Given below are two statements:
Statement I: The correct order in terms of bond dissociation enthalpy is $$Cl_{2} > Br_{2} > F_{2} > I_{2}$$.
Statement II : The correct trend in the covalent character of the metal halides is $$[SnCl_{4} > SnCl_{2}]$$, $$[PbCl_{4}> PbCl_{2}]$$, and $$[UF_{4} > UF_{6}]$$.
In The light oh the above statements, choose the correct answer from the options given below:

We need to evaluate two statements about bond dissociation enthalpies and covalent character of metal halides.

First, consider Statement I, which claims that the correct order in terms of bond dissociation enthalpy is $$Cl_2 > Br_2 > F_2 > I_2$$. The bond dissociation enthalpies (in kJ/mol) are: $$Cl\text{-}Cl$$: 242, $$Br\text{-}Br$$: 193, $$F\text{-}F$$: 159, $$I\text{-}I$$: 151.

One might expect $$F_2$$ to have the highest bond dissociation enthalpy since fluorine is the smallest halogen and a shorter bond tends to be stronger; however, $$F_2$$ has an anomalously low bond dissociation enthalpy due to strong lone pair-lone pair repulsion between the two small fluorine atoms. The three lone pairs on each fluorine atom are very close to each other, causing significant interelectronic repulsion that weakens the bond.

Since the experimental enthalpies follow the order $$Cl_2 > Br_2 > F_2 > I_2$$, Statement I is true.

Statement II asserts that the correct trend in covalent character is $$SnCl_4 > SnCl_2$$, $$PbCl_4 > PbCl_2$$, and $$UF_4 > UF_6$$. According to Fajans’ rules, covalent character increases when the cation has a higher charge (more polarizing power), when the cation has a smaller size, and when the anion is larger and more polarizable.

In $$SnCl_4$$, tin is in the +4 oxidation state, giving it greater polarizing power than in $$SnCl_2$$ where tin is in the +2 oxidation state; accordingly, $$SnCl_4$$ is more covalent than $$SnCl_2$$, which agrees with the statement. Similarly, $$PbCl_4$$ (Pb in +4) is more covalent than $$PbCl_2$$ (Pb in +2), so that part of the trend is also correct.

In contrast, for the uranium fluoride pair, $$UF_6$$ contains U6+ (higher charge and more polarizing) while $$UF_4$$ contains U4+ (lower charge), so by Fajans’ rules $$UF_6$$ should have greater covalent character than $$UF_4$$. The statement claims $$UF_4 > UF_6$$, which is incorrect. Therefore, Statement II is false.

The correct answer is Option (3): Statement I is true but Statement II is false.