Calculate crystal field spliting energy in th following pair. (Fe H²O6]²+ and (Fe H²06]³+
Answers
The splitting of the d-orbitals into different energy levels in transition metal complexes has important consequences for their stability, reactivity, and magnetic properties. Let us first consider the simple case of the octahedral complexes [M(H2O)6]3+, where M = Ti, V, Cr. Because the complexes are octahedral, they all have the same energy level diagram: The Ti3+, V3+, and Cr3+ complexes have one, two and three d-electrons respectively, which fill the degenerate t2g orbitals singly. The spins align parallel according to Hund's rule, which states that the lowest energy state has the highest spin angular momentum.
For each of these complexes we can calculate a crystal field stabilization energy, CFSE, which is the energy difference between the complex in its ground state and in a hypothetical state in which all five d-orbitals are at the energy barycenter.
For Ti3+, there is one electron stabilized by 2/5 ΔO, so CFSE=−(1)(25)(ΔO)=−25ΔO
Similarly, CFSE = -4/5 ΔO and -6/5 ΔO for V3+ and Cr3+, respectively.
For Cr2+ complexes, which have four d-electrons, the situation is more complicated. Now we can have a high spin configuration (t2g)3(eg)1, or a low spin configuration (t2g)4(eg)0 in which two of the electrons are paired.High spin: CFSE=(−3)(25)ΔO+(1)(35)ΔO=−35ΔO
Low spin: CFSE=(−4)(25)ΔO+P=−85ΔO+P , where P is the pairing energy
Energy difference = -ΔO + P
