P Block Elements

NH? + 3Cl? (excess) → NCl? + 3HCl

Zone refining is used to obtain high purity elements which are used in the manufacture of semiconductors. Boron and silicon both are used in semiconductors.

Number of bond between sulphur and oxygen $=8$

Number of bond between sulphur and sulphur $=8$

$6\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+3{\mathrm{C}\mathrm{l}}_2?5\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}+{\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O}$

$2\mathrm{C}\mathrm{a}(\mathrm{O}\mathrm{H}{)}_2+{\mathrm{C}\mathrm{l}}_2?\mathrm{C}\mathrm{a}(\mathrm{O}\mathrm{C}\mathrm{l}{)}_2+{\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$

Compound Oxidation state of P
H? P? O? → 5
H? P? O? → 3
H? P? O? → 4

Hydrogen peroxide reduces iodine to iodide ion is basic medium as;

$H_2{O}_2+2O{H}^{-}+l_2\to O_2+2l^{-}+2H_{2}O$

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

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

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

= (-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