The sum of `IE_(1)` and `IE_(2)` and those of `IE_(2)` and those of `IE_(3)` and `IE_(4)` in `(kJ mol^(-1))` of `Ni` and `Pt` are :
`{:(,(IE_(1)+IE_(2)),(IE_(2)+IE_(4)),"Total",),(Ni,2.5 xx 10^(3),8.8 xx 10^(3),11.3 xx 10^(3),),(Pt,2.7 xx 10^(3),6.7 xx 10^(3),9.4 xx 10^(3),):}`
a. What is the most common oxidation state (O.S.) of `Ni` and `Pt`.
b. Name of the metal (`Ni`or `Pt`) which can more easily form compounds in its `+4` O.S.

Explanation

Thermodynamic stability of the transition elements and their compounds can be evaluated in terms of magnitude of `IE's` of metals. The smaller the `IE` of the metal, the stable is its compound.
a. `Ni = +2` and `Pt = +4`.
Since `(IE_(1) + IE_(2)) (2.5 xx 10^(3))` of `Ni` is less than `(IE_(1) + IE_(2))` of `Pt`, so, the most common oxidation states of `Ni = +2`. Similarly `(1E_(3) + 1E_(4))` of `Pt` is less than `(IE_(3) + IE_(4))` of `Ni`, so, the most common oxidation state of `Pt = +4`.
b. `Pt` forms more stable complexes in `+4` state due to its higher stability than `+2` state.