Step 1: Recall the ideal-solution condition.
A solution is ideal only when the solute-solvent attractions match the attractions inside each pure liquid, giving $\Delta H_{\text{mix}} = 0$ and $\Delta V_{\text{mix}} = 0$.
Step 2: Translate into a structural clue.
This happens when the two liquids are chemically similar in polarity, size and bonding, so mixing changes nothing energetically.
Step 3: Inspect benzene and toluene.
Both are non-polar aromatics of almost the same shape; the new A-B dispersion forces equal the old A-A and B-B forces, so they mix ideally.
Step 4: Inspect phenol and aniline.
Strong O-H to N hydrogen bonds form on mixing, attractions get stronger, so this shows negative deviation.
Step 5: Inspect chloroform and acetone.
A new H-bond forms between chloroform's H and acetone's carbonyl O, again strengthening attractions and giving negative deviation.
Step 6: Inspect ethanol and acetone.
Acetone breaks ethanol's hydrogen-bond network, weakening attractions, so this shows positive deviation.
Step 7: Select the ideal pair.
Only benzene plus toluene meets the no-change requirement, so the answer is option (1).
\[ \boxed{\text{Benzene + toluene (nearly ideal)}} \]