IB Chemistry S2.3: The metallic model worked solutions

Positive cores attract the delocalized electron population.

B treats a metal as molecular; C substitutes the ionic model; D incorrectly makes like charges attract.

Mobile delocalized electrons transfer thermal energy.

A wrongly makes bonding directional and fixed; B mistakes cations for the mobile charge carriers; D reverses the electrostatic force between like charges.

A metal conducts through delocalized electrons, whereas melting frees both ion types in the salt to move. Option D wrongly treats the metal cations as mobile and the anions in the melt as immobile.

Layer displacement can retain strong non-directional attraction, so easy shaping does not imply easy melting.

A confuses displacement with loss of cohesion; C invents a change of bonding model; D mistakes melting for electron removal.

The second metal has more highly charged positive ions and a greater supply of delocalized electrons, increasing electrostatic attraction at equal radius. C ignores the effect of charge. [1]

Smaller radius favours stronger attraction in H, while greater charge favours it in J; no exact cancellation or quantitative comparison is supplied.

A ignores charge; B ignores radius; C invents an exact cancellation.

Melting removes the ordered solid arrangement, while cation-electron attraction and mobile delocalized electrons persist.

B retains a solid lattice in the liquid; C invents neutral molecules; D replaces electrons with anions.

Participating d electrons add to the metallic bonding population and support strong cation-electron attraction.

A substitutes an ionic carrier; B makes like charges attractive; D confuses disrupting the solid structure with electron removal.

(a) Malleability. [1]

(b) Layers of cations move relative to neighbouring layers. [1] Non-directional attraction to the delocalized electron population continues after the displacement, maintaining cohesion. [1]

(c) Metallic bonds can be strong; low bond strength is not required for malleability. [1] Deformation is possible because changing neighbours does not remove the attraction to the shared electron population. [1]

(a) J<L<N. [1]

(b) Both L and N have cation charge +2, but N has the smaller radius. [1] Its higher charge density gives stronger attraction to the delocalized electrons. [1]

(c) J has lower cation charge and supplies fewer delocalized electrons per atom than the other two. [1] It also has the largest radius; these factors reduce the attraction between cations and electrons. [1]

(d) N, which has the strongest predicted bonding. [1]

(a) The 96 pm core gives the strongest attraction. [1] (b) Each group 2 atom supplies two delocalised electrons, leaving the same model core charge, +2. [1] The smaller core brings the electron population closer to the positive centre, strengthening attraction when structural effects are comparable. [1]

(a) The positive core charge stays +1; equivalently, each atom contributes one shared electron. [1] (b) Increasing core radius increases separation from the delocalised electrons and weakens the electrostatic attraction. [1] Less energy is therefore needed to disrupt the ordered solid sufficiently for melting, explaining the broad fall in melting temperature. [1] Mass does not supply this electrostatic explanation.

(a) Neutrality requires N_e=3\times45=135 electrons. [1] (b) The electrons form a population shared throughout the metal rather than isolated pair bonds. [1] Electrostatic attraction between these negative electrons and the positive ion cores provides cohesion. [1] The cores include bound electrons; they are not bare nuclei.

(a) Delocalised electrons carry current through the molten metal. [1] Both positive and negative ions migrate in the molten ionic compound. [1] (b) In the solid metal, delocalised electrons remain mobile and drift under a potential difference although the cores remain at lattice positions. [1] Melting is not a prerequisite for metallic conduction.

(a) Ductility is the property used when drawing a wire. [1] (b) Rows of positive ion cores can shift relative to neighbouring rows. [1] The shared electron population continues to attract the cores non-directionally, so the sample retains cohesion as neighbours change. [1] The electrons are not fixed between particular pairs of cores.

(a) Electrons in the warmer region acquire greater kinetic energy. [1] Through interactions they transfer energy to other particles, producing net energy transfer towards cooler regions without requiring net charge drift. [1] (b) Lattice vibrations also transfer thermal energy. [1] Electrical conduction and thermal conduction involve different net flows.

