4.2 Bonding, structure, and the properties of matter — coverage pack

Answer

• Sodium chloride: ionic bonding.
• Oxygen: covalent bonding.
• Copper: metallic bonding.

Method: Sodium is a metal and chlorine is a non-metal, so sodium chloride is ionic. Oxygen is a non-metallic element whose atoms share electrons, so it is covalent. Copper is a metallic element, so it has metallic bonding.

Answer

• In magnesium chloride, electrons are transferred from magnesium atoms to chlorine atoms.
• The resulting oppositely charged ions attract electrostatically, forming ionic bonds.
• In hydrogen, two atoms share a pair of electrons.
• The shared pair is attracted to both nuclei, forming a covalent bond.

Method: Treat each substance separately. Magnesium is a metal and chlorine is a non-metal, so describe electron transfer and attraction between the ions formed. Hydrogen contains only non-metal atoms, so describe a shared pair of electrons and its attraction to both nuclei.

Answer

• P has ionic bonding: oppositely charged ions attract.
• Q has metallic bonding: positive metal ions are attracted to delocalised electrons.
• R has covalent bonding: each shared electron pair is attracted to the nuclei of the bonded atoms.

Method: Match each particle description to a bonding model. Oppositely charged ions indicate ionic bonding. Mobile delocalised electrons in a giant metal structure indicate metallic bonding. Shared electron pairs between non-metal atoms indicate covalent bonding. Then state the relevant electrostatic attraction rather than stopping at the bond name.

Answer

• K⁺
• S²⁻

Method: A Group 1 atom loses one electron, so potassium forms K⁺. A Group 6 atom gains two electrons to complete its outer shell, so sulfur forms S²⁻.

Answer

• The magnesium atom loses two outer-shell electrons.
• Each of two chlorine atoms gains one electron.
• One Mg²⁺ ion and two Cl⁻ ions form.

Method: Magnesium is in Group 2, so it loses two electrons and forms Mg²⁺. Chlorine is in Group 7, so each atom gains one electron and forms Cl⁻. Two chlorine atoms are therefore needed to receive the two transferred electrons.

Answer

• X loses two electrons to form X²⁺.
• Two Y atoms each gain one electron to form two Y⁻ ions.
• Each ion is shown in brackets with a complete outer shell and its charge.
• Dots and crosses distinguish electrons originally from X and Y, including the transferred electrons.
• The formula is XY₂.

Method: Use the group numbers to obtain X²⁺ and Y⁻. One X atom supplies two electrons, so two Y atoms are needed. In the diagram, use dots and crosses to distinguish the original and transferred outer-shell electrons, bracket every ion, and label the charges. The $1:2$ ion ratio gives XY₂.

Answer

• It has a giant lattice of oppositely charged ions.
• Strong electrostatic attractions act between the ions in all directions.

Method: Name both the scale of the structure and the force holding it together: a giant lattice, not separate molecules, with electrostatic attraction between opposite charges in every direction.

Answer

• MN₂

Method: The ion count ratio is $18:36$. Divide both numbers by $18$ to obtain $1:2$, so the simplest empirical formula is MN₂. The charges also balance: one $2+$ ion is matched by two $1-$ ions.

Answer

• The ion ratio $8:12$ simplifies to $2:3$.
• The empirical formula is A₂B₃.
• The model may not show the ions at their relative sizes or separations.
• Sticks can suggest directional physical bonds that are not present.

Method: Simplify the count ratio $8:12$ by dividing by $4$, giving $2:3$ and hence A₂B₃. The charges balance because two A³⁺ ions give $+6$ and three B²⁻ ions give $-6$. For limitations, compare the simplified representation with a giant lattice whose electrostatic attractions act in all directions.

Answer

• A shared pair of electrons.

Method: Use the definition of a covalent bond: the single bond between the two non-metal atoms represents one pair of electrons shared by both atoms.

Answer

• Oxygen is in the centre with one hydrogen on each side.
• There is one shared electron pair between oxygen and each hydrogen.
• Oxygen also has two unshared pairs of outer-shell electrons.
• Dots and crosses distinguish which atom supplied each electron.

