Question

below is a set of boundary - surface representations of molecular orbitals in homonuclear (second period) diatomic molecules. drag - and - drop the orbital designations to the appropriate molecular orbitals.
see periodic table
see hint

Explanation:

Molecular orbitals in homonuclear diatomic molecules of the second - period are classified based on their symmetry and formation. $\sigma$ orbitals have cylindrical symmetry about the internuclear axis. $\sigma_{2s}$ is formed from the combination of 2s atomic orbitals. $\sigma_{2p}$ is formed from the head - on overlap of 2p atomic orbitals along the internuclear axis. $\sigma_{2p}^*$ (antibonding $\sigma$ orbital) has a node between the nuclei. $\pi$ orbitals have a nodal plane containing the internuclear axis. $\pi_{2p}$ is formed from the side - on overlap of 2p atomic orbitals, and $\pi_{2p}^*$ (antibonding $\pi$ orbital) has an additional node.

Looking at the shapes:

• For $\sigma$ orbitals: They are symmetric about the internuclear axis. The $\sigma_{2s}$ (C) is formed from 2s orbitals (simpler in shape compared to 2p - based $\sigma$ orbitals). The $\sigma_{2p}$ (D) has a more elongated shape due to the 2p orbital contribution. The $\sigma_{2p}^*$ (B) has a node (a region of zero electron density) between the nuclei.
• For $\pi$ orbitals: $\pi_{2p}$ (A) has a shape with two lobes on either side of the internuclear axis (side - on overlap), and $\pi_{2p}^*$ (E) has an additional node. $\sigma_{2s}^*$ (F) is also an antibonding orbital but from 2s - orbital combination.

By analyzing the symmetry and nodal patterns:

• The top - left orbital has the symmetry and nodal pattern of $\sigma_{2p}^*$ (B).
• The top - right orbital has the symmetry of $\pi_{2p}^*$ (E).
• The middle - left orbital has the symmetry of $\sigma_{2s}$ (C).
• The middle - right orbital has the symmetry of $\sigma_{2p}$ (D).
• The bottom - left orbital has the symmetry of $\pi_{2p}$ (A).
• The bottom - right orbital (assuming it's the remaining one) would be $\sigma_{2s}^*$ (F) (but based on standard orbital - shape - to - label mapping in such problems, if we consider the common ones first:
• If we match based on head - on (for $\sigma$) and side - on (for $\pi$) overlap and antibonding (with node) and bonding (without extra node for antibonding):
• The orbital with two lobes along the axis and no node (middle - right) is $\sigma_{2p}$ (D).
• The orbital with two lobes along the axis and a node (top - left) is $\sigma_{2p}^*$ (B).
• The orbital with side - on lobes (bottom - left) is $\pi_{2p}$ (A).
• The orbital with side - on lobes and an extra node (top - right) is $\pi_{2p}^*$ (E).
• The simpler $\sigma$ - like orbital (middle - left) is $\sigma_{2s}$ (C).
• The remaining (if we assume a standard 6 - orbital set) with the appropriate $\sigma_{2s}^*$ shape is F.

Answer:

Top - left: B ($\sigma_{2p}^*$), Top - right: E ($\pi_{2p}^*$), Middle - left: C ($\sigma_{2s}$), Middle - right: D ($\sigma_{2p}$), Bottom - left: A ($\pi_{2p}$), Bottom - right: F ($\sigma_{2s}^*$)