Chemistry · Chemical Bonding

JEE Main 2024 — 9 April, Shift 2 — Question 81

Total number of electron present in (π∗)\left(\pi^{*}\right) molecular orbitals of O2,O2+\mathrm{O}_{2}, \mathrm{O}_{2}^{+}and O2−\mathrm{O}_{2}^{-}is \qquad

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

O2(16e):(σ1  ⁣ ⁣  ⁣ ⁣ s)2(σ1  ⁣ ⁣  ⁣ ⁣ s*)2(σ2  ⁣ ⁣  ⁣ ⁣ s)2(σ2  ⁣ ⁣  ⁣ ⁣ s*)2{{\text{O}}_{2}}\left( 16\text{e} \right):{{\left( {{\sigma }_{1\text{ }\!\!~\!\!\text{ s}}} \right)}^{2}}{{\left( \sigma _{1\text{ }\!\!~\!\!\text{ s}}^{\text{*}} \right)}^{2}}{{\left( {{\sigma }_{2\text{ }\!\!~\!\!\text{ s}}} \right)}^{2}}{{\left( \sigma _{2\text{ }\!\!~\!\!\text{ s}}^{\text{*}} \right)}^{2}}

(σ2p)2[(π2p)2=(π2p)2],[(π2p*)1=(π2p*)1]{{\left( {{\sigma }_{2p}} \right)}^{2}}\left[ {{\left( {{\pi }_{2p}} \right)}^{2}}={{\left( {{\pi }_{2p}} \right)}^{2}} \right],\left[ {{\left( \pi _{2p}^{\text{*}} \right)}^{1}}={{\left( \pi _{2p}^{\text{*}} \right)}^{1}} \right]

Number of e−{{\text{e}}^{-}}present in (π*)\left( {{\pi }^{\text{*}}} \right) of O2=2{{\text{O}}_{2}}=2 Number of e−{{\text{e}}^{-}}present in (π*)\left( {{\pi }^{\text{*}}} \right) of O2+=1\text{O}_{2}^{+}=1 Number of e−{{\text{e}}^{-}}present in (π*)\left( {{\pi }^{\text{*}}} \right) of O2−=3\text{O}_{2}^{-}=3 So total e−{{\text{e}}^{-}}in (π*)=2+1+3=6\left( {{\pi }^{\text{*}}} \right)=2+1+3=6

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Chemistry
Chapter
Chemical Bonding
Topic
Molecular Orbital Theory