Edexcel Jun 2017 (IAL) Paper 2 Q23 (with explained solutions) |

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Edexcel Jun 2017 (IAL) Paper 2 Q23

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SECTION C Answer ALL the questions. Write your answers in the spaces provided. 23 Boron nitride, BN, is a compound first made commercially in the 1940s from boric acid and ammonia, in an atmosphere of nitrogen. It forms structures analogous to graphite and diamond because it is isoelectronic with these corresponding carbon structures. Boron nitride has also been used to form nanotube structures in a similar way to carbon. Just as synthetic diamonds are produced from graphite by using high temperatures and high pressures, the diamond-like cubic boron nitride can also be made from heating the graphite-like hexagonal boron nitride under high pressure. Boron nitride forms ceramic materials with very high thermal and chemical stability and, a wide range of uses. For example, they are stable in air up to 1000(cid:113)C, which is an advantage over similar graphite materials. The hexagonal form of boron nitride is a very effective lubricant and is also used in cosmetics. However, it is an electrical insulator, in contrast to graphite, which is a good electrical conductor. (a) (i) Write the equation for the formation of boron nitride from boric acid, H3BO3, and ammonia. State symbols are not required. A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D (ii) Suggest why this reaction is carried out in an atmosphere of nitrogen. (1) (1)A E R A S I H T N I E T I R W T O N O D *P48383A01924* 19 Turn over<br />
(b) The structure of the cubic boron nitride corresponds to the diamond structure. The boron and nitrogen atoms alternate throughout the structure. (i) In the left hand box, the diagram shows a section of the diamond structure, where each black circle represents a carbon atom. In the right hand box label all the nitrogen and boron atoms in the diagram of cubic boron nitride. Diamond Cubic boron nitride (ii) State the bond angle and shape around the carbon atoms in diamond and fully justify your answer. (1) (4) Bond angleShapeJustification20 *P48383A02024* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />
A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D (c) The equilibrium between graphite and diamond is C(graphite) (cid:85) C(diamond) (cid:507)H = +1.9 kJ mol1 The density of graphite is 2.27 g cm3 and the density of diamond is 3.51 g cm3. *(i) Suggest why a very high temperature and high pressure are needed to convert graphite to diamond. (4)(ii) The use of a catalyst in the conversion of graphite to diamond has been reported. Describe how the addition of a catalyst can lower the temperature required for a reaction. (3)A E R A S I H T N I E T I R W T O N O D*P48383A02124* 21 Turn over<br />
(d) Diamond and graphite are stable in air up to approximately 800(cid:113)C. Identify one of the products if diamond or graphite is heated in air above this temperature. (1)(e) The structure of hexagonal boron nitride corresponds to that of graphite. (i) The simplified diagram in the left hand box shows the bonding in hexagonal boron nitride. In the right hand box, complete the dot and cross diagram showing only the electrons around the nitrogen atom which is labelled with an asterisk (*). Use () for the nitrogen electrons and () for the boron electrons. N* *(ii) Describe how each carbon atom is bonded in the graphite structure and hence explain why graphite is a good conductor of electricity. Suggest why hexagonal boron nitride is an electrical insulator. (1) (3) D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A22 *P48383A02224* D O N O T W R I T E I N T H I S A R E A<br />
(iii) Graphite and the hexagonal boron nitride are both used as lubricants because of the weak intermolecular forces between the layers of hexagonal rings. Identify these intermolecular forces and describe how they arise. (3)(Total for Question 23 = 22 marks) TOTAL FOR SECTION C = 22 MARKS TOTAL FOR PAPER = 80 MARKS A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *P48383A02324* 23<br />

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Section C Acceptable Answers Question Number 23(a)(i) H3BO3 + NH3BN + 3H2O Reject Mark (1) ALLOW multiples IGNORE state symbols even if incorrect Acceptable Answers Question Number 23(a)(ii) Nitrogen is an unreactive gas/ to prevent (nitrogen) oxides from forming / to prevent oxidation (of ammonia) ALLOW inert / wont react with other things/ no oxygen present/prevent combustion IGNORE stable Reject Mark (1) Acceptable Answers Question Number 23(b)(i) Diagram where the boron and nitrogen atoms Reject Mark (1) alternate throughout the structure, e.g. Accept use of alternative ways of labelling e.g key<br />
Acceptable Answers Question Number 23(b)(ii) (Bond angle) 109.5 (1) Reject Mark (4) (shape) Tetrahedral (1) Four bonded electron pairs/ Four areas of electron density (around each carbon) (1) Atom Repulsion between electron pairs to give minimum repulsion repulsion (that results in tetrahedral shape) ALLOW Repulsion to give maximum separation of electron pairs/get as far away from each other as possible (1) Acceptable Answers Question Number *23(c)(i) Marking point 1 Reject Mark (4) Equilibrium shifts to the right/ Equilibrium favours the formation of diamond IGNORE just yield (1) Marking point 2 Because the reaction is endothermic (high temperature favours the formation of diamond) (1) M2 depends on M1 or near miss, but these points can be in any order Marking point 3 Any reference to (High) pressure favours the formation of higher moles or density diamond molecules ALLOW smaller volume for higher density (1) Marking point 4 Intermolecul There are (many) strong covalent bonds to break (and the rate is slow) ar forces ALLOW High temperature and pressure needed because activation energy is high (1)<br />
Acceptable Answers Question Number 23(c)(ii) A catalyst lowers the activation energy for the Reject Mark (3) reaction (1) By providing an alternative pathway for the reaction (1) So the same number of particles can react at a lower temperature /more particles have sufficient energy to react/more particles exceed EA ALLOW Molecules/atoms for particles (1) Question Number 23(d) Acceptable Answers Reject Mark Carbon dioxide/carbon monoxide OR CO2/CO (1) Soot Carbon C<br />
Acceptable Answers Question Number 23(e)(i) Single dot and cross in overlap between top and right-hand boron atoms and the nitrogen atom and Three crosses and one dot in the overlap area between left-hand boron atom and the nitrogen atom Reject Mark (1) Any additional electrons around the N IGNORE Other electrons around the boron atoms. ALLOW The single bond and the dative bonds can be reversed<br />
Acceptable Answers Question Number *23(e)(ii) In graphite each carbon atom has three (covalent) bonds (to carbon atoms) (1) With (one) electron delocalised (between the layers of hexagonal rings which can move and carry charge) ALLOW JUST delocalised electrons (1) The electrons in hexagonal boron nitride are all fixed in position / localised / not delocalised / not free moving / not mobile (1) Reject Mark (3) Just free electrons Acceptable Answers Question Number 23(e)(iii) London forces / dispersion forces Mark (3) ALLOW van der Waals forces (1) Instantaneous dipole due to (asymmetric) electron distribution/movement (1) Induced dipoles (in adjacent layers) ALLOW In molecules/atoms (1) Reject Permanent dipole/ Polar bonds TOTAL FOR SECTION C (QUESTION 23) = 22 MARKS TOTAL FOR PAPER = 80 MARKS<br />

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