$\text{Arrange the following acids in increasing order of their acidic strengths:}$

$\text{<br>}$

$\text{HCOOH, FCH₂COOH, NO₂CH₂COOH, ClCH₂COOH}$

The correct option is:
$\text{HCOOH < ClCH₂COOH < FCH₂COOH < NO₂CH₂COOH}$

Step-by-Step Explanation:

Acidic strength of carboxylic acids depends on the stability of the carboxylate anion formed after losing a proton ($H+$). Electron-withdrawing groups increase acidic strength by dispersing the negative charge on the carboxylate group through the inductive effect (-I effect) or resonance effect. Conversely, electron-donating groups decrease acidic strength by destabilizing the carboxylate anion.

Let us analyze the structures and substituents of the given acids:
1. Formic acid ($\text{HCOOH}$): It has no electron-withdrawing groups attached to the carboxyl carbon. The hydrogen atom has negligible inductive effect.
2. Monochloroacetic acid ($\text{ClCH₂COOH}$): It contains a chlorine ($\text{Cl}$) atom, which is moderately electronegative and exerts a moderate electron-withdrawing inductive effect (-I effect).
3. Monofluoroacetic acid ($\text{FCH₂COOH}$): It contains a fluorine ($\text{F}$) atom. Since fluorine is more electronegative than chlorine, its electron-withdrawing inductive effect (-I effect) is stronger than that of chlorine ($\text{-I of F > -I of Cl}$).
4. Nitroacetic acid ($\text{NO₂CH₂COOH}$): It contains a nitro group ($\text{-NO₂}$). The nitro group is a very strong electron-withdrawing group and exerts a significantly stronger -I effect compared to both halogens ($\text{-NO₂ > -F > -Cl}$).

Comparing the electron-withdrawing strength of the substituents, we have the order:
$\text{-H < -Cl < -F < -NO₂}$

Since a stronger electron-withdrawing group stabilizes the conjugate base more effectively, the acidic strength increases in the same order.

Therefore, the increasing order of acidic strength is:
$\text{HCOOH < ClCH₂COOH < FCH₂COOH < NO₂CH₂COOH}$