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

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1. **Principle:** Bond order is calculated as half the difference between the number of electrons in bonding molecular orbitals ($$N_b$$) and antibonding molecular orbitals ($$N_a$$). 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 $$N_2$$. The configuration fills the bonding orbitals optimally ($$N_b=10, N_a=4$$), resulting in a bond order of **3**.
   - **NO:** Total electrons = 7 (N) + 8 (O) = 15. Compared to $$N_2$$, it has one extra electron in an antibonding orbital ($$\pi^*_{2p}$$). Bond order = 2.5.
   - **CN:** Total electrons = 6 + 7 = 13. Bond order = 2.5.
   - **CN$$^+$$:** Total electrons = 6 + 7 - 1 = 12. Isoelectronic with $$C_2$$. Bond order = 2 .
3. **Conclusion:** The species with the highest bond order is CN$$^-$$ (3).

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