Question

e) Answer the questions using quantum selection rule and transition-types. i. [Mn(OH2)6]** is essentially colorless. Why is this? ii. Besides colorless (Mn(OH2)], the other hexaaqua first-row transition metal complexes are only faintly colored. Why is this? iii. [Cu"(L)4]?* salts are often quite dark in comparison. Why is this? iv. The [Mnacac)3] and [Fe"(bipy)s] - complexes are even darker than all the above complexes. What kind of transitions are responsible?

Answer 1

ci) [MMC 0 [Mn(H2O) J has a ds configuration & has high spin State 47 477 Therefore tramition of electron is both Laporte & forbidden & spin forbidden. Since the probability of election transition is neglt negligible due to sonbidden selection rule (Mn(H₂O), J2t complexes are usually colourless. host of the hexaqua first row complexes are faintly coloured because are octahedral and they have a centow symmetric environment. d-d transitions are Laporte forbidden , however for non-centrosymmetric environment dop mixing of orbitals take place which makes the d-orbital lose its identity as d-orbital and it becomes a mixed orbital making it Laporte allowed. But incaree of hexaaqua complexes this is he less likely because they are highly Symmetric They are slightly coloured out because of the spin selection as allowance (except Ma) & spin orbit coupling,