Properties of Molecular Orbitals Formed by Two 2pz Orbitals

To determine the correct statement(s) regarding the molecular orbitals formed by the overlap of two $2p_z$ orbitals, we analyze the symmetries and nodal properties under both possible orientations of the internuclear (molecular) axis.

---

1. Head-on (Axial) Overlap: $z$-axis as the Molecular Axis

When the $z$-axis is designated as the internuclear axis, two $2p_z$ atomic orbitals overlap head-on along the axis of symmetry, forming a bonding $\sigma_{2p_z}$ orbital and an antibonding $\sigma^*_{2p_z}$ orbital.

Analysis of $\sigma$ (Bonding) Orbital:

• Each isolated $2p_z$ atomic orbital has a nodal plane perpendicular to the $z$-axis passing through its respective nucleus ($z = z_A$ and $z = z_B$).
• In the bonding combination $\sigma_{2p_z} = \psi_{2p_z, A} - \psi_{2p_z, B}$ (where lobe signs facing each other are identical):
  • Constructive interference occurs in the region between the nuclei ($z_A < z < z_B$).
  • The wavefunction changes sign on the outer side of nucleus $A$ ($z < z_A$) and nucleus $B$ ($z > z_B$).
• Consequently, the $\sigma$ molecular orbital possesses two nodal planes perpendicular to the molecular axis near/at the positions of the two nuclei.

$\text{Total number of nodal planes for } \sigma_{2p_z} = 2$

Thus, Option A is correct.

Analysis of $\sigma^*$ (Antibonding) Orbital:

• In the antibonding combination $\sigma^*_{2p_z} = \psi_{2p_z, A} + \psi_{2p_z, B}$, destructive interference occurs between the nuclei.
• A nodal plane exists midway between the two nuclei at $z = 0$ (which is the $xy$-plane, perpendicular to the molecular axis).
• Since $\sigma^*$ is cylindrically symmetric about the internuclear $z$-axis, the $xz$-plane (which contains the molecular axis) is not a nodal plane.

Thus, Option B is incorrect.

---

2. Sideways (Lateral) Overlap: $x$-axis or $y$-axis as the Molecular Axis

When the internuclear axis is perpendicular to the $z$-axis (e.g., along the $x$-axis or $y$-axis), two parallel $2p_z$ atomic orbitals overlap sideways, forming a bonding $\pi_{2p_z}$ orbital and an antibonding $\pi^*_{2p_z}$ orbital.

Analysis of $\pi$ (Bonding) Orbital:

• Both atomic $2p_z$ orbitals have a node at $z = 0$ (the $xy$-plane).
• In the $\pi_{2p_z}$ bonding MO, electron density is concentrated above and below the molecular axis (in $+z$ and $-z$ regions).
• The orbital has a single nodal plane at $z = 0$ (the $xy$-plane), which contains the molecular axis.
• It does not have a nodal plane perpendicular to the molecular axis passing through the center of the molecule.

Thus, Option C is incorrect.

Analysis of $\pi^*$ (Antibonding) Orbital:

• In the $\pi^*_{2p_z}$ antibonding MO, out-of-phase sideways overlap creates two nodal planes:
  1. The $xy$-plane ($z = 0$), which contains the molecular axis.
  2. A plane perpendicular to the molecular axis passing through the center of the molecule (the $yz$-plane if the molecular axis is along $x$, or the $xz$-plane if the molecular axis is along $y$).
• Therefore, the $\pi^*$ orbital indeed has a nodal plane in the $xy$-plane containing the molecular axis.

Thus, Option D is correct.

---

Conclusion

The correct statements are A and D.