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:

For ionic compounds, the greater the difference in ionisation enthalpy and electron gain enthalpy, the greater the ionic character. More negative electron gain enthalpy indicates a stronger ionic bond.

Answer 1

The Correct Option is C

The ionic character of a compound depends on the difference in ionisation enthalpy and electron gain enthalpy. The larger the difference, the greater the ionic character. A compound with a more negative electron gain enthalpy will result in a stronger ionic bond. Given that the electron gain enthalpy values for elements A, B, C, and D are as follows: 
- \( {A} \) has \( -328 \, {kJ/mol} \) 
- \( {B} \) has \( -349 \, {kJ/mol} \) 
- \( {C} \) has \( -325 \, {kJ/mol} \) 
- \( {D} \) has \( -295 \, {kJ/mol} \) 
The order of ionic character is given by the electron gain enthalpy, where \( {D} \) has the highest ionic character and \( {A} \) has the least. Thus, the correct order is (3): \( {ED}>{EC}>{EB}>{EA} \).