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Friday, 9 October

JEE Main 2025 · ChemistryNumerical answerNumerical valueHardMulti-step

JEE Main 8 April 2025, Shift 2, Chemistry Q73

Question 73 of 75 in this shift, Chemistry question 23 of 25, Section B.

Resonance in X2Y\mathrm{X_2Y} can be represented as X¨⊖=X⊕=Y¨¨↔ :X≡X⊕−Y¨¨:⊖\overset{\ominus}{\ddot{X}} = \overset{\oplus}{X} = \ddot{\ddot{Y}} \leftrightarrow\ :X \equiv \overset{\oplus}{X} - \ddot{\ddot{Y}}\overset{\ominus}{:} The enthalpy of formation of X2Y\mathrm{X_2Y} (X≡X(g)+12Y=Y(g)→X2Y(g))\left(\mathrm{X \equiv X}(g) + \frac{1}{2}\mathrm{Y = Y}(g) \rightarrow \mathrm{X_2Y}(g)\right) is 80 kJ mol−1^{-1}. The magnitude of resonance energy of X2Y\mathrm{X_2Y} is __________ kJ mol−1^{-1} (nearest integer value) Given : Bond energies of X≡X\mathrm{X \equiv X}, X=X\mathrm{X = X}, Y=Y\mathrm{Y = Y} and X=Y\mathrm{X = Y} are 940, 410, 500 and 602 kJ mol−1^{-1} respectively. valence X: 3 , Y: 2

The figure, in words

Two resonance structures of X2Y: X=X=Y with formal charge minus on the first X and plus on the middle X (lone pairs on the end atoms), and :X≡X–Y: with plus on the middle X and minus on Y.

Official answer

98

NTA final key.

Chapter
Chemical Bonding and Molecular Structure
Topic
Enthalpy and Hess's Law
Idea tested
Resonance Energy Calculation

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.