Determine the CFSE and OSSE of Co(II) for the weak field case in both Oh and Td

Question

Determine the CFSE and OSSE of Co(II) for the weak field case in both Oh and Td symmetry in Dq units. Based on these calculations, which structure would you expect for the chloride complex?

Explanation / Answer

Tetrahedral complexes are always high spin or weak field => maximum CFSE can only be 12 Dq = > even smaller when converted to octahedral field equivalent by multiplying the value with 4/9 (CFSE Td = 4/9 CFSE Oh ) Therefore the CFSE is rarely important for tetrahedral 7 complexes as it is in octahedral complexes In fact, when comparing a given ion in a tetrahedral field with the same ion in octahedral hole, the ion is at least as stable – usually more so - in the octahedral hole => The energy which always favors the octahedral with respect to tetrahedral coordination is called the octahedral site stabilization energy (OSSE) For d1, d2, d5, d6 and d7 configurations, the advantage of the octahedral in weak field or high spin arrangement is little or nothing. However, d3 and d8 configurations are strongly favored to be octahedral => The OSSE for low spin is cut in half ? Co2+ does not favor strong fields and spin pairing because of its lower charge ? Co2+ complexes are high spin octahedral with ligands such as NH3 and H2O and they tend to be unstable with respect to oxidation ? High spin octahedral complexes are not particularly stabilized compared to tetrahedral coordination for Co3+ ? so the halides and pseudohalides of Co complexes are [CoCl4]-2, [CoBr4]-2 and [Co(NCS)4]-2 tetrahedral

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