Question

discuss molecular orbital diagram for H2 and O7 calculate bond order

Answer 1

In case of H₂ :- H₂ has two electrons .
configuration of H₂ = σ1s²,σ*1s⁰
bond order , B.O = 1/2 [ Nb - Na]
Here, Na = number of electrons in antibonding molecular orbital
Nb = number of electrons in bonding molecular orbital
∴ B.O = 1/2 [ 2 - 0] = 1
there are no unpaired electron exists in molecular orbitals so, H₂ is a diamagnetic molecule.

In case of O₂ :- O₂ has 16 electrons .
configuration of O₂ = σ1s², σ*1s², σ2s² , σ*2s² , σ2Pₓ² , π2Py² ≈ π2Pz² , π*2Py¹≈π*2Pz¹
Nb = 10 and Na = 6
So, B.O = 1/2 [ 10 - 6] = 2
There are two unpaired molecular orbitals in O₂ so, O₂ is a paramagnetic molecule.