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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-

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

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 , p2py, p2pz2, p*2py, 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 , p2py, p2pz2, p*2py, p2pz2 , s*2p2x  s3s2 , s*3s2, s3p2x , p2py, p2pz2, p*2py, 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

Solution:

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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