Which one of the following arrangements does not give the correct picture of the trends indicated against it?

Which one of the following arrangements does not give the correct picture of the trends indicated against it?
  1. $\mathrm{F}_2 > \mathrm{Cl}_2 > \mathrm{Br}_2 > \mathrm{I}_2$ : Oxidising power
  2. $\mathrm{F}_2 > \mathrm{Cl}_2 > \mathrm{Br}_2 > \mathrm{I}_2$ : Electron gain enthalpy
  3. $\mathrm{F}_2 > \mathrm{Cl}_2 > \mathrm{Br}_2 > \mathrm{I}_2$ : Bond dissociation energy
  4. $\mathrm{F}_2 > \mathrm{Cl}_2 > \mathrm{Br}_2 > \mathrm{I}_2$ : 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 :
$\mathrm{Cl}_2 > \mathrm{Br}_2 > \mathrm{F}_2 > \mathrm{I}_2$

Asked in: NEET 2008 (Screening)

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