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In a giant ionic lattice, these attractions act between ions in all directions throughout the structure."},{"text":"Strong covalent bonds in all directions.","correct":false,"explanation":"Covalent bonds involve shared electron pairs, whereas ions in a giant ionic lattice are held together by strong electrostatic attractions between opposite charges in all directions."},{"text":"Weak intermolecular forces in all directions.","correct":false,"explanation":"Intermolecular forces act between molecules and are much weaker. A giant ionic lattice contains ions held together by strong electrostatic attractions in all directions."},{"text":"Metallic attractions in all directions.","correct":false,"explanation":"Metallic bonding involves positive metal ions attracted to delocalised electrons. An ionic lattice is instead held together by strong electrostatic attractions between positive and negative ions."}]},{"id":"32497909-5199-4318-aeb5-b6cdc906610c","cardType":"cloze","points":null,"front":"Hess’s law: the total enthalpy change is independent of the [ ] if the [ ] are the same.","back":"Hess’s law: the total enthalpy change is independent of the route taken if the initial and final states are the same.","mcqOptions":[{"text":"Hess’s law: the total enthalpy change is independent of the route taken if the initial and final states are the same.","correct":true,"explanation":"Enthalpy is a state function, so its overall change depends only on the initial and final states, not on the route taken between them."},{"text":"Hess’s law: the total enthalpy change is independent of the initial state if the route and final state are the same.","correct":false,"explanation":"The initial state cannot be changed without potentially changing the enthalpy difference. Hess’s law states that the route is irrelevant when the initial and final states are unchanged."},{"text":"Hess’s law: the total enthalpy change is independent of the final state if the initial state and route are the same.","correct":false,"explanation":"Changing the final state changes the enthalpy difference. It is the route taken that does not matter, provided the initial and final states remain the same."},{"text":"Hess’s law: the total enthalpy change is independent of the initial and final states if the route taken are the same.","correct":false,"explanation":"This reverses the relationship: enthalpy change depends on the initial and final states. It is independent of the route taken between those fixed states."}]},{"id":"9644ceaf-a298-49e6-af8f-9655cc3c469a","cardType":"cloze","points":null,"front":"\$\\Delta_\\text{LE}H\$ is for forming [ ] from [ ] under standard conditions.","back":"\$\\Delta_\\text{LE}H\$ is for forming 1 mol of ionic lattice from gaseous ions under standard conditions.","mcqOptions":[{"text":"\$\\Delta_\\text{LE}H\$ is for forming 1 mol of ionic lattice from gaseous ions under standard conditions.","correct":true,"explanation":"Lattice enthalpy of formation is the enthalpy change when one mole of an ionic lattice forms from its separated gaseous ions under standard conditions."},{"text":"\$\\Delta_\\text{LE}H\$ is for forming 1 mol of gaseous ions from an ionic lattice under standard conditions.","correct":false,"explanation":"This describes separating the lattice into gaseous ions, which is lattice dissociation enthalpy. Lattice enthalpy of formation forms one mole of ionic lattice from gaseous ions."},{"text":"\$\\Delta_\\text{LE}H\$ is for forming 1 mol of ionic lattice from solid elements under standard conditions.","correct":false,"explanation":"Formation from elements describes standard enthalpy of formation, not lattice enthalpy. Lattice enthalpy specifically forms one mole of ionic lattice from gaseous ions."},{"text":"\$\\Delta_\\text{LE}H\$ is for forming 1 mol of aqueous ions from gaseous ions under standard conditions.","correct":false,"explanation":"Forming aqueous ions from gaseous ions is hydration, not lattice formation. Lattice enthalpy forms one mole of solid ionic lattice from gaseous ions."}]},{"id":"bfc8242a-e451-49d0-a6b7-9f57334be01d","cardType":"basic","points":2,"front":"Why is lattice enthalpy of formation usually negative?","back":"Ion-ion attractions form, so energy is released.","mcqOptions":[{"text":"Ion-ion attractions form, so energy is released.","correct":true,"explanation":"When separated gaseous ions form a lattice, strong electrostatic attractions develop between opposite charges. Formation of these attractions releases energy, giving a negative enthalpy change."},{"text":"Ion-ion attractions break, so energy is released.","correct":false,"explanation":"Breaking ion-ion attractions requires an input of energy rather than releasing it. Lattice formation is negative because electrostatic attractions form between oppositely charged ions."