Consider the bonding $\left(\Phi_{\mathrm{b}}\right)$ and antibonding $\left(\Phi_{\mathrm{a}}\right)$ molecular orbitals of the $\mathrm{H}_2{ }^{+}$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:

• Correct answer
  A

  As the internuclear distance is increased from the equilibrium bond distance, the energy of $\Phi_b$ is increased and that of $\Phi_{\mathrm{a}}$ is lowered.
• B

  For $\Phi_{\mathrm{a}}$, the probability density between the nuclei is increased by $c \Psi_{\mathrm{X}} \Psi_{\mathrm{Y}}$, where $c$ is a positive real number.
• C

  For the function $\Phi_{\mathrm{a}}$, 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 $\Phi_{\mathrm{b}}$ orbital.

Hence, the correct answer is option: 'A'.