Which of the following pictorial diagram most correctly represents the \( \pi^* \) (\( \pi \) - antibonding) molecular orbital between two atoms if the internuclear axis is taken to be in the z-direction ( \( \xrightarrow{\text{z-axis}} \) )?}
Step 1: Understanding the Concept:
Molecular orbitals are formed by the linear combination of atomic orbitals (LCAO). A $\pi$ (pi) bond is formed by the sideways (lateral) overlap of $p$-orbitals. When they combine "out of phase" (subtractive overlap), they form a $\pi^$ antibonding orbital, which features a node directly between the nuclei. Step 2: Key Formula or Approach:
Antibonding Orbital ($\Psi^$) = $\Psi_A - \Psi_B$.
A $\pi^$ orbital is characterized by a nodal plane perpendicular to the internuclear axis (between the atoms) in addition to the nodal plane containing the internuclear axis itself. Step 3: Detailed Explanation:
Let the internuclear axis be the z-axis. The $p_x$ or $p_y$ atomic orbitals are oriented perpendicular to this axis.
When two $p_x$ (or $p_y$) orbitals approach each other out-of-phase (e.g., positive lobe of one approaches the negative lobe of the other), they repel and create a region of zero electron density (a nodal plane) exactly bisecting the internuclear axis.
The resulting shape consists of four distinct lobes pointing away from the center, with alternating phases (signs of the wave function) diagonally.
Looking at standard diagrams:
Image (Option A) shows an s-orbital overlapping out of phase, producing a $\sigma^$ antibonding orbital.
Image (Option B) shows two p-orbitals overlapping sideways in-phase, producing a bonding $\pi$ orbital (two large continuous lobes).
Image (Option C) shows two p-orbitals overlapping sideways out-of-phase, producing four distinct lobes separated by a central vertical nodal plane. This is the hallmark $\pi^$ antibonding orbital.
Image (Option D) shows head-on overlap creating a $\sigma$ bond or represents a d-orbital array, not a sideways overlap.
Thus, the third image accurately reflects the $\pi^$ symmetry. Step 4: Final Answer:
Option (C) correctly represents the $\pi^$ antibonding molecular orbital.
