Question

Match the following

The correct answer is

• A. A-II, B-IV, C-I, D-III
• B. A-V, B-IV, C-II, D-I
• C. A-III, B-IV, C-II, D-I
• D. A-III, B-II, C-IV, D-I

Hint:

The most important exception to remember for electron gain enthalpy is that period 3 p-block elements (like Si, P, S, Cl) have more negative \(\Delta_{eg}H\) than their corresponding period 2 counterparts (C, N, O, F). This single fact is often the key to solving matching questions involving these elements.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question requires knowledge of the trends in electron gain enthalpy (\(\Delta_{eg}H\)) in the periodic table. Electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom to form a negative ion. Generally, it becomes more negative (more exothermic) across a period and less negative down a group. However, there are important exceptions.
Step 2: Analyzing the Elements and Trends:
The elements are from Group 16 (O, S) and Group 17 (F, Cl).
General Trends:
- Halogens (Group 17) have the most negative electron gain enthalpies in their respective periods because they are one electron short of a stable noble gas configuration.
- Chalcogens (Group 16) have less negative \(\Delta_{eg}H\) than halogens.
Anomalies:
- The electron gain enthalpy of a period 2 element (like O, F) is less negative than that of the corresponding period 3 element (S, Cl). This is due to the small size and high electron density of the period 2 atoms, which leads to significant electron-electron repulsion when an extra electron is added.
- Therefore, Cl has a more negative \(\Delta_{eg}H\) than F.
- And S has a more negative \(\Delta_{eg}H\) than O.
Step 3: Matching the Values:
Based on the trends:
1. Chlorine (Cl) vs. Fluorine (F): Both are halogens. Cl is in period 3, F is in period 2. Due to the anomaly, Cl will have the most negative \(\Delta_{eg}H\) of the group. The most negative value in List-2 is -349 kJ/mol. So, C \(\rightarrow\) II (-349).
2. Fluorine (F) will have the next most negative value among the halogens. The value is -328 kJ/mol. So, B \(\rightarrow\) IV (-328).
3. Sulphur (S) vs. Oxygen (O): Both are chalcogens. S is in period 3, O is in period 2. S will have a more negative \(\Delta_{eg}H\) than O. The remaining negative values are -200 and -141. The more negative of these is -200. So, D \(\rightarrow\) I (-200).
4. Oxygen (O) will have the least negative value among these. So, A \(\rightarrow\) III (-141).
The value V (+48) is for elements like noble gases or alkaline earth metals which have a positive (endothermic) electron gain enthalpy, and is not used here.
Summary of Matching:
- A (O) \(\rightarrow\) III (-141)
- B (F) \(\rightarrow\) IV (-328)
- C (Cl) \(\rightarrow\) II (-349)
- D (S) \(\rightarrow\) I (-200)
This corresponds to option (C).
Step 4: Final Answer:
The correct match is A-III, B-IV, C-II, D-I. Therefore, option (C) is correct.