electrostatic forces of attraction / strong ionic bonds M1\n- These forces exist between oppositely charged ions ($F^-$ and $Na^+$) M1\n- A large amount of energy is required to overcome / break these forces/bonds A1","marks":4,"order":3,"markingRubric":null,"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"4e10cea9-3c65-4bab-b1c4-00f0f1e96cce","revisionId":"37e2be75-621f-49b1-b2ca-3530d49bc206","questionNumber":null,"questionType":"LA","specification":"Sodium oxide is an ionic compound.","difficulty":3,"difficultyBand":"medium","totalMarks":9,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"831a2964-e22c-40f6-b0b4-5af92b3506af","label":"a","content":"Complete the table to show the arrangement of electrons in the ions formed when sodium and oxygen react to form sodium oxide.\n\nGive the charge on each of the ions.\n\n| Element | Arrangement of electrons in atom | Arrangement of electrons in ion | Charge on ion |\n| :---: | :---: | :---: | :---: |\n| $\\text{Na}$ | 2.8.1 | | |\n| $\\text{O}$ | 2.6 | | |","markscheme":"- Sodium ion ($\\text{Na}^+$) has electron arrangement $2.8$ B1\n- Sodium ion has charge $1+$ (or $+$) B1\n- Oxide ion ($\\text{O}^{2-}$) has electron arrangement $2.8$ B1\n- Oxide ion has charge $2-$ B1","marks":4,"order":0,"markingRubric":null,"explanation":null,"options":[]},{"id":"1704cf66-6a2f-4372-93d4-c9da227794df","label":"b","content":"Explain why sodium oxide conducts electricity when molten.","markscheme":"- Ions are free to move / mobile (in the molten state) M1\n- (to) carry charge / current A1","marks":2,"order":1,"markingRubric":null,"explanation":null,"options":[]},{"id":"df65bc85-30ad-49af-bbd9-a2add27aa68c","label":"c","content":"Explain why sodium oxide has a high melting point.","markscheme":"- Strong electrostatic forces of attraction M1\n- Between oppositely charged ions (or between $\\text{Na}^+$ and $\\text{O}^{2-}$) M1\n- (Require a) large amount of energy to overcome / break A1","marks":3,"order":2,"markingRubric":null,"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"724f0f82-35f5-4355-83dc-2772dadcab70","revisionId":"c13ce2e3-6890-408c-8601-eb27647a2a01","questionNumber":null,"questionType":"LA","specification":"Calcium and fluorine react to form the ionic compound calcium fluoride.\n\nThe chemical equation for this reaction is:\n\\[\\text{Ca} + \\text{F}_2 \\rightarrow \\text{CaF}_2\\]","difficulty":4,"difficultyBand":"medium","totalMarks":9,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"81edf1a1-fb38-41f6-9f02-5f971e1ea680","label":"a","content":"Write a word equation for this reaction.","markscheme":"- Writes the correct reactants and product: $\\text{calcium} + \\text{fluorine} \\rightarrow \\text{calcium fluoride}$ B1","marks":1,"order":0,"markingRubric":null,"explanation":null,"options":[]},{"id":"707a854d-500a-44b9-b6d7-89938e3d7eef","label":"b","content":"Describe the changes in electronic configurations when calcium reacts with fluorine to form the ionic compound calcium fluoride (\$\\text{CaF}_2\$). Include the electronic configurations of the atoms and the resulting ions, and show the charges on the ions. (Atomic numbers: \$\\text{Ca} = 20\$, \$\\text{F} = 9\$)","markscheme":"- States that the calcium atom has configuration $2.8.8.2$ and loses $2$ electrons M1\n- States that this forms a calcium ion ($\\text{Ca}^{2+}$) with configuration $2.8.8$ A1\n- States that each fluorine atom has configuration $2.7$ and gains $1$ electron M1\n- States that this forms fluoride ions ($\\text{F}^-$) with configuration $2.8$ A1","marks":4,"order":1,"markingRubric":null,"explanation":null,"options":[]},{"id":"173a3544-faed-44b3-adf9-537e86c63a94","label":"c","content":"Calculate the mass of calcium fluoride (\

Ionic bonding

Question 6

Ionic bonding Questions

Ionic bonding

Question 6

Ionic bonding Questions