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.
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.
(a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about 537 K, a class of organosilicon polymers called methylsubstituted chlorosilanes
(MeSiCl3,Me2SiCl2,Me3SiCl,andMe4Si) are formed.
(b) When silicon dioxide(SiO2) is heated with hydrogen fluoride (HF), it forms silicon tetrafluoride (SiF4). Usually, the Si–O bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperature. However, it is attacked by HF.
The SiF4 formed in this reaction can further react with HF to form hydrofluorosilicic acid.
(c) When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.
(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.
(a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about 537 K, a class of organosilicon polymers called methylsubstituted chlorosilanes
(MeSiCl3,Me2SiCl2,Me3SiCl,andMe4Si) are formed.
(b) When silicon dioxide(SiO2) is heated with hydrogen fluoride (HF), it forms silicon tetrafluoride (SiF4). Usually, the Si–O bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperature. However, it is attacked by HF.
The SiF4 formed in this reaction can further react with HF to form hydrofluorosilicic acid.
(c) When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.
(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.
