Chemqueries: Hybridization in outer orbital octahedral complex

Hybridization in outer orbital octahedral complex

Outer orbital octahedral complex:

        Let's take an example   [Fe(H2O)5(NO)]2+
  • In this complex ion oxidation state of Fe is +1, because NO exists in +1 oxidation state in complex of Fe and its valence shell electronic configuration is 3d6, 4s1
  • Magnetic moment measurement indicates that its experimental magnetic moment is 3.89 B.M. which corresponds to three unpaired electron in the d-orbital of complex ion.
  • The single NOstrong ligand has little tendency to pair up only two unpaired electrons.Since  H2O is a weak ligand therefore, it has no tendency to pair up electrons and none of the five 3d orbital is vacant. 
  • Therefore the  4s, 4p and two of the five 4d orbital(i.e.4dx2-y2  and  4dz)combine to give six sp3d2 hybrid orbital.
  • These hybrid orbital form bonds with six ligands by accepting six pairs of electrons, one pair from each of the six ligand.

  • Here as outer d-orbital is involved in hybridization it gives outer orbital octahedral geometry.

Note:

  • The octahedral complex of  d1, d2, and d3 metal cation are always inner orbital octahedral complexes whether the ligands are strong or weak.
  • The octahedral complex of d8, d9, and d10 metal cation are always outer orbital complexes either the lihands are strong or weak.
  • The complexes of d4, d5,d6 and d7 metal cation are outer orbital complex if the ligands are weak.

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