Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: For diatomic molecules, the correct statement(s) about the molecular orbitals formed by the overlap to two $2p_z$ orbitals is(are)

Imagine two atoms approaching each other to form a diatomic molecule. Their atomic orbitals don't just sit there; they interact, overlap, and merge to form entirely new probability landscapes called Molecular Orbitals (MOs). In this problem, we are specifically looking at the overlap of two $2p_z$ orbitals. The geometry of this overlap—and the resulting nodal planes—depends entirely on how these orbitals approach each other relative to the internuclear axis.

The $\sigma$ Orbitals

Let's assume the atoms are approaching each other along the $z$-axis. The $2p_z$ orbitals are aligned perfectly head-to-head.

When they overlap in-phase (positive lobe with positive lobe), they form a $\sigma$ bonding molecular orbital. The electron density builds up strongly between the two nuclei, acting like a glue. But what about the nodes? A single $p_z$ orbital inherently has a nodal plane at the nucleus (the $xy$-plane, where $z=0$). When two such orbitals form a $\sigma$ bond, these two original nodal planes remain intact, situated at each respective nucleus and perpendicular to the internuclear axis. Thus, the $\sigma$ bonding orbital has exactly two nodal planes. This makes statement (A) absolutely correct.

Conversely, if they overlap out-of-phase (positive lobe with negative lobe), they form a ${\sigma }^{\ast }$ antibonding molecular orbital. The electron density between the nuclei cancels out, creating a brand new nodal plane exactly midway between the atoms. Add this to the two original nodal planes, and you get a total of three nodal planes. Crucially, because $\sigma$ orbitals are cylindrically symmetric around the internuclear axis, they never have a nodal plane that contains the axis itself. Therefore, statement (B) is incorrect.

The $\pi$ Orbitals

Now, what if the atoms approach each other along the $x$-axis or $y$-axis? The $2p_z$ orbitals are now parallel to each other and must overlap sideways.

When they overlap in-phase sideways, they form a $\pi$ bonding molecular orbital. The electron density is concentrated above and below the internuclear axis. The original nodal plane of the $p_z$ orbitals (the $xy$-plane) merges into a single, continuous nodal plane that actually contains the internuclear axis. There is no node between the nuclei perpendicular to the axis. Statement (C) incorrectly claims there is a perpendicular node, which is a characteristic of antibonding, not bonding. So, (C) is wrong.

Finally, if they overlap out-of-phase sideways, they form a ${\pi }^{\ast }$ antibonding molecular orbital. Here, the electron density cancels out between the nuclei, creating a new nodal plane perpendicular to the internuclear axis. But the original nodal plane containing the internuclear axis (the $xy$-plane) is still there! So, a ${\pi }^{\ast }$ orbital has two nodal planes: one containing the axis and one perpendicular to it. Statement (D) correctly identifies the presence of the nodal plane containing the molecular axis. Thus, (D) is correct.

Visualizing molecular orbitals is like visualizing 3D standing waves. By tracking where the wave function changes sign, you can easily map out the nodal planes and predict the stability of the bond. The correct statements are indeed (A) and (D).