},{"text":"Ion-ion repulsions form, so energy is released.","correct":false,"explanation":"Repulsions do not explain the stability or energy release of an ionic lattice. Energy is released because strong electrostatic attractions form between oppositely charged ions."},{"text":"Ion-ion attractions form, so energy is absorbed.","correct":false,"explanation":"Forming attractive interactions lowers the system’s energy and releases energy rather than absorbing it. Therefore, lattice enthalpy of formation is usually negative."}]},{"id":"6757d871-85ee-4aba-9114-29d42ed26eff","cardType":"basic","points":1,"front":"What does a more negative \$\\Delta_\\text{LE}H\$ show about the giant lattice?","back":"Stronger ionic bonding.","mcqOptions":[{"text":"Stronger ionic bonding.","correct":true,"explanation":"A more negative lattice enthalpy means more energy is released when the lattice forms, indicating stronger electrostatic attractions and therefore stronger ionic bonding."},{"text":"Weaker ionic bonding.","correct":false,"explanation":"Weaker attractions would release less energy and produce a less negative lattice enthalpy. A more negative value indicates stronger electrostatic attractions and stronger ionic bonding."},{"text":"Stronger covalent bonding.","correct":false,"explanation":"Lattice enthalpy measures electrostatic attraction between ions, not covalent electron-pair sharing. A more negative value therefore shows stronger ionic bonding."},{"text":"Weaker covalent bonding.","correct":false,"explanation":"Covalent bond strength is not measured by lattice enthalpy. A more negative lattice enthalpy shows that the electrostatic attractions within the ionic lattice are stronger."}]},{"id":"10d5ed61-96f8-4503-8604-9416fc6e58f3","cardType":"basic","points":1,"front":"If lattice enthalpy of formation is \$-x\$, what is lattice dissociation enthalpy?","back":"\$+x\$, the opposite sign.","mcqOptions":[{"text":"\$+x\$, the opposite sign.","correct":true,"explanation":"Lattice dissociation is the exact reverse of lattice formation, so its enthalpy change has the same magnitude but the opposite sign, changing from \$-x\$ to \$+x\$."},{"text":"\$-x\$, the same sign.","correct":false,"explanation":"Keeping the negative sign ignores that dissociation reverses the formation process. Reversing a reaction reverses the enthalpy sign, so the dissociation enthalpy is \$+x\$."},{"text":"\$+2x\$, double the magnitude.","correct":false,"explanation":"Reversing the process changes only the sign, not the magnitude. No equation has been multiplied, so the lattice dissociation enthalpy is \$+x\$, not \$+2x\$."},{"text":"\$0\$, no enthalpy change.","correct":false,"explanation":"Breaking the ionic lattice requires the same amount of energy that its formation releases. The reverse process therefore has enthalpy \$+x\$, rather than zero."}]},{"id":"688d34fd-da8a-481d-ba51-d3c03f8398d3","cardType":"cloze","points":null,"front":"One mole of ionic lattice means one mole of [ ], not one [ ].","back":"One mole of ionic lattice means one mole of formula units, not one individual crystal.","mcqOptions":[{"text":"One mole of ionic lattice means one mole of formula units, not one individual crystal.","correct":true,"explanation":"Ionic substances do not contain discrete molecules. One mole refers to Avogadro’s constant of formula units, which may together make up any number of physical crystals."},{"text":"One mole of ionic lattice means one mole of individual ions, not one formula unit.","correct":false,"explanation":"A formula unit gives the simplest whole-number ratio of ions in an ionic compound. One mole of lattice means one mole of formula units, not one mole of individual ions."},{"text":"One mole of ionic lattice means one mole of individual crystals, not one formula unit.","correct":false,"explanation":"A crystal is a macroscopic piece containing an enormous number of formula units, so it is not the counting unit used here. One mole means one mole of formula units."},{"text":"One mole of ionic lattice means one mole of ionic molecules, not one individual crystal.","correct":false,"explanation":"Giant ionic compounds do not consist of discrete molecules. Their amount is counted using formula units, so one mole of lattice means one mole of formula units."}]},{"id":"d01a9e7d-8211-469d-9f10-2b15e0c97b1d","cardType":"basic","points":1,"front":"What equation represents \$\\Delta_\\text{f}H^\\circ\$ for NaCl?","back":"\$\\mathrm{Na}(s)+\\frac{1}{2}\\mathrm{Cl}_2(g)\\to\\mathrm{NaCl}(s)\$","mcqOptions":[{"text":"\$\\mathrm{Na}(s)+\\frac{1}{2}\\mathrm{Cl}_2(g)\\to\\mathrm{NaCl}(s)\$","correct":true,"explanation":"Standard enthalpy of formation produces exactly one mole of NaCl solid from sodium and chlorine in their standard states: solid sodium and gaseous diatomic chlorine."