2019 VJC H2 Chem Prelim P2 Ans

 VJC 2019 9729/02/PRELIM/19 [Turn over 1 Victoria Junior College 2019 H2 Chemistry Prelim Exam 9729/2 Suggested Answers 1 (a) An unknown solid sample, with a mass of 1.50 g, contains three sodium salts, NaC l, NaClO3 and NaNO3. The sample was completely dissolved in water and diluted in a 250 cm3 volumetric flask to obtain solution L. In one experiment, a 50 cm3 portion of solution L was reacted with excess silver nitrate solution. The AgCl precipitate formed was removed by filtration, dried and weighed. The AgCl precipitate was found to have a mass of 0.240 g. In another experiment, a gas was bubbled into a new 50.0 cm3 portion of solution L to convert ClO3 to C l. Excess silver nitrate was then added to the resulting mixture. The AgC l precipitated formed was treated similarly as before and found to have a mass of 0.285 g. (i) Determine the amount of ClO3- ions present in 50 cm3 of solution L. From 1st experiment, nAgCl = 0.240 107.9 + 35.5 = 1.674  10–3 mol = nCl- From 2nd experiment, nAgCl = 0.285 107.9 + 35.5 = 1.987  10–3 mol = nCl- + nClO3 - nClO3 - in 50.0 cm3 of L = 1.987  10–3 – 1.674  10–3 = 3.13  10–4 mol [2] (ii) Hence, determine the percentage by mass of NaClO3 present in the original solid sample. nClO3 - in 250.0 cm3 of L = (250 / 50)  3.13  10–4 = 1.565  10–3 mol = nNaClO3 Percentage by mass of NaClO3 = 1.565  10-3  (23.0 + 35.5 + 3  16.0) 1.50  100% = 11.1% [2] (b) Suggest whether NaCl or NaClO3 has a lower melting point. Explain your answer. NaClO3 has a lower melting point. Both Na Cl and Na ClO3 have giant ionic structure with electrostatic attraction between ions of opposite charges. │Lattice energy│∝ q+ × q- r++ r- . Since ClO3– is larger than Cl–, NaClO3 has a lower magnitude of lattice energy. Hence, less energy is required to break the ionic bonds. [2]

 VJC 2019 9729/02/PRELIM/19 [Turn over 2 (c) (i) Draw a dot-and-cross diagram of the ClO3– anion. [1] (ii) Chlorine forms a number of oxides, one of which is Cl2O7. Cl2O7 is a symmetrical molecule, with a central oxygen atom bonded to two chlorine atoms. Draw the structure of Cl2O7. [1] (iii) The bonds in Cl2O7 exhibit two different bond lengths, namely, 0.141 nm and 0.171 nm. However, all the bonds in the ClO3– ion have the same bond length of 0.149 nm. Suggest why the bond length in ClO3– is intermediate between those present in Cl2O7. Resonance arising from the overlapping of p orbitals of Cl atom with those of the surrounding O atoms. This leads to delocalization of the lone pair on O– into the Cl=O double bonds, causing all bonds to have partial double bond character. [2] (iv) The bond angle formed between the central oxygen atom and its two surrounding chlorine atoms is larger than expected with a value of 118.6o. Suggest a possible reason why. Repulsion between the electron clouds (or steric hindrance) of the two bulky –Cl atoms [1] [Total: 11] 2 (a) Use of the Data Booklet is necessary for this part of the question. The sub–atomic makeup of certain ions formed from isotopes of cobalt and lead is given below: Number of neutrons Number of electrons Co ion 33 25 Pb ion 122

 VJC 2019 9729/02/PRELIM/19 [Turn over 3 In a particular experimental set–up, a beam containing the above ions of cobalt was passed through an electric field and was deflected by an angle of +10.2o. Under identical conditions, another beam containing the above ions of lead was deflected by an angle of +6.0o. What is the overall charge of the lead ions? From the Data Booklet, Proton number: Co = 27, Pb = 82 Charge of Co ion = +27 – 25 = +2 Nucleon number of Co = 33 + 27 = 60 Nucleon number of Pb = 122 + 82 = 204 Angle of deflection of Co ion = +10.2 = k ( 2 60) ⇒ k = 306 Let charge of Pb be x. Angle of deflection of Pb ion = +6.0 = 306 ( x 204) x = +4 (+ve sign must be included) [2] (b) Many transition metals and their complexes are paramagnetic. Paramagnetism is a property of a substance which allows it to be weakly attracted to a magnet. This property is due to the presence of unpaired electrons in the substance. CoF63– and [Co(NH 3)6]3+ are both complexes with cobalt in the +3 oxidation state. However, only CoF 63– displays paramagnetism while [Co(NH 3)6]3+ does not. This is determined by whether the electronic configuration of the Co 3+ ion in the complex displays a “high spin” or a “low spin” state. The following diagram shows how the d –orbitals are split in an octahedral environment. In a ‘high spin’ state, the electrons occupy all the d –orbitals singly, before starting to pair up in the lower energy d–orbitals. In a ‘low spin’ state, the lower energy d –orbitals are filled first, by pairing up if necessary, before the higher energy d–orbitals are used. (i) Draw the shapes of the dxy and dz2 orbitals. Label your drawings clearly.

 VJC 2019 9729/02/PRELIM/19 [Turn over 4 [2] (ii) Using ↑ or ↓ to represent electrons, show, on the two diagrams below, the electronic distribution of a Co3+ ion in a high spin state, and in a low spin state. Hence, identify the cobalt complex that corresponds to each particular spin state. High spin Low spin Complex: CoF63– [Co(NH3)6]3+ [2] (iii) Suggest why electrons usually prefer to occupy orbitals singly rather than in pairs. To minimize repulsion between negatively (OR similarly) charged electrons. [1] (iv) Using the explanation in (b)(iii), together with the information given above, state and explain which of the two cobalt complexes contains the larger energy gap, E, between its d orbitals. The low spin complex, [Co(NH3)6]3+, contains the larger energy gap. Electrons pair up only if the energy gap is larger than the energy required to overcome the interelectronic repulsion. [1] [Total: 8] dz2 dxy

 VJC 2019 9729/02/PRELIM/19 [Turn over 5 3 Frederick Thomas Trouton was an Irish physicist who observed a relationship between boiling poin ts and enthalpy changes of vaporisation after studying many liquids. He published his findings and formulated Trouton’s rule which states that the molar entropy of vaporisation, ΔSvap, for most liquids is about 85 J K1 mol1. The boiling points and enthalpy changes of vaporisation of several organic liquids are as follows: substance boiling point / °C ΔHvap / kJ mol-1 ΔSvap / J K-1 mol-1 propanone, (CH3)2CO 56.1 29.1 88.4 dimethyl ether, (CH3)2O -24.8 21.5 86.6 ethanol, CH3CH2OH 78.4 38.6 109.8 octane, CH3(CH2)6CH3 125.6 34.4 86.3 pyridine, 115.3 35.1 90.4