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Question:1) Stability of the species Li2, Li2- , Li2+ increases in the order of: (IIT JEE 2013)
1. Li2- < Li2+ < Li2
2. Li2< Li2- < Li2+
3. Li2-< Li2 < Li2+
4. Li2< Li2+ < Li2-
Answer: A
Solution:
Stability of species depends on its bond order. Greater the bond order, more stable would be the species.
Electronic configuration of Li2 = σ1s2, σ*1s2, σ2s2
Bond Order = (No. of electrons in bonding orbitals –No. of electrons in antibonding orbitals)/2
= ½ (4-2) =1
Electronic configuration of Li2+ = σ1s2, σ*1s2, σ2s1
Bond Order = ½(3-2) = 0.5
Electronic configuration of Li2- = σ1s2, σ*1s2, σ2s2, σ*2s1
Bond Order = ½(4-3) = 0.5
Bond order of both Li2+ and Li2- is same but Li2+ will be more stable as it has lesser number of antibonding electrons.
So, the correct order is Li2- < Li2+ < Li2
And hence, the correct order is A.
Question: 2 ) Which one of the following molecules is expected to exhibit diamagnetic behaviour? (IIT JEE 2013)
1) N2 2) O2 3) S2 4) C2
Answer:
N2(14) = σ1s2, σ*1s2, σ2s2, σ*2s2, σ2p6
There is no unpaired electron so it is diamagnetic.
O2(16) = s1s2, s* 1s2, s2s2 , s*2s2, s2p2x , p2py2 , p2pz2, p*2py1 , p2pz1
There are two unpaired electrons in antibonding 2p orbitals, hence it is paramagnetic.
C2(12) = σ1s2, σ*1s2, σ2s2, σ*2s2, p2py2 , p2pz2
There are no unpaired electrons, hence it is diamagnetic.
S2(32) = s1s2, s* 1s2, s2s2 , s*2s2, s2p2x , p2py2 , p2pz2, p*2py2 , p2pz2 , s*2p2x s3s2 , s*3s2, s3p2x , p2py2 , p2pz2, p*2py1 , p2pz1
There are two unpaired electrons in antibonding 3p orbitals, hence it is paramagnetic.
Hence, b and c are the correct options.
Question:3) The shape of XeO2F2 molecule is (IIT JEE -2012)
1. trigonal bipyramidal 2. square planar 3. tetrahedral 4. see-saw
Answer: d
XeO2F2 molecule has see-saw geometry with two F atoms on axial positions while two O atoms and a lone pair of electrons on equatorial positions.
Hence, the correct option is D.
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