MCQMediumJEE 2026Enthalpies (Bond, Combustion, Formation…)

JEE Chemistry 2026 Question with Solution

The heat of atomisation of methane and ethane are xkJ mol1x \, \text{kJ mol}^{-1} and ykJ mol1y \, \text{kJ mol}^{-1} respectively. The longest wavelength (λ\lambda) of light capable of breaking the C–C bond can be expressed in SI unit as:

  • A

    hc1000(y6x4)1\dfrac{hc}{1000}\left(\dfrac{y-6x}{4}\right)^{-1}

  • B

    NAhc250(y6x)\dfrac{N_A hc}{250(y-6x)}

  • C

    NAhc(y6x4)1N_A hc\left(\dfrac{y-6x}{4}\right)^{-1}

  • D

    NAhc250(4y6x)\dfrac{N_A hc}{250(4y-6x)}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Heat of atomisation of methane is xkJ mol1x \, \text{kJ mol}^{-1} and that of ethane is ykJ mol1y \, \text{kJ mol}^{-1}.

Find: The longest wavelength λ\lambda of light capable of breaking the C–C bond.

Bond energy concept: Longest wavelength corresponds to minimum energy required to break a bond.

Step 1: Heat of atomisation of methane gives C–H bond energy, while that of ethane includes both C–H bonds and one C–C bond.

Step 2: Ethane has 66 C–H bonds and 11 C–C bond. Hence C–C bond energy is proportional to 4y6x4y - 6x.

Step 3: Use the relation between photon energy and wavelength:

E=hcλ=(4y6x)×103NAE = \frac{hc}{\lambda} = \frac{(4y-6x)\times 10^3}{N_A}

Step 4: Rearranging,

λ=NAhc250(4y6x)\lambda = \frac{N_A hc}{250(4y-6x)}

Therefore, the correct option is D.

Common mistakes

  • Using the shortest wavelength condition instead of the longest wavelength condition is incorrect because bond breaking starts at the minimum required energy. Use the fact that longest wavelength corresponds to minimum photon energy sufficient to break the bond.

  • Treating the heat of atomisation of ethane as involving only C–C bond breaking is wrong because ethane atomisation includes breaking all 6 C–H bonds and 1 C–C bond. First isolate the C–C bond contribution from the total atomisation data.

  • Forgetting to divide by NAN_A is incorrect because the given atomisation values are per mole, while E=hcλE = \frac{hc}{\lambda} is energy per photon. Convert molar bond energy to energy per molecule before applying the wavelength relation.

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