Method: Oxygen has six outer-shell electrons and each hydrogen has one. Oxygen shares one electron with each hydrogen, making two O–H shared pairs. Four oxygen electrons remain as two lone pairs. Use different symbols for electrons from oxygen and hydrogen.

Answer

• The molecular formula is N₂H₄.
• Five covalent bonds are shown.
• The sticks do not show electron pairs, or the model may not show realistic atom sizes, distances or bond angles.

Method: Count atoms rather than bonds to obtain two nitrogen atoms and four hydrogen atoms, so the formula is N₂H₄. Count the one N–N connection and four N–H connections to obtain five bonds. A ball-and-stick model simplifies electron density, scale and geometry.

Answer

• Positive metal ions.
• Delocalised electrons.

Method: The outer-shell electrons are no longer attached to individual atoms. The model therefore shows a regular array of positive metal ions surrounded by mobile, negatively charged delocalised electrons.

Answer

• Metal ions are positively charged.
• Delocalised electrons are negatively charged.
• There is a strong electrostatic attraction between the ions and delocalised electrons throughout the giant structure.

Method: Identify the opposite charges and then state their attraction. Because the ions and delocalised electrons extend through a giant structure, the electrostatic attraction acts throughout the metal.

Answer

• The diagram represents a giant metallic structure, not a simple molecule.
• The circles represent positive metal ions in a regular arrangement.
• The negative symbols represent delocalised electrons.
• The electrons move through and are shared across the whole structure, not only one neighbouring pair.
• Metallic bonding is the electrostatic attraction between the positive ions and delocalised electrons.

Method: Use every feature of the diagram: regular repeated particles show a giant lattice, while many small negative symbols between them show delocalised electrons. Correct both parts of the claim, then give the electrostatic definition of metallic bonding.

Answer

• Liquid to gas: boiling.
• Gas to liquid: condensing.

Method: Use the named changes at the boiling point: heating a liquid produces a gas by boiling, while cooling a gas produces a liquid by condensing.

Answer

• At $-30\,^{\circ}\text{C}$: solid.
• At $20\,^{\circ}\text{C}$: liquid.
• At $100\,^{\circ}\text{C}$: gas.

Method: $-30\,^{\circ}\text{C}$ is below the melting point, so S is solid. $20\,^{\circ}\text{C}$ lies between the melting and boiling points, so S is liquid. $100\,^{\circ}\text{C}$ is above the boiling point, so S is a gas.

Answer

• P is a gas at $100\,^{\circ}\text{C}$.
• Q is a solid at $100\,^{\circ}\text{C}$.
• Q has much stronger forces or bonds between its particles than P.
• More energy is needed to overcome the attractions in Q, giving it higher melting and boiling points.

Method: $100\,^{\circ}\text{C}$ is above P's boiling point, so P is a gas, but it is below Q's melting point, so Q is solid. Higher change-of-state temperatures mean that more energy must be transferred to separate or rearrange the particles, so the attractions in Q are stronger.

Answer

• (l) means liquid.
• (aq) means dissolved in water.

Method: Read each symbol as a physical state. The symbol (aq) is reserved for an aqueous solution, so it must not be treated as another symbol for a pure liquid.

Answer

• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

Method: Balance carbon and hydrogen first, then oxygen. Translate the stated physical state of each substance directly: methane, oxygen and carbon dioxide receive (g), while water receives (l). Coefficients do not change state symbols.

Answer

• Barium chloride: (aq).
• Sodium sulfate: (aq).
• Barium sulfate: (s).
• Sodium chloride: (aq).

Method: Both named reactant solutions receive (aq). The barium sulfate is stated to be a solid precipitate, so it receives (s). Sodium chloride remains dissolved in water, so it receives (aq), not (l).

Answer

• Solid: does not conduct.
• Molten: conducts.

