ACSI Paper 2 Prelim (Student Version) Answer

1 YEAR 6 PRELIMINARY EXAMINATION 2023 INTERNATIONAL BACCALAUREATE DIPLOMA PROGRAMME CHEMISTRY HIGHER LEVEL PAPER 2 Wednesday 13 th September 2023 2 hours 15 minutes INSTRUCTIONS TO CANDIDATES  Do not open this examination paper until instructed to do so.  Write your candidate session number in the box above.  A calculator is required for this paper.  A copy of the Chemistry Data Booklet is required for this paper.  Write your answers in the boxes provided.  If you use additional sheets of paper for your answer, attach them to the booklet. Indicate the question number clearly on these sheets.  All drawings must be in ink. For examiner’s use Qn 1 /6 Qn 2 /22 Qn 3 /8 Qn 4 /10 Qn 5 /13 Qn 6 /9 Qn 7 /15 Qn 8 /7 Wrong s.f. /units Total /90 ________________________________________________________________________ This question paper consists of 31 printed pages including this cover page. Name Candidate Number

2 Answer all questions. Write your answers in the boxes provided. 1. Carbon consists of 4 allotropes in the form of diamond, graphite, fullerene and graphene. (a) State and explain the difference in electrical conductivity of diamond and graphite. [2] Diamond is a non-conductor Graphite is a good conductor Diamond uses all the valence electrons in bonding/ no free electrons Graphite has delocalized electrons to conduct electricity Marker comments Most students can answer the question. Some students misused words like “lone pairs” or lone electrons (b) Explain why the carbon-carbon bonds in graphite are stronger than the carbon- carbon bonds found in diamond. [2] Resonance or delocalization of pi electrons in graphite/ overlap of pi electrons or overlap sp2-sp2 and sp3-sp3 C Resulting in shorter bonds/ partial double bond character in C-C bonds/ 1.333 Bond order Marker comments A few students related this to Diamond having a weaker London Dispersion Force compared to the dipole-dipole forces in graphite, this is a conceptual misunderstanding. Students are to note that the bond resulting from delocalised electrons is not 1.5 bond order and it is not a double bond in graphite but rather a partial double bond. (c) State the type of hybridization around the carbon atoms in diamond and graphite. [2] Diamond: sp3 Graphite: sp2 Marker comments Generally well done

3 2. The energy cycle shown can be used to calculate the enthalpy change of formation of KCl (s). Each arrow indicates a transformation, W, X, and Y. Each transformation consists of one or more steps. (a) (i) Calculate the value of the following transformations using section 8 of the data booklet and the following information. Enthalpy change of atomisation of K (s) +89 kJ mol─1 Enthalpy change of atomisation of Cl2 (g) +121 kJ mol─1 Enthalpy change of solut