Corrections

• This question is about the reactions of Magnesium and its compounds.

  Magnesium is used in one of the stages in the extraction of Titanium.

  Give an equation for the reaction between Titanium (IV) chloride and Magnesium.(2)

  Equation:

  • 2Mg + TiCl4 -> Ti + 2MgCl2

  Role of Magnesium:

  Reducing agent
• A mixture of Magnesium oxide and Magnesium hydroxide has a mass of 3200mg.

  This mixture is reacted with Carbon dioxide to form Magnesium carbonate and water.

  The mass of water produced is 210mg.

  Mg(OH)2 + CO2 -> MgCO3 + H2O

  MgO + CO2 -> MgCO3

  Calculate the percentage by mass of Magnesium oxide in this mixture. (4)

  • m / Mr = (210 / 1000) / 18 = 0.0117mol
  • mass of Mg(OH)2 = 0.0117 x 58.3 (Mr) = 0.680g
  • mass of MgO = (3200 / 1000) - 0.68 = 2.52g
  • % of MgO = (2.52 / 3.2) x 100 = 78.7%
• A student investigates to determine the Mr of unknown diprotic acid(s), H2A.

  H2A + 2NaOH -> Na2A + 2H2O

  250cm3 of solution(aq) are prepared using 1300mg of H2A.

  25.0cm3 of 0.112moldm-3 aqueous Sodium hydroxide added.

  Sodium hydroxide(aq) is titrated with the acid solution.

  Titration results: R - 27.35, 26.40, 26.75, 26.50

  Use the info to calculate Mr of H2A. (5)

  • Average titre = (26.50 + 26.40) / 2 = 26.45
  • n of NaOH = (25 x 0.112) / 1000 = 2.80 x 10-3mol
  • n in titre = (2.80 x 10-3) / 2 = 1.40 x 10-3mol
  • n in 250cm3 = ((1.40 x 10-3) x 250) / 26.45 = 0.0132mol
  • Mr = (1300 / 1000) / 0.0132 = 98.48
• Before adding the solution from the burette in the rough titration, there was an air bubble below the tap.

  At the end of this titration the air bubble was not there.

  Explain why this air bubble increases the final burette reading of the rough titration. (1)

  • Some solution replaces air bubble.
• Calculate the mass in mg, of propanedioic acid (Mr = 104.0) needed to prepare 250cm3 of a 0.00500moldm-3 solution. (2)

  • moles of acid = 0.00500 x (250 / 1000) = 0.00125
  • mass of acid = 0.00125 x 104.0 = 0.130g x 1000 = 130mg
• The mass of MgO obtained in this experiment is slightly less than that expected from the mass of Mg(NO3)2 used.

  Suggest one practical reason for this. (1)

  • Some of the solid is lost in weighing products.
• Some of the liquid injected did not evaporate because it dripped into the gas syringe nozzle outside the oven.

  Explain how this would affect the value of the Mr of Y calculated from the experimental results. (2)

  • Lower volume recorded.
  • Mr would be greater than the real Mr.
• A sample of pure Mg(NO3)2 was decomposed by heating as shown in the equation below.

  2Mg(NO3)2(s) -> 2MgO(s) + 4NO2(g) + O2(g)

  A 3.74 x 10-2g sample of Mg(NO3)2 was completely decomposed by heating.

  Calculate the total volume, in cm3, of gas produced at 60.0°C and 100kPa.

  Give your answer to the appropriate number of significant figures.

  The gas constant R = 8.31JK-1mol-1. (5)

  • n = (3.74 x 10-2) / 148.3 = 2.522 x 10-4
  • Total moles of gas produced = (5 / 2) x (2.522 x 10-4) = 6.305 x 10-4
  • V = ((6.305 x 10-4) x 8.31 x 333) / 100,000 = 1.745 x 10-5m3
  • 1.745 x 10-5 x 1000 = 17.5cm3
• State the meaning of the term electronegativity. (1)

  • Ability of an atom to attract electron density in a covalent bond.
• State and explain the trend in electronegativity values across Period 3 from Sodium to Chlorine. (3)

  Trend:

  • Increases

  Explanation:

  • Nuclear charge increases.
  • Electrons in same shell.
• What is meant by the term first ionisation energy? (1)

