The sum of the first and second ionisation enthalpies and those of third and fourth ionisation enthalpies of nickel and platinum are given below: {:(,,IE_(1)+IE_(2)(kJ mol^(-1)),,IE_(3)+IE_(4)(kJ mol^(-1))),(Ni,,2.49,,8.80),(Pt,,2.66,,6.70):} Taking these values into account write : (i) the most common oxidation state for Ni and Pt and its reason. (ii) the name of the metal (Ni or Pt) which can form compounds in +4 oxidation state more easily and why?

The sum of the first and second ionisation enthalpies and those of thi

1.	The sum of the first and second ionisation enthalpies and those of third and fourth ionisation enthalpies of nickel and platinum are given below: {:(,,IE_(1)+IE_(2)(kJ mol^(-1)),,IE_(3)+IE_(4)(kJ mol^(-1))),(Ni,,2.49,,8.80),(Pt,,2.66,,6.70):} Taking these values into account write : (i) the most common oxidation state for Ni and Pt and its reason. (ii) the name of the metal (Ni or Pt) which can form compounds in +4 oxidation state more easily and why?
Answer» Solution :(i) Ni SHOWS +2 oxidation STATE because of LOWER `(IE_(1)+IE_(2))`. Platinum shows +4 oxidation state because extra ENERGY `IE_(3)+IE_(4)` is balanced by HYDRATION energy. (ii) Pt can form compounds in +4 oxidation state because the energy required to remove four electrons is less in case of Pt than Ni.