Question

The ionic radii in \AA of \( \text{N}^{3-} \), \( \text{O}^{2-} \), and \( \text{F}^- \) are respectively:

• A. \( 1.71, 1.40 \) and \( 1.36 \)
• B. \( 1.71, 1.36 \) and \( 1.40 \)
• C. \( 1.36, 1.40 \) and \( 1.71 \)
• D. \( 1.36, 1.71 \) and \( 1.40 \)

Hint:

For isoelectronic ions, remember: As the negative charge decreases (or positive charge increases), the ionic radius decreases.

Answer 1

The Correct Option is A

Concept: The given species \( \text{N}^{3-} \), \( \text{O}^{2-} \), and \( \text{F}^- \) are isoelectronic species, meaning they all possess the same number of electrons (\( 10 \) electrons each).

Step 1: Understanding the relationship between nuclear charge and ionic radius. For isoelectronic species, the number of electrons is constant, but the number of protons (nuclear charge) increases as we move from Nitrogen (\( Z=7 \)) to Oxygen (\( Z=8 \)) to Fluorine (\( Z=9 \)).

Step 2: Applying the principle of effective nuclear charge. As the atomic number (\( Z \)) increases, the nucleus exerts a stronger attractive force on the same number of electrons, pulling the electron cloud closer to the nucleus. Therefore, the ionic radius decreases as the nuclear charge increases. \[ Z(\text{N}) = 7, Z(\text{O}) = 8, Z(\text{F}) = 9 \] Since \( Z(\text{N}) < Z(\text{O}) < Z(\text{F}) \), the ionic radii follow the order: \[ \text{N}^{3-} > \text{O}^{2-} > \text{F}^- \]

Step 3: Matching the given values. Based on the trend, the radius should be largest for \( \text{N}^{3-} \) (\( 1.71\text{\AA} \)) and smallest for \( \text{F}^- \) (\( 1.36\text{\AA} \)). The correct sequence is: \( \text{N}^{3-} = 1.71\text{\AA} \) \( \text{O}^{2-} = 1.40\text{\AA} \) \( \text{F}^- = 1.36\text{\AA} \)