Concept: Bond order is a measure of the number of chemical bonds between a pair of atoms. It is directly related to the stability of a molecule and inversely related to bond length. For diatomic molecules, bond order can be calculated using Molecular Orbital Theory (MOT) as half the difference between the number of electrons in bonding molecular orbitals and antibonding molecular orbitals.

Why (B) is correct: To determine the bond order, we first need to calculate the total number of electrons for each species and then fill them into the molecular orbitals. The formula for bond order is: Bond Order = 1/2 (Number of electrons in bonding MOs - Number of electrons in antibonding MOs).

• Step 1: Calculate total electrons for each species.
  - ${\mathrm{CN}}^{+}$: C (6) + N (7) - 1 (for +ve charge) = 12 electrons
  - ${\mathrm{CN}}^{-}$: C (6) + N (7) + 1 (for -ve charge) = 14 electrons
  - $\mathrm{NO}$: N (7) + O (8) = 15 electrons
  - $\mathrm{CN}$: C (6) + N (7) = 13 electrons
• Step 2: Determine bond order using MOT electron configuration.
  - For species with 14 electrons (like ${\mathrm{CN}}^{-}$), the electron configuration is ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p}^4{\sigma }_{2p}^2$.
  Bonding electrons = 2+2+4+2 = 10
  Antibonding electrons = 2+2 = 4
  Bond Order = 1/2 (10 - 4) = 3.0
• Step 3: Calculate bond orders for other species.
  - ${\mathrm{CN}}^{+}$ (12 electrons): ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p}^4$.
  Bonding electrons = 2+2+4 = 8
  Antibonding electrons = 2+2 = 4
  Bond Order = 1/2 (8 - 4) = 2.0
  - $\mathrm{NO}$ (15 electrons): ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p}^4{\sigma }_{2p}^2{\pi }_{2p}^{\ast 1}$.
  Bonding electrons = 2+2+4+2 = 10
  Antibonding electrons = 2+2+1 = 5
  Bond Order = 1/2 (10 - 5) = 2.5
  - $\mathrm{CN}$ (13 electrons): ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p}^4{\sigma }_{2p}^1$.
  Bonding electrons = 2+2+4+1 = 9
  Antibonding electrons = 2+2 = 4
  Bond Order = 1/2 (9 - 4) = 2.5
• Step 4: Compare bond orders.
  - ${\mathrm{CN}}^{+}$: 2.0
  - ${\mathrm{CN}}^{-}$: 3.0
  - $\mathrm{NO}$: 2.5
  - $\mathrm{CN}$: 2.5
  ${\mathrm{CN}}^{-}$ has the highest bond order of 3.0.

Option Analysis:

• A) $\mathrm{NO}$: Has 15 electrons, leading to a bond order of 2.5. Incorrect.
• B) ${\mathrm{CN}}^{-}$: Has 14 electrons, leading to a bond order of 3.0. This is the highest. Correct.
• C) ${\mathrm{CN}}^{+}$: Has 12 electrons, leading to a bond order of 2.0. Incorrect.
• D) $\mathrm{CN}$: Has 13 electrons, leading to a bond order of 2.5. Incorrect.

Correct Answer: (B)