Explain why a solution of [MnO4]- is very dark in colour, but a solution of [MnF6]4- is very pale.
In MnO4-, the oxidation no. of MN is +7 while in (MnF6)4-, the oxidation no. is +2. In the second case.. it looses its 4s2 e- and has 3d5 e- in the outermost orbit..[Absorption of visible light by this ion is accomplished only by a spin-forbidden transition in which one of the d electrons must pair with another, to give the atom a change in spin of two units. The unlikeliness of such a transition is seen in the uniformly pale and almost colorless nature of Mn(II) compounds relative to other oxidation states of manganese..] this part is frm wiki
sweeeet, thanks. i'm still a little confused about the spin forbidden transition. do you know if that means it's going from a g -> g (or u->u) orbital?
more the no. of unpaired e- paler is the colour
how does that tie in with the forbidden transition?
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