The elements of 3d transition series are given as :
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following :
(i) Write the elements which shows maximum number of oxidation states. Give reason.
(ii) Which element has the highest m.p. ?
(iii) Which element shows only +3 oxidation state ?
(iv) Which element is a strong oxidising agent in +3 oxidation state and why ?

Explanation

(i) Mn shows the maximum number of oxidation states viz., +2, +3, +4, +5, +6 and +7. This is because Mn has electronic configuration `3d^(5)4s^(2)`. Thus there are 7 electrons that can be shared. If it shares (loses) `4s^(2)` electrons, we get +2 oxidation and with successive sharing of d electrons we get other oxidation states.
(ii) Cr has the highest m.p. This is because of the involvement of unpaired electrons in d and s-orbitals to form strong metallic bonds. This number is maximum in Cr, `3d^(5)4s^(1)(5+1=6)`.
(iii) Sc shows only +3 oxidation state. It has the configuration `3d^(1)4s^(2)`. Thus only 3 electrons can be lost.
(iv) Mn is a strong oxidising agent in +3 oxidation state. The change from `Mn^(3+)` to `Mn^(2+)` results in half-filled `(d^(5))` stable configuration.