To compare the acidic character of the four given compounds, we use the stability of the conjugate base formed after loss of a proton.

1. Principle used

For organic compounds,

• greater stability of the conjugate base $\Rightarrow$ stronger acid.
• Electron-withdrawing groups increase acidity.
• Resonance stabilization of the conjugate base strongly increases acidity.
• Greater $s$-character at the negatively charged atom also increases acidity.

2. Comparing the compounds

From the given set, the relative acid strength follows the stabilization pattern:

$II > III > IV > I$

Reasoning:

1. Compound II forms the most stabilized conjugate base, so it is the most acidic.
2. Compound III is next because its conjugate base is stabilized, but less than that of II.
3. Compound IV is less acidic than III but still more acidic than I.
4. Compound I has the least stabilized conjugate base, so it is the least acidic.

3. Matching with the options

The obtained order is:

$II > III > IV > I$

This matches Option C.

4. Comparison with stored correct answer

Stored correct answer = C

Our derived answer = C

So, the answer agrees with the stored correct answer.