The p-Block Elements

$I_2+10HN{O}_3\left(conc\right)\to 2Hl{O}_3+10N{O}_2+4H_{2}O$

$I_2+10HN{O}_3\left(conc\right)\to 2Hl{O}_3+10N{O}_2+4H_{2}O$

$Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$  

$\begin{array}{l}C{l}_2+2Kl\to 2KCl+I_2\\ I_2+2N{a}_2{S}_2{O}_3\to 2Nal+N{a}_2{S}_4{O}_6\end{array}$               

Here, meq of MnO₂ = meq of Na₂S₄O₆

$\frac{w}{E}*1000=M*n-factor*V$                              

$\frac{w}{87/2}*1000=0.1*1*60$                              

Mass of MnO₂ in sample = 0.261 g

Percentage of MnO₂ in sample =   $\frac{0.261}{2}*100$

= 13.05%

Pentavalent oxides of group – 15 elements, E₂O₅ is more acidic than trivalent oxides, E₂O₃ of the same element.

Acidic strength of trivalent oxides decreases down the group as metallic strength increases.

N₂ gas is obtained by thermal decomposition of Ba (N₃)₂ as,

$Ba{\left(N_3\right)}_2\underset{}{\overset{\Delta }{\to }}Ba+3N_2\uparrow$

 $S=\sqrt[]{Ksp}=\sqrt[]{8*10^{-28}}=2.82*10^{-14}$

Or solubility of PbS = 282 * 10^-16

Fluorine forms only one oxoacid which is hypofluorous acid HOF because it shows only – 1 oxidation state. Which is due to its the smallest size among halogens & the highest electronegativity.

$N{H}_{4}C\mathcal{l}\left(aq\right)+NaN{O}_2\left(aq\right)\to N_2\left(g\right)+2H_{2}O\left(\mathcal{l}\right)+NaC\mathcal{l}\left(aq\right)$  

Hence, N₂ gas is produced.

Structure PCl₅ is trigonal bipyramidal,

Hybridization of P is sp³d

Equatorial bonds lie in a plane

Axial bonds are longer than equatorial bonds so axial bonds are weaker than equatorial bonds.