O-LevelChemistryBonding & Structure5070/22 · Oct/Nov 2023 · Q99 Marks

Table 9.1 shows the melting points and relative electrical conductivities of three elements. Table 9.1 | | calcium | carbon (diamond) | iodine | |---|---|---|---| | melting point /°C | 839 | 3550 | 114 | | relative electrical conductivity of solid | good | poor | poor | Use ideas about structure and bonding to explain:

(a)Table 9.1 shows the melting points and relative electrical conductivities of three elements. Table 9.1 | | calcium | carbon (diamond) | iodine | |---|---|---|---| | melting point /°C | 839 | 3550 | 114 | | relative electrical conductivity of solid | good | poor | poor | Use ideas about structure and bonding to explain:
(a)(i)the difference in the melting points of diamond and iodine[3]
(a)(ii)the difference in the electrical conductivities of calcium and iodine.[2]
(b)Diamond and graphite are different forms of carbon. Explain, in terms of its structure, why graphite is a lubricant.[2]
(c)A compound of sodium, iodine and oxygen contains 11.62% sodium, 64.14% iodine and 24.24% oxygen by mass. Deduce the empirical formula of this compound. empirical formula[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

[1] M1 diamond (has a high melting point because) has a giant structure of covalent bonds (1)
[1] M2 diamond has strong bonding between all the atoms (1)
[1] M3 iodine (has a low melting point because) there are weak forces between the molecules / iodine has weak intermolecular forces (1)
[1] calcium has mobile electrons / electrons which can move (1)
[1] iodine has no mobile electrons (1)
[1] atoms are arranged in layers in graphite (1)
[1] these layers can slip / layers slide (over each other) (1)
[1] division by correct relative atomic mass e.g. I = 64.14/127 O = 24.24/16 Na = 11.62/23 OR 0.505 1.515 0.505 (1)
[1] division by lowest value to get correct answer NaIO3 (1)

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