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NEET CHEMISTRYChemical Bonding and Molecular StructureMedium

Question

Consider the following species: CN+^+, CN−^-, NO and CN. The highest bond order associated with:

A

NO

B

CN−^-

C

CN+^+

D

CN

Step-by-Step Solution

  1. Principle: Bond order is calculated as half the difference between the number of electrons in bonding molecular orbitals (NbN_b) and antibonding molecular orbitals (NaN_a). Generally, isoelectronic species have the same bond order .
  2. Electron Count and Calculation:
  • CN−^-: Total electrons = 6 (C) + 7 (N) + 1 (charge) = 14. This is isoelectronic with N2N_2. The configuration fills the bonding orbitals optimally (Nb=10,Na=4N_b=10, N_a=4), resulting in a bond order of 3.
  • NO: Total electrons = 7 (N) + 8 (O) = 15. Compared to N2N_2, it has one extra electron in an antibonding orbital (π2p∗\pi^*_{2p}). Bond order = 2.5.
  • CN: Total electrons = 6 + 7 = 13. Bond order = 2.5.
  • CN+^+: Total electrons = 6 + 7 - 1 = 12. Isoelectronic with C2C_2. Bond order = 2 .
  1. Conclusion: The species with the highest bond order is CN−^- (3).

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Chemical Bonding and Molecular Structure. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYChemical Bonding and Molecular Structureconsiderfollowingspecieshighestassociated

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