Intermolecular Forces JEE 2025: Why C and D Are Correct

What does the intermolecular forces JEE 2025 question ask?

Select C and D, as supported by the supplied worked solution and standard theory. The supplied material reports that the scan prints D alone. This intermolecular forces JEE 2025 question is from JEE Advanced 2025, Chemistry, Paper 2, under Chemical Bonding and Molecular Structure. It is a multiple-correct question: select every valid claim, not just one.

The question bank tags it medium and gives an expected solving time of 120 seconds. These are question-bank guidance, not an official difficulty rating or time limit.

The four claims, paraphrased, are:

Why is option A incorrect about how quickly energy approaches zero?

Charge–dipole energy approaches zero faster, reversing A’s claim. Compare potential energies, not forces, and hold the charge–dipole orientation fixed while comparing distance dependences.

Define separation and potential energy as:

r = separation, U = potential energy

The relevant electrostatic energy laws are:

U_{q q} \propto \frac{1}{r}, U_{q μ} \propto \frac{1}{r^{2}}

Doubling the separation gives:

\frac{| U_{q q} (2 r) |}{| U_{q q} (r) |} = \frac{1}{2}, \frac{| U_{q μ} (2 r) |}{| U_{q μ} (r) |} = \frac{1}{4}

Charge–dipole energy retains one-quarter of its initial magnitude, while charge–charge energy retains one-half. A larger inverse power approaches zero faster as separation tends to infinity.

This compares decay rates, not absolute energies with different prefactors. Therefore, A is false.

Why does rotation make option B incorrect?

Rotating polar molecules require a thermal orientational average, whose attractive energy follows an inverse-sixth-power law. The inverse-cube law belongs to fixed dipole orientations, with a coefficient that depends on their relative orientation:

U_{μ μ} \propto \frac{1}{r^{3}}

The question asks for the average interaction of rotating molecules, not an instantaneous fixed-orientation interaction. This thermally averaged attraction is the Keesom interaction:

⟨ U_{μ μ} ⟩ \propto - \frac{1}{T r^{6}}, T = absolute temperature

The negative sign denotes attraction relative to zero energy at infinite separation. The distance dependence is inverse sixth power, and raising temperature weakens the average attraction at fixed separation.

This is not a uniform average of equally weighted orientations. Thermal weighting favours lower-energy orientations; increasing temperature reduces that preference.

Option B assigns the fixed-orientation law to a rotating-molecule average. That mismatch makes B false.

How does the induction-energy derivation prove option C?

C is correct because the standard Debye induction energy contains no explicit temperature factor. A permanent dipole’s electric field distorts a neighbouring nonpolar molecule’s electron cloud, inducing a dipole.

Polarizability, denoted by alpha, measures how readily that cloud distorts. The induced dipole moment is: $p_{i n d} = α E$

The induced-dipole energy is:

U_{i n d} = - \frac{1}{2} α E^{2}

The permanent dipole’s electric field has the distance dependence:

E \propto \frac{1}{r^{3}}

Substituting this dependence gives:

E^{2} \propto {(\frac{1}{r^{3}})}^{2} = \frac{1}{r^{6}}

Hence:

U_{i n d} \propto - \frac{1}{r^{6}}

This is Debye induction, not the Keesom interaction in B. Its coefficient depends on polarizability and the permanent dipole moment, but the standard model introduces no explicit temperature factor.

This does not mean all real-material properties are temperature invariant. It describes the interaction law with molecular parameters held fixed, under which C is true.

Why is D correct, and should you select D alone or C and D?

Nonpolar molecules attract through London dispersion, so D is correct, but it is not the only valid statement. Permanent dipole moments are not required.

Electron-density fluctuations create an instantaneous dipole in one molecule, which induces a dipole in a neighbour, producing attraction. The standard London interaction law has no explicit temperature factor:

U_{L o n d o n} \propto - \frac{1}{r^{6}}

A false, B false, C true, D true; therefore select C and D.

