Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : <img alt="\mathrm{N}_2 ; \mathrm{N}_2^{+} ; \mathrm{O}_2 ; \mathrm{O}_2^{+} \text {?

What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : \mathrm{N}_2 ; \mathrm{N}_2^{+} ; \mathrm{O}_2 ; \mathrm{O}_2^{+} \text {? }

Option: 1

2,1,0,1


Option: 2

0, 1, 0, 1


Option: 3

0,1,0,1


Option: 4

2,1,2,1


Answers (1)

best_answer

\mathrm{N}_2=\sigma_{1 \mathrm{~s}^2}, \sigma_{1 \mathrm{~s}^2}^*, \sigma_{2 \mathrm{~s}^2}, \sigma_{2 \mathrm{~s}^2}^*, \pi_{2 \mathrm{Px}^2}=\pi_{2 \mathrm{Py}^2}, \sigma_{2 \mathrm{Pz}^2}

no. of \mathrm{e}^{-} present in Homo =0
$$ \mathrm{N}_2^{+}=\sigma_{1 \mathrm{~s}^2}, \sigma_{1 \mathrm{~s}^2}^*, \sigma_{2 \mathrm{~s}^2}, \sigma_{2 \mathrm{~s}^2}^*, \pi_{2 \mathrm{Px}^2}=\pi_{2 \mathrm{Py}^2}, \sigma_{2 \mathrm{Pz}^1}
no. of unpaired \mathrm{e}^{-}present in \mathrm{HOMO}=1
$$ \mathrm{O}_2=\sigma_{1 \mathrm{~s}^2}, \sigma_{1 \mathrm{~s}^2}^*, \sigma_{2 \mathrm{~s}^2}, \sigma_{2 \mathrm{~s}^2}^*, \sigma_{2 \mathrm{Pz}^2}, \pi_{2 \mathrm{Px}^2}=\pi_{2 \mathrm{Py}^2}, \pi_{2 \mathrm{Px}}^* 1=\pi_{2 \mathrm{Py}} 1
no. of unpaired \mathrm{e}^{-} present in \mathrm{HOMO}=2
$$ \mathrm{O}_2^{+}=\sigma_{1 \mathrm{~s}^2}, \sigma_{1 \mathrm{~s}^*}^*, \sigma_{2 \mathrm{~s}^2}, \sigma_{2 \mathrm{~s}^*}, \sigma_{2 \mathrm{Pz}^2}, \pi_{2 \mathrm{Px}^2}=\pi_{2 \mathrm{Py}}^2, \pi_{2 \mathrm{Px}^*}=\pi_{2 \mathrm{Py}}^* 0
no. of unpaired \mathrm{e}^{-} present in \mathrm{HOMO}=1

Posted by

Kuldeep Maurya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question