Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

According to Fajan's rule,

Covalent character $\propto$ size of anion

$\therefore Ca{F}_2<CaC{l}_2<CaB{r}_2<Ca{l}_2$

 $k=A{e}^{-Ea/RT}$

$lnk=lnA-\frac{Ea}{RT}$

Comparing with y = mx + c

Slope = $+\frac{Ea}{R}=+\frac{20}{5}$

$\therefore Ea=\frac{20}{5}R=4R=4*2cal$

= 8 Cal

First order reaction

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{{\mathrm{a}}_0}{{\mathrm{a}}_0-\mathrm{x}}$

$\mathrm{K}=\frac{2.303}{90}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{{\mathrm{a}}_0}{0.25{\mathrm{a}}_0}$

$=0.0154$

$\begin{array}{rr}& t=60\mathrm{\%}=\frac{2.303}{K}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{a_0}{a_0}\dots .\left(2\right)\\ & =\frac{2.303}{0.0154}*(1-0.602)=59.51\mathrm{m}\mathrm{i}\mathrm{n}s\approx 60\end{array}$

Kindly consider the following figure

A - B has the lowest potential energy and it means it has stronger bond.

Rate constant, k = 5.5 * 10^-14 s^-1.

$t=\frac{2.303}{k}log\frac{{\left[R\right]}_0}{\left[R\right]}$                

= $\frac{2.303}{k}log3-(i)$  

$t_{50\mathrm{\%}}=\frac{2.303}{k}log2-(ii)$                

From (i) & (ii)

$\frac{t_{67\mathrm{\%}}}{t_{50\mathrm{\%}}}=\frac{log3}{log2}$                

= 1.58 t_50%

So; t_67% is 15.8 * 10^-1 times half life.

X = 16 (the nearest integer)

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

$Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$  

$\begin{array}{l}C{l}_2+2Kl\to 2KCl+I_2\\ I_2+2N{a}_2{S}_2{O}_3\to 2Nal+N{a}_2{S}_4{O}_6\end{array}$               

Here, meq of MnO₂ = meq of Na₂S₄O₆

$\frac{w}{E}*1000=M*n-factor*V$                              

$\frac{w}{87/2}*1000=0.1*1*60$                              

Mass of MnO₂ in sample = 0.261 g

Percentage of MnO₂ in sample =   $\frac{0.261}{2}*100$

= 13.05%

According to condition

Rate $\propto {\left[X\right]}^1{\left[Y\right]}^0$

By using experimental data I and II

$\frac{4*10^{-3}}{2*10^{-3}}=\frac{L}{0.1}\therefore L=0.2$

And by using experimental data I and II

$\frac{M*10^{-3}}{2*10^{-3}}=\frac{0.4}{0.1}$

$\therefore$ M = 8

Ratio of $$$\left(\frac{M}{L}\right)=\frac{8}{0.2}=40$

SCl₂ – 2 lone pair of central atom

O₃ – 1 lone pair of central atom

ClF₃ – 2 lone pair of central atom

SF₆ – 0 lone pair of central atom