},{"text":"\$2\\mathrm{Na}(s)+\\mathrm{Cl}_2(g)\\to2\\mathrm{NaCl}(s)\$","correct":false,"explanation":"Although chemically balanced, this equation forms two moles of NaCl. Standard enthalpy of formation must form exactly one mole, requiring half a mole of chlorine."},{"text":"\$\\mathrm{Na}(g)+\\frac{1}{2}\\mathrm{Cl}_2(g)\\to\\mathrm{NaCl}(s)\$","correct":false,"explanation":"Sodium’s standard state is solid, not gaseous. The formation equation must use elements in their standard states and produce exactly one mole of NaCl solid."},{"text":"\$\\mathrm{Na}(s)+\\mathrm{Cl}(g)\\to\\mathrm{NaCl}(s)\$","correct":false,"explanation":"Chlorine’s standard state is diatomic \$\\mathrm{Cl}_2(g)\$, not separate gaseous atoms. The correct equation therefore uses \$\\frac{1}{2}\\mathrm{Cl}_2(g)\$."}]},{"id":"e9c09055-b80d-4bfc-a5fd-3468c7d3d637","cardType":"cloze","points":null,"front":"In Born-Haber data, ionisation energies and atomisation enthalpies are [ ], while first electron affinities are often [ ].","back":"In Born-Haber data, ionisation energies and atomisation enthalpies are positive, while first electron affinities are often negative.","mcqOptions":[{"text":"In Born-Haber data, ionisation energies and atomisation enthalpies are positive, while first electron affinities are often negative.","correct":true,"explanation":"Ionisation and atomisation require energy to overcome attractions, so they are endothermic. First electron affinities are often exothermic because energy is released when an electron joins a gaseous atom."},{"text":"In Born-Haber data, ionisation energies and atomisation enthalpies are negative, while first electron affinities are often positive.","correct":false,"explanation":"This reverses both sign conventions. Ionisation and atomisation require energy and are positive, whereas a first electron affinity commonly releases energy and is therefore negative."},{"text":"In Born-Haber data, ionisation energies and atomisation enthalpies are positive, while first electron affinities are often positive.","correct":false,"explanation":"The ionisation and atomisation signs are correct, but first electron affinity is usually negative because attraction between the nucleus and added electron releases energy."},{"text":"In Born-Haber data, ionisation energies and atomisation enthalpies are negative, while first electron affinities are often negative.","correct":false,"explanation":"The electron-affinity sign is usually negative, but ionising or atomising a substance requires an energy input. Their enthalpy changes must therefore be positive."}]},{"id":"9b55fe92-1f7e-4853-b3b4-040c62c2dfa4","cardType":"basic","points":1,"front":"What is the sodium first ionisation step in a NaCl Born-Haber cycle?","back":"\$\\mathrm{Na}(g)\\to\\mathrm{Na}^+(g)+e^-\$","mcqOptions":[{"text":"\$\\mathrm{Na}(g)\\to\\mathrm{Na}^+(g)+e^-\$","correct":true,"explanation":"First ionisation removes one electron from each gaseous sodium atom, forming a gaseous ion with a single positive charge."},{"text":"\$\\mathrm{Na}(s)\\to\\mathrm{Na}^+(g)+e^-\$","correct":false,"explanation":"Ionisation energy is defined for gaseous atoms, not solid sodium. Sodium must first be atomised, so the correct step begins with \$\\mathrm{Na}(g)\$."},{"text":"\$\\mathrm{Na}(g)+e^-\\to\\mathrm{Na}^-(g)\$","correct":false,"explanation":"This represents electron affinity because sodium gains an electron. First ionisation instead removes an electron from gaseous sodium to form \$\\mathrm{Na}^+(g)\$."},{"text":"\$\\mathrm{Na}^+(g)+e^-\\to\\mathrm{Na}(g)\$","correct":false,"explanation":"This is the reverse of first ionisation and would release the ionisation energy. The Born-Haber ionisation step forms \$\\mathrm{Na}^+(g)\$ and an electron."}]},{"id":"3a7fee0a-bcc5-4b8e-91ac-197460d72d20","cardType":"basic","points":1,"front":"What chlorine atomisation step is used in a NaCl Born-Haber cycle?","back":"\$\\frac{1}{2}\\mathrm{Cl}_2(g)\\to\\mathrm{Cl}(g)\$","mcqOptions":[{"text":"\$\\frac{1}{2}\\mathrm{Cl}_2(g)\\to\\mathrm{Cl}(g)\$","correct":true,"explanation":"One mole of chlorine atoms is required per mole of NaCl, so half a mole of gaseous chlorine molecules is dissociated into one mole of gaseous atoms."},{"text":"\$\\mathrm{Cl}_2(g)\\to2\\mathrm{Cl}(g)\$","correct":false,"explanation":"This produces two moles of chlorine atoms, doubling the required atomisation quantity. For one mole of NaCl, the step starts with half a mole of \$\\mathrm{Cl}_2\$."},{"text":"\$\\frac{1}{2}\\mathrm{Cl}_2(l)\\to\\mathrm{Cl}(g)\$","correct":false,"explanation":"Elemental chlorine is gaseous under standard conditions, so its standard-state reactant is \$\\mathrm{Cl}_2(g)\$, not liquid chlorine."