Rank the indicated bonds in order of increasing bond dissociation energy.

The energy applied to dissociate a bond is called bond dissociation energy.

The bond dissociation energy is directly proportional to the strength of the bond.

Inorganic compounds, alkanes have the least bond dissociation energy. Alkenes have higher bond dissociation energy. In comparison, alkynes have the maximum bond dissociation energy.

There are various factors on which the amount of energy required to break the bonds depends. These factors are described hereunder:

Hence, the more the number of electrons is shared, the number of bonds and bond strength increases between two atoms. Hence, bond dissociation energy increases and vice-versa.

In the given three structures, the three halogens, i.e., chlorine, iodine, and bromine, are bonded with a single bond to the same chemical moiety.

The basis of comparing the Bond Dissociation energy for the three given structures is the atomic size of the halogen atoms attached.

Among $\mathrm{Cl}-{\mathrm{CCl}}_3,\mathrm{Br}-{\mathrm{CCl}}_3\mathrm{and}\mathrm{I}-{\mathrm{CCl}}_3$, chlorine has the smallest size, bromine has the intermediate size, and iodine has the largest size. Hence, the Cl-C bond length is the smallest, Br-C bond length is intermediate, and I-C bond length is the largest. Comparison of bond lengths

Thus, the bond strength increases in the order: I-C < Br-C <Cl-C

Hence, the bond dissociation energy also has the same order as the bond strength as $\mathrm{I}-{\mathrm{CCl}}_3<\mathrm{Br}-{\mathrm{CCl}}_3<\mathrm{Cl}-{\mathrm{CCl}}_3$.