The d orbital electron configuration of octahedral complexes can either be described as high- spin with the maximum possible number of unpaired d-electrons, or low-spin containing one or more paired d-electrons. [Fe(H20)s]2 is a high-spin octahedral complex. What is its spin- state (S-?)? Draw a d-orbital splitting diagram for this complex and fill it with the appropriate number of electrons. Where does the final electron go in this diagram? If you were to oxidize this molecule do you think this would affect the bond lengths? Explain

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Explanation:

Fe(H20)6]2+ is a high spin octahedral complex because water is weak field ligand. A high spin complex has a maximum number of unpaired d electrons.

The spin state of Fe(H20)6]2+ is S=2. The final electron goes into an eg orbital. If the metal is oxidized to Fe^3+, the bond lengths decreases. For an oxidation of M2+ complex to M3+, the M3+L bonds will be shorter due to the higher charge density on the metal. Since the occupation of the eg orbitals in both complexes is the same it then follows that that the difference in the bond lengths must be due to the charge alone.

the answer is c i believe