Ncert Solutions Chemistry Class 11th

$\stackrel{\stackrel{+4}{{\mathrm{N}\mathrm{O}}_2}}{2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{N}\mathrm{O}}_2}+\stackrel{+5}{\mathrm{H}\mathrm{N}\mathrm{O}}{}_3$

$\stackrel{0}{{\mathrm{C}\mathrm{l}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{-1}{\mathrm{H}\mathrm{C}\mathrm{l}}+\mathrm{H}\mathrm{O}\mathrm{C}\mathrm{l}$

$2\stackrel{+4}{{\mathrm{C}\mathrm{l}\mathrm{O}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_2}+\stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}$

$\stackrel{+6}{{\mathrm{C}\mathrm{l}}_2}{\mathrm{O}}_6+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}+\stackrel{+{+}^{+}}{{\mathrm{C}\mathrm{l}\mathrm{O}}_4}$

 Photochemical smog has high concentration of oxidising agent like ${\mathrm{N}\mathrm{O}}_2$ ${}_{}$ and ${}_{}$ ${\mathrm{O}}_3$ and is therefore called as oxidising agent.

Kindly consider the following Image 

$1^{\text{st}}\mathrm{H}$ - atom

$\mathrm{n}=4$

$\mathrm{n}=1$

$4\to \mathrm{3,3}\to \mathrm{2,2}\to 1$

 $4\to 1$ or $4\to \mathrm{3,3}\to 1$

or $4\to \mathrm{2,2}\to 1$

Hence, number of different lines possible $=3+1=4$ $$

Minimally, both can have same transition i.e $4\to 1$ etc.

Initial m equivalent of Cu²? = 200 × 0.5 × 2 = 200 m eq
So electricity passed = (0.965×3600)/96500 = 36 × 10? ³ eq. = 36 m eq
m eq CuBr? remaining = 200 – 36 = 164 ∴ N = meq / V (in ml) = 164/200 = 0.82

From ΔT? = K? × m, and ΔT? = 15°C
∴ m = ΔT? / K? = 15 / 1.86 = 8.06
So the amount of propyl alcohol to be added.
= m × molwt = 8.06 × 60 = 483.6 g

Factual on lyophilic colloids
(A) It is easy to prepare
(B) Stable
(C) Revisable
(D) Viscosity is high and surface tension is low for DP. and DM.

Sm ⇒ +2, +3 (4f? , 6 s²); Gd ⇒ +3 (4f?5 d¹6 s²)
Tm ⇒ +2 + 3 (4f¹³,6 s²); Nd ⇒ +2, +3, +4 (4f? , 6 s²)

  $Zn+2NaOH\to N{a}_{2}Zn{O}_2+H_2\uparrow$

$\begin{array}{l}2AI+2NaOH+H_{2}O\to 2NaAl{O}_2+3H_2\uparrow \\ Sn+2NaOH+H_{2}O\to NaSn{O}_3+2H_2\uparrow \\ P_4+3NaOH+3H_{2}O\to P{H}_3\uparrow +3Na{H}_{2}P{O}_2\end{array}$