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 $$(\xrightarrow{z-axis})$$?

The internuclear axis of the diatomic molecule is chosen as the $$z$$-axis.

Recall the shapes of the four common kinds of molecular orbitals :

• $$\sigma$$ - no nodal plane that contains the $$z$$-axis.

• $$\sigma^{*}$$ - one nodal plane that contains the $$z$$-axis (because the two combining $$p_z$$ orbitals overlap out-of-phase).

• $$\pi$$ - one nodal plane that contains the $$z$$-axis (because the lobes lie above and below the axis) but no nodal plane perpendicular to the axis between the nuclei (the overlap is in-phase).

• $$\pi^{*}$$ - two nodal planes: (i) a plane that contains the $$z$$-axis, and (ii) a plane perpendicular to the $$z$$-axis at the bond centre (because the overlap is out-of-phase).

Therefore a correct diagram of a $$\pi^{*}$$ orbital must show:

1. Lobes located above and below the $$z$$-axis - this gives the nodal plane containing the $$z$$-axis (typical for any $$\pi$$ type orbital).

2. Opposite phases (shaded vs unshaded) on the two atoms - this produces a second nodal plane perpendicular to the $$z$$-axis exactly between the nuclei, characteristic of an antibonding orbital.

3. No electron density exactly on the internuclear axis itself (because $$p_x$$ or $$p_y$$ basis functions are used, not $$p_z$$).

Inspecting the four given sketches:

Option (a) - shows in-phase overlap; it represents a bonding $$\pi$$ orbital, not antibonding.
Option (b) - shows lobes above and below the axis with opposite shading on the two atoms; there is a node at the bond centre and a nodal plane containing the $$z$$-axis. This satisfies both conditions for $$\pi^{*}$$.
Option (c) - lobes are along the internuclear axis, so it would be a $$\sigma$$ type, not $$\pi$$.
Option (d) - again shows in-phase overlap (bonding) rather than antibonding.

Thus only Option (b) has the correct orientation and the two required nodal planes of a $$\pi^{*}$$ molecular orbital.

Option B which is: Option (b)