Question

An element 'E' has the ionisation enthalpy value of 374 kJ mol\(^{-1}\). 'E' reacts with elements A, B, C, and D with electron gain enthalpy values of −328, −349, −325, and −295 kJ mol\(^{-1}\), respectively. The correct order of the products EA, EB, EC, and ED in terms of ionic character is:

• A. EA > EB > EC > ED
• B. ED > EC > EA > EB
• C. ED > EC > EB > EA
• D. EB > EA > EC > ED

Hint:

A larger difference between the ionization enthalpy and electron gain enthalpy generally leads to more ionic bonds.

Answer 1

The Correct Option is C

The ionic character of a bond is determined by the difference in ionization enthalpy of the element and the electron gain enthalpy of the other element. The greater the difference, the more ionic the bond.
- Element D has the least electron gain enthalpy, resulting in the strongest ionic bond with E.
- Thus, the correct order is \( \text{ED} > \text{EC} > \text{EB} > \text{EA} \).