Q.Which of the following have identical bond order? (Note: more than one of the given options may be correct.)

The key idea here is Bond Order Strength, which is determined using Molecular Orbital (MO) theory. Bond order quantifies the number of chemical bonds between a pair of atoms and is directly related to bond strength and stability.

To find the bond order for each species:

1. Count total electrons: Sum the atomic numbers of constituent atoms and adjust for the charge.
2. Fill molecular orbitals: Distribute electrons into bonding ($N_b$) and antibonding ($N_a$) molecular orbitals according to increasing energy, remembering the different MO energy order for species with $\le 14$ electrons versus $> 14$ electrons.
3. Calculate bond order: Use the formula: Bond Order = $\frac{1}{2}(N_b - N_a)$.

Let's apply this to each option:

• (A) CN$^-$: Total electrons = C(6) + N(7) + 1 = 14.

  MO configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p}^4 \sigma_{2p_z}^2$.

  $N_b = 2+2+4+2 = 10$, $N_a = 2+2 = 4$.

  Bond Order = $\frac{1}{2}(10 - 4) = 3$.

• (B) NO$^+$: Total electrons = N(7) + O(8) - 1 = 14.

  MO configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p}^4 \sigma_{2p_z}^2$.

  $N_b = 2+2+4+2 = 10$, $N_a = 2+2 = 4$.

  Bond Order = $\frac{1}{2}(10 - 4) = 3$.

• (C) O$_2^-$: Total electrons = O(8) + O(8) + 1 = 17. …

Understanding the strength and stability of a chemical bond is fundamental in chemistry. For diatomic molecules and ions, Molecular Orbital (MO) theory provides a powerful framework to determine these properties, particularly through the concept of bond order.

What is Bond Order?

Bond order is a measure of the number of chemical bonds between a pair of atoms. It directly correlates with bond strength and inversely with bond length. A higher bond order indicates a stronger and shorter bond. For example, a bond order of 1 corresponds to a single bond, 2 to a double bond, and 3 to a triple bond. Fractional bond orders (like 1.5 or 2.5) are also possible, indicating delocalized bonding or resonance structures.

Calculating Bond Order using Molecular Orbital Theory

Molecular Orbital theory describes how atomic orbitals combine to form molecular orbitals (MOs) that span the entire molecule. Electrons then fill these MOs according to the Aufbau principle, Hund's rule, and Pauli's exclusion principle.

MOs are classified as either bonding (lower energy, stabilize the molecule) or antibonding (higher energy, destabilize the molecule).

The energy ordering of molecular orbitals for diatomic species depends on the total number of electrons:

• For species with $\le 14$ electrons (e.g., N₂, CN⁻, NO⁺):

  The energy order is:

  $\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \pi_{2p} < \sigma_{2p_z} < \pi_{2p}^* < \sigma_{2p_z}^*$

  (Note: $\pi_{2p}$ orbitals are lower in energy than $\sigma_{2p_z}$ orbitals.)

• For species with $> 14$ electrons (e.g., O₂, F₂, O₂⁻, O₂²⁻):

  The energy order is:

  $\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \sigma_{2p_z} < \pi_{2p} < \pi_{2p}^* < \sigma_{2p_z}^*$

  (Note: $\sigma_{2p_z}$ orbital is lower in energy than $\pi_{2p}$ orbitals.)

  In both cases, the $\pi_{2p}$ and $\pi_{2p}^*$ sets each consist of two degenerate orbitals.

Let's apply this to each given option:

1. Calculate the total number of electrons for each species.

   • (A) CN⁻: Carbon (6 electrons) + Nitrogen (7 electrons) + 1 (for the negative charge) = 14 electrons
   • (B) NO⁺: Nitrogen (7 electrons) + Oxygen (8 electrons) - 1 (for the positive charge) = 14 electrons
   • (C) O₂⁻: Oxygen (8 electrons) + Oxygen (8 electrons) + 1 (for the negative charge) = 17 electrons
   • (D) O₂²⁻: Oxygen (8 electrons) + Oxygen (8 electrons) + 2 (for the 2- negative charge) = 18 electrons

2. Determine the Molecular Orbital (MO) electron configuration and calculate bond order for each species.

   • (A) CN⁻ (14 electrons):

     Since it has 14 electrons, we use the MO energy order for $\le 14$ electrons.

     MO configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p}^4 \sigma_{2p_z}^2$

     Number of bonding electrons ($N_b$) = $2 (\sigma_{1s}) + 2 (\sigma_{2s}) + 4 (\pi_{2p}) + 2 (\sigma_{2p_z}) = 10$

     Number of antibonding electrons ($N_a$) = $2 (\sigma_{1s}^*) + 2 (\sigma_{2s}^*) = 4$

     Bond Order = $\frac{1}{2} (10 - 4) = \frac{6}{2} = 3$

   • (B) NO⁺ (14 electrons):

     Since it also has 14 electrons, we use the same MO energy order.

     MO configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p}^4 \sigma_{2p_z}^2$

     Number of bonding electrons ($N_b$) = $2 (\sigma_{1s}) + 2 (\sigma_{2s}) + 4 (\pi_{2p}) + 2 (\sigma_{2p_z}) = 10$ …