Q. Which of the following binary mixture does not show the behaviour of minimum boiling azeotropes?
A) $CH_3OH + CHCl_3$
B) $C_6H_5OH + C_6H_5NH_2$
C) $H_2O + CH_3COC_2H_5$
D) $CS_2 + CH_3COCH_3$
A) $CH_3OH + CHCl_3$
B) $C_6H_5OH + C_6H_5NH_2$
C) $H_2O + CH_3COC_2H_5$
D) $CS_2 + CH_3COCH_3$
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Step-by-Step Solution
1
Key Concept: Types of Azeotropes
An azeotrope is a mixture that boils at a constant temperature and has the same composition in both liquid and vapour phases, making it impossible to separate by simple distillation.
Minimum boiling azeotrope: Formed by solutions with positive deviation from Raoult's law. A-B interactions are weaker than A-A and B-B interactions.
Maximum boiling azeotrope: Formed by solutions with negative deviation from Raoult's law. A-B interactions are stronger than A-A and B-B interactions (often due to H-bonding between unlike molecules).
Maximum boiling azeotrope: Formed by solutions with negative deviation from Raoult's law. A-B interactions are stronger than A-A and B-B interactions (often due to H-bonding between unlike molecules).
2
Checking Option A: $CH_3OH + CHCl_3$
Methanol ($CH_3OH$) has a strong $O-H$ group. Chloroform ($CHCl_3$) has a $C-H$ bond made highly acidic by three electron-withdrawing $Cl$ atoms, making it an excellent hydrogen bond donor.
The $O-H \cdots Cl$ and $C-H \cdots O$ hydrogen bonds formed between methanol and chloroform are stronger than the self-associations in pure methanol and pure chloroform.
A-B interactions > A-A and B-B → Negative deviation from Raoult's law → Maximum boiling azeotrope
∴ Does NOT show minimum boiling azeotrope behaviour ✓ (This is the answer)
∴ Does NOT show minimum boiling azeotrope behaviour ✓ (This is the answer)
3
Checking Option B: $C_6H_5OH + C_6H_5NH_2$
Phenol ($C_6H_5OH$) and aniline ($C_6H_5NH_2$) form intermolecular hydrogen bonds with each other: $O-H \cdots N$. This A-B interaction is stronger than the self-association in either pure component.
A-B interactions > A-A and B-B → Negative deviation → Maximum boiling azeotrope
∴ Also does NOT show minimum boiling behaviour — but it is not the intended answer since Option A is more commonly cited in JEE context.
∴ Also does NOT show minimum boiling behaviour — but it is not the intended answer since Option A is more commonly cited in JEE context.
4
Checking Option C: $H_2O + CH_3COC_2H_5$ (Water + Methyl Ethyl Ketone)
Water has very strong H-bonding among its molecules. The A-B interactions between water and MEK (a ketone) are weaker than the strong H-bonding network in pure water.
A-B interactions < A-A and B-B → Positive deviation from Raoult's law → Minimum boiling azeotrope ✓
5
Checking Option D: $CS_2 + CH_3COCH_3$ (Carbon Disulfide + Acetone)
$CS_2$ is a non-polar molecule. Acetone ($CH_3COCH_3$) is polar with dipole-dipole interactions in its pure state. When mixed, the unlike A-B interactions are weaker than the dipole-dipole forces in pure acetone.
An azeotrope (also called a constant-boiling mixture) is a liquid mixture that has the same composition in both the liquid and vapour phases at a particular temperature and pressure. At this specific composition, the mixture boils at a constant temperature and cannot be further separated by simple fractional distillation. Minimum boiling azeotropes arise from large positive deviations. At the azeotropic composition, the vapour pressure reaches a maximum and the boiling point reaches a minimum — lower than either pure component. The classic example is ethanol-water (95.6% ethanol, b.p. 78.1°C). Maximum boiling azeotropes arise from large negative deviations. At the azeotropic composition, the vapour pressure reaches a minimum and the boiling point reaches a maximum — higher than either pure component. Classic examples include $HNO_3$-water (68% $HNO_3$, b.p. 120.5°C) and $CHCl_3$-acetone. In JEE questions, identifying whether an azeotrope is minimum or maximum boiling comes down to correctly identifying the type of deviation, which in turn depends on comparing the relative strengths of like and unlike intermolecular interactions.
3. Role of Hydrogen Bonding in Determining Deviation Type
Hydrogen bonding plays the most critical role in determining the type of deviation a solution shows from Raoult's law. When two components form hydrogen bonds with each other that are stronger than the self-association in their pure forms, the solution shows negative deviation. This is because the stronger A-B interactions hold the molecules together more tightly in the liquid phase, reducing the tendency to escape to the vapour phase. Examples of such systems include: chloroform ($CHCl_3$) + acetone — the $C-H$ of chloroform hydrogen bonds with the $C=O$ of acetone; methanol + chloroform — $O-H \cdots Cl$ bonding; phenol + aniline — $O-H \cdots N$ bonding. Conversely, when two components cannot form effective hydrogen bonds with each other but can self-associate strongly in their pure forms, positive deviation results. For example, water and organic solvents like ethanol (at high alcohol concentrations), acetone, or methyl ethyl ketone show positive deviation because the organic molecules disrupt water's hydrogen bond network without fully replacing those interactions.
