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JEE Advanced 2017Paper 1
Exam
JEE Advanced
Year
2017
Paper
Paper 1
Subject
Chemistry
Chapter
Chemical Bonding and Molecular Structure

JEE Advanced 2017 · Paper 1 · Chemistry · Chemical Bonding and Molecular Structure

Among H _ 2 , H e _ 2 ^ + , L i _ 2 , B e _ 2 , B _ 2 , C _ 2 , N _...

Among{H}_{2}, H{e}_{2}^{+}, L{i}_{2}, B{e}_{2}, {B}_{2}, {C}_{2}, {N}_{2}, {O}_{2}^{-}, and{F}_{2},thenumberofdiamagneticspeciesis−<br/>(Atomicnumber:the number of diamagnetic species is -<br />(Atomic number :H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, F = 9 )

Correct answer

5

Step-by-step solution

H2⇒σ1s2{H}_{2} ⇒ \sigma1{s}^{2} (Diamagnetic)
He2+⇒σ1s2H{e}_{2}^{+} ⇒ \sigma1{s}^{2}σ∗1s1{\sigma}^{*} 1{s}^{1} (Paramagnetic)
Li2⇒σ1s2σ∗1s2σ2s2L{i}_{2} ⇒ \sigma1{s}^{2}{\sigma}^{*}1{s}^{2}\sigma2{s}^{2} (Diamagnetic)
Be2⇒σ1s2σ∗1s2σ2s2σ∗2s2B{e}_{2} ⇒ \sigma1{s}^{2}{\sigma}^{*}1{s}^{2}\sigma2{s}^{2}{\sigma}^{*}2{s}^{2} (Does not exist)
B2⇒σ1s2σ∗1s2σ2s2σ∗2s2π2px1=π2py1{B}_{2} ⇒ \sigma1{s}^{2}{\sigma}^{*}1{s}^{2}\sigma2{s}^{2}{\sigma}^{*}2{s}^{2}\pi2{p}_{x}^{1} = \pi2{p}_{y}^{1} (Paramagnetic)
C2⇒σ1s2σ∗1s2σ2s2σ∗2s2π2px2=π2py2{C}_{2} ⇒ \sigma1{s}^{2}{\sigma}^{*}1{s}^{2}\sigma2{s}^{2}{\sigma}^{*}2{s}^{2}\pi2{p}_{x}^{2} = \pi2{p}_{y}^{2} (Diamagnetic)
N2⇒σ1s2σ∗1s2σ2s2σ∗2s2π2px2=π2py2σ2pz2{N}_{2} ⇒ \sigma1{s}^{2}{\sigma}^{*}1{s}^{2}\sigma2{s}^{2}{\sigma}^{*}2{s}^{2}\pi2{p}_{x}^{2} = \pi2{p}_{y}^{2}\sigma2{p}_{z}^{2} (Diamagnetic)
O2Θ⇒σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2px2=π2py2π∗2px2=π∗2py1{O}_{2}^{Θ} ⇒ \sigma1{s}^{2}{\sigma}^{*}1{s}^{2}\sigma2{s}^{2}{\sigma}^{*}2{s}^{2}\sigma2{p}_{z}^{2}\pi2{p}_{x}^{2} = \pi2{p}_{y}^{2}{\pi}^{*}2{p}_{x}^{2} = {\pi}^{*}2{p}_{y}^{1} (Paramagnetic)
F2⇒σ1s2σ∗1s2σ2s2σ∗2s2σ2oz2π2px2=π2py2π∗2px2=π∗2py2{F}_{2} ⇒ \sigma1{s}^{2}{\sigma}^{*}1{s}^{2}\sigma2{s}^{2}{\sigma}^{*}2{s}^{2}\sigma2{o}_{z}^{2}\pi2{p}_{x}^{2} = \pi2{p}_{y}^{2}{\pi}^{*}2{p}_{x}^{2} = {\pi}^{*}2{p}_{y}^{2} (Diamagnetic)

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