Question

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

In isoelectronic species, the ionic radius decreases with increasing nuclear charge because the greater positive charge attracts the electrons more strongly, causing the ion to contract.

Answer 1

The Correct Option is A

Step 1: {Isoelectronic Species} 
The ions \( {N}^{3-} \), \( {O}^{2-} \), and \( {F}^- \) are isoelectronic species, meaning they all have the same number of electrons. Specifically, each of these ions has 10 electrons. Despite having the same number of electrons, their ionic radii will differ because of the varying nuclear charges. The nuclear charge increases from nitrogen to oxygen to fluorine as we move across the periodic table: - \( {N} \) has atomic number \( Z = 7 \), - \( {O} \) has atomic number \( Z = 8 \), - \( {F} \) has atomic number \( Z = 9 \).
Step 2: {Ionic Radii Order} 
As the nuclear charge increases, the electrons are attracted more strongly by the nucleus, which results in a decrease in ionic radius. Therefore, the ionic radii will decrease as we move from \( {N}^{3-} \) to \( {O}^{2-} \) to \( {F}^- \). This gives the following order of ionic radii: \[ {N}^{3-}>{O}^{2-}>{F}^- \] Hence, the correct answer is (A). 
 

Answer 2

Step 1: Understand ionic radii and trends in anions
Ionic radius refers to the effective size of an ion in a crystal lattice. When atoms gain electrons to form anions, their radius increases due to enhanced electron-electron repulsion and reduced effective nuclear charge per electron.

Step 2: Analyze the ions N³⁻, O²⁻, and F⁻
These are all isoelectronic species (each has 10 electrons), but their nuclear charges differ:
- N³⁻ has 7 protons
- O²⁻ has 8 protons
- F⁻ has 9 protons
All have the same number of electrons, but different numbers of protons pulling on them.

Step 3: Compare effective nuclear charge
Effective nuclear charge (Zₑff) increases across the period:
Zₑff ∝ (number of protons – inner shell electrons)
Since all three ions are isoelectronic, the ion with fewer protons (N³⁻) will exert the weakest attraction on its electron cloud, resulting in a larger radius.
Hence, the order of ionic radii is:
N³⁻ > O²⁻ > F⁻

Step 4: Provide the ionic radii values
- N³⁻: 1.71 Å
- O²⁻: 1.40 Å
- F⁻: 1.36 Å

Step 5: Final Answer
The ionic radii in (Å) of N³⁻, O²⁻ and F⁻ are respectively:
1.71, 1.40 and 1.36