Question 705070

SCQMEDIUM

The bond dissociation energy of a molecule is defined as the energy required to dissociate it. For H2H_2 and H2+H_2^+ molecules, the bond dissociation energies are 4.4784.478 eV and 2.6512.651 eV respectively. If the equilibrium bond lengths of both H2H_2 and H2+H_2^+ are identical, the value of the ionization potential of hydrogen molecule will be closest to

(A)

15.42715.427 eV

(B)

11.77311.773 eV

(C)

20.72920.729 eV

(D)

6.4716.471 eV

Detailed Solution

We need to calculate the energy for H2→H2++e−H_2 \to H_2^+ + e^-. The process can be divided as: H2→H+HH_2 \to H + H (E1=4.478E_1 = 4.478 eV), H→H++e−H \to H^+ + e^- (E2=13.6E_2 = 13.6 eV), and H++H→H2+H^+ + H \to H_2^+ (E3=−2.651E_3 = -2.651 eV). Total energy ΔE=E1+E2+E3=4.478+13.6−2.651=15.427\Delta E = E_1 + E_2 + E_3 = 4.478 + 13.6 - 2.651 = 15.427 eV.

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