Source note: The supplied material reports that the scanned page prints D alone, whereas the supplied worked solution and standard theory support C and D. This is not an independently checked official answer key or a claim that the examination authority issued a correction.

What formula-selection mistake leads to choosing B?

The error is dropping the formula’s fixed-orientation assumption. A reproducible wrong pathway is: recognise two permanent dipoles, recall the inverse-cube energy law, then select B without processing “rotating” and “average”.

The recalled formula is not wrong. Its fixed-orientation assumption is wrong for this statement:

Fixed dipoles: orientation-dependent energy. $U \propto r^{- 3}$
Thermally rotating polar molecules: Keesom attraction.

⟨ U ⟩ \propto - \frac{1}{T r^{6}}

Permanent dipole inducing another dipole: Debye attraction, with no explicit temperature factor.

U_{i n d} \propto - \frac{1}{r^{6}}

Before choosing a formula, identify the interacting species. Then check whether their dipoles have fixed orientations, are thermally averaged or are induced.

Can you apply the distinction to three related questions?

Use the interaction law to predict changes in energy magnitude, not just name the force. These are original practice questions from Chemical Bonding and Molecular Structure, not additional verified PYQs.

At fixed temperature, what happens when the separation between rotating polar molecules doubles?

The Keesom energy magnitude becomes one-sixty-fourth of its initial value. The inverse-sixth-power dependence gives:

\frac{| ⟨ U (2 r) ⟩ |}{| ⟨ U (r) ⟩ |} = {(\frac{r}{2 r})}^{6} = \frac{1}{64}

Use magnitudes here: the interaction remains attractive, but its energy moves closer to zero.

At fixed separation and molecular parameters, how does doubling temperature affect Keesom and Debye energies?

Keesom’s magnitude halves; Debye’s remains unchanged in the standard model. Their temperature ratios are:

\frac{| U_{K e e s o m} (2 T) |}{| U_{K e e s o m} (T) |} = \frac{1}{2}, \frac{| U_{D e b y e} (2 T) |}{| U_{D e b y e} (T) |} = 1

Only the Keesom law contains the explicit inverse-temperature factor.

What attraction can act between two isolated argon atoms with no permanent dipoles?

London dispersion can act between them. An electron-density fluctuation creates an instantaneous dipole in one atom, which induces a dipole in the other, producing attraction.

Before your next interaction-energy calculation, write down the mechanism and the orientation assumption first.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

Is the answer to the intermolecular forces JEE 2025 question D or C and D?

The supplied worked solution and standard theory support C and D. The supplied material reports that the scan prints D alone, but this is not an independently verified official answer key. C follows from Debye induction having no explicit temperature factor, while D follows from London dispersion between molecules without permanent dipoles.

Why is option A wrong in the intermolecular forces question?

Charge–charge potential energy varies as 1/r, while fixed-orientation charge–dipole potential energy varies as 1/r². The latter therefore approaches zero faster at large separation, contrary to A. This compares energy decay rates, not forces or absolute energy magnitudes.

Does dipole–dipole interaction energy vary as 1/r³ or 1/r⁶?

For fixed dipole orientations, the interaction energy varies as 1/r³, with an orientation-dependent coefficient. For thermally rotating polar molecules, the average Keesom attraction varies as −1/(Tr⁶). Option B is false because it assigns the fixed-orientation law to rotating molecules.

Is Debye induction energy dependent on temperature?

The standard Debye induction energy has no explicit temperature dependence when molecular parameters are held fixed. Using the induced-dipole energy −αE²/2 and the permanent dipole field E proportional to 1/r³ gives an attraction proportional to −1/r⁶. This does not imply that all real-material properties are temperature invariant.

How do nonpolar molecules attract each other?

Nonpolar molecules attract through London dispersion. Electron-density fluctuations create an instantaneous dipole in one molecule, which induces a dipole in a neighbour. The standard attractive energy varies as −1/r⁶ and requires no permanent dipole.