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 Question:
This question asks for the comparative order of bond dissociation energies between a carbon-sulfur double bond ($\text{C}=\text{S}$) and a carbon-tellurium double bond ($\text{C}=\text{Te}$), and between single bonds in chlorine ($\text{Cl}-\text{Cl}$) and fluorine ($\text{F}-\text{F}$).

Step 2: Detailed Explanation:
Let's analyze both pairs of bonds:

• 1. C=S versus C=Te:
- Bond energy is strongly dependent on the extent of orbital overlap.
- Carbon is a second-period element with $2p$ valence orbitals.
- Sulfur is a third-period element ($3p$ valence orbitals), while Tellurium is a fifth-period element ($5p$ valence orbitals).
- The overlap between the small $2p$ orbital of Carbon and the relatively smaller $3p$ orbital of Sulfur is much more effective than the overlap with the large, diffuse $5p$ orbital of Tellurium.
- Consequently, the $\text{C}=\text{S}$ double bond is significantly shorter, more stable, and possesses a higher bond energy than the diffuse $\text{C}=\text{Te}$ double bond.
- Hence: $\text{C}=\text{S} > \text{C}=\text{Te}$.

• 2. Cl–Cl versus F–F:
- Generally, bond energy decreases down a group as atomic size increases. Based on this, one might expect the $\text{F}-\text{F}$ bond to be stronger than the $\text{Cl}-\text{Cl}$ bond.
- However, Fluorine ($\text{F}_2$) is an anomalous case.
- The fluorine atom is extremely small, meaning the non-bonding valence electron lone pairs on the two adjacent fluorine atoms are forced into very close proximity.
- This creates intense, destabilizing lone-pair-lone-pair electrostatic repulsion in the $\text{F}-\text{F}$ single bond.
- In chlorine ($\text{Cl}_2$), the larger atomic size and more diffuse $3p$ orbitals minimize this lone-pair repulsion, making the $\text{Cl}-\text{Cl}$ single bond stronger and more stable than the $\text{F}-\text{F}$ bond.
- Hence: $\text{Cl}-\text{Cl} > \text{F}-\text{F}$.

Step 3: Final Answer:
The correct orders are $\text{C}=\text{S} > \text{C}=\text{Te}$ and $\text{Cl}-\text{Cl} > \text{F}-\text{F}$, which corresponds to option (A).