Chemistry AQA A-level (PPQ)

• Deduce which of Na+ and Mg2+ is the smaller ion. Explain your Answer:

  
  

  Smaller Ion: More protons. Same Shielding Effect.
• In terms of Structure and bonding, explain why the boiling point of Bromine is different from that of Magnesium. Suggest why Magnesium is liquid over a much greater temperature range compared to Bromine (5m)

  • Bromine is Simple Molecular
  • Magnesium contains metallic bonding - requires a lot of energy to break
  • Bromine contains Van Der Wall forces between molecules which are weak and easy to break.
  • Magnesium has a much greater liquid range because of the forces of attraction in the liquid.
• State which of the elements, Magnesium and Aluminum, has the lower first ionization energy

  • Aluminium
  • Outer electron is a 3P subshell
  • Further away from the nucleus -> less energy required to remove electron.
• Give one reason why the bond enthalpy you calculated is different from the mean bond enthalpy quoted in a data book:

  The mean bond enthalpy in data book is an average over several compounds.
• Magnesium metal displaces Zinc from aqueous Zinc Sulfate. Suggest how method and the analysis of results could be improved for more accurate values of enthalpy. Justify. (6m)

  • Use a polystyrene cup and lid -> to reduce heat loss
  • Measure temp at 1 minute intervals for 3 minutes before adding Magnesium -> establishes an accurate initial temperature
  • Record the temperature increase & fall in several increments (eg. ever min)
  • Plot a graph, Time in X axis, Temp on Y axis
  • Extrapolate the graph -> more accurate maximum temperature or temp change calculated
• Give the meaning of the term Electronegativity:

  The tendency of an atom to attract a bonding pair of electrons in a covalent bond.
• Explain how permanent dipole-dipole forces arise between Hydrogen Chloride molecules:

  • Difference in electronegativity leads to bond polarity.
  • Molecule is a permanent dipole - Attraction between Delta positive and Delta negative on opposite molecule.
• Suggest two reasons why the Sulfur Dioxide by-product is removed from exhaust gas:

  • To prevent acid rain
  • Can cause breathing problems
• Identify the Period 2 element, from Carbon to Fluorine, that has the largest atomic radius. Explain your answer:

  • Carbon
  • Contains fewer protons (than Nitrogen onwards).
  • Weaker attraction between valence electrons and nucleus.
  • Outer shell would not be pulled closer to the nucleus like Nitrogen, so Atomic Radius is slightly larger.
• State the general trend in First ionization energy from Carbon to Neon. Deduce the element that deviates from this trend and explain why this element deviates from this trend:

  • First ionization energies increase.
  • Oxygen deviates.
  • Oxygen would have the first paired orbital with 2 electrons.
  • Spin-pair repulsion = less energy required to remove electron.
• Explain why the Second Ionization energy of Carbon is higher than the first ionization energy:

  • Outer shell would be more attracted to positive nucleus.
  • More energy needed to overcome attraction.
• Explain why the first Ionization Energy of Sulfur is less than the first ionization energy of Phosphorus:

  • In Sulfur the outer electron in the 3d orbital begin to pair
  • Spin-pair repulsion makes it easier to remove an electron.
• State the trend in atomic radius from Phosphorus to Chlorine and explain the trend:

  • Atomic Radius decreases.
  • Nuclear charge increases.
  • Greater force of attraction between nucleus and outer shell.
  • Outer shell is drawn closer the nucleus.
• In terms of bonding, explain why Sulfur has a higher melting point than Phosphorus:

  • Sulfur molecules are much larger than Phosphorus.
  • Van Der Wall forces (in Sulfur) are stronger.
  • More energy required to loosen VDW forces between molecules.
• In terms of atomic structure, explain why the Van Der Waals' forces in liquid argon are very weak:

  • Argon particles are single atoms with electrons closer to the nucleus.
  • Cannot easily be polarized.
  • Small Molecule.
• Develop a procedure to prepare a sample of Silver bromide from the mixture of Sodium Chloride and Sodium Bromide using aqueous Silver Nitrate. Explain procedure:

  • Add silver nitrate (to solution)
  • Forms precipitate of AgCl and AgBr
  • Add excess dilute Ammonia to mixture of precipitate.
  • AgCl precipitate dissolves in Ammonia.
  • Filter off remaining AgBr precipitate.
  • Wash to remove soluble compounds and dry
• Explain why the melting point of Calcium Sulphate is so high:

  • Ionically bonded (between oppositely charged ions)
  • More energy needed to overcome strong electrostatic forces of attraction.
• According to VSEPR theory, repulsion between electron pairs in a valence shell decreases in the order?

  lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
• What statement is correct about two elements whose atoms form a covalent bond with eachother?

  The elements are non-metals
• What intermolecular forces are present in gaseous hydrogen?

  Van Der Waal forces
• What is the correct description of Metallic bonding?

