Transition metal with higher oxidation number will acts as:

Arrange the following ions in order of their magnetic moment:

($\mathrm{i}$) ${\mathrm{V}}^{4+}$ ($\mathrm{ii}$) ${\mathrm{Mn}}^{4+}$ ($\mathrm{iii}$) ${\mathrm{Fe}}^{3+}$ ($\mathrm{iv}$) ${\mathrm{Ni}}^{2+}$

(Atomic number of $\mathrm{V}=23, \mathrm{M} \mathrm{n}=25, \mathrm{F} \mathrm{e}=26, \mathrm{N} \mathrm{i}=28)$

Among the following transition elements, pick out the elements with the highest second ionization energy.
($\mathrm{i}$) $\mathrm{V}(\mathrm{At}. \mathrm{no}.=23)$
($\mathrm{ii}$) $\mathrm{Cr}\left(\mathrm{At}, \mathrm{no}.=24\right)$
($\mathrm{iii}$) $\mathrm{Mn}\left(\mathrm{At}. \mathrm{no}.=25\right)$

($\mathrm{iv}$) $\mathrm{Cu}\left(\mathrm{At}. \mathrm{no}.=29\right)$

($\mathrm{v}$) $\mathrm{Zn}\left(\mathrm{At}. \mathrm{no}.=30\right)$