2024 YISHUN INNOVA JC2 H2 Chemistry Group 2 & 17 Tutorial Solutions PDF

Teacher’s Copy YISHUN INNOVA JUNIOR COLLEGE 2024 JC2 H2 CHEMISTRY GROUP 2 & 17 TUTORIAL Learning Strategies 1. Retrieval Practice Put away your class materials, try to recall the trends and variations in...

Teacher’s Copy YISHUN INNOVA JUNIOR COLLEGE 2024 JC2 H2 CHEMISTRY GROUP 2 & 17 TUTORIAL Learning Strategies 1. Retrieval Practice Put away your class materials, try to recall the trends and variations in physical and chemical properties for elements in Group 2 and Group 17. 2. Spaced Practice Review information from each tutorial class, but not immediately after class. Put aside a little bit of time every day to review and assess your understanding in this topic. Topics like atomic structure and Period 3 are linked to this topic. It is important to also review them to help reinforce key ideas in this topic. 3. Dual Coding Take information you are trying to learn and draw visuals to go along with it. E.g Use equations to represent chemical reactions, use answering frames/visuals to help structure your explanations so that you do not miss out on important points. 4. Elaboration Make sure you are answering to what the question is asking. Check for accuracy and keywords needed when describing and explaining trends and/or accounting for the scenario given in the question stem. You may need to refer to the data booklet and extract relevant information to include in your answers. Compulsory qns Tutorial qns on Group 2: Q1,3-6 Tutorial qns on Group 17: Q2, 7-8 Questions to be reviewed in class Tutorial qns on Group 2: Q4-6 Tutorial qns on Group 17: Q7-8 A Physical Properties of Group 2 and 17 elements [Refer to lecture notes pages 3 - 4, 11 - 13] 1 Revision from Atomic Structure N19/2/1(a) (a) State the electronic configuration of a calcium atom and its ion. Ca 1s2 2s2 2p6 3s2 3p6 4s2 Ca2+ 1s2 2s2 2p6 3s2 3p6 (b) Explain the difference in size between the radius of the calcium ion and the radius of a calcium atom. The radius of Ca2+ ion is smaller than the radius of Ca atom. Both Ca2+ ion and Ca atom have the same nuclear charge since they have the same number of protons. However, Ca2+ ion have one electron shell less than Ca atom and hence experience weaker shielding effect. The effective nuclear charge of Ca2+ ion is higher than the Ca atom and valence electrons are more strongly attracted to the nucleus. (c) Explain the trend in ionic radii down Group 2 Down Group 2, the nuclear charge increases due to an increase in the number of protons, the shielding effect also increases due to the increase in the number of inner electronic shells. the increase in the shielding effect OUTWEIGHS the increase in the nuclear charge resulting in a decrease in the effective nuclear charge. valence electrons are less strongly attracted to the nucleus ionic radius increases (d) Table 1.1 shows the densities, at 298 K, of two Group 2 elements which have the same packing arrangement of atoms. Table 1.1 element density / g cm−3 Ca 1.5 Sr 2.6 Suggest why the density of Sr is greater than that of Ca. density = mass ÷ volume Sr has a larger mass and a larger volume than Ca. However, the larger mass of Sr outweighs its larger volume. Hence the density of Sr is higher than Ca. YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 2 of 11 Examiner’s Comments (a) The electronic configuration of a calcium atom and its ion was well known. (b) Correct descriptions of the difference in electronic structure of the calcium ion compared to its atom, in terms of the loss of the outer shell of electrons, were frequently seen. (c) Correct descriptions of the trend in ionic radii down Group 2 were common; the majority of these responses gave explanations involving the increased number of electron shells or increased shielding effect. Stronger responses went on to explain that the ionic radii increased despite the increase in nuclear charge. (d) Responses frequently detailed the difference in atomic mass and atomic or ionic radii for calcium and strontium without making any reference to the difference in volume. Weak responses referred only to the difference in mass or incorrectly stated that the volume of a calcium atom was larger than that of a strontium atom. 2 Revision from Atomic Structure and Chemical Bonding The alkaline earth metals (beryllium, magnesium, calcium, strontium and barium) all have a fixed oxidation number in their compounds. Explain this observation in terms of structure and bonding. Compounds of Group 2 elements all have giant ionic structure with strong electrostatic forces of attraction between oppositely charged ions. They have two electrons in the valence