GR11 Support pack for Matter and materials

Study & Master Support Pack | Grade 11 CAPS Physical Sciences Matter and materials This support pack for the Matter and materials strand of the Physical Sciences Grade 11 CAPS curriculum provides revision summaries on the topic to help prepare for the examinations. Learners can work through these individually at home or these could form the basis of a catch-up class or online lesson. You have permission to print or photocopy this document or distribute it electronically via email or WhatsApp. Cambridge University Press Africa is a proudly South African publisher. For more information on our Study & Master CAPS-approved textbooks and valuable resource materials, visit www.cambridge.org Brighter thinking I Better learning www.cambridge.org

Unit 1: Atomic combinations: molecular structure Summary • A chemical bond in a molecule is the net electrostatic force that two atoms that share electrons exert on each other. • Core electrons are electrons in the inner energy levels. • Valence electrons are electrons in the highest occupied energy level that can interact to form chemical bonds. • Valency is the number of electrons that an atom can gain, lose or share to reach noble gas configuration. • Bond length refers to the distance between the two nuclei of the bonding atoms. • Bond energy is the amount of energy released when a bond forms. • Lewis diagrams are used to show valence electrons during chemical bonding. • Lone pairs are paired non-bonding valence electrons. • A coordinate (dative) covalent bond forms when one atom in a molecule shares a lone pair of electrons with another atom. • A valence-shell electron-pair repulsion (VSEPR) model is used to predict the shape of molecules, e.g. linear, trigonal planar, tetrahedral, trigonal bipyramidal or octahedral. • Multiple covalent bonds can occur, e.g. in O2, CO2 and N2 • Electronegativity is the ability of an atom in a molecule to attract the bonding electron pair(s). • Both the shape of a molecule and the polarity of the bonds determine whether the molecule will be polar or non-polar. MATTER AND MATERIALS 11

Revision exercises: Unit 1 – Atomic combinations: molecular structure 1. Give one term for the following descriptions. c) Calculate the electronegativity difference for each pair of atoms below and predict a) The distance between the centres of two whether the bond between them will be bonded atoms non-polar covalent, slightly polar covalent, b) The structure of an atom that has eight polar covalent or ionic. electrons in its highest energy level i) Carbon and iodine c) An explanation for a scientific ii) Potassium and fluorine phenomenon, based on observation, iii) Hydrogen and bromine experimentation and reasoning, that d) Study the given molecule below. is widely accepted in the scientific community as being true d) A covalent bond, where both the bonding H O pairs are from the same atom H C C e) Electrons in the outer energy level of O H an atom H Classify each bond as non-polar covalent, 2. The orbital diagrams of three atoms, X, Y and slightly polar covalent, polar covalent Z, are given below. or ionic. 3p 4. The following diagram shows the 3s arrangement of electrons in a hydronium ion. Increasing energy 2p 2p 2p + 2s 2s 2s HOH H 1s 1s 1s Atom X Atom Y Atom Z a) Redraw the diagram and indicate the following: i) a coordinate covalent or dative a) Draw Lewis diagrams for atoms X, Y covalent bond and Z. ii) a bonding pair b) Draw Lewis diagrams for the molecule iii) a lone pair. that forms between: b) Use the diagram to explain how the i) two atoms of Y dative covalent bond between oxygen and ii) a diatomic molecule with formula X2 hydrogen is formed. iii) atom X and Y. c) What is the shape of a hydronium ion? c) Why is a molecule with formula Z2 d) Use the VSEPR theory to explain the not possible? shape of a hydronium ion. 3. Electronegativity is useful in determining the nature of a bond. a) Explain the term ‘electronegativity’. b) Distinguish between polar covalent and non-polar covalent bonds. 12 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Unit 1: Answers 1. a) Bond length 4. a) b) Octet structure iii) i) c) Scientific theory ii) d) Dative / coordinate covalent bond e) Valence electrons b) One of the non-bonding electron pairs on the oxygen atom is used to form a covalent 2. a) bond between the hydrogen ion H+ (which X Y Z has no electrons) and the oxygen atom. Atom X Atom Y Atom Z Both bonding electrons in the covalent bond come from the oxygen atom. b) i) ii) iii) c) Trigonal pyramidal YY XX YXY d) In a hydronium ion, the oxygen atom XX Y is surrounded by four pairs of electrons YY (tetrahedral arrangement), but only three YXY of the pairs are used in bonding giving the Y molecule a trigonal pyramidal shape. c) Z has a full outer shell, with eight electrons, making it chemically stable. 3. a) Electronegativity (EN) is the ability of an atom in a molecule to attract the bonding electron pair(s). b) Polar covalent bonds: Bonded electrons are shared unequally between the bonding atoms because the electronegativity difference between participating atoms is greater than zero but smaller than 2,1. Non-polar covalent bond: Electron pairs are shared equally between the participating atoms because the electronegativity difference between participating atoms is close to zero. c) i) DEN = 2,5 − 2,5 = 0 Non-polar covalent ii) DEN = 4,0 − 0,8 = 3,2 Ionic iii) DEN = 2,8 − 2,1 = 0,7 Slightly polar covalent d) C–H: DEN = 2,5 − 2,1 = 0,4 Slightly polar covalent C−O: DEN = 3,5 − 2,5 = 1,0 Polar covalent C–C: DEN = 2,5 − 2,5 = 0 Non-polar covalent O−H: DEN = 3,5 − 2,1 = 1,4 Polar covalent MATTER AND MATERIALS 13

Unit 2: Intermolecular forces Summary • Interatomic forces are strong forces between atoms or ions to form a chemical bond. • Intermolecular forces are forces between ions, atoms or molecules. They are much weaker than interatomic forces. – Forces between ions and atoms or molecules: Ion-dipole force and ion- induced dipole force. – Van der Waals forces: Dipole-dipole force, dipole-induced dipole force and induced dipole force. • Hydrogen bonds form between an H atom on one molecule and a small highly electronegative atom (F, O or N) on another molecule. • Intermolecular forces hold atoms, ions or molecules together in the solid and liquid phase, and their strength determines the density of a liquid or solid. • The size, shape and polarity of molecules influence the strength of the intermolecular forces between them. • Intermolecular forces determine the melting and boiling points. The stronger the forces, the higher the melting and boiling points. • Water: – Polar molecule with O side slightly negative and H sides slightly positive. – Changes phase from solid (ice) to liquid (water) at 0 °C. – Highest density at 4 °C. – Changes phase from liquid to gas (steam) at 100 °C. – High heat capacity moderates Earth’s climate. – High heat of vapourisation results in a high boiling point in comparison to similarly sized molecules. 14 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Revision exercises: Unit 2 – Intermolecular forces 1. Give one term for the following descriptions. 4. Some physical properties of three compounds are given in the table. a) The type of intermolecular force that exists between all molecules Compound Boiling Solubility Density b) The strongest type of intermolecular force point in 100 g of at 20 °C c) The type of intermolecular force between (°C) water at (g·cm−3) permanent polar molecules 20 °C (g) d) The tendency of matter to increase in Ethanol (C2H5OH) 78,4 Infinitely 0,79 volume in response to an increase in soluble temperature Carbon tetrachloride (CCl4) 76,8 0,081 1,60 e) The intermolecular force responsible for Hydrogen chloride (HCl) −84,9 72 1,50 the high heat of evaporation of water a) Convert the boiling point of hydrogen 2. What dominant intermolecular forces must be chloride to Kelvin. overcome when: b) In what phase is hydrogen chloride at 25 °C? a) liquid CS2 turns to vapour c) Explain in terms of molecular polarity why b) ice melts hydrogen chloride is more soluble in water c) liquid Br2 turns to vapour than carbon tetrachloride. d) CuSO4·5H2O is heated and water of d) Explain in terms of intermolecular forces crystallisation is removed. why ethanol has a higher boiling point than hydrogen chloride. 3. A liquid molecular substance is cooled until it e) Ethanol is used in thermometers. Explain solidifies as shown in the diagrams below. why ethanol rises in the narrow column when heated. f) The density of carbon tetrachloride at two different temperatures is given: temperature Y: 1,53 g·cm−3 temperature X: 1,81 g·cm−3 liquid solid Which temperature (X or Y) of CCl4 can possibly be: a) Use the kinetic model of matter to explain i) 30 °C? how the following properties normally ii) –80 °C? change when a substance changes from liquid to solid: i) the movement of the molecules ii) intermolecular forces iii) total volume iv) density of the substance. b) Why can’t this substance be water? MATTER AND MATERIALS 15

