JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
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 \(\mathrm{EA}\gt\mathrm{EB}\gt\mathrm{EC}\gt\mathrm{ED}\)
- C \(\mathrm{EB}\gt\mathrm{EA}\gt\mathrm{EC}\gt\mathrm{ED}\)
- D \(\mathrm{ED}\gt\mathrm{EC}\gt\mathrm{EA}\gt\mathrm{EB}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{EB}\gt\mathrm{EA}\gt\mathrm{EC}\gt\mathrm{ED}\)
Step-by-step Solution
Detailed explanation
The element having high value of Electron gain enthalpy (magnitude) will form a compound having higher ionic character so order of ionic character \(E B>E A>E C>E D\)
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