Question

The correct order of ionic radius of nitrogen family is:

• A. \(N^{3-}<P^{3-}<As^{3-}<Sb^{3-}<Bi^{3-}\)
• B. \(N^{3-}<P^{3-}<Sb^{3-}\)
• C. \(P^{3-}>As^{3-} = Bi^{3-}\)
• D. \(N^{3-}>Bi^{3-}>Sb^{3-}\)

Hint:

Down the group → more shells → larger ionic radius.

Answer 1

The Correct Option is A

The question asks for the correct order of ionic radii for the nitrogen family, which includes the following ions: \( \text{N}^{3-}, \text{P}^{3-}, \text{As}^{3-}, \text{Sb}^{3-}, \text{Bi}^{3-} \). The order of ionic radii can be predicted based on the periodic trends: 

Understanding Periodic Trends: As we move down a group in the periodic table, the size of the atoms increases. This occurs because each subsequent element has an extra electron shell compared to the one above it.

Application to Ionic Radii: In the case of isoelectronic ions (ions with the same number of electrons), the ionic radius increases with an increase in atomic number. Thus, in the nitrogen family:

• \( \text{Bi}^{3-} \) is larger than \( \text{Sb}^{3-} \), which is larger than \( \text{As}^{3-} \), which is larger than \( \text{P}^{3-} \), which is larger than \( \text{N}^{3-} \).

Correct Order: According to the periodic trend, the correct order of ionic radii for the nitrogen family is \( N^{3-} < P^{3-} < As^{3-} < Sb^{3-} < Bi^{3-} \).

Elimination of Incorrect Options:

• The option \( N^{3-} > Bi^{3-} > Sb^{3-} \) is incorrect because it contradicts the consistent increase in ionic radii down the group.
• The option \( P^{3-} > As^{3-} = Bi^{3-} \) is incorrect because, typically, Bi would be larger than As due to the addition of electron shells.
• The option \( N^{3-} < P^{3-} < Sb^{3-} \) is incomplete as it omits describing the position of As and Bi, which are present in the family.

Hence, the correct answer is: \( N^{3-} < P^{3-} < As^{3-} < Sb^{3-} < Bi^{3-} \).