},{"text":"\$\\mathrm{Cl}(g)\\to\\frac{1}{2}\\mathrm{Cl}_2(g)\$","correct":false,"explanation":"This forms chlorine molecules from atoms and is the reverse of atomisation. Atomisation must produce one mole of gaseous chlorine atoms from the element."}]},{"id":"c6536d92-ef0c-4175-ba2e-42a482e8fd0b","cardType":"basic","points":1,"front":"What is the first electron affinity step for chlorine?","back":"\$\\mathrm{Cl}(g)+e^-\\to\\mathrm{Cl}^-(g)\$","mcqOptions":[{"text":"\$\\mathrm{Cl}(g)+e^-\\to\\mathrm{Cl}^-(g)\$","correct":true,"explanation":"First electron affinity is the enthalpy change when each gaseous chlorine atom gains one electron to form a gaseous ion with a single negative charge."},{"text":"\$\\mathrm{Cl}(g)\\to\\mathrm{Cl}^+(g)+e^-\$","correct":false,"explanation":"This is chlorine's first ionisation, involving electron removal. Electron affinity instead involves a gaseous chlorine atom gaining an electron to form \$\\mathrm{Cl}^-(g)\$."},{"text":"\$\\mathrm{Cl}^-(g)\\to\\mathrm{Cl}(g)+e^-\$","correct":false,"explanation":"This is the reverse of the first electron-affinity process. The required step adds an electron to a gaseous chlorine atom rather than removing one from chloride."},{"text":"\$\\mathrm{Cl}(g)+2e^-\\to\\mathrm{Cl}^{2-}(g)\$","correct":false,"explanation":"First electron affinity adds only one electron and forms a singly charged chloride ion. Adding two electrons would involve two separate electron-affinity processes."}]},{"id":"70973fc1-21b6-4677-a6ec-e7b49b2637db","cardType":"basic","points":3,"front":"For NaCl, write the Hess expression for \$\\Delta_\\text{f}H\$ in terms of \$\\Delta_\\text{LE}H\$.","back":"\$\\Delta_\\text{f}H=\\Delta_\\text{at}H(\\mathrm{Na})+\\Delta_\\text{at}H(\\mathrm{Cl})+\\mathrm{IE}_1(\\mathrm{Na})+\\mathrm{EA}_1(\\mathrm{Cl})+\\Delta_\\text{LE}H\$","mcqOptions":[{"text":"\$\\Delta_\\text{f}H=\\Delta_\\text{at}H(\\mathrm{Na})+\\Delta_\\text{at}H(\\mathrm{Cl})+\\mathrm{IE}_1(\\mathrm{Na})+\\mathrm{EA}_1(\\mathrm{Cl})+\\Delta_\\text{LE}H\$","correct":true,"explanation":"Hess's law sums the steps converting standard-state elements into gaseous atoms, then gaseous ions, and finally the ionic lattice. Each term is included with its signed value."},{"text":"\$\\Delta_\\text{f}H=\\Delta_\\text{at}H(\\mathrm{Na})+\\Delta_\\text{at}H(\\mathrm{Cl})-\\mathrm{IE}_1(\\mathrm{Na})+\\mathrm{EA}_1(\\mathrm{Cl})+\\Delta_\\text{LE}H\$","correct":false,"explanation":"Ionising gaseous sodium requires energy, so its ionisation energy is added as a positive term. Subtracting it incorrectly treats electron removal as exothermic."},{"text":"\$\\Delta_\\text{f}H=\\Delta_\\text{at}H(\\mathrm{Na})+\\Delta_\\text{at}H(\\mathrm{Cl})+\\mathrm{IE}_1(\\mathrm{Na})-\\mathrm{EA}_1(\\mathrm{Cl})+\\Delta_\\text{LE}H\$","correct":false,"explanation":"The electron affinity already carries its own negative sign in the data, so it must be added algebraically. Subtracting it would incorrectly make that step endothermic."},{"text":"\$\\Delta_\\text{f}H=\\Delta_\\text{at}H(\\mathrm{Na})+\\Delta_\\text{at}H(\\mathrm{Cl})+\\mathrm{IE}_1(\\mathrm{Na})+\\mathrm{EA}_1(\\mathrm{Cl})-\\Delta_\\text{LE}H\$","correct":false,"explanation":"For lattice enthalpy defined as formation from gaseous ions, the lattice term is added with its negative value. Subtracting it reverses the lattice-formation step."}]},{"id":"01db24e7-3763-416d-8ff9-4ee41be61022","cardType":"basic","points":2,"front":"Using \$-411=108+122+496-349+\\Delta_\\text{LE}H\$, calculate \$\\Delta_\\text{LE}H\$ for NaCl.","back":"\$-788\\ \\mathrm{kJ\\ mol}^{-1}\$","mcqOptions":[{"text":"\$-788\\ \\mathrm{kJ\\ mol}^{-1}\$","correct":true,"explanation":"The non-lattice terms total \$377\\ \\mathrm{kJ\\ mol}^{-1}\$, so \$\\Delta_\\text{LE}H=-411-377=-788\\ \\mathrm{kJ\\ mol}^{-1}\$. The negative sign shows lattice formation is exothermic."},{"text":"\$-34\\ \\mathrm{kJ\\ mol}^{-1}\$","correct":false,"explanation":"This results from combining the terms incorrectly rather than isolating the lattice enthalpy. The other terms total \$377\$, giving \$-411-377=-788\\ \\mathrm{kJ\\ mol}^{-1}\$."},{"text":"\$788\\ \\mathrm{kJ\\ mol}^{-1}\$","correct":false,"explanation":"The magnitude is correct, but the sign has been reversed. Forming solid NaCl from gaseous ions releases energy, so the lattice enthalpy is \$-788\\ \\mathrm{kJ\\ mol}^{-1}\$."},{"text":"\$-377\\ \\mathrm{kJ\\ mol}^{-1}\$","correct":false,"explanation":"This is merely the sum of the atomisation, ionisation and electron-affinity terms. Subtracting that total from \$-411\$ gives \$-788\\ \\mathrm{kJ\\ mol}^{-1}\$."}]},{"id":"4999caaf-df01-45ff-b226-2feadc7c6a60","cardType":"basic","points":2,"front":"In a MgCl₂ Born-Haber cycle, which chlorine terms must be doubled?","back":"

