Q65 / 200 · NEET 2008 (Screening) · Chemistry
Which one of the following arrangements does not give the correct picture of the trends indicated against it?
- A$\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : Oxidising power
- B$\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : Electron gain enthalpy
- Correct answerC$\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : Bond dissociation energy
- D$\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : Electronegativity
Solution
Key Idea : Generally bond dissaciation energies decreases in a group. Bond dissaciation energy also decreases with repulsion.
$\begin{array}{|l|c|c|c|c|}\hline X-X \text { Bond } & \mathrm{F}-\mathrm{F} & \mathrm{Cl}-\mathrm{Cl} & \mathrm{Br}-\mathrm{Br} & \mathrm{I}-\mathrm{I} \\
\hline \text {Bond length }(Å)) & 1.42 & 1.99 & 2.28 & 2.67 \\
\hline \begin{array}{l}\text {Bond dissociation } \\
\text {energy }(\mathrm{kcal} / \mathrm{mol})\end{array} & 38 & 57 & 45.5 & 35.6 \\ \hline \end{array}$
In general the bond dissociation energy decreases as the bond length increases, but the bond dissociation energy of $\mathrm{F}_2$ is less than that of $\mathrm{Cl}_2$. It is due to greater interelectronic repulsions between the lone pair of electrons on the two bonded fluorine atoms. Hence, the order of bond dissociation energy is as :
$\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2$
Hence, the correct answer is option: 'C'.