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:
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:
- Both Statement I and Statement II are false
- Both Statement I and Statement II are true
- Statement I is false but Statement II is true
- Statement I is true but Statement II is false
The Correct Option is C
Approach Solution - 1
To determine the truth of the given statements about molecular orbitals, we need to understand two concepts: \(\pi\) bonding molecular orbitals (MOs) and \(\pi^*\) antibonding molecular orbitals.
- Statement I: A \(\pi\) bonding MO has lower electron density above and below the inter-nuclear axis.
- A \(\pi\) bonding MO is formed by the lateral overlapping of p orbitals. In such MOs, the electron density is concentrated in two lobes above and below the plane of the inter-nuclear axis, not on the axis itself. The statement claims that there is "lower" electron density above and below, which contradicts how \(\pi\) bonding MOs actually behave, where electron density is maximized in these regions. Thus, Statement I is false.
- Statement II: The \(\pi^*\) antibonding MO has a node between the nuclei.
- A \(\pi^*\) antibonding MO is also formed by the lateral overlapping of p orbitals, but in an out-of-phase manner. This results in a nodal plane between the nuclei where there is zero electron density, which is characteristic of antibonding molecular orbitals. Therefore, Statement II is true.
Based on the reasoning above:
- Statement I is false because the \(\pi\) bonding MO should actually have increased electron density above and below the inter-nuclear axis.
- Statement II is true because the \(\pi^*\) antibonding MO correctly has a node between the nuclei.
Thus, the correct answer is: Statement I is false but Statement II is true.
Approach Solution -2
Statement (I) Analysis:
A $\pi$ bonding molecular orbital (MO) is formed by the sideways overlap of $p$-orbitals.
This type of bonding MO has higher electron density above and below the internuclear axis, leading to a bonding interaction between atoms.
Therefore, Statement (I) is false.
Statement (II) Analysis:
The $\pi^*$ antibonding molecular orbital is formed when $p$-orbitals combine in such a way that destructive interference occurs between the wave functions of the atomic orbitals.
This creates a node (a region of zero electron density) between the nuclei.
Therefore, Statement (II) is true.
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