Question

In which of the following ionic pairs, the second ion is smaller in size than the first ion?

• A. \( Al^{3+}, Mg^{2+} \)
• B. \( F^-, Na^+ \)
• C. \( O^{2-}, N^{3-} \)
• D. \( Mg^{2+}, Na^+ \)
  \bigskip

Hint:

Cations are always smaller than their neutral atoms, while anions are larger. The greater the positive charge, the smaller the ion size.

Answer 1

The Correct Option is B

Step 1: Understanding Ionic Radii Trends - Ionic size depends on the nuclear charge and the number of electrons. - Cations (\(+\) ions) are smaller than their parent atoms due to increased nuclear attraction. - Anions (\(-\) ions) are larger than their parent atoms due to increased electron repulsion. Step 2: Comparing the Given Pairs - (1) \( Al^{3+}, Mg^{2+} \): \( Al^{3+} \) has a higher charge and a smaller size than \( Mg^{2+} \), so the second ion is not smaller. - (2) \( F^-, Na^+ \): \( F^- \) (anion) is larger than \( Na^+ \) (cation) because anions are larger due to extra electrons, whereas cations shrink. - (3) \( O^{2-}, N^{3-} \): Both are anions, but \( O^{2-} \) has fewer electrons than \( N^{3-} \), making it smaller. - (4) \( Mg^{2+}, Na^+ \): \( Mg^{2+} \) is smaller than \( Na^+ \) since a higher positive charge leads to a greater nuclear pull. Thus, the correct answer is \( F^-, Na^+ \) where the second ion \( Na^+ \) is smaller than the first ion \( F^- \). \bigskip