Method: In the solid, the ions are held in fixed lattice positions and cannot carry charge through the material. Melting frees the ions to move, so the liquid conducts.

Answer

• Magnesium oxide has a giant ionic lattice.
• There are strong electrostatic attractions between oppositely charged ions in all directions.
• A large amount of energy is needed to overcome the many strong ionic bonds.

Method: Build the explanation from structure to bonding to energy: identify the giant lattice, state the attraction between opposite charges, then connect the many strong attractions to the large energy transfer required for melting.

Answer

• T has a giant ionic lattice.
• Strong electrostatic attractions act between oppositely charged ions in all directions.
• Much energy is needed to overcome these attractions, so the melting point is high.
• In the solid, ions are fixed and cannot carry charge through the lattice.
• When molten or dissolved, ions are free to move and carry charge.

Method: The combined pattern is diagnostic of an ionic compound. Use the giant lattice and strong attractions to explain the melting point. Then compare mobility: ions are fixed in the solid but mobile in the melt and aqueous solution. Charge is carried by ions, not by delocalised electrons.

Answer

• Only weak intermolecular forces need to be overcome when it boils.
• The strong covalent bonds within the molecules are not broken.

Method: Separate the two scales of attraction. Covalent bonds hold atoms together inside each molecule, but boiling moves whole molecules apart by overcoming the much weaker forces between them.

Answer

• V usually has the higher boiling point.
• Larger molecules have stronger intermolecular forces.
• More energy is needed to overcome these forces between V molecules.

Method: Apply the specified trend: increasing molecular size strengthens the forces between molecules. Stronger intermolecular forces require a larger energy transfer to overcome, so the boiling point is higher.

Answer

• B is made of larger molecules than A.
• B therefore has stronger intermolecular forces.
• More energy is needed to overcome the forces between B molecules, so B has the higher boiling point.
• Neither pure substance conducts electricity because the molecules have no overall charge.

Method: Use relative molecular mass as evidence that B has larger molecules. Link larger molecules to stronger intermolecular forces and hence a greater energy requirement for boiling. For conductivity, use the stated neutral molecular particles: there are no mobile charged particles in either pure substance.

Answer

• Polymer.
• Covalent bonds join the atoms within a chain.

Method: Use the phrase ‘separate, very long chains’ to identify a polymer. Then distinguish the bonds inside a molecule from the forces between molecules: the atoms along each chain are linked by strong covalent bonds.

Answer

• P has very large polymer molecules.
• The intermolecular forces between its molecules are relatively strong.
• At $20\,^\circ\text{C}$ there is insufficient energy to overcome these forces, so P remains solid.

Method: Build a structure-to-property chain. Long covalent chains mean very large molecules; these have relatively strong intermolecular forces. Link that explicitly to the state: room-temperature energy does not overcome enough of those forces for the molecules to move past one another.

Answer

• R is the polymer.
• R consists of separate very large molecules or chains.
• Forces between R's chains are intermolecular forces rather than covalent bonds.
• S is a giant covalent structure in which covalent bonds connect atoms throughout one continuous network.

Method: First decide whether the covalent bonding stops at the edge of a molecule. It does in R, leaving separate chains, so R is the polymer. In S the covalent bonds continue throughout the structure, so S is a giant covalent network rather than a collection of polymer molecules.

Answer

• A giant covalent structure.

Method: Look for a network with no boundary around an individual molecule. Because the covalent bonding continues throughout the sample, the structure is giant covalent.

Answer

• Silicon dioxide has a giant covalent structure.
• There are many strong covalent bonds between its atoms.
• A large amount of energy is needed to overcome these bonds, giving it a very high melting point.

Method: Name the structure, identify the strong bonds that extend throughout it, and then link bond strength to energy. Since $1500\,^\circ\text{C}$ is still below the melting point, the structure remains solid.

Answer

• V has a giant covalent structure.
• Strong covalent bonds extend throughout V's network.
• Many strong covalent bonds must be overcome to melt V, requiring a large amount of energy.
• Melting U only overcomes intermolecular forces between its separate molecules, so U melts at a much lower temperature.

Method: Use ‘no separate molecules’ and the extreme temperature to infer a giant network for V. Compare what must be overcome: covalent bonds throughout V but only intermolecular forces between U's molecules. This accounts for the large difference in energy needed.

Answer

• Its atoms are arranged in layers.
• The layers can slide over one another when a force is applied.

Method: Identify the relevant structural feature, which is the layered arrangement of atoms, and link movement of those layers to the observed change of shape.

Answer

• The different-sized atoms distort the regular layers in M's structure.
• The distorted layers cannot slide over one another as easily.
• More force is therefore needed to change the alloy's shape, so it is harder.

Method: Follow the causal sequence demanded by AQA: different atom sizes, then distortion of the layers, then reduced sliding. Finish by connecting reduced sliding to increased hardness.

Answer

• Both materials have giant metallic structures with strong metallic bonding.
• A large amount of energy is needed to overcome the strong metallic bonding, so their melting points are high.
• Pure Q has regular layers of atoms that can slide over one another.
• Different-sized atoms in the alloy distort these layers, making sliding more difficult.
• Therefore the alloy needs the larger force to bend it.

Method: Treat the two properties separately. Explain melting using strong metallic bonding in a giant structure. Explain bending using layer movement: regular layers slide in pure Q, while different-sized alloy atoms distort the layers and oppose sliding, consistent with $47\,\text{N}>18\,\text{N}$.

Answer

• Delocalised electrons.

Method: In a solid metal the positive ions do not travel along the wire. The mobile charge carriers are the delocalised electrons.

Answer

• The metal contains delocalised electrons.
• These electrons are free to move through the metallic structure.
• Their movement carries electrical charge through the strip.

Method: State which charged particles are present, establish that they are mobile, and then connect their motion to charge transfer. The comparison does not require claiming that the polymer has mobile ions or electrons.

Answer

• Copper contains delocalised electrons.
• The delocalised electrons can move through the metallic structure.
• They carry electrical charge through a wire.
• They also transfer thermal energy rapidly through the saucepan base.

Method: Identify the shared cause before treating the two uses. Mobile delocalised electrons account for both properties: charge transport explains electrical conduction, while energy transfer by those electrons explains thermal conduction.

Answer

• Four covalent bonds.

Method: Recall that diamond uses all four outer electrons of each carbon atom to make four covalent bonds in its giant structure.

Answer

• Diamond has a giant covalent structure.
• Each carbon atom is joined by four strong covalent bonds in a rigid network.
• A large force is needed to deform or break this network, so diamond is very hard.

Method: Translate the use into the required property, hardness. Then link hardness to the rigid three-dimensional network of four strong covalent bonds around each carbon atom.

Answer

• Diamond has a giant covalent structure with four covalent bonds from each carbon atom.
• Many strong covalent bonds must be overcome to melt it, so its melting point is very high.
• The rigid network of strong bonds makes diamond hard.
• All four outer electrons of each carbon atom are used in covalent bonds, so there are no delocalised electrons.
• Diamond cannot carry the heating current because it does not conduct electricity.

Method: Separate the three requested properties. Use strong covalent bonds throughout the network for melting point, rigidity for hardness, and the absence of mobile delocalised electrons for electrical conduction. The last point rules out diamond as the current-carrying element.

Answer

• Three covalent bonds.

Method: Each graphite carbon bonds to three neighbouring carbon atoms in a hexagonal layer, leaving one electron delocalised.

Answer

• One electron from each carbon atom is delocalised.
• The delocalised electrons can move through the structure.
• They carry electrical charge, so graphite conducts electricity.

Method: An electrode must conduct. Identify the delocalised electron supplied by each carbon atom, state that these electrons are mobile, and connect their movement to charge transfer.

Answer

• Graphite has strong covalent bonds between carbon atoms within each layer.
• Many strong bonds require a large amount of energy to overcome, so graphite has a high melting point.
• One electron per carbon atom is delocalised.
• The delocalised electrons move and carry charge, so graphite conducts electricity.
• There are no covalent bonds between adjacent layers, only weak forces.
• The layers can slide over one another, so the contact can move with low friction.

Method: Match one structural feature to each demand. Strong bonds within layers give thermal stability; delocalised electrons give electrical conduction; the absence of covalent bonds between layers allows them to slide. Keep the within-layer and between-layer bonding explanations distinct.

Answer

• Graphene is a single layer of graphite.

Method: Treat graphite as a stack of carbon layers. One isolated layer from that stack is graphene.

Answer

• Graphene is a single layer of graphite and is only one atom thick.
• It therefore adds very little thickness to the sensor.
• Graphene contains delocalised electrons.
• The electrons can move and carry charge, so graphene conducts electricity.

Method: Extract the two design requirements: small thickness and electrical conduction. Link the first to graphene being a single atomic layer, and the second to its mobile delocalised electrons.

Answer

• A is graphene.
• B is Buckminsterfullerene or C₆₀.
• C is a carbon nanotube.
• A is suitable for electronics or as part of a composite.
• C is suitable for nanotechnology, electronics or materials.

Method: Use the defining shapes first: one layer identifies graphene, the 60-atom hollow sphere identifies Buckminsterfullerene, and the high-aspect-ratio cylinder identifies a nanotube. Then select applications from the specified families rather than inventing a shape-dependent use that has not been supported.

Answer

• T is a nanoparticle.
• $300$ atom diameters span T.

Method: The size $72\,\text{nm}$ lies inside the nano interval $1$–$100\,\text{nm}$. Because both dimensions use nanometres, compare them directly: $72/0.24=300$, so T spans about $300$ atom diameters.

Answer

• Surface area $=150\,\text{nm}^2$.
• Volume $=125\,\text{nm}^3$.
• Surface-area-to-volume ratio $=1.2:1$.

Method: A cube has six square faces, so $A=6a^2=6\times5^2=150\,\text{nm}^2$. Its volume is $V=a^3=5^3=125\,\text{nm}^3$. Therefore $A:V=150:125=1.2:1$.

Answer

• The nano cubes' surface-area-to-volume ratio is $20$ times larger.
• Estimated nano mass $=0.60\,\text{g}$.

Method: For similar cubes, $A:V=6/a$, so the ratio increases by the inverse side-length factor: $800/40=20$. Under the stated proportional-effectiveness assumption, divide the required mass by $20$: $12/20=0.60\,\text{g}$.

Answer

• Any two from medicine, electronics, cosmetics, sun creams, deodorants and catalysts.

Method: Select two distinct application areas from the specification list. Do not give two products from the same area if the question asks for two areas.

Answer

• The nanoparticles have a high surface-area-to-volume ratio, so less catalyst is needed for the same effect.
• Using less material may reduce raw-material use or cost.
• Less catalyst may also reduce waste or the mass that must be handled.
• A possible disadvantage is uncertain health or environmental risk if nanoparticles enter organisms or the environment.

Method: Use the supplied equal-performance data to justify the benefits rather than assuming better performance. Credit the smaller mass through its high surface-area-to-volume ratio, then state a resource, cost or waste advantage. Balance this with a possible risk caused by the particles' small size.

Answer

• J blocks slightly more ultraviolet radiation: $96\%$ compared with $94\%$.
• K is transparent, which may make it more acceptable to users.
• K needs only one quarter as much active material, so it may reduce resource use, cost or waste.
• Detection inside some cells shows exposure is possible and creates a potential health concern.
• The data do not show that K causes harm, so the risk remains uncertain rather than proven.
• A justified conclusion may choose either product, but should weigh performance and user/resource benefits against the uncertainty and recommend further safety testing before K is widely used.

Method: Compare every relevant datum: protection, appearance, quantity required and cell exposure. Distinguish evidence of entry into cells from evidence of harm. Reach a conditional judgement; for example, K has strong practical and resource advantages, but development should include further safety testing because the long-term risk is unresolved.