Chemistry · JEE Advanced only

Intermolecular Interactions: Types and Distance Dependence (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 3-4 hours: 1 hour to understand each interaction type and its origin, 1 hour to memorize distance dependences and formulas, 1-2 hours practicing comparison and application problems..

Last updated: 26 October 2026

Direct answer

Intermolecular interactions are the forces between molecules that determine physical properties like boiling point, viscosity, and solubility.

Intermolecular interactions are the forces between molecules that determine physical properties like boiling point, viscosity, and solubility. These include ion-dipole, dipole-dipole, hydrogen bonding, and London dispersion forces, each with a specific dependence on distance. Understanding how these forces weaken with distance helps you predict and compare molecular behavior.

Official syllabus

What the JEE Advanced syllabus says

“Intermolecular interactions: types, distance dependence, and their effect on properties”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Elementary idea of metallic bonding, hydrogen bonding and its applications.

JEE Main covers only basic hydrogen bonding and metallic bonding concepts. Advanced expects you to know the mathematical distance dependence of each interaction type and apply it to compare force strengths and predict properties quantitatively.

Key ideas

What you must understand

Ion-dipole interaction: between an ion and a polar molecule, strongest among intermolecular forces, energy ∝ 1/r²
Dipole-dipole interaction: between two polar molecules with permanent dipoles, energy ∝ 1/r³
Hydrogen bonding: special strong dipole-dipole interaction when H is bonded to F, O, or N and interacts with another electronegative atom
London dispersion forces: present in all molecules due to instantaneous dipoles, energy ∝ 1/r⁶, increases with molecular size and polarizability
Ion-induced dipole and dipole-induced dipole forces exist when a charged or polar species induces a dipole in a nonpolar molecule
The steeper the distance dependence (higher power of 1/r), the more rapidly the force weakens with distance

Formulas

Formulas and key facts

Ion-dipole energy: U ∝ −qμ/r²
Dipole-dipole energy: U ∝ −μ₁μ₂/r³ (for fixed orientation)
Keesom interaction (rotating dipoles): U ∝ −μ₁²μ₂²/r⁶T
London dispersion energy: U ∝ −α₁α₂/r⁶ (α = polarizability)
Ion-induced dipole energy: U ∝ −q²α/r⁴
Dipole-induced dipole energy: U ∝ −μ²α/r⁶
Hydrogen bond energy: typically 10-40 kJ/mol
Total van der Waals interaction: sum of Keesom + Debye + London contributions

Avoid

Common mistakes

Confusing intramolecular bonds (covalent, ionic) with intermolecular forces. Intermolecular forces are always weaker.
Thinking hydrogen bonding occurs whenever H is present. It requires H bonded to F, O, or N specifically.
Ignoring London forces for polar molecules. All molecules have dispersion forces, and for large polar molecules, dispersion can dominate.
Forgetting that distance dependence matters: a 1/r⁶ force drops off much faster than 1/r² as molecules move apart.

Worked example

A JEE Advanced-level example

Question. Arrange the following interactions in order of how quickly their energy decreases as intermolecular distance doubles: (i) ion-dipole, (ii) dipole-dipole (fixed orientation), (iii) London dispersion. Also calculate the factor by which each interaction energy decreases when r becomes 2r.

Show solution

Step 1: Write the distance dependence for each. Ion-dipole: U ∝ 1/r². Dipole-dipole (fixed): U ∝ 1/r³. London: U ∝ 1/r⁶.
Step 2: When r → 2r, the new energy U' = U × (1/2)ⁿ where n is the power of r in the denominator.
Step 3: For ion-dipole (n=2): U' = U × (1/2)² = U/4. Energy becomes 1/4 of original.
Step 4: For dipole-dipole (n=3): U' = U × (1/2)³ = U/8. Energy becomes 1/8 of original.
Step 5: For London (n=6): U' = U × (1/2)⁶ = U/64. Energy becomes 1/64 of original.
Step 6: Order of fastest decrease: London (64×) > dipole-dipole (8×) > ion-dipole (4×).

Answer: London dispersion decreases fastest (to 1/64), then dipole-dipole (to 1/8), then ion-dipole (to 1/4). Order: London > dipole-dipole > ion-dipole.

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Intermolecular Interactions: Types and Distance Dependence in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is Elementary idea of metallic bonding, hydrogen bonding and its applications.

How long does Intermolecular Interactions: Types and Distance Dependence take to prepare for JEE Advanced?

About 3-4 hours: 1 hour to understand each interaction type and its origin, 1 hour to memorize distance dependences and formulas, 1-2 hours practicing comparison and application problems., if you already know the related JEE Main topic.