  Positively charged metal ions are attracted to the sea of delocalized electrons = strong electrostatic forces of attraction between opposite charges.
• Sodium and Sodium Iodide can both conduct electricity when molten, but only Sodium can conduct electricity when solid. Explain this difference in conductivity in terms of the structures of Sodium and Sodium Iodide:

  
  

  • Sodium Iodide involves Ionic bonding
  • When solid, the Sodium and Iodine ions cannot move freely so it cannot carry an electrical charge. Ions can only move freely when molten.
  • Sodium itself involves metallic bonding.
  • The metal ions are within a "sea" of delocalized electrons which can more around the solid and can carry an electrical charge.
• Explain why an Ammonia molecule may be polar?

  The molecule is asymmetrical.
• Diamond and Graphite are allotropes of the element Carbon.

  State the structures of and the bonding in Diamond and Graphite:

  • Diamond: Tetrahedral shape of Carbon atoms. Only has covalent bonds.
  • Graphite: Trigonal Planar shape of Carbon atoms. Forms 3 Covalent bonds and contains Van Der Waal forces between layers.
• Compare and explain the hardness and electrical conductivity of Diamond and Graphite (4m):

  • Diamond is hard - all Carbon atoms are covalently bonded together.
  • Diamond does not conduct electricity as all electrons are localised and involved in bonding.
  • Graphite is soft -> arranged in planes with weak bonds. Planes can slide over each other.
  • Graphite can conduct electricity as Carbon only forms 3 covalent bonds -> delocalized electrons are free to carry an electrical charge.
• Explain how lone pairs of electrons on an Oxygen atom can influence the bond angle in Oxygen diflouride:

  • Lone pairs experience the most / maximum repulsion
  • Lone pairs want to be as far away from each other as possible -> takes up more space.
  • The bond angle of bonding pairs is therefore reduced by 2.5 degrees.
• Deduce the intermolecular forces in SiF4. Explain how this type of intermolecular force arises and why no other type of intermolecular force exists in a sample of SiF4:

  • Van Der Waals
  • Temporary induced dipoles are created from the random movement of electrons.
  • SiF4 is symmetrical. Bonding pairs cancel out so there cannot be a permanent dipole.
• State the meaning of the term First Ionization Energy:

  The amount of energy required to remove electrons from 1 mol of gaseous atoms

  to form 1 mol of gaseous ions.
• State how the element Sulfur deviates from the general trend in First Ionisation energy across period 3. Explain your answer:

  • Ionization energy is lower than Phosphorus.
  • Sulfur 3p sub shell has 4 electrons and the first paired orbital.
  • Spin-pair repulsion.
  • Makes it easier to remove electron.
• Explain why the Second Ionization Energy of Aluminum is greater than the first ionization energy of Aluminum:

  • Electrons are being removed from a positive ion.
• Iodine and Diamond are both crystalline solids at Room Temp. Identify one similarity in bonding and one different in structure. Explain why these two solids have very different melting points:

  • SIMILARITY: Both contain strong covalent bonds
  • DIFFERENCE: Iodine is simple molecular whereas Diamond is macromolecular.
  • Iodine has a lower melting point as it contains weak Van Der Waal forces which require little energy to break.
  • Whereas in Diamond the covalent bonds between the Carbon have to be broken - requires a lot of energy.
• Define the term electronegativity:

  • The tendency of an atom to attract a bonding pair of electrons in a covalent bond.
• State and explain the trend in melting point of Group 2 elements Ca-Ba:

  Trend: Decreases

  Explanation: Greater atomic radius and shielding reduces electrostatic attraction between positive metal ions and "sea" of delocalized electrons.
• Name three types of intermolecular forces:

  • Van Der Waal
  • Permanent Dipole-Dipole
  • Hydrogen Bonds
• Explain in terms of intermolecular forces present in each compound why HF has a higher boiling point than HCL:

  • HF contains hydrogen Bonds.
  • HCL is made of permanent dipole-dipole bonds.
  • Hydrogen bonds are stronger.
  • More energy is required to loosen the Hydrogen bonds between the HF molecules.
• State the meaning of Enthalpy Change:

  • Heat energy change in a reaction under constant pressure.
• A data book value for the H-H bond enthalpy is 436 kJ mol-1. Suggest one reason why this value may be different to an experimental value calculated for enthalpy change:

  • Mean bond enthalpies are not the same as actual bond enthalpies as it is an average over several compounds.
• Define the term standard Enthalpy of Combustion:

  • Enthalpy change when 1 mol of a substance in burned excess Oxygen under standard conditions
  • 100 KPa & 298K
• Define the term Standard Enthalpy of Formation:

  • Enthalpy change when 1 mol of a substance is formed from its elements.
  • In their standard states.
  • Under standard conditions (100KPa & 298K)
• Define Hess's Law:

  • Enthalpy change is independent of the reaction route / pathway taken.