Given below are two statements :

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

Statement (II) : The $\pi^*$ antibonding MO has a node between the nuclei.

In the light of the above statements, choose the most appropriate answer from the options given below :

<p>Let's analyze both statements:</p> <p><b>Statement (I)</b>: "A $\pi$ bonding MO has lower electron density above and below the inter-nuclear axis."</p> <p>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.</p> <p><b>Statement (II)</b>: "The $\pi^*$ antibonding MO has a node between the nuclei."</p> <p>This statement is true. In a $\pi^*$ (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 $\pi^*$ orbital antibonding—the electrons in this orbital actually serve to destabilize the bond between the two atoms.</p> <p>Therefore, the correct answer is:</p> <p>Option D: Statement I is false but Statement II is true.</p>