For the compounds CH$$_3$$Cl, CH$$_3$$Br, CH$$_3$$I and CH$$_3$$F, the correct order of increasing C-halogen bond length is:

To determine the correct order of increasing carbon-halogen bond length in the compounds CH$$_3$$F, CH$$_3$$Cl, CH$$_3$$Br, and CH$$_3$$I, we need to understand how bond length relates to the halogen atoms. Bond length is the distance between the nuclei of two bonded atoms. Since the carbon atom is the same in all these compounds, the bond length depends primarily on the size of the halogen atom bonded to it.

Halogens belong to Group 17 of the periodic table. As we move down the group, the atomic size increases due to the addition of electron shells. The atomic radii increase in the order: fluorine (F) < chlorine (Cl) < bromine (Br) < iodine (I). Therefore, the covalent radius of fluorine is the smallest, and iodine is the largest.

In a covalent bond, the bond length is approximately the sum of the covalent radii of the bonded atoms. Since carbon has a fixed covalent radius, the carbon-halogen bond length will increase as the covalent radius of the halogen increases. Thus, the C-F bond will be the shortest, and the C-I bond will be the longest.

Let us list the bonds:

• CH$$_3$$F has a C-F bond
• CH$$_3$$Cl has a C-Cl bond
• CH$$_3$$Br has a C-Br bond
• CH$$_3$$I has a C-I bond

Based on the halogen sizes, the bond lengths should follow the order: C-F < C-Cl < C-Br < C-I. Therefore, the compounds should be ordered as: CH$$_3$$F < CH$$_3$$Cl < CH$$_3$$Br < CH$$_3$$I.

Now, comparing with the given options:

• Option A: CH$$_3$$F < CH$$_3$$Br < CH$$_3$$Cl < CH$$_3$$I — This is incorrect because C-Br bond length is greater than C-Cl, so CH$$_3$$Br should come after CH$$_3$$Cl.
• Option B: CH$$_3$$F < CH$$_3$$I < CH$$_3$$Br < CH$$_3$$Cl — This is incorrect because C-I bond is longer than C-Br, so CH$$_3$$I should be last, not second.
• Option C: CH$$_3$$F < CH$$_3$$Cl < CH$$_3$$Br < CH$$_3$$I — This matches our predicted order.
• Option D: CH$$_3$$Cl < CH$$_3$$Br < CH$$_3$$F < CH$$_3$$I — This is incorrect because CH$$_3$$F has the shortest bond and should be first, not third.

This order is also supported by the trend in bond strength. The C-F bond is the strongest due to better orbital overlap with the small fluorine atom, and bond strength decreases down the group (C-F > C-Cl > C-Br > C-I). Stronger bonds are shorter, confirming the bond length order.

Hence, the correct answer is Option C.