Major product of the given reaction is

(CH₃)₃CBr + CH₃–CH₂–ONa → product

 

For  overlap, the lobes of the atomic orbitals are perpendicular to the line joining the nuclei.

Bond strength $\propto$  Bond order

$N_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^2$  

$O_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^{*1}$

$C_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2$  

$B_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^1\equiv {\pi }_{2py}^1$  

NO ® Number of electron = 7 + 8 = 15

B.O. Similar to  $N_2^{-}$  

$N_2^{-}\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^{*}$  

B.O. of N₂ = 3     B.O of C₂ = $\frac{8-4}{2}=2$  

Removal of e^- form antibonding molecular orbital increases bond order.

In NO & O₂ has valance e^- in p orbital.

O? (15) will have configuration σ1s²σ1s²σ2s²σ2s²σ2p? ² (π2p? ²=π2p? ²) (π*2p? ¹). This ion is paramagnetic.

B.O. of CO = 3
B.O. of NO? = 3
Both are isoelectronic
So difference = 0
∴ x = 0

$O_2^{-2}=8\times 2+2=18e^{-}$

${\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2={\pi }_{2py}^2{\pi }_{2px}^{*2}={\pi }_{2py}^{*2}$      

Number of unpaired e^- = 0

Ans. = 0