Question

The correct sequence of bond enthalpy of ' $C - X$' bond is:

• A. $CH _{3}- F < CH _{3}- Cl < CH _{3}- Br < CH _{3}- I$
• B. $CH _{3}- F > CH _{3}- Cl > CH _{3}- Br > CH _{3}- I$
• C. $CH _{3}- F < CH _{3}- Cl > CH _{3}- Br > CH _{3}- I$
• D. $CH _{3}- Cl > CH _{3}- F > CH _{3}- Br > CH _{3}- I$

Answer 1

The Correct Option is B

In order to solve the question about the correct sequence of the bond enthalpy for the \(C - X\) bond in various methyl halides, let's analyze the concept of bond enthalpy and how it relates to halogen atoms. 

Bond enthalpy (or bond energy) is the energy required to break a bond between two atoms in a molecule. When comparing bond enthalpies of similar types of molecules, such as CH₃-X (where X is a halogen), the bond enthalpy generally decreases as the size of the halogen atom increases. This is because larger atoms have their valence electrons farther from the nucleus, which results in weaker bond strength.

Let's examine the bond enthalpy sequence for methyl halides:

1. Methyl Fluoride (CH₃-F): Fluorine is the smallest halogen. Due to its small size, the C-F bond is very strong, resulting in high bond enthalpy.
2. Methyl Chloride (CH₃-Cl): Chlorine is larger than fluorine, which means the C-Cl bond is weaker than the C-F bond.
3. Methyl Bromide (CH₃-Br): Bromine is even larger than chlorine, leading to a further decrease in bond strength and bond enthalpy.
4. Methyl Iodide (CH₃-I): Iodine is the largest among the halogens listed, thus the C-I bond is the weakest and has the lowest bond enthalpy.

Based on this understanding, the correct sequence of bond enthalpy for the \(C - X\) bond is:

\(CH _{3}- F > CH _{3}- Cl > CH _{3}- Br > CH _{3}- I\)

This option correctly reflects the trend in bond strengths due to the increasing size and thus decreasing bond enthalpy as you move from fluorine to iodine.

Therefore, the correct answer is the option: \(CH _{3}- F > CH _{3}- Cl > CH _{3}- Br > CH _{3}- I\).