(i) Name the element of 3d transition series which shows maximum number of oxidation states. Why does it show so?
(ii) Which transition metal of 3d series has positive `E^(@)(M^(2+)//M)` value and why?
(iii) Out of `Cr^(3+)` and `Mn^(3+)`, which is a stronger oxidising agent and why?
(iv) Name a member of the lanthanoid series which is well known to exhibit +2 oxidation state.
(v) Complete the following equation :
`MnO_(4)^(-)+8H^(+)+5e^(-)to`

Explanation

(i) Mn shows the maximum number of oxidation states, from +2 to +7. This is because it can lose or share maximum number of electrons `(3d^(5)4s^(2))`.
(ii) Copper has the positive value for `E^(@)(M^(2+)//M)`.
This is because the high energy to transform Cu (s) to `Cu^(2+)` is not balanced by its hydration energy.
(iii) Out of `Cr^(3+)` and `Mn^(3+),Mn^(3+)` is a stronger oxidising agent. That is `Mn^(3+)` has a tendency to get reduced to `Mn^(2+)`. This is because `Mn^(3+)` has d-orbital configuration `3d^(4)` while `Mn^(2+)` has a more stable half-filled configuration `3d^(5)`.
On the other hand, `Cr^(3+)` has no tendency to act as oxidising agent. In other words, it has no tendency to change to `Cr^(2+).Cr^(3+)` has d-orbital configuration as `d^(3)`. These three electrons occupy the three `t_(2g)` orbitals (splitting of d-orbitals takes place). It is again half-filled stable configuration. Thus, `Cr^(3+)` has no tendency to act as oxidising agent.
(iv) Sm and Eu exhibit +2 oxidation state.
(v) `MnO_(4)^(-)+8H^(+)+5e^(-)to Mn^(2+)+4H_(2)O`