Question

Given below are two statements: Statement-I: The percent compositions of $\mathrm{Ni}^{2+}$ and $\mathrm{Ni}^{3+}$ in $\mathrm{Ni}_{0.98}\mathrm{O}$ respectively is 96% and 4%. Statement-II: The fraction of $\mathrm{Fe}^{3+}$ and $\mathrm{Fe}^{2+}$ ions in 1 mole of $\mathrm{Fe}_{0.93}\mathrm{O}$ is 0.14 and 0.79 respectively. Choose the correct option.

• A. Both statements I and II are correct
• B. Statement I is correct, but statement II is not correct
• C. Statement I is not correct, but statement II is correct
• D. Both statements I and II are not correct

Hint:

For non-stoichiometric oxides, always apply electrical neutrality. The total positive charge must equal the total negative charge contributed by oxide ions.

Answer 1

The Correct Option is A

Concept: Non-stoichiometric metal oxides often contain metal ion vacancies. To maintain electrical neutrality, some metal ions are oxidized to a higher oxidation state. The fractions of different oxidation states can be calculated using charge balance.

Step 1: Verify Statement-I for $\mathrm{Ni}_{0.98}\mathrm{O}$. Let the fraction of $\mathrm{Ni}^{3+}$ ions be $x$. Total nickel ions present per mole of oxide: \[ 0.98 \] Charge neutrality requires \[ 2(0.98-x)+3x=2. \] Simplifying, \[ 1.96-2x+3x=2 \] \[ x=0.04. \] Thus, \[ \mathrm{Ni}^{3+}=0.04 \] and \[ \mathrm{Ni}^{2+}=0.98-0.04=0.94. \] Percentage among total nickel ions: \[ \frac{0.94}{0.98}\times100\approx95.92%\approx96% \] \[ \frac{0.04}{0.98}\times100\approx4.08%\approx4%. \] Hence Statement-I is correct.

Step 2: Verify Statement-II for $\mathrm{Fe}_{0.93}\mathrm{O}$. Let moles of $\mathrm{Fe}^{3+}$ be $x$. Then moles of $\mathrm{Fe}^{2+}$ are \[ 0.93-x. \] Applying charge neutrality, \[ 3x+2(0.93-x)=2. \] \[ 3x+1.86-2x=2. \] \[ x=0.14. \] Hence \[ \mathrm{Fe}^{3+}=0.14 \] and \[ \mathrm{Fe}^{2+}=0.93-0.14=0.79. \] Therefore Statement-II is also correct.

Step 3: Conclusion. Both statements are correct. \[ \boxed{\text{Option (1)}} \]