Question:medium

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:

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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.
Updated On: Jan 14, 2026
  • \( {EA}>{EB}>{EC}>{ED} \)
  • \( {ED}>{EC}>{EA}>{EB} \)
  • \( {ED}>{EC}>{EB}>{EA} \)
  • \( {EB}>{EA}>{EC}>{ED} \)
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The Correct Option is C

Solution and Explanation

Ionic character is determined by the ionization enthalpy and electron gain enthalpy difference; a larger difference signifies greater ionic character. Compounds with more negative electron gain enthalpies exhibit stronger ionic bonds. The electron gain enthalpies for elements A, B, C, and D are: - \( {A} \) = \( -328 \, {kJ/mol} \), - \( {B} \) = \( -349 \, {kJ/mol} \), - \( {C} \) = \( -325 \, {kJ/mol} \), and - \( {D} \) = \( -295 \, {kJ/mol} \). Based on electron gain enthalpy, element \( {D} \) possesses the highest ionic character, while element \( {A} \) has the least. Therefore, the order of ionic character is \( {ED}>{EC}>{EB}>{EA} \), as indicated by option (3).

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