Arrange the following bonds according to their average bond energies in descending order: $\mathrm{C}-\mathrm{Cl}, \mathrm{C}-\mathrm{Br}, \mathrm{C}-\mathrm{F}, \mathrm{C}-\mathrm{I}$

(A) $\mathrm{C}-\mathrm{F}>\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}$

(B) $\mathrm{C}-\mathrm{I}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{F}$

(C) $\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}>\mathrm{C}-\mathrm{F}$

(D) $\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}>\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{F}$

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Chemical Bonding and Molecular Structure

Arrange the following bonds according to their average bond energies in descending order:

C - Cl, C - Br, C - F, C - I

Answer: (A)

Bond energy falls as bond length grows. Down the halogens from fluorine to iodine the atom gets larger and the

C - X

bond lengthens and weakens, so the shortest, strongest bond is

C - F

and the longest, weakest is

C - I

. The descending order is

C - F > C - Cl > C - Br > C - I

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option