Chemical Bonding and Molecular Structure Part 3

Concept:Bond dissociation enthalpy trends and Fajans’ rule for covalent character are tested.Explanation:Statement I: Bond dissociation enthalpy generally decreases down the group due to increasing bond length.But $F_2$ is an exception because of strong lone pair–lone pair repulsion on small fluorine atoms.Hence, the correct order is $Cl_2 > Br_2 > F_2 > I_2$.So, Statement I is true.Statement II: According to Fajans’ rule, higher cation charge and smaller size increase covalent character.For $\mathrm{SnCl_4}$ vs $\mathrm{SnCl_2}$: $Sn^{4+}$ has higher charge and smaller size than $Sn^{2+}$, so $\mathrm{SnCl_4}$ is more covalent. This is true.For $\mathrm{PbCl_4}$ vs $\mathrm{PbCl_2}$: $Pb^{4+}$ polarises more than $Pb^{2+}$, so $\mathrm{PbCl_4}$ has more covalent character. This is true.For $\mathrm{UF_4}$ vs $\mathrm{UF_6}$: $U^{6+}$ has higher charge than $U^{4+}$, so $\mathrm{UF_6}$ should be more covalent.But the statement claims $\mathrm{UF_4} > \mathrm{UF_6}$, which is incorrect.Thus, Statement II is false.Shortcut:For bond dissociation enthalpy, remember the mnemonic “Cl > Br > F > I”; for Fajans’ rule, higher charge = more covalent character.Answer:Statement I is true but Statement II is false.Correct option: D.