Question

Consider the bonding ($\Phi_b$) and antibonding ($\Phi_a$) molecular orbitals of the $H_2^+$ molecule formed by the linear combination of the atomic orbitals $\Psi_X$ and $\Psi_Y$ centered on the two H atoms X and Y. The correct statement is:

• A. As the internuclear distance is increased from the equilibrium bond distance, the energy of $\Phi_b$ is increased and that of $\Phi_a$ is lowered.
• B. For $\Phi_a$, the probability density between the nuclei is increased by $c\Psi_X\Psi_Y$, where $c$ is a positive real number.
• C. For the function $\Phi_a$, the probability density at any point on a plane midway between the nuclei and perpendicular to the internuclear axis is 0.5.
• D. An electron present in the $\Phi_a$ orbital feels more electron-nuclei attraction compared to the one present in the $\Phi_b$ orbital.

Answer 1

Answer: (A)

As the internuclear distance is increased from the equilibrium bond distance, the energy of $\Phi_b$ is increased and that of $\Phi_a$ is lowered.

Answer 2

The $H_2^+$ molecule is formed from the linear combination of two 1s atomic orbitals, $\Psi_X$ and $\Psi_Y$, centered on atoms X and Y. The bonding ($\Phi_b$) and antibonding ($\Phi_a$) molecular orbitals are given by:

$\Phi_b = N_b (\Psi_X + \Psi_Y)$ (bonding molecular orbital, $\sigma_{1s}$)

$\Phi_a = N_a (\Psi_X - \Psi_Y)$ (antibonding molecular orbital, $\sigma^*_{1s}$)

where $N_b$ and $N_a$ are normalization constants.

• Energy of $\Phi_b$: The potential energy curve for a bonding molecular orbital ($\Phi_b$) has a minimum at the equilibrium bond distance. If the internuclear distance is increased from this equilibrium distance, the bond weakens, and the energy of the system increases, approaching the energy of the separated atoms. So, the energy of $\Phi_b$ is increased.
• Energy of $\Phi_a$: The potential energy curve for an antibonding molecular orbital ($\Phi_a$) is purely repulsive. Its energy is always higher than that of the separated atomic orbitals. As the internuclear distance increases, the repulsive interactions decrease. Consequently, the energy of $\Phi_a$ continuously decreases, approaching the energy of the separated atomic orbitals from above. So, the energy of $\Phi_a$ is lowered.

Therefore, statement (A) is correct.