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, s2p²_{x ,} p2p_y²  , p2p_z², p*2p_y¹  , p2p_z¹

There are two unpaired electrons in antibonding 2p orbitals, hence it is paramagnetic.

C₂(12) = σ1s², σ*1s², σ2s², σ*2s², p2p_y² , p2p_z²

There are no unpaired electrons, hence it is diamagnetic.

S₂(32) = s1s², s^* 1s², s2s² , s*2s², s2p²_{x ,} p2p_y²  , p2p_z², p*2p_y²  , p2p_z² , s*2p²_x  s3s² , s*3s², s3p²_{x ,} p2p_y²  , p2p_z², p*2p_y¹  , p2p_z¹

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

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.