An ideal solution is one in which the enthalpy of mixing is zero — option 3.

The two conditions. A solution is ideal when it obeys Raoult’s law over the whole range of composition, and that is equivalent to :

\(\Delta H_{\text{mix}}=0\qquad\text{and}\qquad\Delta V_{\text{mix}}=0\)

No heat is absorbed or given out on mixing, and the volumes are exactly additive.

Why. Mixing costs or releases energy only if the A–B interactions differ in strength from the A–A and B–B interactions already present. In an ideal solution the three are essentially the same, so a molecule is as comfortable beside an unlike neighbour as beside a like one. Nothing is gained or lost, and no volume change occurs.

Why options 1 and 2 must be wrong on principle. If the entropy of mixing were zero, mixing would have no driving force and the liquids would not mix at all. And since ΔH is zero while ΔS is positive, the Gibbs free energy of mixing is necessarily negative — which is precisely why mixing is spontaneous. A zero ΔG would mean the mixed and unmixed states were equally favoured.

Examples. Ideal, or very nearly so: benzene and toluene; n-hexane and n-heptane; bromoethane and chloroethane — pairs of chemically similar molecules. Positive deviations from Raoult’s law (ΔH positive, mixture boils more easily) occur with ethanol and water; negative deviations (ΔH negative, heat given out) with chloroform and acetone, where hydrogen bonding between unlike molecules is stronger than within either liquid.

Hence, the answer is that the enthalpy of mixing should be zero.