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

The correct sequence of bond enthalpy for the carbon‑halogen bonds in methyl halides is:

CH₃−F > CH₃−Cl > CH₃−Br > CH₃−I

To understand why this order holds, consider two main factors that govern bond dissociation energy (BDE):

1. Electronegativity of the halogen. A larger difference in electronegativity between carbon (2.55) and the halogen leads to a more polar, stronger covalent bond. Fluorine is the most electronegative element (3.98), followed by chlorine (3.16), bromine (2.96), and iodine (2.66). Thus the C‑F bond benefits from the greatest polarity and strongest attraction.

2. Atomic size and bond length. As the halogen gets larger, the C‑X bond length increases, reducing orbital overlap and weakening the bond. The covalent radii increase from F (≈ 60 pm) to Cl (≈ 100 pm), Br (≈ 115 pm), and I (≈ 133 pm). Longer bonds are weaker.

Combining these trends, the bond dissociation energies (typical experimental values) follow the same order:

$\Delta H\_\{C‑F\}\; \approx \; 485 kJ mol⁻¹$
$\Delta H\_\{C‑Cl\}\; \approx \; 350 kJ mol⁻¹$
$\Delta H\_\{C‑Br\}\; \approx \; 285 kJ mol⁻¹$
$\Delta H\_\{C‑I\}\; \approx \; 213 kJ mol⁻¹$

Because the C‑F bond is the most energetic (strongest) and the C‑I bond is the least energetic (weakest), the sequence of bond enthalpies is exactly as stated above.

Therefore, the decreasing order of bond strength—and hence bond enthalpy—is:

CH₃−F > CH₃−Cl > CH₃−Br > CH₃−I