(a) (i) Mn exhibits highest oxidation states (upto +7 ) in the oxides, for example Mn2O7 in comparison to Mn fluorides, MnF4 as oxygen is able to form multiple bond with metal. And also fluorine does not have d-orbital.
(ii) Cr2+ is a strong reducing agent as its electrode potential is more negative. Also, it can loose on electron to become Cr3+ which has half-filled d-orbital which is its stable state.
(iii) Electronic configuration :
‌‌29‌Cu=[Ar]3d104 s1Cu2+ =[Ar]3d9
‌‌30‌Zn=[Ar]3d104 s2Zn2+ =[Ar]3d10
(i)

2MnO2

Manganese‌dioxide

+

4‌KOH

Potassium‌hydroxide

+O2

∆

───▸

2K2MnO4

Potassium‌manganate

+2H2O
(ii) Cr2O72−+14H++6I−→2Cr3+ +3I2+7H2O
OR
(i) Manganese exhibits greatest number of oxidation states as it is able to acquire half filled to completely filled configurations. Thus Mn can loose or share electrons from both the orbitals.
(ii) Cr has the highest melting point. As the number unpaired electrons increases upto d5 configuration, it results in the increase in the strength of metallic bonds. To break the metallic bond, significant energy is required thus Cr with highest number of unpaired electrons (6) has the highest melting point.
(iii) Sc shows only +3 oxidation state as if has too few electrons to lose thus, looses one electron from 3d-electron to attain +3 oxidation state.
(iv) Co is the strongest oxidising agent as the E value for the redox couple M3+∕M2+ is highest.