Class 12th

Order of limiting molar conductivity

$O{H}^{-}>S{O}_4^{2-}>C{l}^{-}>C{H}_{3}CO{O}^{-}$

Higher the acidic strength, lower will be the value of pK_a.

Correct order of pK_a is

$C{H}_{3}COOH>C_6{H}_{5}COOH>HCOOH>O_{2}NC{H}_{2}COOH$  

$Xe{F}_2+P{F}_5\to {\left[XeF\right]}^{+}{\left[P{F}_6\right]}^{-}$

  ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+},N{H}_3$ becomes strong ligand due to +3 oxidation state of cobalt ion. Hence electronic configuration of CO³⁺ is $t_{2g}^6{e}_g^0$  

Haemoglobin is a positive colloid. Hence greater is the charge of anion, more effective will be the coagulation of haemoglobin.

$$ Correct order of coagulating power is $P{O}_4^{3-}>S{O}_4^{2-}>C{l}^{-}$

Electrophoresis : The migration of colloidal particles under the influence of an electric field is known as electrophoresis.

$r_1=k{\left[x\right]}^n$ … (i)

$$ $2r_1=k{\left[4x\right]}^n$  … (ii)

$\left(ii\right)÷\left(i\right)$

$\Rightarrow \frac{2r_1}{r_1}=\frac{k{\left[4x\right]}^n}{k{\left[x\right]}^n}$

$\Rightarrow 2={\left(2\right)}^{2n}$

$\Rightarrow n=\frac{1}{2}$

XeF2 : Hybrid orbital = sigma bond + L.P

= 2 + 3

= 5 = sp3d

Shape will be linear

For a first order reaction.

$k=\frac{0.693}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{0.693}{231}=3*10^{-3}{s}^{-1}$   

$5C_2{O}_4^{2-}+2Mn{O}_4^{-}+16H^{+}\to 2M{n}^{+}+8H_{2}O+10C{O}_2$

5 moles of $C_2{O}_4^{2-}$ ${}_{}{}_{}^{}$ is needed to reduce 2 moles of $Mn{O}_4^{2-}$ ${}_{}^{}$

1 mole of $C_2{O}_4^{2-}$  is needed to reduce $\frac{2}{5}$  moles of $Mn{O}_4^{2-}$