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Chemical Bonding and Molecular Structure question

2018 · 15 Apr · Shift 2 · Q6
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Chemical Bonding and Molecular Structure question

2018 · 15 Apr · Shift 2 · Q6

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Which of the following best describes the diagram below of a molecular orbital ? JEE Main 2018 (Online) 15th April Evening Slot Chemistry - Chemical Bonding & Molecular Structure Question 201 English
  1. A
    A non-bonding orbital
  2. B
    An antibonding σ\sigmaσ orbital
  3. C
    A bonding π\piπ orbital
  4. D
    An antibonding π\piπ orbital
View written solutionFree

Correct answer: D

  1. Identify the key features of the molecular orbital diagram

    To classify a molecular orbital, we look for:

    • whether the overlap is head-on or sidewise
    • whether there is electron density between the nuclei
    • whether there is a node between the nuclei
  2. Distinguish between σ\sigmaσ and π\piπ orbitals

    • A σ\sigmaσ orbital is formed by head-on overlap along the internuclear axis.
    • A π\piπ orbital is formed by sidewise overlap of orbitals, with electron density above and below the internuclear axis.

    Since the diagram is of a molecular orbital with sidewise character, it must be a π\piπ-type orbital, not a σ\sigmaσ orbital.

  3. Distinguish between bonding and antibonding

    • In a bonding orbital, electron density is concentrated between the two nuclei.
    • In an antibonding orbital, there is a node between the nuclei, so electron density is absent from the internuclear region.

    The diagram indicates destructive overlap with a node between the nuclei, so it is antibonding.

  4. Combine both observations

    • Type of overlap: π\piπ
    • Nature: antibonding

    Therefore, the orbital is an antibonding π\piπ orbital, written as π∗\pi^*π∗.

  5. Check options

    • A: A non-bonding orbital →\rightarrow→ Incorrect
    • B: An antibonding σ\sigmaσ orbital →\rightarrow→ Incorrect
    • C: A bonding π\piπ orbital →\rightarrow→ Incorrect
    • D: An antibonding π\piπ orbital →\rightarrow→ Correct

Hence, the correct answer is D.

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