Question

Which of the following is the correct decreasing order of bond dissociation enthalpy of halogens ?

• A. \(\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2\)
• B. \(\text{I}_2 > \text{Br}_2 > \text{Cl}_2 > \text{F}_2\)
• C. \(\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2\)
• D. \(\text{Br}_2 > \text{I}_2 > \text{F}_2 > \text{Cl}_2\)

Hint:

Fluorine is an exception: its small size causes strong lone pair repulsion that weakens the F-F bond.

Answer 1

The Correct Option is C

Step 1: Recall the data
Bond dissociation enthalpies (kJ/mol): \(\text{Cl}_2\) = 242, \(\text{Br}_2\) = 193, \(\text{F}_2\) = 158, \(\text{I}_2\) = 151.

Step 2: Understand the trend
From chlorine to iodine the bond lengthens as the atoms grow, so the bond weakens: \(\text{Cl}_2 > \text{Br}_2 > \text{I}_2\).

Step 3: Why fluorine is unusual
The F-F bond is very short, so the lone pairs on the two small F atoms repel each other strongly. This weakens the bond, and its enthalpy falls below that of chlorine and bromine.

Step 4: Final order
Combining these gives \(\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2\). Option (A) assumes a smooth decrease down the group, and (B) the reverse, while (D) does not fit the data.

Final Answer:
Chlorine has the highest and iodine the lowest bond enthalpy. This is option (C). \[ \boxed{\text{(C) }\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2} \]