Chlorine atomisation: \$2\\Delta_\\text{at}H(\\mathrm{Cl})\$
Chlorine electron affinity: \$2\\mathrm{EA}_1(\\mathrm{Cl})\$

","mcqOptions":[{"text":"

Chlorine atomisation: \$2\\Delta_\\text{at}H(\\mathrm{Cl})\$
Chlorine electron affinity: \$2\\mathrm{EA}_1(\\mathrm{Cl})\$

","correct":true,"explanation":"Each formula unit contains two chloride ions, requiring two chlorine atoms and two gained electrons. Both chlorine atomisation and first electron affinity terms are therefore doubled."},{"text":"

Chlorine atomisation: \$\\Delta_\\text{at}H(\\mathrm{Cl})\$
Chlorine electron affinity: \$2\\mathrm{EA}_1(\\mathrm{Cl})\$

","correct":false,"explanation":"Two electron-affinity steps require two gaseous chlorine atoms, not one. Both the chlorine atomisation and electron-affinity terms must be doubled for \$\\mathrm{MgCl}_2\$."},{"text":"

Chlorine atomisation: \$2\\Delta_\\text{at}H(\\mathrm{Cl})\$
Chlorine electron affinity: \$\\mathrm{EA}_1(\\mathrm{Cl})\$

","correct":false,"explanation":"Producing two chlorine atoms is correctly represented, but each atom must gain an electron. The first electron affinity must also be counted twice."},{"text":"

Chlorine atomisation: \$\\frac{1}{2}\\Delta_\\text{at}H(\\mathrm{Cl})\$
Chlorine electron affinity: \$\\frac{1}{2}\\mathrm{EA}_1(\\mathrm{Cl})\$

