46 and 49

Dear Student,


46) The correct answer is (3)No. of electrons in O22- = 18                                           =σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2px2π2py2π*2px2π*2py2Bond order=Nb-Na2=10-82=22=1No. of electrons in B2 =10                                      = σ1s2σ*1s2σ2s2σ*2s2π2px1π2py1Bond order=Nb-Na2=6-42=22=1Kindly ask the remaining query in the next thread. 

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