The sums of first and second ionization enthalpies and those of third and fourth ionization enthalpies of nickel and platinum are given below:

Metal	${\mathrm{E}}_1+ {\mathrm{E}}_2$	${\mathrm{E}}_3+{\mathrm{E}}_4$
$\mathrm{Ni}$	$2.49$	$8.80$
$\mathrm{Pt}$	$2.66$	$6.70$

 The most common oxidation state for $\mathrm{Ni}$ and $\mathrm{Pt}$ and its reason.

The sums of first and second ionization enthalpies and those of third and fourth ionization enthalpies of nickel and platinum are given below:

Taking these values into account write the name of the metal ($\mathrm{Ni}$ or $\mathrm{Pt}$ ) which can form compounds in $+4$ oxidation state more easily and why?

Among ionic species,${\mathrm{Sc}}^{3+},{\mathrm{Ce}}^{4+}$,  and ${\mathrm{Eu}}^{2+}$ , which one is a good oxidising agent? Give a suitable reason for your answer.

Of the ions ${\mathrm{Co}}^{2+},{\mathrm{Sc}}^{3+}$ and ${\mathrm{Cr}}^{3+}$ which one will give coloured aqueous solutions and how will each of them respond to a  magnetic field and why ? 

$\text{ (Atomic numbers: }\mathrm{Co},=27,\mathrm{Sc}=21,\mathrm{Cr}=24)$