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Analysis shows that nickel oxide has the formula
\(\mathrm{Ni}_{0.98} \mathrm{O}_{1.00^*}\). What fractions of nickel exist as \(\mathrm{Ni}^{2+}\) and \(\mathrm{Ni}^{3+}\) ions.
\(\mathrm{Ni}_{0.98} \mathrm{O}_{1.00^*}\). What fractions of nickel exist as \(\mathrm{Ni}^{2+}\) and \(\mathrm{Ni}^{3+}\) ions.
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The formula \(\mathrm{Ni}_{0.98} \mathrm{O}_{1.00}\) implies that \(98 \mathrm{Ni}\) atoms are associated with \(100 \mathrm{O}\) atoms.
Let out of \(98 \mathrm{Ni}\) atoms present, \(x\) atoms are present as \(\mathrm{Ni}^{2+}\)
ions and \((98-x)\) are present as \(\mathrm{Ni}^{3+}\) ions.
\(\therefore\) Total charge on \(x \mathrm{Ni}^{2+}\) ions and \((100-x) \mathrm{Ni}^{3+}\) ions should balance the charge on \(100 \mathrm{O}^{2-}\) ions.
Hence, \(x \times 2+(98-x) \times 3=100 \times 2\)
\(2 x+294-3 x=200 \quad x=94\).
\(\therefore\) Fraction of \(\mathrm{Ni}\) atoms present as
\(\mathrm{Ni}^{2+}=\frac{94}{98} \times 100=95.92 \%\)
and fraction of \(\mathrm{Ni}\) atoms present as
\(\mathrm{Ni}^{3+}=\frac{98-94}{98} \times 100=\frac{4}{98} \times 100=4.081 \%\).
Let out of \(98 \mathrm{Ni}\) atoms present, \(x\) atoms are present as \(\mathrm{Ni}^{2+}\)
ions and \((98-x)\) are present as \(\mathrm{Ni}^{3+}\) ions.
\(\therefore\) Total charge on \(x \mathrm{Ni}^{2+}\) ions and \((100-x) \mathrm{Ni}^{3+}\) ions should balance the charge on \(100 \mathrm{O}^{2-}\) ions.
Hence, \(x \times 2+(98-x) \times 3=100 \times 2\)
\(2 x+294-3 x=200 \quad x=94\).
\(\therefore\) Fraction of \(\mathrm{Ni}\) atoms present as
\(\mathrm{Ni}^{2+}=\frac{94}{98} \times 100=95.92 \%\)
and fraction of \(\mathrm{Ni}\) atoms present as
\(\mathrm{Ni}^{3+}=\frac{98-94}{98} \times 100=\frac{4}{98} \times 100=4.081 \%\).
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