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

To determine the correct sequence of bond enthalpy for the \(C - X\) bond in halomethanes (\(CH_3 - X\)), we need to understand the concept of bond enthalpy and how it relates to different halogens (F, Cl, Br, I). Bond enthalpy is the energy required to break a bond in one mole of gaseous molecules. It is a measure of bond strength; higher bond enthalpy indicates a stronger bond.

The strength of the \(C - X\) bond decreases as the size of the halogen increases. This is because as the atomic size increases down the group (from F to I), the bond length increases and the bond becomes weaker.

For the halogens bonded to the methyl group (\(CH_3\)), this order of size is:

• Fluorine (F) is the smallest
• Chlorine (Cl)
• Bromine (Br)
• Iodine (I) is the largest

The bond enthalpy follows the reverse order of atomic size due to increased bond length and decreased bond strength with larger halogens. Therefore, the correct bond enthalpy sequence is:

• \(CH_3-F\) has the highest bond enthalpy because F is the smallest and forms the strongest bond with carbon.
• \(CH_3-Cl\) has less bond enthalpy compared to \(CH_3-F\).
• \(CH_3-Br\) has less bond enthalpy compared to \(CH_3-Cl\).
• \(CH_3-I\) has the lowest bond enthalpy because I is the largest and forms the weakest bond with carbon.

Hence, the correct order is \(CH_3-F > CH_3-Cl > CH_3-Br > CH_3-I\).

This reasoning matches with the given correct answer option:

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