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 is false but Statement II is true

Both Statement I and Statement II are true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Chemical Bonding and Molecular Structure PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q17 with Solution

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Correct answer: D

Analyze Statement (I)

A $\pi$ molecular orbital is formed by sideways overlap of atomic orbitals.
- In a bonding $\pi$ MO, electron density is concentrated above and below the internuclear axis.
- In fact, this increased electron density in the region above and below the axis is what makes it a bonding MO.
Therefore, the statement

"A $\pi$ bonding MO has lower electron density above and below the inter-nuclear axis" is false.
Analyze Statement (II)

A $\pi^*$ antibonding molecular orbital is formed by out-of-phase sideways overlap.
- Antibonding orbitals always have a node between the nuclei due to destructive overlap.
- For $\pi^*$ , there is a nodal plane containing the internuclear axis, and also a node between the nuclei because of antibonding character.
Therefore, the statement

"The $\pi^*$ antibonding MO has a node between the nuclei" is true.
Choose the correct option
- Statement (I): False
- Statement (II): True
Hence, the correct option is:

$\boxed{\text{D}}$
Comparison with stored answer

Stored correct answer: $\text{D}$

My derived answer also is $\text{D}$ , so they agree.

View written solutionFree

Correct answer: D

Analyze Statement (I)

A $\pi$ molecular orbital is formed by sideways overlap of atomic orbitals.
- In a bonding $\pi$ MO, electron density is concentrated above and below the internuclear axis.
- In fact, this increased electron density in the region above and below the axis is what makes it a bonding MO.
Therefore, the statement

"A $\pi$ bonding MO has lower electron density above and below the inter-nuclear axis" is false.
Analyze Statement (II)

A $\pi^*$ antibonding molecular orbital is formed by out-of-phase sideways overlap.
- Antibonding orbitals always have a node between the nuclei due to destructive overlap.
- For $\pi^*$ , there is a nodal plane containing the internuclear axis, and also a node between the nuclei because of antibonding character.
Therefore, the statement

"The $\pi^*$ antibonding MO has a node between the nuclei" is true.
Choose the correct option
- Statement (I): False
- Statement (II): True
Hence, the correct option is:

$\boxed{\text{D}}$
Comparison with stored answer

Stored correct answer: $\text{D}$

My derived answer also is $\text{D}$ , so they agree.