Question:

The electronic configuration of four elements are given below:

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- Elements in Group 17 (Halogens) have the highest negative electron gain enthalpies. - Fluorine has a lower enthalpy than Chlorine due to electron repulsions in its small size. - Oxygen and Sulfur have lower values due to electronic repulsions.
Updated On: Mar 18, 2025
  • \( II > III > IV > I \)
  • \( II > I > III > IV \)
  • \( I > IV > III \)
  • \( II > I > IV > III \) 
     

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The Correct Option is D

Solution and Explanation

Understanding Electron Gain Enthalpy
Electron gain enthalpy is the energy change when an electron is added to a neutral atom in its gaseous state. The trend of electron gain enthalpy is influenced by:
- Atomic size (smaller atoms have higher negative values).
- Effective nuclear charge (more nuclear attraction increases the enthalpy).
- Electronic configuration stability (half-filled and fully-filled orbitals resist electron gain).
Analyzing the Given Configurations
- Element I ([He] 2s\(^2\) 2p\(^5\)): This corresponds to Fluorine (F), which has a very high electron gain enthalpy due to its high nuclear attraction and small size.
- Element II ([Ne] 3s\(^2\) 3p\(^5\)): This corresponds to Chlorine (Cl), which has slightly lower but still high electron gain enthalpy compared to Fluorine.
- Element III ([He] 2s\(^2\) 2p\(^4\)): This corresponds to Oxygen (O), which has lower electron gain enthalpy than Fluorine due to higher repulsion in a smaller size.
- Element IV ([Ne] 3s\(^2\) 3p\(^4\)): This corresponds to Sulfur (S), which has a lower electron gain enthalpy compared to Fluorine and Chlorine.
Determining the Correct Order From known values: \[ \text{Electron Gain Enthalpy: } \text{Cl}>\text{F}>\text{S}>\text{O} \] Thus, the correct order is: \[ II>I>IV>III \]
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