NJC H2 CHEM P3 Solutions Prelim

1 [Turn Over NATIONAL JUNIOR COLLEGE PRELIMINARY EXAMINATIONS Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 3 Free response 9647/03 Wed 17 Sep 2014 2 hours READ THESE INSTRUCTIONS FIRST Answer any four questions. Start your answer to each question on a fresh piece of paper. A Data Booklet is provided. You are reminded of the need for good English and clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely behind the cover page. This paper consists of 11 printed pages and 1 cover page.

2 [Turn Over 1 Iodine monochloride, ICl, is an interhalogen compound that is formed when chlorine gas is passed through iodine crystals. Generally, interhalogen compounds have chemical properties similar to that of halogens. (a) (i) Predict the physical state of ICl at room temperature. Liquid / solid. Mr of ICl : 162.5 Mr of Br2 : 159.8 Strength of td-id expected to be similar to that of Br2, hence liquid state. Td-id stronger than Cl2 but weaker than I2. (ii) By considering the partial charges in ICl, explain how water reacts with ICl. The lone pair of electrons on the oxygen of H2O attacks the + I, breaking the I-Cl bond, forming HOI. The H+ then combines with the Cl¯. (iii) Hence construct an equation for the reaction of ICl with water. ICl + H2O  HOI + HCl [3] (b) Methylbenzene reacts with ICl in the presence of FeCl3 to form compound A. (i) Draw the structure of A. 2-iodo methylbenzene (ii) Describe the mechanism of this reaction, clearly indicating the role of FeCl3. Electrophilic substitution reaction FeCl3 + ICl  [FeCl4]¯ + I+ H+ + [FeCl4]¯  FeCl3 + HCl [4]

3 [Turn Over (c) FeCl3 and FeF3 are both formed from a metal and a non -metal. However, in molten state, only FeF3 is able to conduct electricity and FeCl3 does not. (i) Suggest the structures of FeC l3 and FeF 3 and explain why iron forms two different types of compounds when reacted with the different hal ogens – fluorine and chlorine. FeCl3 : Simple covalent molecules FeF3 : Ionic lattice structure Fe3+ has a high charge density and is hence able to polarise the larger electron cloud of the chlorine atoms resulting in more extensive sharing of electrons hence forming a covalent bond. Fluoride ions due to their small size have low polarisability. (ii) Using the information given below as well as relevant data from the Data Booklet, construct an energy level diagram for the formation of FeF 3 from its elements to determine its lattice energy. enthalpy change of atomisation of Fe +340 kJ mol–1 first electron affinity of F –328 kJ mol–1 standard enthalpy change of formation of FeF3 –989 kJ mol–1 +340 + 762 + 1560 + 2960 + 3/2 x 158 + 3 x -328 + LE = -989 LE = -5860 kJmol-1 [7] Suggest an equation for the Fe(s) + 3/2F2(g) +577 +5282 Fe(g) + 3F(g) Fe3+ (g) + 3e + 3F(g) Fe3+ (g) + 3F¯(g) -984 LE 0 Enthaply/ kJmol-1 -989 FeF3(s)

4 [Turn Over (d) Iron is a transition metal, a d -block element that varies from s-block metals in terms of physical properties. The table below gives data about some physical properties of the elements – iron and calcium. Property calcium Iron relative atomic mass 40.1 55.8 atomic radius/ nm 0.197 0.126 density/ g cm–3 1.54 7.86 electrical conductivity Good Good (i) Write down the full electronic configurations of calcium and iron. Ca : 1s22s22p63s23p64s2 Fe : 1s22s22p63s23p63d64s2 (ii) Explain why the atomic radius of iron is smaller than that of calcium. Iron has more proton s and hence has a higher nuclear charge than calcium. The additional d electrons in iron are poor shielding electrons and hence the shielding effect is negligible. Iron hence has a higher effective nuclear charge. The valence electrons in iron are held more tightly resulting in a smaller atomic radius. (iii) Suggest why iron has a higher density than calcium. Iron has a larger atomic mass, but a smaller atomic radius that results in a smaller volume than calcium. Since density is mass/volume, iron has a higher density. (iv) Explain which of the 2 metals would be expected to have a higher electrical conductivity. Both the 3d and 4s electrons in iron can be delocalised into the sea of electrons as they are of similar energy. Hence iron is expected to have a higher electrical conductivity due to the larger number of charge carriers. [6] [Total:20]

5 [Turn Over 2 (a) In the qualitative analysis of organic compou nds, the presence of a carbonyl functional group can be confirmed by using 2,4dinitrophenylhydrazine. When 2 cm 3 of 2,4dinitrophenylhydrazine solution is added to separate test tubes containing 2 cm 3 propanone and 2 cm 3 propanal respectively, a bright orange precipitate is observed in both test tubes. Write a balanced equation for the reaction of 2,4 dinitrophenylhydrazine with propanal. Hence, identify the type of reaction. [2] C O H CH3CH2 + NN H H H N NO2 O2 NN H C N NO2 H CH3C O2 H2 + H2O Type of reaction: Condensation reaction (b) Fehling's solution, an alkaline copper(II) tartrate complex, can be used to distinguish between propanal and propanone. A positive test would show the formation of a brick red solid, Cu2O. (i) Identify which of the two organic compounds above would give a positive test with the Fehling’s solution . G ive the structure of the organic product for this reaction. Propanal will react with Fehling’s solution. Organic product is C O-CH3CH2 O The brick red solid, Cu 2O, is isolated through filtration. It is soluble in excess ammonia to give a colourless solution B. When left exposed to air, the colourless solution B turns into a deep blue solution, containing the complex ion [Cu(NH3)4]2+. (ii) Explain why the solution containing the complex ion [Cu(NH 3)4]2+ is deep blue in colour. In the presence of NH 3 ligands, the d orbitals of Cu 2+ split into two different energy levels with an energy gap, E. he electron from the lower energy d orbital can be promoted to the higher energy d orbital by absorbing wavelength of light with energy corresponding to the energy gap, E. Wavelength of light not absorbed will be reflected and seen as the complementary colour, deep blue.

6 [Turn Over (iii) Identify the oxidation state of copper in the copper containing species in the colourless solution B, and write its electronic configura tion. Hence explain why solution B is colourless. Oxidation state of Cu in B = +1 Electronic configuration: [Ar] 3d10 In Cu+ complex, the d orbitals are fully filled. He nce, it is not possible to have dd transition and all the wavelengths of visible light spectrum are reflected. The solution appears colourless. (c) [Cu(NH3)4]2+ can also be prepared by adding excess aqueous NH3 to aqueous copper(II) sulfate. Three reactions involving aqueous copper( II) sulfate are illustrated below. (c) (i) Explain fully , with the aid of equations, how the pale blue precipitate C is formed when NH3 (aq) is added to CuSO4 (aq). NH3 + H2O NH4 + + OH Cu2+ + 2 OH  Cu(OH)2 (ii) Write an equation for the formation of the yellow solution D. Hence , explain what will be observed when water is added to solution D. [Cu(H2O)6]2+ + 4Cl [CuCl4]2 + 6H2O Blue Yellow By Le