shell, and they tend to lose both electrons to form cations of +2 charge and form ionic bonds. The 3rd ionisation energy of the Group 2 elements is much higher than the 2nd ionisation energy as the 3rd electron is in an inner quantum shell which is more strongly held by the nucleus. Hence ionic compounds having the group 2 metals cations of +3 charge would not be feasible. YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 3 of 11 3 Revision from Chemical Bonding State the physical states and the colours of chlorine, bromine and iodine at room temperature, and explain the observed trend in their volatilities. chlorine bromine iodine physical state at rtp gas liquid solid colour greenish-yellow reddish-brown black Halogens have simple molecular structure and are non-polar with instantaneous dipole- induced dipole interactions (id-id) between their molecules. From Cl2 to I2, the electron cloud becomes larger and more polarisable. Hence the strength of id-id between molecules increases. Thus, more energy is required to overcome the stronger id-id and boiling point increases. Thus, volatility decreases down the group. (Note: volatility and boiling point has an inverse relationship. i.e. if a substance has a low boiling point, it is highly volatile.) B Reducing Power of Group 2 elements [Refer to lecture notes page: 5 - 6] 4 Revision from Atomic Structure and Electrochemistry (a) Describe and explain the following aspects of the chemistry of Group 2 elements (from Be to Ba) (i) the variation in the first ionisation energy of the elements, Down Group 2, the nuclear charge increases due to an increase in the number of protons, the shielding effect also increases because of the increase in the number inner electronic shells. the increase in the shielding effect OUTWEIGHS the increase in the nuclear charge resulting in a decrease in the effective nuclear charge. valence electrons are less strongly attracted to the nucleus amount of energy needed to remove valence electron decreases and first ionisation energy decreases. YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 4 of 11  (ii) the reducing power of the elements. Mg2+ + 2e− ⇌ Mg Eo = −2.38 V Ca2+ + 2e− ⇌ Ca Eo = −2.87 V Ba2+ + 2e− ⇌ Ba Eo = −2.91 V Down group 2, the Eo(X2+/X) values become more negative. Equilibrium position lies more to the left, hence higher the tendency for oxidation process. Reducing power of the Group 2 elements increases down the group. (b) By using the Data Booklet, predict the approximate magnitude of the standard electrode potential of radium (Ra). Ra is below Ba. Hence it is expected to be a stronger reducing agent than Ba and thus have a Eө value more negative than −2.90 V, ie −2.93 V. C Thermal Stability of Group 2 compounds [Refer to lecture notes page: 7-9] 5. New content in this topic. Thermal stability of group 2 compounds is the most important and most commonly tested concept in this topic. N18/3/1(a) Describe and explain the trend in the thermal stabilities of the Group 2 carbonates. Going down the group, while the charge of the cations is the same, the ionic radii of cations increase. Thus, the charge density and hence the polarising power of the cations decreases. As such, the electron cloud of the large CO32− anion is polarised to a lesser extent and the C−O bonds within the CO32− is weakened to a smaller extent by the M2+. Hence, more energy is required for the decomposition to occur. Thermal stabilities of the Group 2 carbonates increase down the group. Examiner’s Comments Most candidates correctly recalled that stabilities increase down the group due to less polarisation of the C–O bond in the carbonate ion by the Group 2 cation. Most candidates explained that this was due to the decreasing ionic charge density. YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 5 of 11 6 N19/2/1(b) The minimum temperature required to decompose the Group 2 carbonates increases down Group 2, as shown in Table 5.1. Table 5.1 minimum temperature of Group 2 carbonate decomposition/ °C MgCO3 350 CaCO3 832 SrCO3 1340 BaCO3 1450 (a) When solid copper (II) carbonate is heated, it behaves in a similar way to the Group 2 carbonates. Write an equation for the reaction that occurs when copper (II) carbonate is heated. CuCO3(s) → CuO(s) + CO2(g) (b) Use data from the Data Booklet and Table 5.1 to suggest a minimum temperature at which pure copper(II) carbonate will decompose. Explain your reasoning. [Calculations are not required for this question] ionic radii of Cu2+ = 0.073 nm; Mg2+ = 0.065 nm; Ca2+ = 0.099 nm While the charge of Cu2+ is the same as that of Mg2+ and Ca2+, the ionic radii of Cu2+ is bigger than that of Mg2+ but smaller than that of