I understand that the complex [Re2Cl8]2- has a bond order of 4 between the ruthenium ions. The bond comprises of sigma interactions between Dz2 orbitals, two sets of pi interactions between dxz and d yz, and delta interactions between the dxy orbitals of the two ruthenium ions. My textbook says that the dx2-y2 orbitals remain for the metal ligand bonding. How do you fit 16 electrons from 8 ligands into two orbitals, I thought you could only fit two electrons in an orbital. Are the dx2-y2 orbitals forming hybrid orbitals with the empty 4s orbitals ?enter image description here

  • $\begingroup$ Eight of the 16 electrons are in the quadruple bond. $\endgroup$ – gsurfer04 Dec 2 '17 at 14:03
  • $\begingroup$ If you understand how a normal triple bond between 2 carbons has a bond order of 3, then it should be pretty clear why adding a $\delta$ bond in addition to this creates a bond order of 4. $\endgroup$ – Zhe Dec 2 '17 at 15:19
  • $\begingroup$ Zhe, sure I understand that part, but its not quite what im stuck on $\endgroup$ – George Sandle Dec 3 '17 at 0:03
  • $\begingroup$ gsurfer04, yeah I agree that 8 electrons are in the quadruple bond, but we have to account for 24 electrons, 4 from each rhenium ion and 2 from each of the 8 ligands. If the dz2, dxz, dyz and dxy are used to form the quadruple bond, where are the remaining 16 electrons going, as all we have left over are two dx2-y2 orbitals which can hold a total of 4 electrons. So where do we put the other 8? $\endgroup$ – George Sandle Dec 3 '17 at 0:07

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