2023 ASRJC H2 Chem Prelim P3 MS

ASRJC JC2 PRELIM 2023 9729/03/H2 [Turn over Anderson Serangoon Junior College 2023 JC2 H2 Chemistry Preliminary Examination Paper 3 Suggested Solutions 1 (a) (i) Describe and explain the relative basicities of methylamine, dimethylamine and trimethylamine in the gas phase. [3] Order of increasing basicity in the gas phase: methylamine < dimethylamine < trimethylamine The increasing number of electron-donating methyl / alkyl group from methylamine to dimethylamine to trimethylamine increases the availability of the lone pair of electrons on N atom to accept a proton. [1]: trend [1]: more electron donating alkyl group [1]: greater availability of lone pair (ii) Explain why amides are neutral. [1] Amides are neutral because the lone pair on nitrogen atom is delocalised into the  bond of the adjacent C=O by resonance and hence not available for donation to a proton. [1] (b) Deuterium (symbol D or 2 1H ) was discovered in 1931. Deuterium accounts for 0.0156% of all the naturally occurring hydrogen in the oceans, while the most common isotope 1 1H accounts for 99.98%. Tritium (symbol T or 3 1H ), a rare and radioactive isotope of hydrogen account for only 0.0044%. Chemically, deuterium behaves similarly to ordinary hydrogen. (i) Calculate the average Ar of hydrogen. Give your answer to four decimal places. [1] Ar = 0.0156 99.98 0.0044( 2) ( 1) ( 3)100 100 100 +  +  = 0.000312 + 0.9998 + 0.000132 = 1.000244 = 1.0002 (4 decimal places) [1] (ii) On the same diagram, sketch how a beam of singly positively charged deuterium ions and a beam of hydrogen ions will behave in an electric field. In your diagram, indicate clearly the relative angle of deflection for each beam. (You may let the angle of deflection of hydrogen ions be x°) [2] Diagram should show charged deuterium ion deflecting less than hydrogen ion. Deuterium ion and hydrogen ion must deflect in the same direction and towards negative plate. Source xo 1 1H+ + ‒ x/2o 2 1H+

2 ASRJC JC2 PRELIM 2023 9729/03/H2 Above two points [1] Since angle of deflection  e/m ratio:  of Deuterium ion is half of  of hydrogen ion [1] (iii) Explain the difference in the thermal stability of DCl, DBr and DI. [2] Down Group 17 from chlorine to bromine to iodine, the size of valence orbitals increases and become more diffused. This causes a decrease in effectiveness of orbital overlap between the valence orbital of halogen and s–orbital of hydrogen. / Electronegativity difference between the halogen and deuterium decreases, resulting in a decrease in bond polarity. [1] D–X bond becomes weaker and hence the D–X bond energy decreases from DCl to DBr to DI. So, thermal stability decreases from DCl to DBr to DI. [1] (c) Deuterium can replace the normal hydrogen in water molecules to form heavy water, D2O. Some data of light water and heavy water are given in Table 1.1. Table 1.1 Property D2O (Heavy water) H2O (Light water) Freezing point (oC) 3.82 0.00 Boiling point (oC) 101.4 100.0 Density at standard temperature and pressure (g cm–3) 1.106 (solid) 0.998 (solid) (i) Suggest if distillation is effective in separating heavy water from light water. [1] The boiling point of light water and heavy water is too close for distillation to take place effectively. [1] (ii) Using Table 1.1, suggest with reasoning, how a scientist can differentiate the two types of water without the use of a temperature measuring device. (Density of liquid light water is 1.0 g cm–3) [2] Freeze the heavy water and light water separately. [1] Drop the heavy water ice cube and light water ice cube into a glass of light water. Heavy water having a higher density than light water will sink in the glass of light water.[1]

3 ASRJC JC2 PRELIM 2023 9729/03/H2 [Turn over (d) Deuterated solvents (such as D2O) are a group of compounds where one or more hydrogen atoms are substituted by deuterium atoms. It may be assumed that they have similar chemical reactivity as their hydrogen analogues. (i) D2O is added to 3-chloropropionyl chloride. C1 C Cl H H H H C2 O Cl 3-chloropropionyl chloride Comment on the reactivity of C 1 and C2 and write a balanced chemical equation for the reaction. [2] C2 is more reactive than C1 C2 is highly electron –deficient because it is bonded to TWO highly electronegative atoms, oxygen and chlorine. [1] This makes the carbon very susceptible to reaction with nucleophiles. C1 C Cl H H H H C2 O Cl + D2O C1 C Cl H H H H C2 O OD + DCl [1] (ii) Construct a balanced chemical equation to show how deuterated ethanol, C 2D5OD, reacts with ethanoic acid, CH3CO2H in the presence of acid catalyst. [1] CH3COOH + C2D5OD CH3COOC2D5 + DOH [1] [Total: 15]

4 ASRJC JC2 PRELIM 2023 9729/03/H2 2 (a) Iron is a transition metal. The following scheme illustrates a series of reactions involving various oxidation states of iron. I II (i) Explain why [Fe(H2O)5SCN]2+(aq) is blood-red. [3] In the presence of water and SCN - ligands, the partially filled 3d orbitals of Fe3+ are split into two levels with a small energy gap (that falls within the visible light spectrum). [1] When white light passes though the solution, 3d electron absorbs light energy that is equal to the energy gap and gets excited from the lower energy 3d orbital to the higher energy 3d orbital. [1] Red colour of Fe3+(aq) observed is complementary to the green colour absorbed. [1] (ii) State the formula of the cation present in A and identify B. [1] A: [Fe(H2O)6]3+ ; B: Fe(OH)3 [1] (iii) State the type of reaction that occurred in I, II and III. [3] I − Redox reaction [1] II − Acid-base [1] III − Ligand exchange [1] (iv) With the aid of relevant equations, explain why • solution A is acidic. • effervescence was observed from reaction II. [3] [Fe(H2O)6]3+(aq) + H2O(l) [Fe(H2O)5(OH)]2+(aq) + H3O+(aq) [1] Why acidic? Fe3+ has a high charge size and small cationic radius , giving rise to high charge density. Thus, it polarises water ligands to a large extent, hence weakening the O-H bond and liberating H3O+ ions readily. [1] Why effervescence? H3O+ ions will react with carbonate to give effervescence of CO2. H3O+ + CO32- → CO2 + H2O [1] FeO42–(aq) dark red A yellow solution (acidic) C2O42– Na2CO3 B red-brown ppt + effervescence [Fe(H2O)5SCN]2+(aq) blood-red SCN– III

5 ASRJC JC2 PRELIM 2023 9729/03/H2 [Turn over (b) The electrolysis of an aqueous solution of potassium hydroxide was carried out using an iron anode and a platinum cathode. After a current was passed through the cell for some time, 360 cm 3 of gas was collected at the cathode (measured at r.t.p.) while the