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

Step 1 : Understanding the Question:
This question belongs to the topic of periodic properties, specifically dealing with the ionic radii of isoelectronic species. We are asked to identify the correct set of ionic radii for the ions \( \text{N}^{3-} \), \( \text{O}^{2-} \), and \( \text{F}^- \).
Step 2 : Key Formulas and Approach:
Isoelectronic species are atoms or ions that possess the same number of electrons. For these species, the size of the ion is determined by the effective nuclear charge.
The nuclear charge is given by the atomic number \( Z \). As \( Z \) increases, the ratio of protons to electrons increases:
\[ \text{Ionic Radius} \propto \frac{1}{Z} \]
A higher nuclear charge exerts a stronger attractive pull on the electron cloud, pulling it closer to the nucleus and decreasing the ionic radius.
Step 3 : Detailed Explanation:

Step 1: Count the total number of electrons in each given ion.
\( \text{N}^{3-} \) has \( 7 + 3 = 10 \) electrons.
\( \text{O}^{2-} \) has \( 8 + 2 = 10 \) electrons.
\( \text{F}^- \) has \( 9 + 1 = 10 \) electrons. Since all three have 10 electrons, they are isoelectronic.

Step 2: Compare their atomic numbers (nuclear charge, \( Z \)).
The atomic numbers are: \( Z(\text{N}) = 7 \), \( Z(\text{O}) = 8 \), and \( Z(\text{F}) = 9 \).

Step 3: Analyze the trend in ionic radii. Since nitrogen has the lowest nuclear charge (\( Z = 7 \)), its nucleus exerts the weakest pull on the 10 electrons, making \( \text{N}^{3-} \) the largest ion.

Conversely, fluorine has the highest nuclear charge (\( Z = 9 \)), exerting the strongest pull on the electron cloud, making \( \text{F}^- \) the smallest ion.

The descending order of their ionic radii is: \( \text{N}^{3-}>\text{O}^{2-}>\text{F}^- \).

Step 4: Match this trend with the values given in the options. The values must follow a decreasing sequence: \( 1.71\text{ \AA}>1.40\text{ \AA}>1.36\text{ \AA} \).

Step 4 : Final Answer:
The ionic radii of \( \text{N}^{3-} \), \( \text{O}^{2-} \), and \( \text{F}^- \) are \( 1.71 \), \( 1.40 \), and \( 1.36\text{ \AA} \) respectively, which corresponds to option (A).