Question

Analysis shows nickel oxide has the formula $N{i}_{0.98}{O}_{1.00}$. What fraction of $Ni$ exists as $N{i}^{2+}$ and $N{i}^{3+}$ ions in given oxide?

Answer 1

Formula is ${Ni}_{0.98}{O}_{1.00}$

Ration of $Ni$ to $O=0.98:1$, i.e.,

$0.98$ atom of $Ni$ combines with $1$ atom of $O$.

Let $x$ be the number of atoms of ${Ni}^{+2}$

Therefore,

Number of atoms of ${Ni}^{+3}=0.98-x$

Total charge on $Ni=-$ Total charge on $O$

$\because 1$ oxygen atom has charge $-2$.

$\therefore 3(0.98-x)+2x=2$

$2.94-3x+2x=2$

$-x=-0.94$

$\Rightarrow x=0.94$

Therefore,

Number of atoms of ${Ni}^{+2}=x=0.94$

Number of atoms of ${Ni}^{+3}=0.98-x=0.98-0.94=0.04$

Now

Percentage of ${Ni}^{+2}=\frac{94}{98}\times 100\approx 95.92\%$

Percentage of ${Ni}^{+3}=100-95.92=4.08\%$