O-LevelChemistryBonding & Structure5070/21 · Oct/Nov 2017 · Q710 Marks

B7 Some properties of the Group IV elements are shown in the table. element | melting point /°C | relative electrical conductivity carbon (diamond) | 3550 | non-conductor silicon | 1410 | poor conductor germanium | 937 | poor conductor tin | 232 | conductor lead | 328 | conductor

(a)(i)Explain in terms of structure and bonding why diamond has such a high melting point.[2]
(a)(ii)Use the information in the table to suggest how the type of structure and bonding in carbon (diamond) differs from the type of structure and bonding in tin. Explain your answer.[2]
(a)(iii)Lead oxide is an amphoteric oxide. What is the meaning of the term amphoteric oxide?[1]
(b)(i)A sample containing 64.5 g of a chloride of germanium contains 42.6g of chlorine. Deduce the empirical formula of this chloride. empirical formula[3]
(b)(ii)This chloride of germanium has a boiling point of 87 °C. Predict the structure and bonding of this chloride.[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

[1] B7(a)(i) giant (molecular) structure / many covalent bonds
[1] B7(a)(i) takes a lot of energy to break the bonds / needs a high temperature to break the bonds
[1] B7(a)(ii) Diamond has a giant covalent structure whereas tin has a metallic structure
[1] B7(a)(ii) because diamond or carbon has a much higher melting point OR diamond does not conduct electricity but tin does
[1] B7(a)(iii) (oxide which) reacts with acids or bases
[1] B7(b)(i) mass of germanium = 21.9 (g)
[1] B7(b)(i) mol Ge = 21.9 / 73 and mol Cl = 42.6 / 35.5 OR mol ratio Ge : Cl is 0.3 to 1.2
[1] B7(b)(i) GeCl4
[1] B7(b)(ii) (simple) molecular
[1] B7(b)(ii) covalent bonding

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