Question

The correct sequence of bond enthalpy of ‘C–X’ bond is:

• A. CH₃–F>CH₃–Cl>CH₃–Br>CH₃–I
• B. CH₃–F<CH₃–Cl>CH₃–Br>CH₃–I
• C. CH₃–Cl>CH₃–F>CH₃–Br>CH₃–I
• D. CH₃–F<CH₃–Cl<CH₃–Br<CH₃–I

Hint:

Small atoms form strong bonds because they can get closer to each other, resulting in better orbital overlap. Fluorine is the smallest halogen, so C–F is the strongest bond.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Bond enthalpy is the energy required to break a chemical bond. It is inversely proportional to the bond length. As the size of the halogen atom increases down the group, the bond length increases, making the bond weaker and easier to break.
Step 2: Key Formula or Approach:
Bond Strength $\propto$ $\frac{1}{\text{Bond Length}}$ $\propto$ $\frac{1}{\text{Atomic Size of X}}$.
Step 3: Detailed Explanation:
1. Atomic size order: F<Cl<Br<I.
2. Bond length order: C–F<C–Cl<C–Br<C–I.
3. Because the C–F bond is the shortest, the overlap of orbitals is most effective, resulting in the highest bond enthalpy.
4. Consequently, bond enthalpy decreases as we move from Fluorine to Iodine.
Step 4: Final Answer:
The correct sequence is CH₃–F>CH₃–Cl>CH₃–Br>CH₃–I.