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Chemical Bonding and Molecular Structure question

2012 · Q115
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Chemical Bonding and Molecular Structure question

2012 · Q115

NEETChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Bond order of 1.5 is shown by
  1. A
    O2+
  2. B
    O2−-−
  3. C
    O22−-−
  4. D
    O2
View written solutionFree

Correct answer: B

Configuration of O2

σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py2π∗2px1π∗2py1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2{\pi ^*}2p_x^1{\pi ^*}2p_y^1σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​π2py2​π∗2px1​π∗2py1​

Bond order = No. of e− in bonding M.O−No. of e− in antibonding M.O2{{No.\,of\,{e^ - }\,in\,bonding\,M.O - No.\,of\,{e^ - }\,in\,antibonding\,M.O} \over 2}2No.ofe−inbondingM.O−No.ofe−inantibondingM.O​

Bond order of O2+ = 10−52{{10 - 5} \over 2}210−5​ = 2.5

Bond order of O2- = 10−72{{10 - 7} \over 2}210−7​ = 1.5

Bond order of O22- = 10−82{{10 - 8} \over 2}210−8​ = 1

Bond order of O2 = 10−62{{10 - 6} \over 2}210−6​ = 2

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