InterviewSolution
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InterviewSolution
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Explain why the bond order of N2 is greater than N2+, but the bond order of O2 is less than that of O2+. |
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Answer» N2 = (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2px)2 (π2py)2 (π2pz)2 Bond order = 1/2 (Nb - Na) = 1/2 (10 - 4) = 3 N2+ = (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (π2px)2 [π2py]2 [σ2pz]1 Bond order = 1/2 (7 - 2) = 2.5 Hence bond order of N2 is greater than N2+ O2+ = (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2px)2 (π 2py)2 (2π 2pz)2 (π* 2py)1 Bond order = 1/2 (Nb - Na) = 1/2 (10 - 5) = 2.5 Hence Bond order of O2 is less than of O2+. |
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