Question

A, B, C, D and E are elements with atomic numbers 13, 11, 9, 7 and 16 respectively. Among these elements, ion of an element X has largest size and ion of an element Y has smallest size. X and Y are respectively (Assume that all ions have nearest inert gas configuration)

• A. D, A
• B. A, D
• C. E, A
• D. D, E

Hint:

For isoelectronic ions, the radius is inversely proportional to the atomic number \( Z \). Lower \( Z \) \(\rightarrow\) Larger Radius. \( \text{Anion} \textgreater \text{Neutral} \textgreater \text{Cation} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

First, identify the elements from their atomic numbers and write their most stable ions. Then compare their ionic sizes.

Important rules:
1. In an isoelectronic series, ionic size decreases as nuclear charge increases.
2. Anions are generally larger than cations.
3. If ions belong to different shells, the ion with more shells is usually larger.

Step 2: Key Formula or Approach:

Identify the ions formed by each element to attain the nearest noble gas configuration, then compare their sizes.

Step 3: Detailed Explanation:

Identify the elements and their ions:

A: Z = 13 → Al
Stable ion: Al3+ → 10 electrons

B: Z = 11 → Na
Stable ion: Na+ → 10 electrons

C: Z = 9 → F
Stable ion: F- → 10 electrons

D: Z = 7 → N
Stable ion: N3- → 10 electrons

E: Z = 16 → S
Stable ion: S2- → 18 electrons

Now compare the isoelectronic ions with 10 electrons:
Al3+, Na+, F-, N3-

For isoelectronic ions, size decreases as nuclear charge increases.
So, the order of ionic size is:

N3- > F- > Na+ > Al3+

Thus, among these isoelectronic ions:
Largest = N3- → D
Smallest = Al3+ → A

Element E forms S2-, which has 18 electrons and belongs to a different electronic shell. In many such questions, the intended comparison is within the isoelectronic set A, B, C, D.

Step 4: Final Answer:

X = D, Y = A.