The descending order of acidity for the following carboxylic acid is-
(A) CH$$_3$$COOH
(B) F$$_3$$C-COOH
(C) ClCH$$_2$$-COOH
(D) FCH$$_2$$-COOH
(E) BrCH$$_2$$-COOH
Choose the correct answer from the options given below:

We need to arrange the given carboxylic acids in descending order of acidity: (A) CH$$_3$$COOH, (B) F$$_3$$C-COOH, (C) ClCH$$_2$$-COOH, (D) FCH$$_2$$-COOH, and (E) BrCH$$_2$$-COOH.

The acidity of carboxylic acids is enhanced by electron-withdrawing groups through the negative inductive effect (-I effect). The stronger the -I effect, the more the electron density is pulled away from the O-H bond, making it easier to release the proton.

Trifluoroacetic acid (B) has three fluorine atoms, providing a very strong cumulative -I effect, making it the strongest acid in the group.

Now, among the monohaloacetic acids, we compare based on electronegativity: F (3.98) > Cl (3.16) > Br (2.96). Since all three halogens are at the same position (alpha carbon), the -I effect follows the electronegativity order, giving FCH$$_2$$COOH (D) > ClCH$$_2$$COOH (C) > BrCH$$_2$$COOH (E).

CH$$_3$$COOH (A) has no electron-withdrawing substituent, making it the weakest acid.

So the descending order of acidity is:

$$\text{B} > \text{D} > \text{C} > \text{E} > \text{A}$$

Hence, the correct answer is Option C.