Chemistry · 2025
JEE Main · 8 April 2025, Shift 2 · Q73
Resonance in X_2 Y can be represented as The enthalpy of formation of X_2 Y ( X ≡ X (g)+1/2 Y = Y (g) → X_2 Y (g)) is 80 kJ mol^-1. The magnitude of…
Resonance in $\displaystyle \mathrm{X}_2 \mathrm{Y}$ can be represented as
The enthalpy of formation of $\displaystyle \mathrm{X}_2 \mathrm{Y}\left(\mathrm{X} \equiv \mathrm{X}(g)+\frac{1}{2} \mathrm{Y}=\mathrm{Y}(g) \rightarrow \mathrm{X}_2 \mathrm{Y}(g)\right)$ is $\displaystyle 80 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The magnitude of resonance energy of $\displaystyle \mathrm{X}_2 \mathrm{Y}$ is $\displaystyle \_\_\_\_$ $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)Given : Bond energies of $\displaystyle \mathrm{X} \equiv \mathrm{X}, \mathrm{X}=\mathrm{X}, \mathrm{Y}=\mathrm{Y}$ and $\displaystyle \mathrm{X}=\mathrm{Y}$ are $\displaystyle 940,410,500$ and $\displaystyle 602 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
valence X: $\displaystyle 3$ , Y: $\displaystyle 2$
Official answer
From NTA’s final answer key for this paper.
98
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JEE Main 2025 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.