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Question : 67 of 180
Marks:
+1,
-0
Solution:
He2+=σ1s2,σ*1s1 Bond order
=(Nb−Na) Nb is number of bonding electrons
Na is number of anti - bonding electrons
=(2−1)−=0.5 C22−=σ1s2,σ*1s2,σ2s2,σ*2s2,(π2px2=π2py2),σ2pz2 BO=(10−4)==3.0 O2+=σ1s2,σ*1s2,σ2s2,σ*2s2,σ2pz2,(π2px2=π2py2),π*2px1 BO=(10−5)==2.5 O2−=σ1s2,σ*1s2,σ2s2,σ*2s2,σ2pz2, (π2ps2=π2py2),(π*2px2=π*2py1) BO=(10−7)==1.5 Thus, the correct order of increasing bond order is
$He^{+}_2
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