  • Enthalpy for removal of 1 electron from a gaseous atom.
• What evidence from the diagram of first ionisation energy across Period 3 supports that the maximum number of electrons that can be accommodated in an s sub-level? (1)

  • 2 elements (Na and Mg) before the drop in energy to Al.
• What evidence from the diagram of first ionisation energy across Period 3 supports the fact that the 3p sub-level is higher in energy than the 3s? (1)

  • Ionisation energy of Al is less than that for Mg.
• What evidence from the diagram of first ionisation energy across Period 3 supports the fact that no more than three unpaired electrons can be accommodated in the 3p sub-level? (2)

  • Fall in energy from P to S.
  • From Al to P, there are 3 additional electrons.
• Explain why certain elements in the Periodic Table are classified as p-block elements.

  Illustrate your answer with an example of a p-block element and give its electronic configuration. (3)

  • Elements in the p block have their outer electrons in p sub-shells.
  • Al
  • 1s2 2s2 2p6 3s2 3p1
• Explain periodicity applied to properties of rows of elements in the PT. Describe and explain ar, electronegativity and conductivity (Na to Ar). (13)

  • Pattern in the change in the properties of a row of elements repeated in the next row.
  • Atomic radius decreases across the row and nuclear charge increases. More attraction for e- in the same shell.
  • Electronegativity increases across the row and nuclear charge increases but atomic radius decreases. More attraction for shared e-.
  • Conductivity decreases row. Na - Al metals and 2 of Si - Ar non-metals. Delocalised e- free to move in metals.
• State the general trend in the first ionisation energy of the Period 3 elements from Na to Ar. (1)

  • Increases.
• State how, and explain why, the first ionisation energy of Aluminium does not follow this general trend. (3)

  • Lower than Mg.
  • Less energy needed for electron removed from 3p sub-shell.
  • Shielded by 3s electrons.
• State and explain the trend in the melting points of the Period 3 metals, Na, Mg and Al. (3)

  Trend:

  • Increases.

  Explanation:

  • Smaller atomic radius.
  • Stronger attraction between cations and delocalised electrons.
  • Stronger metallic bonding.
• State the trend in atomic radius from Phosphorous to Chlorine and explain the trend. (3)

  Trend:

  • Decreases.

  Explanation:

  • Number of protons increases.
  • Attracting outer electrons in the same shell.
• In terms of structure and bonding, explain why Sulfur has a higher melting point than Phosphorous. (3)

  • Sulfur molecules (S8) are larger than Phosphorous (P4).
  • Therefore, van der Waals' forces between molecules are stronger.
  • Therefore, more energy needed to loosen forces between molecules.
• In terms of atomic structure, explain why the van der Waals' forces in liquid Argon are very weak. (2)

  • Argon particles are single atoms with electrons closer to nucleus.
  • Cannot easily be polarised.
• Which elements are shown in increasing order of the stated property? (1)

  1. Atomic radius: Phosphorous, Sulfur, Chlorine.
  2. First ionisation energy: Sodium, Magnesium, Aluminium.
  3. Electronegativity: Sulfur, Phosphorous, Silicon.
  4. Melting point: Argon, Chlorine, Sulfur.

  • 4
  • The more delocalised electrons present and the smaller the radius of the atom, the higher the melting point of the metal.
  • The metallic bonds increase in strength.
• Which of these elements has the highest second ionisation energy? (1)

  1. Na
  2. Mg
  3. Ne
  4. Ar

  • 1
  • More energy is needed to eliminate 1 electron from the outermost shell as the outer shell is full so it is difficult to take out 1 electron.
• There are many uses for Group 2 metals and their compounds.

  State a medical use of Barium sulfate.

  State why this use of Barium sulfate is safe, given that solutions containing Barium ions are poisonous. (2)

  Use:

  • Used in a Barium meal.

  Why this use is safe?:

  • Barium sulfate is insoluble.
• Magnesium hydroxide is used in antacid preparations to neutralise excess stomach acid.

  Write an equation for the reaction of Magnesium hydroxide with hydrochloric acid. (1)

  • Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
• Magnesium burns with a bright. white light and is used in flares and fireworks.

  Use your knowledge of the reactions of Group 2 metals with water to explain why water should not be used to put out a fire, in which Magnesium metal is burning. (2)

  • Hydrogen produced.
  • Risk of explosion.
• Group 2 elements and their compounds have a wide range of uses.