Ca2+. Hence the charge density and polarising power of Cu2+ is in between that of Mg2+ and Ca2+, resulting in a minimum temperature for it to decompose to be between 350 oC and 832 oC. minimum temperature for CuCO3 to decompose is 590 oC. Examiner’s Comments (a) The equation representing the thermal decomposition of copper(II) carbonate was well known. A small number of answers showed oxygen gas present as a reactant and/or product and some equations included copper as a product. (b) There were many strong answers. The majority of answers stated the difference in size of the cations in Group 2 and copper(II) ions. Weaker responses gave explanations based on the difference in atomic mass instead and referred only to the difference in polarising power of the cations. Occasionally, answers incorrectly stated that smaller cations had the least polarising effect. YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 6 of 11 D Oxidising Power of Group 17 elements [Refer to lecture notes page: 15 - 17] 7. Bromine has been obtained from sea salt for a number of years. For simplification, sea salt can be considered to be sodium chloride containing some sodium bromide. An early method of obtaining bromine is outlined in Fig. 6.1. Fig 6.1 (a) (i) What causes the yellow colour produced in A? bromine (ii) Write an equation for the reaction in A. 2Br– + Cl2 → 2Cl– + Br2 (b) Ether is an organic solvent immiscible with water. Explain why the yellow substance dissolves in the ether in B but the salt does not dissolve in ether. Br2 is non-polar, so it dissolves better in a non-polar solvent like ether as it forms instantaneous dipole-induced dipole interactions with ether. However, the ionic salt is insoluble in a non-polar solvent as it cannot form favorable interactions with ether. (c) Given that bromine undergoes disproportionation in the presence of KOH to give BrO − as one of the product, suggest an equation for reaction C. Br2 + 2OH– → BrO– + Br – + H2O (d) Suggest an equation for reaction D, the reaction of acid on the products of reaction C. BrO– + Br- + 2H+ → Br2 + H2O YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 7 of 11  New content in this topic! E Thermal Stability of Group 17 Hydrides [Refer to lecture notes page: 18-19] 8 HF HCl HBr HI o boiling point / C +20 −85 −67 −35 bond energy / kJ mol−1 562 431 366 299 Hfo / kJ mol−1 −269 −92 −36 +26 Using the data given above, describe and explain how the thermal stability of the hydrides varies down Group 17. The decrease in the thermal stability from HCl to HBr to HI is because: the size of the halogen atom, X, increases from Cl to I the effectiveness of orbital overlap between the H and X atom decreases, resulting in weaker covalent bond formed between the H and X atom; lesser amount of energy is required to break the H–X bond as reflected in the decreasing bond energies down the group hence, thermal stability of HX decreases down the group. F Additional Questions [Refer to lecture notes page: 19 − 20] 9 Revision from solubility equilibria Ammonia is a reagent commonly used to react with copper and silver ions. The metal ions can be precipitated from their solutions in some instances. With the aid of relevant equations, explain the following observations as fully as you can. When aqueous ammonia is added to AgCl, the precipitate dissolves. However, if aqueous ammonia is added to AgBr, the precipitate remains insoluble. Ag+(aq) + Cl−(aq) ⇌ AgCl(s) ---- (1) When NH3(aq) is added, it reacts with Ag+ to form [Ag(NH3)2 ]+(aq). Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)2 ]+(aq) The concentration of Ag+(aq) decreases and the equilibrium position (1) shifts left and ionic product [Ag+][Cl−] becomes lower than Ksp (AgCl). As the Ksp of AgBr is lower than Ksp of AgCl, the ionic product [Ag+][Br−] remains higher than Ksp of AgBr upon addition of NH3(aq). Hence, the precipitate remains insoluble. YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 8 of 11 10 N03/3/5 (a) Describe what would you see if chlorine water is added to aqueous sodium bromide. Write an equation for the reaction. Suggest why this reaction is spontaneous. Pale yellow solution turns orange. Cl2 + 2Br− → 2Cl− + Br2 The reaction is spontaneous as Eocell = +1.36 V – (+1.07V) = +0.29 V (>0) (b) When chlorine is bubbled through an aqueous solution containing sodium hydroxide and sodium bromide, bromide ions are oxidised to a bromine-containing oxyanion. Careful addition of aqueous silver nitrate to the resulting solution precipitates 4.31 g of AgCl. When this precipitate is filtered off, and more silver nitrate solution added, 1.18 g of a cream solid is produced. The solid had the following