If you are predicting Cr(III)-aquo complex to form a tetrahedral complex, you would need to justify that. Especially, when the OSSE table shows d3 has one of the highest OSSE. 6. Make distinctions between Oh field and Oh point group. Not all molecules in an Oh field will have center of symmetry as the point group is not Oh. As you did the Character Table problem for [Ru(en)3]2+ complex and identified that the point group there is D3, same analysis can be made for the en complexes of Cu2+, Cr3+, and Mn2+. In these cases, the complex can still be Oh geometry but will not have Oh point group and Laporte rule will not apply. Any point group that lacks center of symmetry will relax Laporte rule (the complex does not need to become Td from Oh shape for that, especially, if OSSE is very high, this geometry change is unlikely. Moreover, en being bidentate ligand, trying to make Td complex will cause a lot of strain to the CH2-CH2 in en and I will presume it will not make Td complex). 7. d-d bands appear in the 400-800 nm range. There can be more than one d-d bands