Question

List-I contains some group-I elements and List-II contains their electron gain enthalpy values (in kJ mol\(^{-1}\))
List-I: (i) H, (ii) Li, (iii) Na, (iv) K
List-II: (a) -48, (b) -53, (c) -73, (d) -60
Choose the correct match.

• A. (i)-(a), (ii)-(c), (iii)-(d), (iv)-(b)
• B. (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a)
• C. (i)-(d), (ii)-(b), (iii)-(c), (iv)-(a)
• D. (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d)

Hint:

For Group 1, just remember the trend: "Bigger the atom, smaller (less negative) the electron gain enthalpy." H is smallest, so it releases the most energy upon gaining an electron.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Electron gain enthalpy (\(\Delta_{eg}H\)) for Group 1 elements generally becomes less negative as we move down the group. This is because the atomic size increases, and the incoming electron is less strongly attracted to the larger nucleus.

Step 2: Detailed Explanation:
1. The elements are: Hydrogen (H), Lithium (Li), Sodium (Na), and Potassium (K).
2. Trend: From top to bottom, \(\Delta_{eg}H\) becomes less negative.
3. Order of elements: H (top) \(>\) Li \(>\) Na \(>\) K (bottom).
4. Matching with values:
- H: Most negative \(\to\) -73 (c)
- Li: Next \(\to\) -60 (d)
- Na: Next \(\to\) -53 (b)
- K: Least negative \(\to\) -48 (a)
5. Match: (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a).

Step 3: Final Answer:
The correct match is Option B.