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Calculate the resonance energy of nitrous oxide from the following data
${ΔH}_{f}^{0}$ of $N_{2} O (N = N = O)$ is $82 k J {mol}^{- 1}$ ; $N \equiv N : 946 k J {mol}^{- 1}$ ;
$N = N : 418 k J {mol}^{- 1}$ ; $O = O : 498 k J {mol}^{- 1}$ ;
$N = O : 607 k J {mol}^{- 1}$

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$- 54 k J {mol}^{- 1}$

$- 88 k J {mol}^{- 1}$

$- 22 k J {mol}^{- 1}$

$- 44 k J {mol}^{- 1}$

answer is B.

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Detailed Solution

$N_{2} O$ structure N = N = O

Given $N_{2} + \frac{1}{2} O_{2} \to N_{2} O, ∆ H_{f}^{0} = 82 {KJmol}^{- 1}$

$Δ H_{f}^{0} = [N \equiv N] + \frac{1}{2} (O = O) - [1 N = N] - 1 [N = O] = [946 + \frac{1}{2} (498)] - [418 + 607] = + 170 kJ$ $Resonance energy = ExP . {ΔH}_{f} (N_{2} O) - cal . {ΔH}_{f} (N_{2} O) = 82 - 170 = - 88 k J {mol}^{- 1}$

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