AP EAPCET 05-Jul-2022 Shift 1 Paper

Bond order (BO)=‌1∕2 (number of electron in
bonding MO's - number of electron in anti-bonding MO′ s )
C22−: Total number of electrons =6+6+2=14 Electronic configuration
‌BO=‌1∕2×(10−4)=‌1∕2×6=3
N2 : Total number of electrons =7.+7=14
BO=‌1∕2(10−4)=‌6∕2=3
‌O22−: Total number of electron "‌=8+8+2=18
‌π*2px2=π22py2
BO‌=‌1∕2(10−8)
‌=‌1∕2×2=1
Option (a) is incorrect.
‌O2+
Total number of electrons =8+8−1=15
‌=π(2py2),π*2px1
BO=‌

1

2

(10−5)=‌

1

2

×5=2.5
‌O2− Total number of electrons =8+8+1=17
=π(2py2),π*(2px2)=π*(2py1)
r‌BO=‌

1

2

(10−7)=‌

1

2

×3=1.5
‌N2+ Total number of electron"‌=7+7−1=13
‌BO=‌

1

2

(9−4)=‌

1

2

×5=2.5
Option (b) is correct.
Similarly, bond order of O2=‌

1

2

(10−6)=‌

1

2

×4=2
Bond order of Li2=‌

1

2

(4−2)=1
Bond order of H2+=‌

1

2

(1−0)=0.5
Bond order of C2=‌

1

2

(8−4)=‌

1

2

×4=2
Rest other options are incorrect, as they don't have bond order in fraction.