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NEET CHEMISTRYMedium

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).
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