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Questions on the four crystal structures and their properties
The four crystal structures, ionic, metallic, macromolecular and molecular, with sodium chloride, magnesium, diamond, graphite, silicon dioxide, iodine and ice as examples; the particles and forces in each; the melting point, electrical conductivity and solubility that follow; graphite's layers and conduction; graphene; buckminsterfullerene, C₆₀, as a molecular solid; why ice is less dense than water; enthalpies of fusion and vaporisation and the energy changes at changes of state; drawing the structures; and a full comparison table.
21 original questions · 65 marks · the four crystal structures and their properties notes · Bonding and structure · Flashcards for this unit
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Explain the order of the melting points given in the table in terms of structure and bonding.
Substance Melting point / °C Iodine 114 Magnesium 650 Sodium chloride 801 Silicon dioxide 1710 Mark scheme
Indicative content
- iodine is molecular; melting overcomes only weak induced dipole–dipole forces between I₂ molecules, so its melting point is lowest
- magnesium is a giant metallic lattice of Mg²⁺ ions in delocalised electrons, a strong attraction
- melting a metal only loosens the lattice, as the ions stay attracted to the delocalised electrons in the liquid
- sodium chloride is a giant ionic lattice with strong attractions between oppositely charged ions in all directions
- silicon dioxide is macromolecular, each Si bonded to four O, so melting requires breaking many strong covalent bonds and its melting point is highest
- the three giant structures all melt far higher than iodine because strong bonding must be overcome, not intermolecular forces
Levels
- Level 3 (5–6 marks) all four substances are placed in the right order with the structure and the force or bond overcome on melting correctly identified for each, from the induced dipole forces of iodine to the covalent bonds of silicon dioxide; the answer separates intermolecular forces from bonds and explains why the giant structures melt so much higher
- Level 2 (3–4 marks) three of the four substances are explained with the structure and the force overcome, or all four are named and one explanation is thin
- Level 1 (1–2 marks) one or two structures are named, typically with covalent bonds said to break when iodine melts, and no account of the order
Explain why graphite conducts electricity but diamond does not.
Mark scheme
- B1 in graphite each carbon atom forms only three covalent bonds
- B1 the fourth outer electron of each atom is delocalised across the layer
- B1 the delocalised electrons can move along the layers and carry charge
- B1 in diamond all four outer electrons of each carbon are used in covalent bonds, so no electrons are free to move
Explain why ice is less dense than liquid water.
Mark scheme
- B1 in ice each water molecule is hydrogen bonded to four others
- B1 the hydrogen bonds hold the molecules in a rigid, open tetrahedral lattice with empty space inside it
- B1 on melting some hydrogen bonds break and the open lattice collapses
- B1 the molecules in the liquid pack more closely, so the same mass occupies a smaller volume
Suggest the type of crystal structure of each solid. P melts at 1538 °C and conducts as a solid. Q melts at 44 °C and does not conduct when molten. R melts at 2072 °C, does not conduct as a solid, but conducts when molten. S does not melt below 3000 °C and does not conduct in any state.
Mark scheme
- B1 P is metallic, because it conducts as a solid, so it has delocalised electrons
- B1 Q is molecular, because its low melting point shows only weak intermolecular forces are overcome
- B1 R is ionic, because it conducts only when molten, when its ions are free to move
- B1 S is macromolecular, because its very high melting point and lack of conduction show a network of covalent bonds with no delocalised electrons
Compare the structures of diamond and graphite.
Mark scheme
- B1 both are macromolecular forms of carbon with strong covalent bonds between the atoms
- B1 in diamond each carbon is bonded to four others tetrahedrally in a three-dimensional network
- B1 in graphite each carbon is bonded to three others in flat hexagonal layers held together by weak forces
- B1 graphite has one delocalised electron per atom; diamond has none
Describe the structure of silicon dioxide.
Mark scheme
- B1 a macromolecular, giant covalent, structure
- B1 each silicon atom is covalently bonded to four oxygen atoms arranged tetrahedrally
- B1 each oxygen atom is bonded to two silicon atoms, so the network extends throughout the crystal in a 1:2 ratio
Explain why graphite is soft enough to be used as a lubricant but has a very high melting point.
Mark scheme
- B1 the forces between the layers are weak
- B1 so the layers slide over each other easily, which makes graphite soft
- B1 melting requires breaking the strong covalent bonds within the layers, which needs a great deal of energy
Explain why iodine has a low melting point, and state what happens to the I–I bonds when iodine melts.
Mark scheme
- B1 iodine is a molecular crystal of I₂ molecules
- B1 melting only overcomes the weak induced dipole–dipole forces between molecules, which needs little energy
- B1 the covalent I–I bonds within the molecules are not broken
Explain why diamond has a very high melting point and is very hard.
