Chemistry Chemical Bonding and Molecular Structure

Bond order of $C_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-4}{2}=3$  

 Bond order of $N_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-6}{2}=2$   

Bond order of  $O_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-8}{2}=1$  

N_B = No. of electrons in bonding molecular orbitals.

N_A = No. of electron is Anti bonding molecular orbitals

Bond order in N $N_2$ and O $O_2^{2+}$ is 3 (calculated by energy level diagram)

Kindly consider the following figure

As bond order is increasing it implies bond strength is increasing. As bond order is increasing it implies bond strength is increasing.

Total number of electron in Ti = 22
Total number of electron in Ti? = 22 – 4 = 18 So EAN value of Ti = 18 + 12 + 4 = 34

$N_{2}O,C_2{H}_2,C{O}_2,C_3{O}_2,Be{F}_2$

If hunds rule is followed

${N_2}^{-2}:\sigma 1s^2{\sigma }^{*}1s^2\sigma 2s^2{\sigma }^{*}2s^2\sigma 2{p_z}^2\pi 2{p^2}_x\pi 2{p_y}^2{\pi }^{*}2{p_x}^1{\pi }^{*}2{p_y}^1$            

If Hunds rule is not followed

${N_2}^{-2}:\sigma 1s^2{\sigma }^{*}1s^2\sigma 2s^2{\sigma }^{*}2s^2\sigma 2{p_z}^2\pi 2{p^2}_x\pi 2{p_y}^2{\pi }^{*}2{p_x}^2$            

Number of electrons in

$P{H}_3=15+3=18$  

$B_2{H}_6=5*2+6=16$  

$\begin{array}{l}CC{l}_4=6+17*4=6+68=74\\ N{H}_3=7+1*3=10\\ LiH=3+1=4\\ BC{l}_3=5+17*3=56\end{array}$  

$B_2{H}_6\&BC{l}_3$  are e^- deficient molecules. B₂H₆ is dimer of BH₃, both compound has 6e^- only.

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.