Question

Assertion (A): The ionic radii of \( Na^+ \) and \( F^- \) are the same. 
Reason (R): Both \( Na^+ \) and \( F^- \) are isoelectronic species. 

• (A) and (R) are correct. (R) is the correct explanation of (A)
• (A) and (R) are correct, but (R) is not the correct explanation of (A)
• (A) is correct but (R) is not correct
• (A) is not correct but (R) is correct

Hint:

Isoelectronic species have the same number of electrons but different sizes due to varying nuclear charges. Cations (\( Na^+ \)) are smaller than their neutral atoms, while anions (\( F^- \)) are larger.

Answer 1

The Correct Option is D

Step 1: Understanding Isoelectronic Species 
Isoelectronic species have the same number of electrons but different nuclear charges. Both \( Na^+ \) (11 protons, 10 electrons) and \( F^- \) (9 protons, 10 electrons) are isoelectronic. Thus, Reason (R) is correct. 

Step 2: Comparing Ionic Radii of \( Na^+ \) and \( F^- \) 
- \( Na^+ \) is a cation formed by losing one electron, reducing electron-electron repulsion and making it smaller than its parent atom.
- \( F^- \) is an anion formed by gaining one electron, increasing electron-electron repulsion and making it larger than its parent atom.
Since \( Na^+ \) has a smaller radius than \( F^- \), the assertion that their radii are the same is incorrect.
Step 3: Verify the Correct Answer 
Since (A) is incorrect but (R) is correct, the correct answer is Option (4).