JEE Main 2025 Apr 8 Shift 2, Chemistry Q21: Resonance & Mesomeric Effect

JEE Main2025Apr 8, Shift 2Chemistry

Resonance in $X_{2} Y$ can be represented as

The enthalpy of formation of $X_{2} Y (X \equiv X (g) + \frac{1}{2} Y = Y (g) \to X_{2} Y (g))$ is 80 kJ mol $^{- 1}$ .

The magnitude of resonance energy of $X_{2} Y$ is ________ kJ mol $^{- 1}$ (nearest integer value)

Given: Bond energies of $X \equiv X$ , $X = X$ , $Y = Y$ and $X = Y$ are 940, 410, 500 and 602 kJ mol $^{- 1}$ respectively.

valence X : 3, Y : 2

Your answer

Solution

Step 1: Theoretical Enthalpy of Formation

The theoretical (non-resonating) enthalpy of formation is calculated by treating the product as having one X=X double bond and one X=Y double bond.

Reaction:

$X_{2} (g) + \frac{1}{2} Y_{2} (g) \to X_{2} Y (g)$

Step 2: Bond Energies of Reactants

Bonds broken:

One $X \equiv X$ bond $= 940 kJ mo l^{- 1}$
Half of one $Y = Y$ bond $= \frac{1}{2} \times 500 = 250 kJ mo l^{- 1}$

Total bond energy of reactants:

$940 + 250 = 1190 kJ mo l^{- 1}$

Step 3: Bond Energies of Products

Bonds formed:

One $X = X$ bond $= 410 kJ mo l^{- 1}$
One $X = Y$ bond $= 602 kJ mo l^{- 1}$

Total bond energy of products:

$410 + 602 = 1012 kJ mo l^{- 1}$

Step 4: Calculate Theoretical Enthalpy

$\Delta H_f(\text{theo}) = 1190-1012 = \boxed{178\ \mathrm{kJ\,mol^{-1}}}$

Step 5: Resonance Energy

$\text{Resonance Energy} = \Delta H_f(\text{exp}) - \Delta H_f(\text{theo})$

$= 80 - 178 = - 98 kJ mo l^{- 1}$

Hence, the magnitude of resonance energy is

$98 kJ mo l^{- 1}$

Concept behind this question

Resonance & Mesomeric EffectNotes, formulas and examples →

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →