Question

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

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

Answer 1

The Correct Option is C

To determine the correct sequence of the bond enthalpy of the 'C–X' bond (where X is a halogen), we need to understand how bond enthalpy varies with the size and electronegativity of the halogen involved. 

Bond enthalpy is a measure of the strength of a chemical bond, defined as the amount of energy required to break one mole of bonds in a gaseous substance. The bond enthalpy of C–X bonds is influenced by two main factors:

1. Bond Length: The smaller the halogen atom, the shorter the bond length. As bond length decreases, bond enthalpy generally increases because the atoms are more tightly held together.
2. Electronegativity: The greater the electronegativity difference between carbon and the halogen, the stronger the bond due to increased polarity.

Considering these factors, the bond enthalpy order from strongest to weakest is as follows:

• CH₃–F: Fluorine is the smallest and most electronegative of the halogens, resulting in a very strong bond.
• CH₃–Cl: Chlorine is larger than fluorine, and although it has a significant electronegativity, the bond length is longer, leading to a weaker bond compared to C–F.
• CH₃–Br: Bromine is larger than chlorine, and therefore the bond is even weaker.
• CH₃–I: Iodine is the largest halogen, resulting in the weakest bond.

Thus, the correct sequence of bond enthalpy from highest to lowest is: CH₃–F > CH₃–Cl > CH₃–Br > CH₃–I.

This corresponds to the following option:

CH₃−F > CH₃−Cl > CH₃−Br > CH₃−I