Sulphur in vapour state exhibits paramagnetic behaviour . Give reasons
Sulphur in vapour state exhibits paramagnetic behaviour . Give reasons
The paramagnetism in Sulphur in vapour phase can be explained from the fact that in vapour phase, S2 is the dominant species and is therefore paramagnetic like O2, which has two unpaired electrons. So like O2 molecule, S2 also has two unpaired electrons in the π*px and π*py orbital.
Let us now see how does it happen.
The electronic configuration of sulphur atom is 1s2 2s2 2p6 3s2 3p4. Each sulphur atom has 16 electrons, hence, in S2 molecule there are 32 electrons. The electronic configuration of S2 molecule, will be
σ1s2 < σ*1s2 < σ2s2 < σ*2s2<σ2pz 2 <(π2px 2= π2py 2)< (π*2px 2= π*2py 2)< σ*2pz 2 < σ3s2 < σ*3s2 < σ3pz 2 <(π3px 2= π3py 2)< (π*3px1= π*3py 1)
In vapour phase sulphur exists as S 2 molecule and S 2 molecule has two unpaired electrons in the antibonding 3p orbital and hence exhibits paramagnetism.
The paramagnetism in Sulphur in vapour phase can be explained from the fact that in vapour phase, S2 is the dominant species and is therefore paramagnetic like O2, which has two unpaired electrons. So like O2 molecule, S2 also has two unpaired electrons in the π*px and π*py orbital.
Let us now see how does it happen.
The electronic configuration of sulphur atom is 1s2 2s2 2p6 3s2 3p4. Each sulphur atom has 16 electrons, hence, in S2 molecule there are 32 electrons. The electronic configuration of S2 molecule, will be
σ1s2 < σ*1s2 < σ2s2 < σ*2s2<σ2pz 2 <(π2px 2= π2py 2)< (π*2px 2= π*2py 2)< σ*2pz 2 < σ3s2 < σ*3s2 < σ3pz 2 <(π3px 2= π3py 2)< (π*3px1= π*3py 1)
In vapour phase sulphur exists as S 2 molecule and S 2 molecule has two unpaired electrons in the antibonding 3p orbital and hence exhibits paramagnetism.
