I recently encountered a question asking whether [PtCl4]2- has square planar geometry. I know that Pt has an electronic configuration of [Xe] 5d9 6s1, in the given compound Pt2+ is there, so the E.C will be [Xe] 5d8 . But I'm confused whether Pt2+ will have four pairs of electrons or three pairs of electrons and two unpaired electrons in the 5d orbital. It seems like the latter should be true because paired electrons may be harder to take out. If that is true, is it always true as a general rule that unpaired electrons are always taken out instead of paired ones if both types are there (during the formation of the oxidised form of the element)?

  • 1
    $\begingroup$ If you're trying to draw a Lewis structure, you're going to have a bad time. You need a different model to be able to deal with transition metal complexes. $\endgroup$ – Zhe Mar 25 '19 at 13:52
  • $\begingroup$ No I'm using valence bond theory to model this structure. If the electronic configuration of Pt2+ was such that there were 4 pairs of electrons in the 5d orbitals. I know that there would be dsp2 hybridization and square planar geometry. What I'm not sure about is how the electronic configuration of Pt2+ will be (to use VBT). $\endgroup$ – karun mathews Mar 25 '19 at 17:25
  • 1
    $\begingroup$ How would you know what the hybridization is? Generally, it's not a great idea to use hybridization with transition metals. $\endgroup$ – Zhe Mar 25 '19 at 18:11
  • 1
    $\begingroup$ Your intuition that the electrons are all paired is correct, but the explanation is somewhat beyond what is provided by VBT. In VBT terms, the best way to explain it is that the one d orbital that is involved in bonding results in a low energy bonding orbital that is filled by electrons from Cl- and a high energy antibonding orbital (as happens with all bonding interactions). This antibonding orbital is higher in energy than the remaining four d orbitals, so the lowest energy arrangement is to put pairs in those 4 and leave the antibonding orbital empty. $\endgroup$ – Andrew Mar 26 '19 at 12:41
  • 2
    $\begingroup$ Hybridisation is useless for predicting geometries of transition metal complexes, and it also "predicts" some scientifically unjustifiable rubbish, so it's worse than just being useless - it's actually misleading. To predict the geometry or the electronic configuration you need more accurate theories. $\endgroup$ – orthocresol Mar 26 '19 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.