Chemistry Chemical Bonding and Molecular Structure

Kindly consider the following figure

Species               B.O

Ne₂                        0

N₂                           3

F₂                 1

O₂                          2

During the electrolysis of dilute H₂SO₄

$2H_{2}S{O}_4\stackrel{\left(1\right)electrolysis}{\to }2HS{O}_4^{-}+2H^{+}$  

$H_2{S}_2{O}_8+2H_{2}O\stackrel{hydrolysis}{\to }2H_{2}S{O}_4+\underset{\left(A\right)}{H_2{O}_2}$  

$2H^{+}+2e\to H_2\uparrow$  

In the solid form of  $H_2{O}_2$ dihedral angle is equal to 90.2°.

$Bondorder=\frac{e^{-}ofBondingMO-e^{-}ofABMO}{2}$  

Hence Bond order for  $O_2^{+}=\frac{10-5}{2}=2.5$  

$O_2=\frac{10-6}{2}=2.0$

$O_2^{-}=\frac{10-7}{2}=1.5$

$O_2^{--}=\frac{10-8}{2}=1.0$  

Hybridization of P in PF₅ is sp³d, so value of y = 1

Species =            CH₄                       $N{H}_4^{+}$                     $B{H}_4^{?}$  

Electrons =         10                        10

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

$\begin{array}{l}L{i}_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2\to Diamagnetic\\ C_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2\to Diamagnetic\\ C_2^{-}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^1\to Paramagnetic\end{array}$

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

$O_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^0\to$ Paramagnetic

$H{e}_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*1}\to$ Paramagnetic

$\therefore$ Paramagnetic molecules are $=B_2,C_2^{-},O_2^{+},H{e}_2^{+}$