Q.

Bond order of O2 is

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a

2

b

1.5

c

3

d

3.5

answer is A.

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Detailed Solution

Electronic configuration of  O2 isO2=(σ1s)2σ∗1s2(σ2s)2σ∗2s2σ+2s2σ2pz2π2px2≡π2py2π∗2px1≡π∗2py1Hence bond order  =12Nb−Na=12[10−6]=2
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