","correct":false,"explanation":"This confuses coefficients based on chlorine molecules with tabulated atomisation values per mole of atoms. Two chloride ions are formed, so both chlorine terms are doubled."}]},{"id":"11c2f6ce-55a1-4046-9cf4-1e2b276e8d37","cardType":"basic","points":2,"front":"Which magnesium ionisation energies are used in a MgCl₂ Born-Haber cycle?","back":"\$\\mathrm{IE}_1\$ and \$\\mathrm{IE}_2\$.","mcqOptions":[{"text":"\$\\mathrm{IE}_1\$ and \$\\mathrm{IE}_2\$.","correct":true,"explanation":"Magnesium forms \$\\mathrm{Mg}^{2+}\$ by losing two electrons successively. The cycle must therefore include both the first and second ionisation energies."},{"text":"\$\\mathrm{IE}_1\$ and \$\\mathrm{IE}_3\$.","correct":false,"explanation":"Ionisation energies must be applied successively, so the second electron cannot be represented by the third ionisation energy. Magnesium requires \$\\mathrm{IE}_1\$ and \$\\mathrm{IE}_2\$."},{"text":"\$\\mathrm{IE}_2\$ and \$\\mathrm{IE}_3\$.","correct":false,"explanation":"This omits removal of the first electron and includes removal of a third. Forming \$\\mathrm{Mg}^{2+}\$ requires exactly the first and second ionisation energies."},{"text":"\$\\mathrm{IE}_1\$ and \$\\mathrm{EA}_1\$.","correct":false,"explanation":"Electron affinity describes electron gain, whereas magnesium forms its ion by losing two electrons. Both \$\\mathrm{IE}_1\$ and \$\\mathrm{IE}_2\$ are required."}]},{"id":"4ea414d9-3a42-43bc-9af6-390af9e2f0a2","cardType":"cloze","points":null,"front":"MgCl₂ lattice formation: \$\\mathrm{Mg}^{2+}(g)+[ ]\\to[ ]\$.","back":"MgCl₂ lattice formation: \$\\mathrm{Mg}^{2+}(g)+2\\mathrm{Cl}^-(g)\\to\\mathrm{MgCl}_2(s)\$.","mcqOptions":[{"text":"MgCl₂ lattice formation: \$\\mathrm{Mg}^{2+}(g)+2\\mathrm{Cl}^-(g)\\to\\mathrm{MgCl}_2(s)\$.","correct":true,"explanation":"Lattice formation is the formation of one mole of solid ionic compound from its gaseous ions. Two chloride ions are needed to balance the 2+ charge on magnesium."},{"text":"MgCl₂ lattice formation: \$\\mathrm{Mg}^{2+}(g)+\\mathrm{Cl}^-(g)\\to\\mathrm{MgCl}_2(s)\$.","correct":false,"explanation":"The formula requires two chloride ions to balance Mg²⁺ and form neutral MgCl₂. The correct equation uses 2Cl^-(g) before the arrow."},{"text":"MgCl₂ lattice formation: \$\\mathrm{Mg}^{2+}(g)+2\\mathrm{Cl}^-(aq)\\to\\mathrm{MgCl}_2(s)\$.","correct":false,"explanation":"Lattice formation starts with gaseous ions, not aqueous ions. The correct reactant is 2Cl^-(g), because lattice enthalpy involves forming the solid from separated gaseous ions."},{"text":"MgCl₂ lattice formation: \$\\mathrm{Mg}^{2+}(g)+2\\mathrm{Cl}^-(g)\\to\\mathrm{MgCl}_2(aq)\$.","correct":false,"explanation":"An aqueous product would describe ions dissolved in water, not a lattice forming. The correct product is MgCl₂(s), the ionic solid."}]},{"id":"b28f7b6a-3274-4ef8-93c2-803b5b5aed0a","cardType":"cloze","points":null,"front":"\$\\Delta_\\text{sol}H\$ is the enthalpy change when [ ] [ ].","back":"\$\\Delta_\\text{sol}H\$ is the enthalpy change when 1 mol of solute dissolves in water.","mcqOptions":[{"text":"\$\\Delta_\\text{sol}H\$ is the enthalpy change when 1 mol of solute dissolves in water.","correct":true,"explanation":"Solution enthalpy is the enthalpy change when one mole of solute dissolves in water, producing a solution under the specified standard conditions."},{"text":"\$\\Delta_\\text{sol}H\$ is the enthalpy change when 1 mol of solvent dissolves in water.","correct":false,"explanation":"The definition refers to the solute, not the solvent. The correct wording is 1 mol of solute dissolves in water, under the stated standard conditions."},{"text":"\$\\Delta_\\text{sol}H\$ is the enthalpy change when 1 mol of solute evaporates in water.","correct":false,"explanation":"Dissolving means becoming dispersed in the solvent, whereas evaporation changes a liquid into a gas. The correct process is when 1 mol of solute dissolves in water."},{"text":"\$\\Delta_\\text{sol}H\$ is the enthalpy change when 1 mol of solute dissolves in its own liquid.","correct":false,"explanation":"The solvent specified in the definition is water, so using the solute’s own liquid changes the process being described. The correct phrase is dissolves in water."}]},{"id":"34d19fa9-c288-49a3-b0fc-f25ec51ac5d5","cardType":"cloze","points":null,"front":"\$\\Delta_\\text{hyd}H\$ is for [ ] dissolving in water to form [ ].","back":"\$\\Delta_\\text{hyd}H\$ is for gaseous ions dissolving in water to form aqueous ions.","mcqOptions":[{"text":"\$\\Delta_\\text{hyd}H\$ is for gaseous ions dissolving in water to form aqueous ions.","correct":true,"explanation":"Hydration enthalpy is the enthalpy change when one mole of gaseous ions becomes surrounded by water molecules to form aqueous ions."},{"text":"\$\\Delta_\\text{hyd}H\$ is for aqueous ions dissolving in water to form gaseous ions.","correct":false,"explanation":"This reverses hydration: hydration takes gaseous ions into aqueous solution, not aqueous ions into the gas phase. The correct terms are gaseous ions and aqueous ions."},{"text":"\$\\Delta_\\text{hyd}H\$ is for gaseous atoms dissolving in water to form aqueous ions.","correct":false,"explanation":"Hydration enthalpy starts with ions, not neutral atoms. The correct definition describes gaseous ions dissolving in water to form aqueous ions."