Ncert Solutions Chemistry Class 12th

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

Effective no. of octahedral voids in a lattice = n

Effective no. of lattice point in a lattice = n

$$ $\therefore$  Ratio = n/n = 1

$S\left(g\right)+6F\left(g\right)\to S{F}_6\left(g\right)$

$\Delta {H^0}_R=\Delta {H^0}_f\left(S{F}_6\right)-\Delta {H^0}_f\left(S\right)-6*\Delta {H^0}_f\left(F\right)$

= (-1100) (275) 6 (80) = 1855

$\Delta {H^0}_{S-F}=\frac{1855}{6}=309.16KJ/mol\cong 309$

Beo & Be (OH)₂ are amphoteric in nature, because they react with both acid and base

Kindly consider the following figure