Ncert Solutions Chemistry Class 12th

Here A is CH₃ - CH₂ – CH₂ – COO – CH₂ – CH₃

The given sequence of reaction is:

$C{H}_3-C{H}_2-C{H}_2-C{H}_2-OH\underset{}{\overset{Oxidation}{\to }}C{H}_3-C{H}_2-C{H}_2-COOH$

Ce, Pr, Nd, Tb and Dy are the only lanthanoids which shows + 4 O.S, so can form MO₂, but Yb does not shows +4 O.S so can't form MO₂.

Only 1° aromatic amines give stable diazonium salt on reaction with nitrous acid

Here 1° aromatic amine is  which gives most stable diazonium salt

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products

Calamine is the ore of zinc i.e. ZnCO₃.

Malachite is the ore of copper i.e. CuCO₃.Cu (OH)₂.

Coordination no. of an atom in BCC is 8.

Using $\lambda =\frac{h}{\sqrt[]{2qVm}}$  

${\lambda }_{Li}=\frac{h}{\sqrt[]{2*3e*V*8.3m}}$                 m = mass of proton

${\lambda }_p=\frac{h}{\sqrt[]{2*e*V*m}}$  

$\frac{{\lambda }_{Li}}{{\lambda }_p}=\sqrt[]{\frac{2eVm}{2*24.9eVm}}$            

= 0.2 = 2 * 10^-1

So; x = 2

The balanced equation is :

$S_8\left(s\right)+12O{H}^{-}\left(aq\right)\to 4S^{2-}\left(aq\right)+2S_2{O}_3^{2-}\left(aq\right)+6H_{2}O\left(\mathcal{l}\right)$           

Here ; a = 12

$Mn{O}_4^{-}+S_2{O}_3^{2-}\stackrel{O{H}^{-}}{\to }Mn{O}_2+S{O}_4^{2-}$

O.S of S in $S{O}_4^{2-}=+6$