composition by mass: Ag, 45.8 %; Br, 33.9 %; O, 20.3 %. (i) Calculate the formula of the cream solid. Ag Br O % by mass 45.8 33.9 20.3 Ar 107.9 79.9 16.0 moles 0.42447 0.42428 1.2688 simplest ratio 1 1 3 AgBrO3 (ii) Calculate the number of moles of cream solid and silver chloride formed. Amount of AgBrO3 = 1.18  235.8 = 5.00 × 10−3 mol Amount of AgCl = 4.31  143.4 = 3.01 × 10−2 mol (iii) Construct a balanced equation for the reaction between chlorine, sodium bromide and sodium hydroxide. [R]: Cl2 + 2e− → 2Cl− [O]: Br− + 6OH− → BrO3− + 3H2O + 6e− overall: 3Cl2 + Br− + 6OH− → 6Cl− + BrO3− + 3H2O CHEM~IS~TRY YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 9 of 11 Guiding questions Dear students, you may refer to the following guiding questions to scaffold your thinking process in answering questions 1 to 8. 1 (a) Electronic configuration refers to a particular arrangement of electrons in the orbitals of an atom or ion and is determined by the principal quantum shell, subshell and orbital which the electron occupies. (b) In your answer, you should make reference to nuclear charge and shielding effect (and hence, effective nuclear charge) of calcium ion vs calcium atom. (c) Ionic radius depends on how strongly the valence electrons are attracted to the nucleus. Discuss: Effective Nuclear Charge = Nuclear Charge – Shielding Effect (d) Recall: Density = mass / volume Discuss differences in mass and volume of Sr and Ca in your answer. 2 Recall the oxidation state of alkaline earth metals (ie. Group 2 elements) and consider why they do not exhibit other oxidation states where more electrons are removed. 3 What do you understand by the term ‘volatility’? What does it mean when we say ‘a compound is volatile’? What affects volatility? 4 (a) (i) First ionisation energy of an element is the energy required to remove one mole of the outermost electron from one mole of gaseous atoms of the element to form one mole of gaseous ions with a single positive charge. When an element has a higher 1st IE, it requires more energy to remove its valence electron. Recall: What affects 1st IE down the Group? Discuss: Effective Nuclear Charge = Nuclear Charge – Shielding Effect (ii) Recall: How does reducing power changes down Group 2? When an element has a higher reducing power, it also has a higher tendency to undergo oxidation (=lose electrons) to form its +2 ion. Hint: the strength of reducing power is reflected by the increasing negative Eө value down the group. (b) Given the position of Ra in the periodic table, suggest a suitable Eө value based on your understanding of the trend and variation in Eө value down Group 2. 5 (a) To describe and explain the trend in thermal stabilities of the Group 2 carbonates, consider the polarizing power of the Group 2 cations and its impact on the electron cloud of the carbonate ion. Hint: polarizing power depends on the charge density (ie. compare charges and ionic radii) of the cation. 6 (a) For thermal decomposition of metal carbonates, what are the possible products from this reaction? YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 10 of 11  (b) Recall: How does thermal stability of Group 2 carbonates vary down the group? Using your understanding of why thermal stability vary down Group 2, consider the charge density of copper (II) ions vs those of Group 2 cations and suggest an appropriate decomposition temperature. 7 (a) Read the preamble given to find out the ions present in sea water. Which ions (in the sea water) will react with the chlorine? Use the clue in the flowchart (eg. colour of product formed) to determine the reaction that has taken place and to confirm your answer. (b) Recall that ether is a colourless non-polar organic solvent. Bearing in mind the product formed from Reaction 1, what do you think has occurred in Reaction 2? Since the colour formed is yellow, this means that the product from Reaction 1 has dissolved in the ether. What has dissolved in the ether? Recall on what determines the solubility of compounds in the topic of Chemical Bonding. (c) What do you understand by the term ‘disproportionation’? What reaction has taken place when Br2 becomes BrO− when it reacts with OH-? (Has Br2 been oxidized or reduced?) Subsequently what other reaction does Br2 undergo in this disproportionation reaction? (d) Given that the final product formed is bromine. Use the products formed from (c) to form half-equations in acidic medium. Subsequently get the overall equation. 8 What do you understand by ‘thermal stability’? What happens during this process? What bonds are being broken? What factors affect the strength of bonds? YIJC / JC2 H2 Chemistry / 2024 / Teacher’s Copy Pg 11 of 11