Note:
(1) Bond length
∝(2) Bond order
=[Nb−Na]Nb= No of electrons in bonding molecular orbital
Na= No of electrons in anti bonding molecular orbital
(4) upto 14 electrons, molecular orbital configuration is
Here
Na= Anti bonding electron
=4 and
Nb=10(5) After 14 electrons to 20 electrons molecular orbital configuration is
Here
Na=10and
Nb=10In 0 atom 8 electrons present, so in
O2,8×2=16 electrons present.
Then in
O2+no of electrons
=15in
O2−no of electrons
=17in
O22− no of electrons
=18 ∴ Molecular orbital configuration of
O2 (16 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px1*=π2py1*
∴Na=6Nb=10∴BO=[10−6]=2 Molecular orbital configuration of
O2+(15 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px1*=π2pya*
∴Nb=10Na=5∴BO=[10−5]=2.5 Molecular orbital configuration of
O22+ (14 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px0*=π2pyo*
∴Nb=10Na=4∴BO=[10−4]=3 Molecular orbital configuration of
O2−(17 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px2*=π2py1*
∴Nb=10Na=7∴BO=[10−7]=1.5 Molecular orbital configuration of
O22− (18 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px2*=π2py2*
∴Nb=10Na=8∴B0=[10−8]=1 As Bond length
∝So
O22+ has the shortest bond length.