Chemistry Chemical Bonding and Molecular Structure

(1) $H{e}_2^{-}{\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^{1}B.O=\frac{3-2}{2}=\frac{1}{2}$  

(2)    $H{e}_2^{+}{\sigma }_{1s}^2{\sigma }_{1s}^{*1}B.O=\frac{2-1}{2}=\frac{1}{2}$

(3) $B{e}_2{\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}B.O=\frac{4-4}{2}=0$  

So Be₂ does not exits

SF₄ has sp³d hybridization and one type is axial while other type is equatorial

SiF₄ has sp³ (all bonds are equal)

$B{F}_4^{-}$ has sp³ (all bonds are equal)

XeF₄ has sp³d² and shape is sq. planar (All bonds are equal)

(A) Ethane                   one $\left(C-C\right)\sigma$ bond

(B) Ethene                         one $\left(C-C\right)\sigma$ and one $\left(C-C\right)\pi$ bond

(C) $C_2$                                                       two $\left(C-C\right)\pi$ bonds

(D) Ethyne $H-C\equiv C-H$                       two $\left(C-C\right)\pi$ bonds and one $\left(C-C\right)\sigma$ bond

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

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

Ans 4.

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

$2NaB{H}_4+l_2\to 2Nal+H_2+B_2{H}_6$

Diborane is produced when NaBH₄ is reacted with I₂

Both boron atoms are sp³ hybridised & molecule is non-planar. Diborane as two bridged 3 centre-2-electron bonds.

$Zn{\left(OH\right)}_2?Z{n}^{2+}+2O{H}^{-}$  

S                           -            -

NaOHNa⁺ + OH^--

0.1 M    -            -

$\left[Z{n}^{2+}\right]=S$                

Here ; 2 * 10^-20 = S (0.1)²

So; S = 2 * 10^-18 M