Q4 / 75 · JEE Main 2025 (29 Jan Shift 1) · Chemistry

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 :

  • A
    \(\mathrm{ED}\gt\mathrm{EC}\gt\mathrm{EB}\gt\mathrm{EA}\)
  • B
    \(\mathrm{EA}\gt\mathrm{EB}\gt\mathrm{EC}\gt\mathrm{ED}\)
  • Correct answer
    C
    \(\mathrm{EB}\gt\mathrm{EA}\gt\mathrm{EC}\gt\mathrm{ED}\)
  • D
    \(\mathrm{ED}\gt\mathrm{EC}\gt\mathrm{EA}\gt\mathrm{EB}\)

Solution

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

Hence, the correct answer is option: 'C'.

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