(a) Use q=mc_{\mathrm w}\Delta T, hence \Delta T=q/(mc_{\mathrm w}). [1] \Delta T=1260/(60.0\times4.18)=5.02\,\mathrm{K}. [1] The final temperature is 18.5+5.0239=23.5\,{}^{\circ}\mathrm{C} to one decimal place. [1]

(b) Delocalized electrons move through the metal and gain kinetic energy in hotter regions. [1] They transfer energy through interactions with other particles, helping distribute thermal energy rapidly. [1]

(c) Negligible energy is absorbed by the water container. [1] Accept negligible energy loss from the water to the surroundings during heating.

(a) D is predicted to melt at the higher temperature. [1] Its d-electrons increase the population of electrons participating in metallic bonding. [1] Stronger attraction between the ion cores and bonding electrons requires more energy to overcome during melting. [1]

(b) The prediction is incorrect: s-block metal E also has delocalized electrons. [1] Those electrons move and carry charge under a potential difference without any d-electron contribution. [1]

(c) No; melting point concerns the strength of cohesion, whereas conductivity concerns charge transport, so the melting-point order alone does not determine a conductivity order. [1]

(a) The cores contribute 75\times(+2)=+150 elementary-charge units, while 144 electrons contribute -144, leaving +6. [1] Add six electrons, giving 150-150=0. [1] (b) The second patch contains 150 cores and therefore 300 electrons. [1] Its volume is 2V, so 300/(2V)=150/V; the concentrations are equal. [1] (c) Reject the claim: doubling the size doubles the total electron count without increasing concentration. [1] Conductivity depends on mobile charge transport and material properties; these particle counts alone provide no evidence of a higher conductivity. [1] A larger sample need not have a greater intrinsic conductivity.

(a) R is predicted to have the strongest attraction. [1] Relative to Q it has a greater positive charge, more participating electrons per atom and a smaller core. [1] Relative to S it has the same charge but a smaller radius, strengthening core–electron attraction. [1] (b) S has greater core charge, which favours stronger attraction. [1] Its larger radius favours weaker attraction, and the qualitative model assigns no numerical balance between these effects. [1] (c) Reject the proposed temperature calculation: the model gives qualitative comparisons, not a calibrated equation; packing and the solid-to-liquid energy balance also matter. [1]

(a) The proposed solid has neither mobile ions nor delocalised electrons. [1] It therefore lacks mobile charge carriers and cannot explain the observed current. [1] (b) Positive metal cores can move relative to neighbouring rows during rolling. [1] Non-directional attraction to a shared electron population persists, so the sheet retains cohesion. [1] (c) An ionic lattice can have strong electrostatic attraction between its ions. [1] A metallic solid can also have strong core–electron attraction, so a high melting temperature alone does not identify which structure is present. [1]

(a) Rows of positive metal cores can change neighbours during bending. [1] Attraction to the shared electrons persists, retaining cohesion. [1] (b) The metal conducts by mobile delocalised electrons. [1] The molten spacer conducts by mobile cations and anions. [1] (c) The coolant consists of neutral molecules and, as specified, does not ionise. [1] It therefore has no population of mobile charged particles to transport current. [1] Liquidity alone does not establish electrical conduction.

(a) Use Q=mc\Delta T and rearrange to \Delta T=Q/(mc). [1] \Delta T=836/(40.0\times4.18)=5.00\,\mathrm K. [1] Thus T_f=21.0+5.00=26.0\,{}^{\circ}\mathrm C. [1] The supplied energy is explicitly the transfer to the water. (b) Mobile electrons gain kinetic energy in warmer regions. [1] Their interactions transfer energy to other particles towards cooler regions; lattice vibrations also contribute. [1] (c) If 836 J instead meant total heater input, some would warm the spreader, leaving less for the water and giving a smaller rise. [1] The interpretation of the energy input must be changed as stated in this part.

(a) Participating d electrons add to the delocalised bonding population. [1] With the supplied comparable sizes and packing, stronger core–electron cohesion increases the energy needed to disrupt the ordered solid for melting. [1] (b) E still has delocalised outer s electrons. [1] They are mobile and carry charge under an applied potential difference. [1] (c) Different-sized atoms distort the regular arrangement and obstruct relative layer motion. [1] This makes plastic deformation more difficult without requiring every core–electron interaction to become stronger. [1] Hardness and melting behaviour require distinct explanations.