What is the Bond Order of O2 (Oxygen Gas)?

An oxygen atom has 6 valence electrons. To achieve a stable octet (8 electrons), two oxygen atoms must share two pairs of electrons with each other. This creates a double bond (O=O). Since there is a double bond connecting the two atoms, the bond order is simply 2.

For competitive exams (like JEE or NEET), you must calculate bond order using MO Theory. Formula: Bond Order = (Number of Bonding Electrons - Number of Antibonding Electrons) / 2

1. A single Oxygen atom (atomic number 8) has 8 electrons. Therefore, an O₂ molecule has a total of 16 electrons.
2. We fill these 16 electrons into the molecular orbitals: σ1s², σ1s², σ2s², σ2s², σ2pz², π2px² = π2py², π2px¹ = π2py¹.
3. Counting the electrons:
   • Bonding electrons (Nb) = 10
   • Antibonding electrons (Na) = 6 (those with the * symbol)
4. Calculation: Bond Order = (10 - 6) / 2 = 4 / 2 = 2.

The MO Theory configuration reveals something the Lewis structure cannot. In the highest energy orbitals (π2px¹ and π2py¹), there are two unpaired electrons. Because of these unpaired electrons, liquid oxygen is actually attracted to a magnetic field. This property is called paramagnetism.