Describe the bonding in both the superoxide ion (O2-) and the peroxide ion (O22-) in terms of molecular orbital theory, and compare their bond order to that of O2. (10)
The Correct Answer and Explanation is:
Bonding in Superoxide (O₂⁻), Peroxide (O₂²⁻), and Dioxygen (O₂) Using Molecular Orbital Theory
In molecular orbital (MO) theory, the bonding in diatomic oxygen and its ions can be explained by the filling of molecular orbitals formed by the overlap of atomic orbitals from each oxygen atom. The relevant orbitals for O₂ and its ions arise mainly from the 2s and 2p atomic orbitals, resulting in the following MO configuration for molecules with 12–16 electrons (such as O₂):
σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p_z)², π(2p_x)² = π(2p_y)², π*(2p_x)¹ = π*(2p_y)¹
Dioxygen (O₂):
O₂ has 16 electrons. The MO configuration leads to a bond order of:
Bond order=12(Nb−Na)=12(10−6)=2\text{Bond order} = \frac{1}{2}(N_b – N_a) = \frac{1}{2}(10 – 6) = 2Bond order=21(Nb−Na)=21(10−6)=2
O₂ thus has a double bond and two unpaired electrons in the π* orbitals, making it paramagnetic.
Superoxide (O₂⁻):
The superoxide ion has 17 electrons. The extra electron enters a degenerate π* orbital:
Bond order=12(10−7)=1.5\text{Bond order} = \frac{1}{2}(10 – 7) = 1.5Bond order=21(10−7)=1.5
This results in a bond weaker than in O₂. O₂⁻ is paramagnetic due to one unpaired electron.
Peroxide (O₂²⁻):
Peroxide contains 18 electrons. The two additional electrons fully occupy the π* orbitals:
Bond order=12(10−8)=1\text{Bond order} = \frac{1}{2}(10 – 8) = 1Bond order=21(10−8)=1
This yields a single bond. All electrons are paired, so O₂²⁻ is diamagnetic.
Comparison:
- O₂ bond order = 2 (double bond)
- O₂⁻ bond order = 1.5
- O₂²⁻ bond order = 1 (single bond)
The trend shows that adding electrons to antibonding orbitals lowers the bond order, weakening the bond and increasing bond length.
