CHEM-6 Midterm: Chem 6 Dartmouth Spring Exam 4S
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Sample final exam answers: for both these compounds, the formal oxidation state is mn(ii), which is d5. Since ncs is a weak-field, high-spin ligand (from the spectrochemical series), the electron configuration for the octahedral complex will be (t2g)3(eg)2, and it will have 5 unpaired electrons. The tetrahedral complex, which is also high-spin, will have electron configuration (e)2(t2)3 and it will also have 5 unpaired electrons. Since the magnetic susceptibility is a function of the number of unpaired electrons, you can"t tell these compounds apart in this way: (a) the cu(i) complex is d10 and colorless; the cu(ii) complex is d9 and blue. The colors are a result of d-d electronic transitions; none of these are possible in the d10 case, since there are no empty orbitals into which an electron can go. However, for the d9 case, such an excitation is possible. (b) for 1 febr63 , this is fe(iii), d5.
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