ionisation energies of `Ni` and `Pt` in `Kjmol^(-1)` are given below
`{:(,ubrace((IE)_(1)+(IE)_(2)),ubrace((IE)_(3)+(IE)_(4))),(Ni,2.49,8.80),(Pt,2.60,6.70):}`
So, (select the correct statement)

For Ni and Pt different I.P values are given below : {:(,,ul((IP_(1))+(IP)_(2)),,(IP)_(3)+(IP)_(4)),(Ni,,2.49,,8.80),(Pt,,2.60,,6.70):} hence :

Ionisation energies of Ni and Pt in Kj/mol are given below {:("",IE_1 + IF_2,IE_3 + IE_4),(Ni,2.49,8.80),(Pt,2.60,6.70):} So the correct statment is

The sums of first and second ionization energies and those of third and fourth ionization energies ("in kJ mol"^(-1)) of nickel and platinum are: {:(,(IE_(1)),+(IE)_(2)),(,Ni,2.49),(,Pt,2.66):} {:(,(IE_(3)),+(IE)_(4)),(,Ni,8.80),(,Pt,6.70):} Justify using above data why Ni^(+2) and Pt^(+4) are more stable than their corresponding counter part ions ?

The sum of IE_(1) and IE_(2), IE_(3) and IE_(4) for element P and Q are given below: {:(,IE_(1)+IE_(2),IE_(2)+IE_(4)),((P),2.45,8.82),((Q),2.85,6.11):} Then according to the given information the correct statement (s) is/are

The sum of IE_(1) and IE_(2), IE_(3) and IE_(4) for element P and Q are given below: {:(,IE_(1)+IE_(2),IE_(2)+IE_(4)),((P),2.45,8.82),((Q),2.85,6.11):} Then according to the given information the correct statement (s) is/are

First three ionisation energies (in kJ/mol) of three representative elements are given below: {:("Element",IE_(1),IE_(2),IE_(3)),(P,495.8,4582,6910),(Q,737.7,1451,7733),(R,577.5,1817.2745):} then incorrect option is:

First three ionisation energies (in kJ/mol) of three representative elements are given below: {:("Element",IE_(1),IE_(2),IE_(3)),(P,495.8,4582,6910),(Q,737.7,1451,7733),(R,577.5,1817.2745):} then incorrect option is: