Question

What is the order of bond energy between C=S & C=Te, and between Cl–Cl & F–F?

• A. C=S $>$ C=Te and Cl–Cl $>$ F–F
• B. C=Te $>$ C=S and Cl–Cl $>$ F–F
• C. C=Te $>$ C=S and F–F $>$ Cl–Cl
• D. C=S $>$ C=Te and F–F $>$ Cl–Cl

Hint:

The weak $\text{F}-\text{F}$ bond is a classic periodic exception:
Because of high lone-pair-lone-pair repulsion in the tiny $\text{F}_2$ molecule, the halogen bond energy order is: $\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2$.
Keep this exception in mind, as it is tested frequently!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Bond energy depends on the effectiveness of orbital overlap and the presence of destabilizing factors like lone-pair repulsions.

Step 2: Detailed Explanation:
$\bullet$ C=S vs C=Te: Carbon is in period 2. Sulfur is period 3, while Tellurium is period 5. The overlap between $2p$ (Carbon) and $3p$ (Sulfur) is much more effective than between $2p$ and the large, diffuse $5p$ (Tellurium). Smaller atoms and better size matching lead to stronger bonds. Thus, C=S $>$ C=Te.
$\bullet$ Cl–Cl vs F–F: This is a famous exception. Usually, bond energy increases as atoms get smaller. However, Fluorine is so small that the lone-pair-lone-pair repulsions on the two F atoms are extremely intense, weakening the bond significantly. Chlorine, being larger, minimizes these repulsions. Thus, Cl–Cl $>$ F–F.

Step 3: Final Answer:
The orders are C=S $>$ C=Te and Cl–Cl $>$ F–F.
This corresponds to option (A).