I have encountered with the following question:

The complex having zero crystal field stabilisation energy is:

(A) $\ce{[Mn(H_2O)6]^3+}$ (B) $\ce{[Fe(H_2O)6]^3+}$ (C) $\ce{[Co(H_2O)6]^2+}$ (D) $\ce{[Co(H_2O)6]^3+}$

Zero crystal field stabilisation energy will be the case when the metal has either $d^{10}$ configuration or $d^5$ configuration with every orbital having one electron. For Iron($\ce{Fe^3+}$) the d orbital is half filled. Now the problem is, water is in the middle of the spectrochemical series, i.e. it is moderately strong ligand. So, how to decide whether the d orbital will have the 'every orbital half filled' condition or not ?

  • $\begingroup$ In principle crystal field theory is an obsolete, empirically derived, incomplete concept/model, which gives tempting explanations that are not necessary right. It was combined with molecular orbital theory to form ligand field theory. In none of these complexes the $\ce{M-OH2}$ bonds will be of significant strength to cause the metal to adopt a low spin configuration. $\ce{Fe^3+}$ is your only choice because it had $d^5$ configuration. $\endgroup$ Nov 25, 2016 at 6:58
  • $\begingroup$ Trivalent hexa Aqua complexes of the first row transition metals don't go low spin until Cobalt 3+ (where the highly favourable t2g 6 configuration is reached). So the answer to your question is the iron complex. $\endgroup$
    – RobChem
    Nov 25, 2016 at 13:10
  • $\begingroup$ @RobChem Yes, that explains it. $\endgroup$
    – S R Maiti
    Nov 26, 2016 at 6:13
  • $\begingroup$ @Martin-マーチン I have read that the theory is now changed with the molecular orbital concept to give a more reasonable explanation. But it is in my syllabus. So, I have to read it nevertheless. $\endgroup$
    – S R Maiti
    Nov 26, 2016 at 6:15
  • $\begingroup$ By the way, I am wondering why Martin-マーチン and RobChem did not answer the question, but instead wrote it by comment. $\endgroup$
    – S R Maiti
    Nov 26, 2016 at 6:17

1 Answer 1


In the given example it is pretty simple to chose B as the correct answer, as it is the only option that could possibly create a spin forbidden transition $d^5$ or a $d^{10}$ configuration. For most situatons I have encountered it is sufficient to assume high-spin filling, until you encounter very stable configurations: completely full or half full orbitals (like $d^6$)


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