Unit 2: Answers 1. a) London forces / dispersion forces / Strong intermolecular forces require induced dipole forces more energy to break and result in higher b) Hydrogen bond boiling points. c) Dipole–dipole force e) As the temperature of ethanol in a d) Thermal expansion thermometer increases, the kinetic e) Hydrogen bond energy of the molecules increases. The molecules move further apart (stretching 2. a) Van der Waals (dispersion) forces (CS2 the bonds between them), making them are linear, non-polar molecules); due to occupy more space. The process is called its linear, and thus symmetrical, shape the thermal expansion. molecules are non-polar f) i) X b) Hydrogen bonds ii) Y c) Van der Waals (dispersion) forces (Br2 are linear, non-polar molecules) d) Ion–dipole (polar water molecules are attracted to the Cu2+ and SO42 −-ions) 3. a) i) The molecules move around more slowly until they only vibrate in one position (form a solid). ii) The intermolecular forces become stronger. iii) The intermolecular forces pull the molecules closer together, the substance contracts and the total volume decreases. iv) Solids are more dense than liquids (a solid will sink in liquid). b) When water freezes, the volume increases and it expands. The density of water therefore decreases when it freezes. 4. a) 188,25 K b) Gaseous phase c) Hydrogen chloride and water both consist of polar molecules (dissolve). Carbon tetrachloride is a non-polar molecule and water is polar (will not dissolve). d) Ethanol consists of polar molecules attracted to each other by strong hydrogen bonds (in addition to Van der Waals forces). Hydrogen chloride consists of polar molecules with weaker dipole–dipole interaction (no hydrogen bonds). 16 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Unit 3: Ideal gases and thermal properties Summary • Properties of an ideal gas: – The particles of a gas are identical and in constant motion. – Gases fill the whole space and their particles are so small that their volume is negligible. – Gases increase their volume when heated. – There are no intermolecular forces between gas particles. – The average speed of gas particles is determined by the temperature of the gas. – Gas particles exert pressure when they collide with the walls of a container. Collisions are elastic with no energy loss. – The pressure increases when the gas in a closed container is heated. – The large open spaces between gas particles allow compressibility of gases. – The diffusion rate in gases is high. • Boyle’s Law: The volume of a fixed amount of gas is inversely proportional to 1 the pressure on the gas if the temperature remains constant (V ∝ ) P • Charles’ Law: The volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas (V ∝ T) • Gay-Lussac’s Law: The pressure of a fixed amount of gas is directly proportional to the absolute temperature (P ∝ T) • Avogadro’s Law: At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present (V ∝ n) pV pV • General gas equation: 1 1 = 2 2 T1 T2 • Ideal gas equation: PV = nRT • Absolute zero: 0 K = –273,15 °C. • Real gases deviate from ideal gas behaviour at high pressure and low temperature because the gas particles have volume and exert intermolecular forces on each other. • The gases closest to ideal gases are H2 and He. • STP: standard temperature = 273 K or 0 °C; standard pressure = 1,013 × 105 Pa or 101,3 kPa. MATTER AND MATERIALS 17

Revision exercises: Unit 3 – Ideal gases and thermal properties 1. Give one term for the following descriptions. d) On microscopic level, the amount of pressure that the gas particles exert on a) The force applied to a unit area of surface the container wall depends mainly on two b) A gas that obeys all gas laws under any factors. Name them. condition of temperature and pressure c) This theory explains the behaviour of gases based upon a set of assumptions that 3. The following questions are relatively simple are much simpler than what actually may applications of the gas laws. When answering be taking place on microscopic level the questions write down the equation, d) The gas law that describes the relationship substitute values and determine the answer. between pressure and volume a) A sample of carbon dioxide gas at e) The physical quantity that directly 100 °C and 150 kPa occupies a volume of determines the average kinetic energy of a 300 dm3. If the volume is increased to gas sample 350 dm3 at 100 °C, what will the gas pressure be? 2. Study the presentation of gas particles in a closed b) The initial pressure in a balloon filled container and answer the questions that follow. with an ideal gas is P. The temperature is now doubled, the volume tripled, while the number of particles remains constant. What is the final pressure in the balloon? c) Nitrous oxide (N2O), or laughing gas, is best known for its use as an anaesthetic in dental surgery. A sample of this gas occupies a volume of 0,025 m3 at 22 °C and 123,55 kPa. How many moles of N2O gas are in the sample? d) A balloon filled with an ideal gas has a volume of 1,22 dm3 at sea level where the pressure is 1 atm and the temperature is a) Write down one fact from the kinetic 32 °C. If the balloon is taken to the top of theory of gases that explains each the a high mountain where the pressure is following observations made from the 0,85 atm, the volume decreases to diagram above. 1,12 dm3. What is the temperature in °C i) An arrow points from each particle. on the mountain? ii) The arrows have different lengths. e) The temperature of a gas sample is iii) The particles do not stick to each other decreased from 200 °C to 100 °C. If the but spread out in the container. initial pressure is 160 kPa and the volume iv) Particles constantly change direction of the container is fixed, what will the new and speed. pressure be at the lower temperature? b) Provide one sentence to compare the total f) The volume of an enclosed mass of gas at volume of the container to the volume 300 kPa is V. If the pressure is decreased occupied by the gas particles themselves. by 100 kPa at constant temperature, what c) Use the kinetic theory to explain why the will the new volume of the gas be? gas particles exert pressure on the walls of the container. 18 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

