Test Index

NCERT Class XI Chemistry The p-Block Elements Solutions

© examsnet.com
Question : 21 of 38
Marks: +1, -0
Explain the following reactions :
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper;
(b) Silicon dioxide is treated with hydrogen fluoride;
(c) CO is heated with ZnO;
(d) Hydrated alumina is treated with aqueous NaOH solution.
Solution:  
(a) When methyl chloride reacts with silicon in the presence of copper as a catalyst at a temperature 573 K various types of methyl substituted chlorosilane of formula MeSiCl3,Me2SiCl2,Me3SiCl\mathrm{MeSiCl}_3, \mathrm{Me}_2\mathrm{SiCl}_2, \mathrm{Me}_3\mathrm{SiCl} with small amount of Me4Si\mathrm{Me}_4\mathrm{Si} are formed.
2CH3Cl+Si2\mathrm{CH}_3\mathrm{Cl} + \mathrm{Si} →573 KCu powder\xrightarrow[573\,\text{K}]{\text{Cu powder}} (CH3)SiCl3+(CH3)2SiCl2(\mathrm{CH}_3)\mathrm{SiCl}_3+(\mathrm{CH}_3)_2\mathrm{SiCl}_2 + (CH3)3SiCl+(CH3)4Si(\mathrm{CH}_3)_3\mathrm{SiCl} + (\mathrm{CH}_3)_4\mathrm{Si}
(b) When silicon dioxide is treated with HF, silicon tetrafluoride is formed.
SiO2+4HF\mathrm{SiO}_2 + 4\mathrm{HF} → SiF4+2H2O\mathrm{SiF}_4 + 2\mathrm{H}_2\mathrm{O}
(c) CO is a powerful reducing agent, it reduces ZnO to Zn.
CO(g)+ZnO(s)\mathrm{CO}_{\text{(g)}} + \mathrm{ZnO}_{\text{(s)}} →Δ\xrightarrow[]{{\Delta}} Zn(s)+CO2(g)\mathrm{Zn}_{\text{(s)}} + \mathrm{CO}_{2\text{(g)}} ↑
(d) Aluminium reacts with aqueous alkali and liberates hydrogen gas.
2Al(s)+2NaOH(aq)+6H2O(l)2\mathrm{Al}_{\text{(s)}} + 2\mathrm{NaOH}_{\text{(aq)}} + 6\mathrm{H}_2\mathrm{O}_{\text{(l)}} → 2Na+[Al(OH)4](aq)−+3H2(g)2\mathrm{Na}^{+}[\mathrm{Al}(\mathrm{OH})_4]^{-}_{\text{(aq)}}+3\mathrm{H}_{2\text{(g)}} ↑
© examsnet.com
Go to Question: