JEE Main 26 June 2022 Shift 1 Solved Paper

(1) Bond strength ∝ Bond order
(2) Bond length ∝‌

1

‌ Bond order ‌

(3) Bond order =‌

1

2

[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 isHere Na= Anti bonding electron =4 and Nb=10
(5) After 14 electrons to 20 electrons molecular orbital configuration is - - -Here Na=10
and Nb=10
In O atom 8 electrons present, so in O2,8×2=16 electrons present.
Then in O2+no of electrons =15
in O2−no of electrons =17
in O22− no of electrons =18
∴ Molecular orbital configuration of O2 (16 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px1*=π2py1*
∴Na=6
Nb=10
∴‌‌BO=‌

1

2

[10−6]=2
Molecular orbital configuration of O2+( 15 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px1*=π2pyo*
∴‌‌Nb=10
Na=5
∴‌‌BO=‌

1

2

[10−5]=2.5
Molecular orbital configuration of O2−(17 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px2*=π2py1*
∴‌‌Nb=10
Na=7

∴‌‌BO=‌

1

2

[10−7]=1.5
Molecular orbital configuration of O22− (18 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px2*=π2py2*
∴‌‌Nb=10
Na=8
∴‌‌BO=‌

1

2

[10−8]=1
So, correct order of Bond order is
O22−<O2−<O2<O2+