A LEVEL 化学 AGroup IV 第四主族元素worksheet with answers

1、A2 Chemistry Group IV (1) answer1. Write balanced equations for the following reactions. (a)  the reaction of germanium (IV) chloride with water  GeCl4 + 2H2O GeO2 + 4HCl(b)  the thermal decomposition of lead (

2、IV) oxide  2PbO2 2PbO + O2 (c)  the reaction of germanium (II) oxide with hydrochloric acid  GeO + 2HCl GeCl2 + H2O (d) the reaction of lead(II) oxide with aqueous hydroxide ions to form the plum

3、bate(II) ionPbO + 2OH PbO22 + H2O (e)  the reaction of carbon dioxide with water to form the hydrogencarbonate ion  CO2 + H2O HCO3 + H+(f)  the reaction of carbon dioxide with aqueous hydroxide ions&#

4、160;  CO2 + 2OH CO32 + H2O (g)  the reaction of silicon (IV) oxide with aqueous sodium hydroxide  SiO2 + 2NaOH Na2SiO3 + H2O(h) the disproportionation of germanium (II) oxide on heating to form germanium(IV) oxide and germanium 2GeO GeO2 + Ge2. The m

5、elting points of the Group IV elements are given in the table.ElementMelting point / °CCarbon (diamond)3550Silicon 1410Germanium937Tin232Lead327(a) Describe and explain how the trend in melting point is related to the structure and bonding in these elements. Melting point decreases down the gro

6、upDiamond, Si and Ge have giant covalent structures; strength of bonds in order of C-C > Si-Si > Ge-Ge;Sn and Pb have giant metallic structures; but metallic bonding is not very strong because of the large ions(b) Carbon dioxide is a gas but tin (VI) oxide is a solid. Explain this difference i

7、n terms of structure and bonding.CO2 has simple molecular structure;Weak V.D.W. forces between molecules;SnO2 has a giant covalent structure;Requires a lot of energy to break the strong bonds(c) Explain in terms of structure and bonding why silicon (IV) oxide is used as a lining in furnaces. It has

8、a giant covalent structure; It takes a lot of energy to break strong covalent bonds; 3. Describe, and explain, the differences in electrical conductivity between carbon (as diamond and as graphite), silicon, germanium, tin and lead.Diamond is insulator because all the electrons are involved in coval

9、ent bonds.Graphite is conductor because some of its electrons are delocalized.Si and Ge are semi-conductors because some electrons are free to move.Sn and Pb are good conductors because they have metallic structure with delocalized electrons.4. All the Group IV elements form chlorides with the formu

10、la MCl4.(a) Describe the bonding in, and the shape of, these chlorides.(i) bonding covalent(ii) shape tetrahedralThe boiling point of lead (IV) chloride cannot be measured directly because it decomposes on heating. The following table lists the boiling points of three Group IV chlorides.(b) (i) Plot

11、 these data on the following axes and extrapolate your graph to predict what the boiling point of PbCl4 would be if it did not decompose.(ii) Suggest why the boiling points vary in this way.b. p. increases due to larger van der Waals(c) SiCl4 reacts vigorously with water whereas CCl4 is inert.(i) Su

12、ggest a reason for this difference in reactivity.Si has empty low-lying d-orbitals(ii) Write an equation for the reaction between SiCl4 and water.SiCl4 + 2H2O SiO2 + 4HCl(iii) Suggest, with a reason, whether you would expect GeCl4 to react with water.Yes, because Ge also has empty low-lying d-orbitals(d) SiCl4 is used to make high-purity silicon for the semiconductor industry. After it has been purified by several fractional distillations, it is reduced to silicon by heating with pure zinc.(i) Suggest an equation for the