The bond dissociation energy of a molecule is defined as the energy required to dissociate it. For H₂ and H₂⁺

molecules, the bond dissociation energies are 4.478 eV and 2.651 eV respectively. If the equilibrium bond lengths of both H^​₂ and H₂⁺

are identical, the value of the ionization potential of hydrogen molecule will be closest to

Correct option is A

Given:

• Bond dissociation energy of H₂: 4.478 eV.
• Bond dissociation energy of H₂+: 2.651 eV.
• Ionization energy of a single hydrogen atom: 13.6 eV.
• We need to calculate the ionization potential of the hydrogen molecule (H₂).

Solution:

1. Step 1: Energy required to remove an electron from H₂:

   • This step involves converting H₂ into H₂+ and one free electron.
   • The energy required for this step is the bond dissociation energy of H₂, which is 4.478 eV.

2. Step 2: Energy required to dissociate H₂+:

   • After ionization, H₂+ dissociates into H and H+.
   • The energy required for this step is the bond dissociation energy of H₂+, which is 2.651 eV.

3. Step 3: Total energy (ionization potential):

   • To find the ionization potential, sum the energy required for both steps:
     Total Energy = 4.478 eV + 2.651 eV + 8.298 = 15.427 eV.

Conclusion:
The ionization potential of H₂ is (a) 15.427 eV.