JEE Main 2019ChemistryChemical Bonding and Molecular StructureHardMCQ

JEE Main 2019Chemical Bonding and Molecular Structure Question with Solution

JEE Main 2019 (09 Apr Shift 1)

Question

Among the following, the molecule expected to be stabilized by anion formation is:
C2, O2, NO, F2

Choose an option

Show full solutionCorrect option: B
Correct answer
BC2

Step-by-step explanation

According to molecular orbital theory the configuration is like that
C2σ1s2<σ*1s<σ2s<σ*2s  π2Px2=π2Py2< σ2Pz0 (BMO)

BO=8-42=2
C2-=σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2<2pz1=9-42=2.5
C2C2- bond order increases so it is more favourable.
For  F2, O2 and NO
F2=σ1s2<σ*1s2<σ2s2<σ*2s2<σ2Pz2<π2Px2=π2Py2<π*2Px2= π*2Py2<σ*2Pz BO=10-82=1
F2-=σ1s2<σ*1s2<σ2s2<σ*2s2<σ2Pz2<π2Px2=π2Py2<π*2Px2= π*2Py2<σ*2Pz1 BO=10-92=0.5
Bond order decreases F2F2- so not favourable.
O2=σ1s2<σ*1s2<σ2s2<σ*2s2<σ2Pz2<π2Py2=π2Px2<π*2Py1= π*2Pz1
BO=10-62=2
O2-=σ1s2<σ*1s2<σ2s2<σ*2s2<σ2Pz2<π2Py2=π2Px2<π*2Py2= π2Pz1
BO=10-72=1.5
Bond order decreases O2O2- so not favourable
NO=σ1s2<σ*1s2<σ2s2<σ*2s2<σ2Pz2<π2Px2=π2Py2<π*2Px1= π*2Py0
BO=10-52=2.5
NO-=σ1s2<σ*1s2<σ2s2<σ*2s2<σ2Pz2<π2Px2=π2Py2<π*2Px1= π*2Py1
BO=10-62=2
BO NONO- decreases so not favourable


In case of only C2, incoming electron will enter the boding molecular orbital which increases the bond order and stability.

Whereas rest of all takes incoming electron in their antibonding molecular orbital which decreased stability.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Chemical Bonding and Molecular Structure chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, covering the Chemical Bonding and Molecular Structure chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.