Chemistry - Mole Concept Question with Solution | TestHub

$\mathrm{N}\mathrm{i}=3{\mathrm{d}}^{8}4{\mathrm{s}}^2$(Atomic number = 28)
So$\mathrm{N}{\mathrm{i}}^{2+}=3{\mathrm{d}}^8$and$\mathrm{N}{\mathrm{i}}^{3+}=3{\mathrm{d}}^7$
Hence, 96% ion of$\mathrm{N}{\mathrm{i}}^{2+}$and 4% ions of$\mathrm{N}{\mathrm{i}}^{3+}$are present.
Let number of${\mathrm{O}}^{2-}$ion present in the crystal = x
Using electroneutrality rule,
Total positive charge = Total negative charge
$96\times 2(+\mathrm{v}\mathrm{e}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e})+4\times 3(+\mathrm{v}\mathrm{e}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e})$
$=\mathrm{x}\times 2(-\mathrm{v}\mathrm{e}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e})$
So$96\times 2+4\times 3-2\mathrm{x}=0$
So x = 1.02
Hence formula of the crystal is$\mathrm{N}{\mathrm{i}}_{1.00}{\mathrm{O}}_{1.02}$or$\mathrm{N}{\mathrm{i}}_{0.98}{\mathrm{O}}_{1.00}$.