JEE MAIN - Chemistry (2024 - 1st February Evening Shift - No. 17)
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 :
Explanation
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 $\pi^*$ antibonding MO has a node between the nuclei."
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.
Therefore, the correct answer is:
Option D: Statement I is false but Statement II is true.
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