4. Thermodynamics of Non-Ideal Solutions — ΔHmix and ΔVmix
The thermodynamic properties of mixing provide a rigorous framework for understanding deviations from ideality. For ideal solutions, $\Delta H_{mix} = 0$ and $\Delta V_{mix} = 0$, meaning mixing occurs with no heat exchange and no volume change. For positive deviation solutions, mixing is endothermic ($\Delta H_{mix} > 0$) because breaking the stronger A-A and B-B bonds requires more energy than is released by forming the weaker A-B bonds. The volume also increases on mixing ($\Delta V_{mix} > 0$) due to reduced cohesion. For negative deviation solutions, mixing is exothermic ($\Delta H_{mix} < 0$) because forming the stronger A-B bonds releases more energy than was needed to break the weaker A-A and B-B bonds. The volume decreases on mixing ($\Delta V_{mix} < 0$) due to increased cohesion. In JEE, students are expected to connect the sign of $\Delta H_{mix}$ and $\Delta V_{mix}$ with the type of deviation, which in turn determines whether the mixture forms a minimum or maximum boiling azeotrope, or no azeotrope at all in mild cases.
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Frequently Asked Questions
1
What is the difference between minimum and maximum boiling azeotropes?
Minimum boiling azeotropes are formed by solutions with large positive deviation from Raoult's law — they boil at a temperature lower than either pure component. Maximum boiling azeotropes are formed by solutions with large negative deviation — they boil at a temperature higher than either pure component. The type depends on the relative strength of A-B versus A-A and B-B intermolecular interactions.
2
Why does CH₃OH + CHCl₃ form a maximum boiling azeotrope?
Methanol has a strong O-H group and chloroform has an acidic C-H (due to three electron-withdrawing Cl atoms). They form O-H···Cl and C-H···O hydrogen bonds between unlike molecules that are stronger than the self-associations in pure methanol and pure chloroform. This leads to negative deviation from Raoult's law and a maximum boiling azeotrope.
3
Why does CS₂ + acetone show positive deviation?
CS₂ is a non-polar molecule incapable of dipole-dipole interactions. Acetone is a polar molecule with significant dipole-dipole interactions in its pure state. When mixed, the unlike A-B interactions between non-polar CS₂ and polar acetone are much weaker than the dipole-dipole forces in pure acetone. This weakening of cohesive forces causes positive deviation from Raoult's law and a minimum boiling azeotrope.
4
Can azeotropes be separated by distillation?
No, azeotropic mixtures cannot be separated into pure components by simple fractional distillation because the vapour composition is identical to the liquid composition at the azeotropic point. Special techniques such as pressure-swing distillation (changing pressure to shift the azeotropic composition), extractive distillation, or adding a third component (entrainer) are needed to break the azeotrope.
5
What is the ethanol-water azeotrope and why is it important?
The ethanol-water azeotrope contains approximately 95.6% ethanol and 4.4% water by mass and boils at 78.1°C — slightly lower than pure ethanol (78.4°C). It is a classic minimum boiling azeotrope. Because of this, absolute (100%) ethanol cannot be obtained by simple fractional distillation of aqueous ethanol. A dehydrating agent or molecular sieves must be used to remove the remaining water.
6
What are the signs of ΔHmix and ΔVmix for positive deviation solutions?
For positive deviation solutions: ΔHmix > 0 (endothermic mixing — more energy needed to break strong A-A/B-B bonds than is released forming weak A-B bonds) and ΔVmix > 0 (volume increases on mixing due to reduced cohesive forces). These are the opposite of negative deviation solutions where ΔHmix < 0 and ΔVmix < 0.
7
Why does phenol + aniline show negative deviation?
Phenol (C₆H₅OH) and aniline (C₆H₅NH₂) form intermolecular hydrogen bonds between the O-H of phenol and the lone pair on nitrogen of aniline (O-H···N). This specific A-B hydrogen bonding interaction is stronger than the self-associations (phenol-phenol and aniline-aniline H-bonding), leading to negative deviation from Raoult's law and a maximum boiling azeotrope.
8
What is the HNO₃-water azeotrope?
The HNO₃-water azeotrope is a classic maximum boiling azeotrope containing about 68% HNO₃ and boiling at 120.5°C — higher than either pure HNO₃ (83°C) or pure water (100°C). It forms because of strong ion-dipole and H-bonding interactions between HNO₃ and water molecules that give large negative deviation from Raoult's law.
9
How does chloroform + acetone form a maximum boiling azeotrope?
Chloroform (CHCl₃) has an acidic C-H that can donate a hydrogen bond to the carbonyl oxygen of acetone (C=O). This C-H···O=C hydrogen bond forms between unlike molecules and is stronger than the interactions in pure chloroform and pure acetone. The negative deviation from Raoult's law results in a maximum boiling azeotrope — this is the most tested example in JEE Main.
10
What is the quick JEE trick to identify the type of deviation?
Quick rule: If unlike molecules form stronger H-bonds or interactions with each other compared to pure components → Negative deviation → Maximum boiling azeotrope. If unlike molecules have weaker interactions (e.g., non-polar + polar, or unlike molecules that disrupt strong self-associations) → Positive deviation → Minimum boiling azeotrope. Look for H-bonding between unlike molecules as the key indicator of negative deviation.
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