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 :

Solution

Chemistry ma’am · 3 steps

E is the same in all four, so compare how eagerly A, B, C and D grab its electron.

1. What decides ionic character. An ionic bond forms when E gives up an electron (374 kJ mol⁻¹ spent) and the other atom takes it. The more energy released when the electron is accepted, the more negative the electron gain enthalpy, and the more easily the ions form.
2. Same donor every time. The ionisation enthalpy of E is the same in all four products, so only the electron gain enthalpy of the partner changes. A more negative value means a more ionic product.

   Product	Electron gain enthalpy of partner	Net energy to form ions, Δᵢ H + Δ_(eg) H	Ionic character
   EB	-349 kJ mol⁻¹	374 - 349 = 25 kJ mol⁻¹	highest
   EA	-328 kJ mol⁻¹	374 - 328 = 46 kJ mol⁻¹	second
   EC	-325 kJ mol⁻¹	374 - 325 = 49 kJ mol⁻¹	third
   ED	-295 kJ mol⁻¹	374 - 295 = 79 kJ mol⁻¹	lowest

3. So. The less energy it costs to make the ions, the more ionic the product. The order is EB > EA > EC > ED, which is option (a).

The idea to keep. With a fixed metal, ionic character follows the partner's electron gain enthalpy: the more negative it is, the cheaper ion formation is and the more ionic the bond.

Answer from the NTA final answer key. Solution by PrepWiser.