4. A 500 cm3 container with 8 g of neon gas has a total mass of 507 g. An identical container is filled with argon gas and the gas is subjected to the same conditions of pressure and temperature. a) State Avogadro’s principle for gases. b) Use this principle to make an important assumption about the neon and argon gas in the two identical containers. c) How many moles of neon are in the container? d) Find the total mass of the container filled with argon gas. PV 5. The graph gives the relationship between nRT against pressure (P) at various temperatures for the gas nitrogen. 2,0 173 K 298 K 873 K 1,5 pV/nRT ideal gas 1,0 0,5 0 200 400 600 800 1 000 Pressure (atm) PV a) What is the value of nRT for an ideal gas on the graph? Explain the reasoning behind your answer. b) Under normal conditions of temperature and pressure, nitrogen is treated as if it is an ideal gas. Use the graphs to explain why this is allowed. c) Consider the graph at 173 K. Explain the deviation from ideal gas behaviour at pressures: i) between 0 atm and 200 atm ii) higher than 300 atm. d) What conclusion can be made from the graphs in terms of temperature and ideal gas behaviour for real gases? MATTER AND MATERIALS 19

Unit 3: Answers P P 1. a) Pressure e) T 1 5 T 2 1 2 b) Ideal gas 160 kPa P2 c) Kinetic theory of gases 473 K 5 373 K d) Boyle’s Law ⇒ P2 5 126,17 kPa e) Temperature f) P1V1 5 P2V2 300 kPa 3 V 5 200 kPa 3 xV 2. a) i) Particles are in constant motion and 1 1 move in straight lines. (They have high x 5 1 2 , therefore 1 2 V kinetic energy.) 4. a) At constant pressure and temperature, the ii) Particles move at different speeds. / volume of a gas is directly proportional to They have different kinetic energies. the number of moles of the gas present. iii) Particles have high kinetic energies. / b) Because the temperature and pressure Weak intermolecular forces exist conditions in the containers are the same between particles. for the two gases, equal volumes (in this iv) Particles collide frequently with case the same container) hold the same each other and with the walls of the number of moles of neon and argon. container. m 8g b) The volume of the particles is small c) n 5 M 5 20 g·mol−1 5 0,4 mol compared to the volume of the container. d) The mass of the empty container 5 507 − 8 c) The particles are in constant motion and 5 499 g. collide against the walls of the container, The container has room for 0,4 moles of exerting a force on the inside area of the argon gas. (Avogadro’s Principle) force container wall. Pressure 5 area mAr 5 nM 5 0,4 mol 3 40 g·mol−1 5 16 g d) The number of collisions per second So, when filled with argon, the container The intensity of the collisions will weigh 499 1 16 5 515 g 3. a) P1V1 5 P2V2 5. a) 1 mol. For ideal gases: PV 5 nRT 150 kPa 3 300 dm3 5 P2 3 350 dm3 PV Therefore nRT should give an answer of 1. ⇒ P2 5 128,57 kPa b) At room temperature (298 K) and pressures PV PV b) T1 1 5 T2 2 around 100 atm (which can be seen as normal 1 2 conditions) the graph line is very close to PV xP3V that of an ideal gas. Under these conditions T 5 2T 2 2 nitrogen obeys the ideal gas equation. x 5 3 , therefore 3 P c) i) Increasing the pressure from 0 atm c) PV 5 nRT brings the molecules closer together. PV Intermolecular attraction forces will ⇒ n 5 RT become greater. Intermolecular forces 123,55 3 103 Pa 3 0,025 m3 become significant enough to affect the 5 8,31 J·K−1·mol−1 3 295 K motion of the molecules. The impact 5 1,26 mol of the molecules striking the wall of PV PV d) T1 1 5 T2 2 the container will be less, therefore 1 2 decreasing the pressure. Substituting a 1 atm 3 1,22 dm3 0,85 atm 3 1,12 dm3 PV 305 K 5 T2 lower pressure into the nRT term will give a lower value than expected. T2 5 238 K 5 −35 °C 20 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

ii) According to the kinetic theory, ideal gas molecules have negligible volume compared with the volume of the container. The molecules of a real gas, however, have small, but measurable, volumes. As the pressure of the gas increases, the intermolecular distances decrease. As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. The total volume occupied by the gas is now greater than the volume predicted by the ideal gas law. Substituting a larger volume PV into the formula nRT will result in a larger value than predicted by the ideal gas law. d) Real gases approach ideal gas behaviour more closely at higher temperatures. MATTER AND MATERIALS 21