Resonance in can be represented as The enthalpy of formation of is The magnitude of resonance energy of is ____ kJ (nearest integer value). Given: Bond energies of , , and are 940, 410, 500 and 602 kJ respectively. Valence

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Solution

Q.
Resonance in $X_{2} Y$ can be represented as

The enthalpy of formation of $X_{2} Y (X \equiv X (g) + \frac{1}{2} Y = Y (g) \to X_{2} Y (g))$ is $80 {kJ mol}^{- 1} .$

The magnitude of resonance energy of $X_{2} Y$ is ____ kJ ${mol}^{- 1}$ (nearest integer value).

Given: Bond energies of $X \equiv X$ , $X = X$ , $Y = Y$ and $X = Y$ are 940, 410, 500 and 602 kJ ${mol}^{- 1}$ respectively.

Valence $X : 3, Y : 2$ [2025]

Ans.
(98)

$X \equiv X (g) + \frac{1}{2} Y = Y (g) ⟶ \overset{(-)}{X} = \overset{(+)}{X} = Y, Δ H_{X_{2} Y} (theoretical)$

$Δ H_{X_{2} Y} (theoretical) = \sum (B.E(reactants)) - \sum (B.E(products))$

$= [B.E (X \equiv X) + \frac{1}{2} B.E (Y = Y)] - [B.E (X = X) + B.E (X = Y)]$

$= [940 + \frac{1}{2} \times 500] - [410 + 602]$

$= 178 kJ/mol$

$Δ H_{X_{2} Y} (exp) = 80 kJ/mol$

$Δ H_{resonance energy} = Δ H_{X_{2} Y} (theoretical) - Δ H_{X_{2} Y} (exp)$

$= (178 - 80) kJ/mol = 98 kJ/mol$

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