$\begin{array}{1 1}95.918,4.081\\75.918,6.081\\65.918,3.081\\45.918,8.081\end{array} $

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The formula $Ni_{0.98}O_{1.00}$ shows that

$Ni : O=0.98 : 1.00=98 : 100$

Thus if there are 100-0 atoms then Ni atoms =98

Charge on $O^{2-}$ ions =$100\times (-2)=-200$

Suppose Ni atoms as $Ni^{2+}=x$

Suppose Ni atoms as $Ni^{3+}=98-x$

Total charge on $Ni^{2+}$ and $Ni^{3+}=(+2) x+(+3)(98-x)$

$\Rightarrow +2x+294-3x$

$\Rightarrow 294-x$

As metal oxide is neutral,total charge on cations =Total charge on anions

$294-x=200$

$-x=-294+200$

$-x=-94$

% of Ni as $Ni^{2+}=\large\frac{94}{98}$$\times 100=95.918$

% of Ni as $Ni^{3+}=\large\frac{98-x}{98}$$\times 100$

$\Rightarrow \large\frac{98-94}{98}$$\times 100$

$\Rightarrow \large\frac{4}{98}$$\times 100$

$\Rightarrow 4.081$

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