Q.
Consider the following statement.I. When O2 is converted into O22+ bond order decreases.II. O2 molecule is paramagnetic because it contains two unpaired electrons in π*2px and π*2py molecular orbitals.III. The bond length in NO is greater than NO+.Choose the correct option.
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a
Only I
b
I and II
c
II and III
d
II and IV
answer is C.
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Detailed Solution
Statement II and III are correct.The molecular orbital configuration of O2 molecule is as follows: σ 1s2, σ *1s2, σ 2s2, σ *2s2, σ 2p2z,π2px2 =π2py2 ,π*2px1=π*2py1There are 10 bonding and 6 nonbonding electrons in the orbitals according to the molecular orbital configuration.Therefore, Bond order =12[Bonding-antibonding]= 12[10−6]=12(4)=2Thus, the bond order O2 is 2. The molecular orbital configuration of O2+2 is σ1s2,σ*1s2,σ2s2,σ*2s2,σ2pz2,π2px2=π2py2There are 10 bonding and 4 nonbonding electrons in the orbitals according to the molecular orbital configuration.Therefore, Bond order =12[Bonding-antibonding]=12[10−4]=3.Hence statement I is incorrect.
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