(1) Bond order
=‌[Nb−Na]Nb= No of electrons in bonding molecular orbital
Na= No of electrons in anti bonding molecular orbital
(2) upto 14 electrons, molecular orbital configuration is
Here
Na= Anti bonding electron
=4 and
Nb=10(3) After 14 electrons to 20 electrons molecular orbital configuration is -
Here
Na=10and
Nb=10In
O atom 8 electrons present, so in
O2,8×2=16 electrons present.
in
O22− no of electrons
=18(A) Molecular orbital configuration of
O22−(18 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px2*=π2py2*∴‌‌Nb=10Na=8∴‌‌BO=‌[10−8]=1 (B)
C22− has 14 electrons.
Moleculer orbital configuration of
C22− is
σ1s2σ1s2*σ2s2σ2s2*π2px2=π2py2σ2pz2∴Na=4Nb=10∴‌‌BO=‌[10−4]=3(C)
N22− has 16 electrons.
Moleculer orbital configuration of
N22− is
σ1s2σ1s2*σ2s2σ2s2*π2px2=π2py2σ2pz2π2px1*=π2py1*∴Na=6 Nb=10∴‌‌BO=‌[10−6]=2The correct order of bond orders of
C22−,N22− and
O2‌2−O2‌2−<N2‌2−<C2‌2−