An element ' \(E^{\prime}\) has the ionisation enthalpy value of \(374 \mathrm{~kJ} \mathrm{~mol}^{-1}\). '…

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

Solution

The element having high value of Electron gain enthalpy (magnitude) will form a compound having higher ionic character so order of ionic character
$E B>E A>E C>E D$

Asked in: JEE Main 2025 (29 Jan Shift 1)

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