Recall approximate $\mathrm{p}K_a$ values of the given compounds.

• Ethanol is a simple alcohol, so it is much less acidic than phenols. $\mathrm{p}K_a(\text{ethanol}) \approx 15.9$

• Phenol is more acidic than ethanol because the phenoxide ion is resonance-stabilized. $\mathrm{p}K_a(\text{phenol}) \approx 10.0$

• Nitrophenols are more acidic than phenol because the $-\mathrm{NO_2}$ group is electron-withdrawing and stabilizes the phenoxide ion.

Compare $m$-nitrophenol and $p$-nitrophenol.

• In $p$-nitrophenol, the $-\mathrm{NO_2}$ group exerts both $-I$ and $-R$ effects, giving stronger stabilization of the conjugate base.
• In $m$-nitrophenol, only the $-I$ effect operates significantly; resonance stabilization of the negative charge is not effective at meta position.

Therefore, $\text{$p$-nitrophenol is more acidic than $m$-nitrophenol}$ so it must have the lower $\mathrm{p}K_a$.

Match with the given values:

Given values are: $\text{(I)}\;10.0,\quad \text{(II)}\;15.9,\quad \text{(III)}\;7.1,\quad \text{(IV)}\;8.3$

So,

• Ethanol $\to 15.9$ $\Rightarrow$ A-II
• Phenol $\to 10.0$ $\Rightarrow$ B-I
• $m$-Nitrophenol $\to 8.3$ $\Rightarrow$ C-IV
• $p$-Nitrophenol $\to 7.1$ $\Rightarrow$ D-III

Final matching: $A-\mathrm{II},\;B-\mathrm{I},\;C-\mathrm{IV},\;D-\mathrm{III}$

Check options:

This corresponds to Option C.