup$ You have false assumptions: just because something is d6 LS it doesn't mean inert. $\endgroup$ – Greg May 22 '16 at 14:03
  • $\begingroup$ Perhaps not, but surely the products, both with an unpaired electron are still less stable than the reactants. Perhaps the driving force is entropy? $\endgroup$ – Jack May 22 '16 at 14:11
  • $\begingroup$ Cobalt(II) is most certainly high-spin in this environment. Cobalt(III) is likely to be. However, I am at a loss for the answer … $\endgroup$ – Jan May 26 '16 at 11:44

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