Question

The enthalpy change for the exothermic reaction between BeI\(_2\) and HgF\(_2\) is ........... kJ mol\(^{-1}\) (rounded off to the nearest integer).
(Given bond dissociation energies in kJ mol\(^{-1}\): Be–F = 632, Be–I = 289, Hg–F = 268, Hg–I = 145)

Hint:

For bond-energy estimates: \(\Delta H \approx \sum D(\text{bonds broken})-\sum D(\text{bonds formed})\). Halogen exchange tends to form the stronger metal–halogen bonds (here, Be–F and Hg–I), driving the reaction exothermic.

Answer 1

Step 1: Write the reaction using halide exchange. \[ \mathrm{BeI_2 + HgF_2 ⇒ BeF_2 + HgI_2} \] Step 2: Apply the bond-energy method. Break: \(2\times\)Be–I \(=2(289)=578\) kJ; \(2\times\)Hg–F \(=2(268)=536\) kJ. \[ E_{\text{broken}}=578+536=1114\ \text{kJ} \] Form: \(2\times\)Be–F \(=2(632)=1264\) kJ; \(2\times\)Hg–I \(=2(145)=290\) kJ. \[ E_{\text{formed}}=1264+290=1554\ \text{kJ} \] Step 3: Compute \(\Delta H\). \[ \Delta H \approx E_{\text{broken}}-E_{\text{formed}} =1114-1554=-440\ \text{kJ mol}^{-1}. \] Thus the reaction is exothermic by approximately \(-440\ \text{kJ mol}^{-1}\).