ASRJC 2021 J2Prelims H2Chem P3 soln

ASRJC JC2 PRELIMS 2021 9729/03/H2 [Turn over Anderson Serangoon Junior College 2021 JC2 Preliminary Examination H2 Chemistry (9729) Paper 3 Suggested Solutions 1 (a) SiH4, PH3 and H2S are three compounds. (i) State and explain the bond angles in the molecules of these three compounds. [4] To minimise repulsion, the valence electron pairs around a central atom are arranged as far apart as possible. Lone pairs exert greater repulsion than bond pairs. [1] There are 4 bond pairs and (0 lone pair) around Si. Hence, SiH4, is tetrahedral with a bond angle of 109.5°. [1] There are 3 bond pairs and 1 lone pair around P. Since lone pair–bond pair repulsion is greater than bond pair–bond pair repulsion, PH3 is trigonal pyramidal with a bond angle of 107°. [1] There are 2 bond pairs and 2 lone pairs around S. Since lone pair–lone pair repulsion is greater than lone pair–bond pair repulsion which is greater than bond pair–bond pair repulsion, H2S is bent with a bond angle of 104.5°. [1] (ii) Phosphorus and hydrogen have same electronegativity. Draw the three molecules SiH4, PH3 and H2S and indicate for each one the polarity of each of the bonds it contains and the overall polarity of the molecule. [4] overall: non polar overall: polar overall: polar *Note: must indicate the polarity of each bond as question mentioned it. [1]: all 3 correct drawings [1]: polarity of each of the bonds [1]: overall polarity of each molecule [1]: correct direction of polarity for PH3 and H2S Thinking process Electronegativity increases across a Period. As P and H have same electronegativity, Si will be less electronegative than H while S is more electronegative than H. (iii) State the properties of a gas necessary for it to approach ideal behavior. [2] Intermolecular forces of attraction between gas molecules are negligible. [1] The volume of the gas particles/molecules is negligible compared to the volume of the container. [1] Si d d d d d H H H H P H H H S H H d d d

2 ASRJC JC2 PRELIMS 2021 9729/03/H2 [Turn over (iv) Of the three gases at 400 K, the behaviour of SiH4 is closest to ideal gas behavior. Suggest why H2S deviate more from ideal behaviour than SiH4. [1] H2S is polar and has stronger permanent dipole-permanent dipole attractions between its molecules. Hence, it has more significant intermolecular forces and deviate more from ideal behaviour. [1] (b) (i) Define the term lattice energy. [1] The amount of heat evolved when one mole of the solid ionic compound is formed from its constituent free gaseous ions under standard conditions of 298 K and 1 bar. [1] (ii) Use the data in Table 1.1, together with data from the Data Booklet, draw an energy cycle to calculate a value for the lattice energy of magnesium iodide, MgI2(s). Show your working. Table 1.1 value / kJ mol–1 electron affinity of iodine, I(g) + e- ® I–(g) –295.4 enthalpy change of sublimation of iodine molecules, I2(s) ® I2(g) +62.4 standard enthalpy change of atomisation of Mg(s) +148 standard enthalpy change of formation of MgI2(s) –364 [3] [1] Hess Law: LE = ∆Hf™ – [∆Hat™(Mg) + 1st IE(Mg) + 2nd IE(Mg) + ∆Hsub™ + BE (I-I) + 2EA(I)] [1] = (-364) – [+148 + 736 + 1450 + 62.4 + 151 + 2 x (-295.4)] = -2320kJ mol−1 [1] Mg(s) + I2(s) Mg(g) + I2(s) Mg(g) + I2(g) Mg(g) + 2I(g) Mg2+(g) + 2I(g) + 2e– Mg2+(g) + 2I–(g) MgI2(s) DHato(Mg) = +148 DHsubo = +62.4 2DHato(I) = B.E.(I-I) = +151 1st + 2nd I.E.(Mg) = +736 + 1450 2EA(I) = 2(–295.4) L.E.(CoI2(s)) DHfo = –364

3 ASRJC JC2 PRELIMS 2021 9729/03/H2 [Turn over (c) Magnesium forms an important group of covalent compounds which are known as Grignard reagents. Many Grignard reagents, with different alkyl or aryl groups, have now been prepared and are widely used in organic synthesis. A typical example of the use of a Grignard reagent is the two-step reaction of C2H5MgBr with propanone, CH3COCH3 to form 2-methylbutan-2-ol. (i) Suggest the type of reaction which occurs in step 2. [1] Hydrolysis [1] (ii) Suggest the structural formula of the final organic product formed when is reacted with ethanal, CH3CHO, in a similar two-step process. [1] [1] (iii) The Grignard reagent CH3CH2MgBr can be readily converted into a carboxylic acid by using carbon dioxide. Given the two-step reaction sequence for this conversion Draw the structural formula of compound A and B. [2] Compound A [1] Compound B [1] C2H5MgBr + CH3COCH3 step 1CH3C OMgBr CH3 C2H5 step 2CH3CCH3 + Mg(OH)Br C2H5 OH MgBr C OH CH3 H CH3CH2MgBr + CO2 step 1 step 2 compound A compound B + Mg(OH)Br C CH3CH2 O OMgBr C CH3CH2 O OH

4 ASRJC JC2 PRELIMS 2021 9729/03/H2 [Turn over CLT (c) F and G are ionic compounds which contain six-coordinate Co3+ complex ions, whose ligands are either NH3 or Cl–. The complexes of E and F are octahedral in shape. octahedral shape When excess aqueous AgNO3 is added to 1 mole of F and 1 mole of G separately, 2 moles of a white precipitate is formed from F while 1 mole of the same white precipitate is formed from G. (i) Deduce the structural formulae of complexes F and G. [2] F has 2 moles of free Cl – ion and one Cl – ion datively bonded to the Co3+ . G has 1 moles of free Cl – ion and two Cl – ion datively bonded to the Co3+. F: [Co(NH3)5Cl]2+. 2Cl– [1] G: [Co(NH3)4Cl2]+ . Cl– [1] (ii) The Co3+ complex ion in G exhibits cis-trans isomerism like alkenes. Suggest the structure of the trans isomer, showing clearly how the ligands are bonded to the central metal ion. [2] Complex G [1] Correct overall charge and dative bonds from nitrogen atom. [1] Correct trans configuration (either structure) [Total: 23] M n(AgCl) 2 n(complex F) 1 n(AgCl) 1 n(complex G) 1 = = NH3 CoIII NH3 Cl NH3 Cl H3N + . Cl– . NH3 Cl CoIII NH3 Cl H3N H3N 2+ . Cl–

5 ASRJC JC2 PRELIMS 2021 9729/03/H2 [Turn over 2 Chlorine is a greenish yellow gas that was first synthesised around 1630. Chlorine forms a variety of oxides such as chlorate(VII), ClO4– and chlorate(V), ClO3–. (a) Chlorine gas can be produced from the electrolysis of brine (concentrated sodium chloride solution) using a diaphragm cell with inert electrodes. As seen in Fig 2.1, a diaphragm is used to keep the products of electrolysis from mixing and hence, prevent the formation of undesirable by-products. Fig 2.1 (i) Given the cathode reaction: 2H2O + 2e ® H2 + 2OH– Write the ion-electron half-equation for the reactions taking place at the anode. Hence, construct the overall equation. You are to provide state symbols for all the equations.