Question

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 \( SnCl_2>SnCl_4 \), \( PbCl_2>PbCl_4 \) and \( UF_4>UF_6 \).
In the light of the above statements, choose the correct answer from the options given below :

• A. Both Statement I and Statement II are true
• B. Statement I is false but Statement II is true
• C. Statement I is true but Statement II is false
• D. Both Statement I and Statement II are false

Hint:

Fluorine's anomalous low bond enthalpy is a classic exam question—always attribute it to inter-electronic repulsion of lone pairs due to its tiny size.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Bond dissociation enthalpy in halogens is affected by bond length and lone pair-lone pair repulsions.
Covalent character in metal halides is determined by Fajan's Rules.

Step 3: Detailed Explanation:
Statement I: In general, bond strength decreases down a group as size increases. However, \( F_2 \) has a very small bond length, leading to significant repulsion between the non-bonding lone pairs on adjacent fluorine atoms. This weakens the F-F bond, making it weaker than Cl-Cl and Br-Br bonds. The correct experimental order is \( Cl_2 (242.6)>Br_2 (192.8)>F_2 (158.8)>I_2 (151.1) \text{ kJ/mol} \). This statement is true.
Statement II: According to Fajan's Rules, smaller cations with higher positive charges have greater polarizing power and thus more covalent character. In pairs like \( SnCl_2/SnCl_4 \), \( Sn^{4+} \) is smaller and more highly charged than \( Sn^{2+} \), so \( SnCl_4 \) is more covalent than \( SnCl_2 \). Similarly, \( PbCl_4>PbCl_2 \) and \( UF_6>UF_4 \) in terms of covalent character. The statement provides the reverse trend. This statement is false.

Step 4: Final Answer:
Statement I is true, but Statement II is false.