Mark scheme
- B1 diamond is macromolecular: each carbon is covalently bonded to four others in a network throughout the crystal
- B1 melting requires breaking many strong covalent bonds, which needs a great deal of energy
- B1 the rigid three-dimensional network of strong bonds resists any change of shape, so diamond is hard
Calculate the number of C–C bonds in the layers of 0.600 g of graphite. Ar of C = 12.0; L = 6.022 × 10²³ mol⁻¹.
Mark scheme
- M1 amount of carbon = 0.600 ÷ 12.0 = 0.0500 mol, which is 0.0500 × 6.022 × 10²³ = 3.011 × 10²² atoms
- M1 each atom forms three bonds, each shared by two atoms, so 1.5 bonds per atom: 1.5 × 3.011 × 10²²
- A1 4.52 × 10²² C–C bonds, to three significant figures
Calculate the increase in volume when 1.00 kg of liquid water at 0 °C freezes. Density of water = 1.00 g cm⁻³; density of ice = 0.917 g cm⁻³.
Mark scheme
- M1 volume of water = 1000 ÷ 1.00 = 1000 cm³
- M1 volume of ice = 1000 ÷ 0.917 = 1090.5 cm³
- A1 increase = 90.5 cm³, to three significant figures
Suggest why hexagonal boron nitride, BN, which has flat layers of hexagons like graphite, is an electrical insulator.
Mark scheme
- B1 like graphite it is macromolecular, with strong covalent bonds within the layers
- B1 nitrogen is more electronegative than boron, so the remaining electrons are held on the nitrogen atoms rather than delocalised across the layer
- B1 with no delocalised electrons there is nothing free to move and carry charge
Evaluate the claim that metallic bonding is always stronger than ionic bonding, using these melting points: magnesium 650 °C, sodium chloride 801 °C, tungsten 3422 °C.
Mark scheme
- B1 the claim is not supported: sodium chloride, which is ionic, melts higher than magnesium, which is metallic
- B1 tungsten shows some metallic bonding is stronger than the ionic bonding in sodium chloride, so neither type is always stronger
- B1 the strength of each depends on the charges and sizes of the particles, so comparisons must be made compound by compound rather than by bonding type
Explain why buckminsterfullerene, C₆₀, turns to vapour at a far lower temperature than diamond, although both are pure carbon.
Mark scheme
- B1 C₆₀ is molecular: separate C₆₀ molecules, with covalent bonds only within each cage
- B1 turning it to vapour overcomes only the induced dipole–dipole forces between the molecules
- B1 diamond is macromolecular, so turning it to vapour means breaking many strong covalent bonds, which needs far more energy
Calculate the energy needed to boil 45.0 g of water at 100 °C. Enthalpy of vaporisation of water = 40.7 kJ mol⁻¹; Mr of H₂O = 18.0.
Mark scheme
- M1 amount of water = 45.0 ÷ 18.0 = 2.50 mol
- M1 energy = 2.50 × 40.7
- A1 102 kJ, to three significant figures
Explain why the enthalpy of vaporisation of water, 40.7 kJ mol⁻¹, is much larger than its enthalpy of fusion, 6.01 kJ mol⁻¹.
Mark scheme
- B1 on melting the molecules stay in contact and only some of the hydrogen bonds are broken, enough to collapse the open lattice of ice
- B1 on boiling the molecules are separated completely, so all the remaining hydrogen bonds must be broken
- B1 far more intermolecular attractions are overcome per mole on boiling, so more energy is needed; no O–H bonds are broken in either change
Name the type of crystal structure in magnesium, sodium chloride, ice and silicon dioxide.
Mark scheme
- B1 magnesium is metallic and sodium chloride is ionic
- B1 ice is molecular and silicon dioxide is macromolecular, also called giant covalent
Give the number of atoms each carbon atom is covalently bonded to in diamond and in graphite, and the bond angle in each.
Mark scheme
- B1 in diamond each carbon is bonded to four others, at 109.5°
- B1 in graphite each carbon is bonded to three others, at 120°
State two physical properties typical of a molecular crystal such as iodine, with a reason for each.
Mark scheme
- B1 a low melting point, because only weak intermolecular forces are overcome on melting
- B1 it does not conduct electricity, because its molecules are uncharged and there are no delocalised electrons
Suggest why iodine dissolves readily in hexane but only slightly in water.
Mark scheme
- B1 I₂ and hexane are both non-polar and form induced dipole–dipole forces with each other of similar strength to those broken
- B1 iodine can form only weak induced dipole–dipole forces with water, which cannot make up for the hydrogen bonds between water molecules that would have to be broken
Suggest why solid C₆₀ is a very poor conductor of electricity, although it has delocalised electrons.
Mark scheme
- B1 the delocalised electrons can move only round their own C₆₀ cage
- B1 there are no bonds between neighbouring molecules, only weak intermolecular forces, so electrons cannot pass from one molecule to the next through the solid
Practise the four crystal structures and their properties one question at a time