Question: 72. The correct sequence of bond enthalpy of ‘ \mathrm{C}-\mathrm{X} ‘ bond is :
(1) \mathrm{CH}{3}-\mathrm{F}<\mathrm{CH}{3}-\mathrm{Cl}<\mathrm{CH}{3}-\mathrm{Br}<\mathrm{CH}{3}-\mathrm{I}
(2) \mathrm{CH}_{3}-\mathrm{F}>\mathrm{CH}_{3}-\mathrm{Cl}>\mathrm{CH}_{3}-\mathrm{Br}>\mathrm{CH}_{3}-\mathrm{I}
(3) \mathrm{CH}{3}-\mathrm{F}<\mathrm{CH}{3}-\mathrm{Cl}>\mathrm{CH}{3}-\mathrm{Br}>\mathrm{CH}{3}-\mathrm{I}
(4) \mathrm{CH}_{3}-\mathrm{Cl}>\mathrm{CH}_{3}-\mathrm{F}>\mathrm{CH}_{3}-\mathrm{Br}>\mathrm{CH}_{3}-\mathrm{I}
Answer: Option (2)
Explanation:
Bond enthalpy is the energy required to break a bond between two atoms in a molecule.
The bond enthalpy of the \mathrm{C}-\mathrm{X} bond depends mainly on the bond length and size of the halogen atom.
As we move from fluorine to iodine, the atomic size of halogen increases, which increases the bond length.
Increase in bond length leads to decrease in bond strength and hence lower bond enthalpy.
Fluorine is the smallest halogen, so the \mathrm{C}-\mathrm{F} bond is the shortest and strongest.
Iodine is the largest halogen, so the \mathrm{C}-\mathrm{I} bond is the longest and weakest.
Therefore, the correct order of bond enthalpy is:
\mathrm{CH}_{3}-\mathrm{F}>\mathrm{CH}_{3}-\mathrm{Cl}>\mathrm{CH}_{3}-\mathrm{Br}>\mathrm{CH}_{3}-\mathrm{I}