1. Evaluate Statement (I): Acidity of nitrophenols

The acidity of nitrophenols depends on the stability of the corresponding phenoxide ion.

• The $-NO_2$ group is a strong electron-withdrawing group.
• At the para and ortho positions, it shows both:
  • $-I$ effect (inductive withdrawal)
  • $-R$ effect / resonance withdrawal
• At the meta position, only the $-I$ effect operates effectively; resonance interaction with the phenoxide ion is not possible.

So, compared to $m$-nitrophenol, both $o$- and $p$-nitrophenol should be more acidic due to extra resonance stabilization.

However, between $o$-nitrophenol and $p$-nitrophenol:

• $o$-nitrophenol shows intramolecular hydrogen bonding, which stabilizes the neutral molecule and reduces its tendency to lose $H^+$ in water.
• Hence, $p$-nitrophenol is more acidic than $o$-nitrophenol.

Therefore, the acidity order is: $p\text{-nitrophenol} > o\text{-nitrophenol} > m\text{-nitrophenol}$

Thus, the statement $p\text{-nitrophenol is more acidic than } m\text{-nitrophenol and } o\text{-nitrophenol}$ is true.

2. Evaluate Statement (II): Ethanol with Lucas reagent

Lucas reagent is a mixture of: $\text{conc. HCl} + \text{anhyd. } ZnCl_2$

It is used to distinguish alcohols:

• Tertiary alcohols: immediate turbidity
• Secondary alcohols: turbidity in a few minutes
• Primary alcohols: no immediate turbidity at room temperature

Ethanol is a primary alcohol. So ethanol does not give immediate turbidity with Lucas reagent.

Hence, Statement (II) is false.

3. Conclusion

• Statement (I): True
• Statement (II): False

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