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Question:

Among H2, He+2, Li2, Be2, B2, C2, N2, O2, and F2, the number of diamagnetic species is (Atomic numbers: H=1, He=2, Li=3, Be=4, B=5, C=6, N=7, O=8, F=9)

Solution:

Six species are diamagnetic as all electrons are paired. These are H2(σ1s)2, Li2 KK(σ2s)2, Be2 KK(σ2s)2(σ2s)2, C2 KK(σ2s)2(σ2s)2(π2py)2(π2pz)2, N2 KK(σ2s)2(σ2s)2(π2py)2(π2pz)2(σ2px)2 F2 KK(σ2s)2(σ2s)2(σ2px)2(π2py)2(π2pz)2(π2py)2(π2pz)2. Three species are paramagnetic as unpaired electrons are present. These are He+2(σ1s)2(σ1s)1, B2 KK(σ2s)2(σ2s)2(π2py)1(π2pz)1, O2 KK(σ2s)2(σ2s)2(σ2px)2(π2py)2(π2pz)2(π2py)1(π*2pz)1.