JEE Main 2024 — Chemical Bonding And Molecular Structure Question with Solution

Let's analyze both statements:

Statement (I): "A $\pi$ bonding MO has lower electron density above and below the inter-nuclear axis."

This statement is false. In molecular orbital (MO) theory, a pi bond ($\pi$ bond) is formed by the sideways overlap of p-orbitals from two adjacent atoms. The characteristic electron density of a $\pi$ bonding molecular orbital is concentrated above and below the inter-nuclear axis, not lower. These regions of electron density are where the p-orbitals overlap and electrons are likely to be found. This concentration of electron density above and below the inter-nuclear axis is what allows the $\pi$ bond to provide additional stability to the molecule, alongside any sigma ($\sigma$) bonds that may be present.

Statement (II): "The ${\pi }^{\ast }$ antibonding MO has a node between the nuclei."

This statement is true. In a ${\pi }^{\ast }$ (pi-star) antibonding molecular orbital, there is indeed a nodal plane between the nuclei where the probability of finding electrons is essentially zero. This node arises from the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive interference and a region of zero electron density in the bonding region. The presence of this nodal plane is what makes the ${\pi }^{\ast }$ orbital antibonding—the electrons in this orbital actually serve to destabilize the bond between the two atoms.

Therefore, the correct answer is:

Option D: Statement I is false but Statement II is true.