To determine the correct order of acidity, we compare the stability of the conjugate bases formed after loss of $H^+$.

1. General principle

For carboxylic acids and related compounds:

• Greater stabilization of the conjugate base $\Rightarrow$ stronger acid.
• Electron-withdrawing groups increase acidity by stabilizing the carboxylate ion.
• Electron-donating groups decrease acidity.
• The effect is stronger when the substituent is closer to the acidic group.

2. Comparison of the given compounds

From the structures (I, II, III, IV), we compare the substituent effects near the acidic group.

• Compound I has the strongest electron-withdrawing influence near the acidic proton, so it is the most acidic.
• Compound II has less electron-withdrawing effect than I, but still more than III and IV.
• Compound III has still weaker stabilization of the conjugate base.
• Compound IV has the least stabilization / greatest electron-donating influence, so it is the least acidic.

Thus the acidity order is: $I > II > III > IV$

3. Match with options

• A: $I > II > III > IV$ ✅
• B: $III > I > II > IV$ ❌
• C: $III > IV > II > I$ ❌
• D: $I > III > IV > II$ ❌

Therefore, the correct option is: $\boxed{A}$

4. Verification with stored answer

Stored correct answer: A

My derived answer: A

They agree.