Class 12th

  $Mn{O}_4^{-2}\to A+B$

Oxidation state of Mn in B < A

  $Mn{O}_4^{-2}+H^{+}\to Mn{O}_4^{-}+Mn{O}_2$             

B is MnO₂

$\therefore$ Oxidation state of Mn = +4

    ${\therefore }_{25}M{n}^{+4}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^3$          

$\therefore$ unpaired electron = 3

Spin only magnetic moment  $\left(\mu \right)$

$=\sqrt[]{n\left(n+2\right)}=\sqrt[]{3\left(3+2\right)}=\sqrt[]{15}$

$=3.87\approx 4$

  $C\mathcal{l}{F}_3$ -> T-shaped (sp³d)

 IF₇ -> Pentagonal bipyramidal (sp³d³)

BrF₅ -> Square pyramidal (sp³d²)

BrF₃ -> T-Shaped (sp³d)

I₂Cl₆ -> Triangular bipyramidal (sp³d)

$I{F}_5\to$ Square Pyramidal (sp³d²)

ClF -> (sp³)

ClF₅ -> Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ -> Square Pyramidal

t1/2 = 0.301 min

t = 2 min

$K=\frac{2.303}{t}\mathcal{l}og\left(\frac{Co}{Ct}\right)$

$\frac{0.693}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$

$\frac{2.303*0.301}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$

$\therefore 2=\mathcal{l}og\left(\frac{Co}{Ct}\right)$

$\frac{Co}{Ct}=10^2=100$

Ans. 100

$m=\frac{w*1000}{molecularwt*w_{solvent}}=\frac{10.2*1000}{176*150}$

$=\frac{1020\cancel{0}}{176*15\cancel{0}}=0.386$

$\Delta T_f=K_f*m$

3.9 * 0.386

$\therefore x*10^{-1}=15.05*10^{-1}$

Ans =15

B.C.C structure

a = 300 pm = 300 * 10^-12 m

d = 6g/cm³

z = 2

$d=\frac{Z*M}{a^3}$            

$6=\frac{2*A}{{\left(300*10^{-10}\right)}^3}=\frac{2*A}{27*10^{-24}}$            

$\therefore AtomsofM$ = 3.69 * 6.022 * 10²³

= 22.22 * 10²³

the nearest integer = 22

$P_1=1.2*10^{-30}C.m{P}_2=2.4*10^{-30}C.m$

$E_1=5*10^{4}N{C}^{-1}{E}_2=15*10^{4}N{C}^{-1}$

So, $\frac{{\tau }_1}{{\tau }_2}=\frac{P_1{E}_{1}sin90°}{P_2{E}_{2}sin90°}=\frac{\left(1.2*10^{-30}\right)*\left(5*10^4\right)}{\left(2.4*10^{-30}\right)*\left(15*10^4\right)}$

$=\frac{1}{2}*\frac{1}{3}=\frac{1}{6}\Rightarrow x=6$

$R_{AB}=20\Omega$

L = 300 cm

Null point is at 60 cm from A, so

$20mV=\frac{4}{R+20}*20*\left(\frac{60}{300}\right)*10^3$

$\Rightarrow 20=\frac{16}{R+20}*10^3$

$R=\frac{15600}{20}=780\Omega$

 $\frac{{}_{}}{\text{exception 25:}}$ $i=\frac{{\mathcal{l}}_0}{\sqrt[]{2}}$  at ${\omega }_1$ ${}_{}$ = 212 rad/s $R=5\Omega$ $$

${}_{}$  ${\omega }_2$ = 232 rad/s $\Delta \omega ={\omega }_2-{\omega }_1=20rad/s$ ${}_{}{}_{}$

$P=\frac{P_{max}}{2}at{\omega }_{1}and{\omega }_2$

So, $\Delta \omega ={\omega }_2-{\omega }_1=\frac{R}{L}$ ${}_{}{}_{}\frac{}{}$

$\Rightarrow L=\frac{R}{\Delta \omega }=\frac{5}{20}=.25H$

= 250 mH

${\lambda }_1=560nm,$ fringe width, B = 7.2 mm

$\Rightarrow B_1=\frac{D{\lambda }_1}{d}=7.2mm$

For ${\lambda }_2\to B_2=\frac{D{\lambda }_2}{d}=8.1mm$ ${}_{}{}_{}\frac{{}_{}}{}$

$\frac{B_2}{B_1}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{8.1}{7.2}\Rightarrow {\lambda }_2\frac{81}{72}{\lambda }_1$

$=\frac{9}{8}*560nm$

= 630 nm

t_1/2 = 2hr 30 min

Radiation intensity a disintegrations / sec (Activity)

As,   $A=\frac{A_0}{{2^t}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

For safe working, $A=\frac{A_0}{64}\Rightarrow \frac{A_0}{64}=\frac{A_0}{2^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$}\right.}}$  

$\Rightarrow t=6t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=6*2.5hr$                            

= 15