},{"text":"\$\\Delta_\\text{hyd}H\$ is for gaseous ions dissolving in water to form solid ions.","correct":false,"explanation":"Dissolved ions are surrounded by water molecules and are described as aqueous, not solid. The correct product is aqueous ions."}]},{"id":"8caa28c5-a679-4f37-bd5a-42e04b91d4d9","cardType":"basic","points":2,"front":"Why are hydration enthalpies usually negative?","back":"Ion-water attractions form, releasing energy.","mcqOptions":[{"text":"Ion-water attractions form, releasing energy.","correct":true,"explanation":"Attractive forces form between the ions and polar water molecules during hydration. Forming these interactions releases energy, so hydration enthalpies are usually negative."},{"text":"Ion-ion attractions break, releasing energy.","correct":false,"explanation":"Breaking attractions requires energy rather than releasing it. Hydration is usually exothermic because ion-water attractions form, releasing energy."},{"text":"Water-water attractions form, absorbing energy.","correct":false,"explanation":"Although water molecules reorganise, the key favourable process is formation of ion-water attractions. This releases energy, giving the usually negative hydration enthalpy."},{"text":"Ion-water attractions break, releasing energy.","correct":false,"explanation":"Breaking ion-water attractions would require energy, not release it. The negative value comes from ion-water attractions forming around the gaseous ions."}]},{"id":"4adf9c25-062d-4e07-b720-1e6faaa8881e","cardType":"basic","points":2,"front":"When \$\\Delta_\\text{LE}H\$ is lattice formation, what is the solution enthalpy equation?","back":"\$\\Delta_\\text{sol}H=-\\Delta_\\text{LE}H+\\sum\\Delta_\\text{hyd}H\$","mcqOptions":[{"text":"\$\\Delta_\\text{sol}H=-\\Delta_\\text{LE}H+\\sum\\Delta_\\text{hyd}H\$","correct":true,"explanation":"Dissolution first reverses lattice formation, giving -Δ_LEH, then hydrates all the ions. Adding the hydration enthalpies gives the solution enthalpy."},{"text":"\$\\Delta_\\text{sol}H=\\Delta_\\text{LE}H+\\sum\\Delta_\\text{hyd}H\$","correct":false,"explanation":"This fails to reverse the lattice enthalpy because Δ_LEH here is defined for formation. The correct equation uses -Δ_LEH before adding the hydration terms."},{"text":"\$\\Delta_\\text{sol}H=-\\Delta_\\text{LE}H-\\sum\\Delta_\\text{hyd}H\$","correct":false,"explanation":"Hydration enthalpies are already signed values, usually negative, so they must be added algebraically rather than given another minus sign. The correct expression is -Δ_LEH + ΣΔ_hydH."},{"text":"\$\\Delta_\\text{sol}H=\\Delta_\\text{LE}H-\\sum\\Delta_\\text{hyd}H\$","correct":false,"explanation":"Both signs are reversed in this expression. Lattice formation must be reversed, while the signed hydration enthalpies are added, giving -Δ_LEH + ΣΔ_hydH."}]},{"id":"a37f7b2e-0cc9-4ab8-9bf4-693c9bbd30e6","cardType":"basic","points":2,"front":"Why is there a minus sign before \$\\Delta_\\text{LE}H\$ in the solution enthalpy equation?","back":"Dissolving uses lattice dissociation, the reverse of formation.","mcqOptions":[{"text":"Dissolving uses lattice dissociation, the reverse of formation.","correct":true,"explanation":"The given lattice enthalpy is for forming the lattice from gaseous ions. Dissolving requires breaking that lattice, so the process is reversed and its enthalpy changes sign."},{"text":"Dissolving uses lattice formation, the same as formation.","correct":false,"explanation":"Dissolving breaks the ionic lattice rather than creating it. Therefore it uses lattice dissociation, which is the reverse of lattice formation and requires the opposite sign."},{"text":"Dissolving uses hydration, the reverse of formation.","correct":false,"explanation":"Hydration is a separate step involving gaseous ions entering water. The minus sign specifically arises because dissolving uses lattice dissociation, the reverse of lattice formation."},{"text":"Dissolving uses lattice dissociation, the same as formation.","correct":false,"explanation":"Lattice dissociation is not the same process as lattice formation: it breaks the solid into gaseous ions. Its enthalpy is opposite in sign, which explains the minus sign."}]},{"id":"2230b39a-3e3c-4fd0-a892-7bf969851263","cardType":"basic","points":2,"front":"Using \$\\Delta_\\text{LE}H=-771\$, \$\\Delta_\\text{hyd}H(\\mathrm{Na}^+)=-406\$, \$\\Delta_\\text{hyd}H(\\mathrm{Cl}^-)=-363\$, calculate \$\\Delta_\\text{sol}H\$ for NaCl.","back":"\$+2\\ \\mathrm{kJ\\ mol}^{-1}\$","mcqOptions":[{"text":"\$+2\\ \\mathrm{kJ\\ mol}^{-1}\$","correct":true,"explanation":"Reverse the formation value to obtain +771 kJ mol^-1, then add -406 and -363 kJ mol^-1. The total is +2 kJ mol^-1."},{"text":"\$-2\\ \\mathrm{kJ\\ mol}^{-1}\$","correct":false,"explanation":"The magnitudes nearly cancel, but reversing lattice formation gives +771, not -771. Thus +771 - 406 - 363 = +2 kJ mol^-1."},{"text":"\$+740\\ \\mathrm{kJ\\ mol}^{-1}\$","correct":false,"explanation":"This treats only one hydration enthalpy as negative or omits the chloride contribution. Both hydration values must be added: +771 - 406 - 363 = +2 kJ mol^-1."},{"text":"\$-1540\\ \\mathrm{kJ\\ mol}^{-1}\$","correct":false,"explanation":"This results from adding the lattice formation value with excessive sign reversal and ignoring the correct Hess cycle. Use +771 - 406 - 363 to obtain +2 kJ mol^-1."}]},{"id":"488087e4-fcf0-44f6-b638-e227820b54af","cardType":"basic","points":3,"front":"Why is MgO’s lattice enthalpy more exothermic than NaCl’s?","back":"

