Why is an oxygen molecule a radical ? There are no unpaired electrons

There are no unpaired electrons in an oxygen molecule because

originally oxygen has a configuration of 1s2 2s2 2p4 ,so now each oxygen atom becomes 1s2 2s2 2p6 ( from the overlapping of the 2p orbitals) .

So there are now no unpaired eelctrons.


Is that right?

Why is oxygen molecule deemed as a radical?

Thanks

It is better than that O2 is a diradical with two unpaired e⁻ and is rationalized by molecular orbital theory :
Simple MO diagram with no sp mixing
O2 = 12 valence e⁻ = σs(2 e⁻) σs(2e⁻) σp(2 e⁻) πp(4 e⁻) πp*(2 e⁻) σp*(0): πpx*(1e⁻)πpy*(1e⁻)
Bond Order = ½[Σ (bonding e⁻) - Σ (antibonding e⁻)]
bo = ½[ σ1(2e⁻) σ3(2e⁻) π1(4 e⁻) - σ2*(2e⁻) π2*(2 e⁻) ] = ½[8-4] = 2
There are two e⁻s in the πp* (two equal energy MOs) and the e⁻s have their spins parallel (unpaired) (Hund's Rule). The O2 molecule is paramagnetic. Lewis would predict O2 to be diamagnetic
with no un p e-.
See http://www.ch.ic.ac.uk/vchemlib/course/m… last example

look up radical in wikipedia, its all there.