Which of the following is a stronger acid?

The correct answer is option (3)

First, let us look at the structures of all the given aromatic compounds one by one. Then we will figure out the substituent and its position in the aromatic compound. So the given aromatic compounds are as follows.

In (1), (2) and (3) the substituent is nitro-group (−NO₂) and is present at ortho, Meta, and para positions. In option (B) substituent (−Cl) is present at the para position.

The nitro group (−NO₂) and (−Cl) both are electron-withdrawing groups which means it withdraws electrons from the electro center. There is one relation you should keep in mind that the electron-withdrawing group increases the acidic strength.

Both the substituents are electron-withdrawing groups but nitrophenol is more acidic than chlorophenol due to resonance. The resonance structure above indicates that delocalization is only feasible at ortho and para positions and the meta position is not delocalized.
Now out of ortho and para-nitrophenol, ortho-nitrophenol takes part in hydrogen bonding which makes it less acidic than the para-nitrophenol. The strongest acid among the following aromatic compounds is therefore para-nitrophenol.

Note: Electron withdrawing effect is represented as \(−I\) inductive effect. When all the substituents are electron withdrawing groups we can use a priority order.

The priority order is as follows:

\(R_3N \text{ }>\text{ } NO_2\text{ } >\text{ } CN\text{ } >\text{ } COOH\text{ } >\text{ } F\text{ } >\text{ } Cl\text{ } >\text{ } Br\text{ } >\text{ } I\text{ } >\text{ } OH\text{ } >\text{ } C_6H_5\text{ } >\text{ } CH=CH_2\text{ } >\text{ } H\)