Higher charges: \$\\mathrm{Mg}^{2+}\$ and \$\\mathrm{O}^{2-}\$.
Smaller ions: ions get closer together.

","mcqOptions":[{"text":"

Higher charges: \$\\mathrm{Mg}^{2+}\$ and \$\\mathrm{O}^{2-}\$.
Smaller ions: ions get closer together.

","correct":true,"explanation":"MgO contains ions with larger charges than NaCl, increasing electrostatic attraction, and its ions are smaller, reducing their separation. Both effects make MgO’s lattice enthalpy more exothermic."},{"text":"

Lower charges: Mg²⁺ and O^2-.
Smaller ions: ions get closer together.

","correct":false,"explanation":"Mg²⁺ and O^2- have higher, not lower, charges than Na⁺ and Cl^-. Higher charges plus smaller ions explain MgO’s more exothermic lattice enthalpy."},{"text":"

Higher charges: Mg²⁺ and O^2-.
Larger ions: ions get farther apart.

","correct":false,"explanation":"The charge comparison is correct, but Mg²⁺ and O^2- are smaller than the corresponding Na⁺ and Cl^- ions. Smaller separation strengthens attraction."},{"text":"

Lower charges: Mg²⁺ and O^2-.
Larger ions: ions get farther apart.

","correct":false,"explanation":"Both trends are reversed: MgO has higher ionic charges and smaller ions than NaCl. These produce stronger electrostatic attractions and therefore a more exothermic lattice enthalpy."}]},{"id":"e709f930-60b9-4bff-b80b-88feae6e51d2","cardType":"basic","points":2,"front":"Why is \$\\mathrm{Mg}^{2+}\$ hydrated more exothermically than \$\\mathrm{Na}^+\$?","back":"Higher charge density, so stronger ion-water attractions.","mcqOptions":[{"text":"Higher charge density, so stronger ion-water attractions.","correct":true,"explanation":"Mg$^{2+}$ has twice the charge of Na$^+$ and a smaller ionic radius, giving it a higher charge density. Its stronger attraction to polar water molecules releases more energy during hydration."},{"text":"Lower charge density, so weaker ion-water attractions.","correct":false,"explanation":"A lower charge density would weaken attraction to water and make hydration less exothermic. Mg$^{2+}$ is more exothermic because its higher charge density produces stronger ion-water attractions."},{"text":"Larger ionic radius, so stronger ion-water attractions.","correct":false,"explanation":"A larger ionic radius spreads the charge over a greater distance, reducing attraction to water. Mg$^{2+}$ is smaller than Na$^+$, and its smaller size contributes to its higher charge density."},{"text":"More electron shells, so stronger ion-water attractions.","correct":false,"explanation":"The number of electron shells is not the key factor controlling hydration enthalpy here. Mg$^{2+}$ attracts water more strongly because its higher charge density creates stronger ion-water attractions."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Lattice enthalpy","title":"Lattice enthalpy Flashcards","topbar":{"backHref":"/a-level/chemistry/ocr/a/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"a5c06767-5605-55b5-bc3c-16f194540d8c","walled":"$L39"}],"$L3a","$L3b"]