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 ?

Explanation

To justify why \( \text{Ni}^{2+} \) and \( \text{Pt}^{4+} \) are more stable than their corresponding counter ions \( \text{Ni}^{+} \) and \( \text{Pt}^{2+} \), we will analyze the given ionization energy data and the electronic configurations of nickel and platinum.

### Step-by-Step Solution:

1. **Understand the Ionization Energies**:
- The first and second ionization energies (IE1 + IE2) for Nickel (Ni) is 2.49 kJ/mol.
- The first and second ionization energies (IE1 + IE2) for Platinum (Pt) is 2.66 kJ/mol.
- The third and fourth ionization energies (IE3 + IE4) for Nickel is 8.80 kJ/mol.
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