JEE Main 24 June 2022 Shift 2 Solved Paper

(1) Bond order =‌

1

2

[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 isHere Na= Anti bonding electron =4 and Nb=10
(3) 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.
in O22− no of electrons =18
(A) Molecular orbital configuration of O22−(18 electrons) is
σ1s2σ1s2*σ2s2σ2s2*σ2pz2π2px2=π2py2π2px2*=π2py2*
∴‌‌Nb=10
Na=8
∴‌‌BO=‌

1

2

[10−8]=1
(B) C22− has 14 electrons.
Moleculer orbital configuration of C22− is
σ1s2σ1s2*σ2s2σ2s2*π2px2=π2py2σ2pz2
∴Na=4
Nb=10
∴‌‌BO=‌

1

2

[10−4]=3
(C) N22− has 16 electrons.
Moleculer orbital configuration of N22− is
σ1s2σ1s2*σ2s2σ2s2*π2px2=π2py2σ2pz2π2px1*=π2py1*
∴Na=6
Nb=10
∴‌‌BO=‌

1

2

[10−6]=2
The correct order of bond orders of C22−,N22− and O2‌2−
O2‌2−<N2‌2−<C2‌2−