JEE Main 2025 Jan 29 Shift 1, Chemistry Q17: Enthalpies for Different Types of Reaction

JEE Main2025Jan 29, Shift 1Chemistry

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 $- 295 kJ mo l^{- 1}$ , respectively.

The correct order of the products EA, EB, EC and ED in terms of ionic character is :

A
EB $>$ EA $>$ EC $>$ ED
B
ED $>$ EC $>$ EA $>$ EB
C
EA $>$ EB $>$ EC $>$ ED
D
ED $>$ EC $>$ EB $>$ EA

Solution

Ionic character of $E - X$ increases when the energy cost of $E \to E^{+}$ (its IE) is closely compensated by the energy released in $X + e^{-} \to X^{-}$ (its $Δ H_{e g}$ ). The smaller the net energy required, $∣ IE + Δ H_{e g} ∣$ , the more favourable the ionic salt.

Compute $I E_{E} + Δ H_{e g}$ in kJ mol $^{- 1}$ (smaller is more ionic):

EA: $374 - 328 = 46$ ; EB: $374 - 349 = 25$ ; EC: $374 - 325 = 49$ ; ED: $374 - 295 = 79$ .

Smaller net value $\Rightarrow$ greater ionic character: EB (25) $>$ EA (46) $>$ EC (49) $>$ ED (79).

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JEE Main 2025 Jan 29 Shift 1, Chemistry Q17: Enthalpies for Different Types of Reaction

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