Question

According to MO theory, number of species/ions from the following having identical bond order is ___.
CN-, NO+, O2, O+2, O2+2

Answer 1

Correct Answer: 3

To solve this problem, we need to determine the bond order of each given species using Molecular Orbital (MO) theory and identify how many of them have identical bond orders. The bond order is calculated using the formula: Bond Order = (Number of electrons in bonding MOs - Number of electrons in antibonding MOs)/2.

1. CN^-: Electron configuration: σ_1s² σ_*1s² σ_2s² σ_*2s² π_2p⁴ σ_2p².
Bonding electrons = 10, Antibonding electrons = 4
Bond Order = (10 - 4)/2 = 3

2. NO⁺: Electron configuration: σ_1s² σ_*1s² σ_2s² σ_*2s² π_2p⁴ σ_2p².
Bonding electrons = 10, Antibonding electrons = 4
Bond Order = (10 - 4)/2 = 3

3. O₂: Electron configuration: σ_1s² σ_*1s² σ_2s² σ_*2s² σ_2p² π_2p⁴ π_*2p².
Bonding electrons = 10, Antibonding electrons = 6
Bond Order = (10 - 6)/2 = 2

4. O₂⁺: Electron configuration: σ_1s² σ_*1s² σ_2s² σ_*2s² σ_2p² π_2p⁴ π_*2p¹.
Bonding electrons = 10, Antibonding electrons = 5
Bond Order = (10 - 5)/2 = 2.5

5. O₂²⁺: Electron configuration: σ_1s² σ_*1s² σ_2s² σ_*2s² σ_2p² π_2p⁴.
Bonding electrons = 10, Antibonding electrons = 4
Bond Order = (10 - 4)/2 = 3

From the calculations above, CN^-, NO⁺, and O₂²⁺ all have a bond order of 3. Therefore, the number of species with identical bond order is 3, which fits the given range of 3, 3.