10..explain the following: (i) O^2- is paramagnetic but O₂ ^2-  is not.(ii) N₂  has higher bond order then NO.

Dear student,

(a) In the electronic configuration of oxygen it has $1s^2 2s^2 2p^4$ , when $O^{2-}$ is given it means 2 extra electrons are added in the oxygen and now electronic configuration becomes $1s^2 2s^2 2p^6$ . All the orbitals are completely filled hence, it is diamagnetic not paramagnetic .In $O_2^{2-}$ the 2 extra electrons are added in the MO diagram of oxygen . Therefore, all orbitals contain 2 electrons in each hence it is diamagnetic. 
The electronic configuration of $O_2^{2-}$ :$(\sigma 2S{)}^2 (\sigma 2S^{*}{)}^2 (\sigma 2p{)}^2 (\mathrm{\pi }2\mathrm{p}{)}^4 (\mathrm{\pi }2{\mathrm{p}}^{*}{)}^4$

This is MO diagram of $O_2$ in this we added 2 extra electrons in $(\pi 2p^{*}{)}^2$ 
Therefore, it is not paramagnetic and it is diamagnetic.

(b) 

The molecular orbital electronic configuration for Nitrogen and NO can be written as:

N₂: (σ 1 s )² < (σ*1 s )² < (σ2 s )² <( σ*2 s )² < (π2 p _x ² = π2p _y ² ) <( σ2p _z ² )

Bond order = no. of bonding - no of antibonding electrons / 2
= (10-4)/2 = 6/2 = 3

NO : (σ 1 s )² < (σ*1 s )² < (σ2 s )² <( σ*2 s )² < (σ2 p _z ² ) (π2 p _x ² = π2p _y ² ) <(π*2p _x ¹ = π*2p _y )

Bond order = (10-5)/2 = 5/2 = 2.5

The higher the bond order, greater is the stability of the molecule therefore N ₂ is more stable than NO.
Hence, we can say that $N_2$ has higher bond order than NO

Regards