Why the Mn2+ gives pale pink colour where as MnO4(-) gives intensed colour in water, although both share same metal? About electron transition, there are certain selection rules. 1. Δn = ±1, ±2, ±3 … (n is the principle quantum number) 2. Δl = ±1 (l is the azimuthal quantum number) this explains that s – s , p – p, d – d, and d – s transitions are not allowed. Allowed transitions are s – p, p –s, d – p, p – d. 3. Δs = 0 this explains that the total spin value of the electron should not change during transition 4. Δml = 0, ±1 (ml = magnetic quantum number) • If obey all the selection rules, the colour will be high in intensity. • Number of rules obey α intensity of the colour In [Mn(H2O)]2+, it’s an octahedral complex. Total spin in the ground state = ½ + ½ + ½ + ½ + ½ = 5/2 Total spin in the exited state = ½ + ½ + ½ + 0 = 3/2 During the excitation, spin has been changed. So it violate important selection rule. i.e. Δs = 0 And it also violate the selection rule, Δl = ±1 because the colour arises due to the d – d transition. Therefore the intensity of the colour is very low. But in MnO4(-), Colour is not due to d – d transition Mn has +7 oxidation state in MnO4(-). All the d electrons are shared with Oxygen and it’s highly oxidized state Instead of d – d transition, charge transfer is occurred here. It is occurred by an electron come from the p orbital of Oxygen is gone to d orbital in Mn. So the energy is absorbed from the visible light. P – d transitions are occurred and is fully allowed. In MnO4(-) no violations of selection rules. Such transitions will have very high probability and the colour is high in intensity.
Posted on: Tue, 23 Jul 2013 06:50:15 +0000
Trending Topics
Recently Viewed Topics
© 2015