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The ground states of high-spin octahedral and tetrahedral Co(II) complexes are, respectively

  • A.\(^{4}\)T\(_{2g}\) and \(^{4}\)A\(_{2}\)
  • B.\(^{4}\)T\(_{1g}\) and \(^{4}\)A\(_{2}\)
  • C.\(^{3}\)T\(_{1g}\) and \(^{4}\)A\(_{2}\)
  • D.\(^{4}\)T\(_{1g}\) and \(^{3}\)T\(_{1}\)

Answer: B

The ground terms are 4T1g (octahedral) and 4A2 (tetrahedral).

Co(II) is d7, and the high-spin free ion has the ground term 4F (three unpaired electrons, quartet). In an octahedral field a d7 F term splits into T1g, T2g and A2g, with T1g lowest (t2g5eg2). In a tetrahedral field the splitting order is reversed, as for d3 in an octahedron, so A2 is lowest (e4t23); there is no g label without a centre of symmetry.

4T2g is an excited state of the octahedral ion. Any triplet (3T) term implies two unpaired electrons, which high-spin d7 does not have.

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