Question

For which of the following liquid mixtures $\Delta_{mixH = 0$ and $\Delta_{mix}V = 0$?}

• A. ethyl chloride, ethyl bromide
• B. ethanol, acetone
• C. phenol, aniline
• D. chloroform, acetone

Hint:

\textbf{Ideal Solution:} $\Delta_{mix}H = 0$, $\Delta_{mix}V = 0$. Obeys Raoult's law. Formed by components with very similar chemical nature and intermolecular forces.
\textbf{Non-ideal solutions with positive deviation:} $\Delta_{mix}H>0$, $\Delta_{mix}V>0$. A-B interactions are weaker than A-A and B-B interactions (e.g., ethanol + acetone, ethanol + water, acetone + CS$_2$).
\textbf{Non-ideal solutions with negative deviation:} $\Delta_{mix}H<0$, $\Delta_{mix}V<0$. A-B interactions are stronger than A-A and B-B interactions (e.g., chloroform + acetone, nitric acid + water, phenol + aniline).
Look for pairs of compounds that are very similar in structure, size, and polarity (e.g., n-hexane and n-heptane; benzene and toluene; ethyl chloride and ethyl bromide).

Answer 1

The Correct Option is A

The conditions $\Delta_{mix}H = 0$ (enthalpy of mixing is zero) and $\Delta_{mix}V = 0$ (volume change on mixing is zero) are characteristic of an ideal solution. Ideal solutions are formed when the components are chemically similar and the intermolecular forces between unlike molecules (A-B) are of the same magnitude as those between like molecules (A-A and B-B). Raoult's law is obeyed by ideal solutions over the entire range of concentration. Let's analyze the options: \begin{itemize} \item (a) ethyl chloride, ethyl bromide (CH$_3$CH$_2$Cl, CH$_3$CH$_2$Br): These are both haloalkanes. They are structurally very similar (ethyl group with a halogen). The halogens Cl and Br are adjacent in the same group, leading to similar electronegativities and polarities of the C-X bond. The intermolecular forces (mainly dipole-dipole and London dispersion forces) are expected to be very similar between ethyl chloride molecules, ethyl bromide molecules, and between ethyl chloride and ethyl bromide molecules. This mixture is likely to behave as an ideal or nearly ideal solution. \item (b) ethanol, acetone (CH$_3$CH$_2$OH, CH$_3$COCH$_3$): Ethanol has strong hydrogen bonding between its molecules. Acetone is a polar molecule (ketone) but does not form hydrogen bonds with itself as effectively as ethanol. When mixed, the hydrogen bonding network of ethanol is disrupted. The interactions between ethanol and acetone molecules (some hydrogen bonding possible: O-H of ethanol with C=O of acetone) are generally weaker than the original ethanol-ethanol hydrogen bonds. This usually leads to a positive deviation from Raoult's law, with $\Delta_{mix}H>0$ (endothermic mixing) and $\Delta_{mix}V>0$. \item (c) phenol, aniline (C$_6$H$_5$OH, C$_6$H$_5$NH$_2$): Phenol is acidic and aniline is basic. They can interact strongly, possibly forming hydrogen bonds or even undergoing a weak acid-base interaction. Phenol molecules have hydrogen bonding among themselves. Aniline molecules also have hydrogen bonding. When mixed, new, stronger interactions (e.g., H-bond between phenol's -OH and aniline's -NH$_2$) can form. This often leads to a negative deviation from Raoult's law, with $\Delta_{mix}H<0$ (exothermic mixing) and $\Delta_{mix}V<0$. \item (d) chloroform, acetone (CHCl$_3$, CH$_3$COCH$_3$): This is a classic example of a mixture showing negative deviation from Raoult's law. Chloroform (CHCl$_3$) has a slightly acidic hydrogen atom. Acetone (CH$_3$COCH$_3$) has an oxygen atom with lone pairs. A hydrogen bond can form between the H of CHCl$_3$ and the O of acetone. This new A-B interaction is stronger than the average of A-A and B-B interactions. This leads to $\Delta_{mix}H<0$ (exothermic) and $\Delta_{mix}V<0$. \end{itemize} Based on this analysis, the mixture of ethyl chloride and ethyl bromide is the most likely to form an ideal solution, satisfying $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$. \[ \boxed{\text{ethyl chloride, ethyl bromide}} \]