Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Option (A): $\frac{1}{2}C{l}_{2(g)}\to C{l}_{(g)}^{-}$ represents the sum of $\frac{1}{2}\Delta H_{BE}$ and $\Delta H_{eg}$ for Chlorine. From the table, this value is $-228kJ/mol$.

Option (B): $\frac{1}{2}B{r}_{2(g)}\to B{r}_{(g)}^{-}$ for Bromine is calculated as $112+(-325)=-213kJ/mol$.

Option (C): $B{r}_{(g)}\to \frac{1}{2}B{r}_{2(g)}$ is the reverse of bond dissociation, equaling $-112kJ/mol$.

Option (D): $C{l}_{(g)}\to C{l}_{(g)}^{-}$ is simply the electron gain enthalpy ($\Delta H_{eg}$) of Chlorine, which is $-349kJ/mol$.

Since $-349kJ/mol$ is the most negative value, Option (D) is the most exothermic change.