(a) Cu is the only metal in the first transition series (3d series) that exhibits +1 oxidation state more frequently.
The electronic configuration of Cu is 3d10 4s1. So, when it loses one ‘s’ electron, it acquires a stable 3d10 configuration. Thus, it shows +1 oxidation state more frequently.
(b) We know that the colour of cations depend upon the number of unpaired electrons present in the d-orbital.
So, the electronic configuration of the following cations are:-
1. Sc (Atomic number 21) = [Ar]3d14s2 and Sc3+ = 3d04s0. As discussed, since d-orbital is empty, it is colourless.
2. V (atomic number 23) = [Ar]3d34s2 and V3+ = 3d24s0. As d-orbital is having 2 unpaired electrons, it undergoes d-d transition and gives green colour.
3. Ti (Atomic number 22) = [Ar]3d24s2 and Ti4+ = 3d04s0. As ‘d’ orbital is empty, it is colourless.
4. Mn (Atomic number 25) = [Ar]3d54s2 and Mn2+ = 3d54s0. As ‘d’ orbital has 5 unpaired electrons, it undergoes d-d transition and gives pink colour.
Thus, V3+ and Mn2+ ions are coloured in their aqueous solution due to presence of unpaired ‘d’ electrons.
Rate this question :