Given below are two statements: Statement I: The correct order in terms of bond dissociation enthalpy is textCl_2 > textBr_2 > textF_2 > textI_2$\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2$. Statement II: The correct trend in the covalent character of the metal halides is [textSnCl_4 > textSnCl_2]$[\text{SnCl}_4 > \text{SnCl}_2]$, [textPbCl_4 > textPbCl_2]$[\text{PbCl}_4 > \text{PbCl}_2]$ and [textUF_4 > textUF_6]$[\text{UF}_4 > \text{UF}_6]$ (or similar Fajan's rule trend). In the light of the above statements, choose the correct answer from the options given below:

### Core Logic - Statement I: Bond dissociation energy order for halogens is textCl_2 > textBr_2 > textF_2 > textI_2$\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2$ due to small size and lone-pair repulsions in fluorine weakening its bond. Statement I is true. - Statement II: According to Fajan's rules, higher charge on cation increases covalent character, so textUF_6 > textUF_4$\text{UF}_6 > \text{UF}_4$ (higher oxidation state has greater covalent character), making the statement II claim regarding textUF_4 > textUF_6$\text{UF}_4 > \text{UF}_6$ false. ### Step 1: Final Conclusion Statement I is true but Statement II is false, corresponding to option (1). ### Pattern Recognition Sees: halogen bond dissociation energy anomalies and Fajan's rules for covalent character. Trap: Assuming fluorine has the highest bond dissociation energy among halogens. ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure