1. Identify element $E$ from ionisation enthalpy

The given ionisation enthalpy of $E$ is $374\ \text{kJ mol}^{-1}$, which is very low and corresponds to an alkali metal. This value is closest to potassium (K).

So, compounds $EA, EB, EC, ED$ are essentially ionic compounds of an alkali metal with different anions.

2. Use Fajan's rule / electronegativity tendency

Since the cation is the same in all four compounds ($E^+$), the ionic character mainly depends on the nature of the anion.

Greater tendency to gain electron (more negative electron gain enthalpy) means the element is more non-metallic and forms a more ionic bond with an alkali metal.

Given electron gain enthalpies:

• $A = -328\ \text{kJ mol}^{-1}$
• $B = -349\ \text{kJ mol}^{-1}$
• $C = -325\ \text{kJ mol}^{-1}$
• $D = -295\ \text{kJ mol}^{-1}$

More negative value $\Rightarrow$ greater electron-accepting tendency $\Rightarrow$ more ionic compound with $E$.

Thus the order of ionic character follows:

Hence,

3. Check options

• A: $EB > EA > EC > ED$ ✅
• B: $EA > EB > EC > ED$ ❌
• C: $ED > EC > EA > EB$ ❌
• D: $ED > EC > EB > EA$ ❌

4. Final Answer

The correct order is:

So, Option A is correct.