  From Mg to Ba, the first ionisation energy... (1)

  • Decreases.
• Explain why Calcium has a higher melting point than Strontium. (2)

  • For Ca, delocalised electrons are closer to nucleus.
  • Ca has stronger attraction between the nucleus and the delocalised electrons.
• Acidified Barium chloride solution is used as a reagent to test for Sulfate ions.

  State why Sulfuric acid should not be used to acidify the Barium chloride. (1)

  • Sulfuric acid contains Sulfate ions.
• Write the simplest ionic equation for the reaction that occurs when acidified Barium chloride solution is added to a solution containing Sulfate ions. (1)

  • Ba2+ + SO42- -> BaSO4
• Samples of solid Sodium fluoride, Sodium chloride, Sodium Bromide and Sodium Iodide are each warmed separately with concentrated Sulphuric acid.

  All four compounds react with concentrated Sulphuric acid but only two can reduce it.

  Identify the two halides, which do not reduce concentrated Sulphuric acid.

  Write an equation for the reaction, which does occur with one of these two halides.(3)

  Halides:

  • Fluoride
  • Chloride

  Equation:

  • H+ + F- -> HF
• Identify the two halides, which reduce concentrated Sulphuric acid to Sulphur dioxide.

  Using half-equations for the oxidation and reduction processes, deduce an overall equation for the formation of Sulphur dioxide when concentrated Sulphuric acid reacts with one of these halides. (3)

  • Bromide
  • Iodide
  • H2SO4 + 2H+ + 2e- -> SO2 + 2H2O
  • 2Br- -> Br2 + 2e-
• In addition to Sulphur dioxide, two further reduction products are formed when one of these two halides reacts with concentrated Sulphuric acid.

  Identify the two reduction products and write a half-equation to show the formation of one of them from concentrated Sulphuric acid. (3)

  • Sulphur
  • Hydrogen sulphide
  • H2SO4 + 6H+ + 6e- -> S + 4H2O
• How would you distinguish between separate solutions of Sodium chloride, Sodium bromide and Sodium iodide, suing solutions of Silver nitrate and ammonia. (6)

  Addition of Silver nitrate:

  • Chloride gives white precipitate.
  • Bromide gives cream precipitate.
  • Iodide gives yellow precipitate.

  Addition of Ammonia:

  • Chloride precipitate soluble in dilute.
  • Bromide precipitate soluble in concentrated.
  • Iodide precipitate insoluble.
• Describe and explain the trend in the boiling points of the elements down Group 7 from Fluorine to Iodine. (4)

  • Increases from Fluorine to Iodine.
  • Size of molecules increase
  • Magnitude of van der Waals' forces increase.
  • More energy required to separate molecules.
• Describe what you would observe when aqueous Silver nitrate, followed by dilute aqueous Ammonia, is added to separate aqueous solutions of Sodium chloride and Sodium bromide. (4)

  With NaCl:

  • Cream precipitate.
  • Partially soluble in ammonia.

  With NaBr:

  • Cream precipitate.
  • Partially soluble in ammonia.
• State the trend in the oxidising abilities of the elements down Group 7 from Chlorine to Iodine.

  Explain how this trend can be shown by displacement reactions between halogens and halide ions in aqueous solutions.

  Illustrate your answer with appropriate observations and equations. (7)

  • Oxidising ability decreases from Chlorine to Iodine.
  • Cl2 + 2Br -> 2Cl- + Br2
  • Br2 red brown liquid.
  • Cl2 + 2I- -> 2Cl- + I2
  • I2 brown solution.
  • Br2 + 2I- -> 2Br- + I2
  • Yellow solution goes brown solution.
• The addition of Silver nitrate solution followed by dilute aqueous ammonia can be used as a test to distinguish between Chloride and Bromide ions.

  For each ion, state what you would observe if an aqueous solution containing the ion was tested in this way. (4)

  Observations with Chloride ions:

  • White precipitate.
  • Soluble in ammonia.

  Observations with Bromide ions:

  • Cream precipitate.
  • Partially soluble in ammonia.
• Write an equation for the reaction between Chlorine and cold, dilute aqueous Sodium hydroxide.

  Give two uses of the resulting solution. (3)

  Equation:

  • Cl2 + 2NaOH -> NaCl + NaOCl + H2O

  Uses:

  • Bleach
  • Steriliser