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CIE Jun 2015 v3 Paper 2 Q1
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![Answer all the questions in the spaces provided. Neon is a noble gas. (a) Complete the full electronic con guration of neon. 1s2[1] (b) (i) Explain what is meant by the termrst ionisation energy.[3] (ii) Explain why therst ionisation energy of neon is greater than that ofuorine.[2] (c) Neon has three stable isotopes. isotope mass number percentage abundance 20 21 9.25 90.48 0.27 (i) De ne the term relative atomic mass.[2] (ii) Use the relative atomic mass of neon, 20.2, to calculate the mass number of isotope 1. mass number =[2]UCLES 2015 9701/23/M/J/15<br />
(d) A mixture of neon and argon has a mass of 0.275 g. The mixture was placed in a gas syringe at a temperature of 25 C and a pressure of 100 kPa. Under these conditions the mixture was found to occupy a volume of 200 cm3. (i) Calculate the average Mr of the mixture. (ii) Use your answer to (i) to calculate the percentage of neon in the mixture. Give your answer to three signi cantgures. average Mr =[2] percentage of neon =% [1] (e) Neon and argon can both be obtained by fractional distillation of liquid air as they have different boiling points. Neon has a boiling point of 27.3 K. The boiling point of argon is 87.4 K. (i) Name the force that has to be overcome in order to boil neon or argon and explain what causes it.[3] (ii) Explain why argon has a higher boiling point than neon.[2]UCLES 2015 9701/23/M/J/15 [Total: 18] [Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2015_v3_QP_-_Paper_2_CIE_Chemistry_A-level_1-0.jpg)
![Page 2 Question Cambridge International AS/A LevelMay/June 2015 9701 23 Mark Scheme Syllabus Paper Mark Scheme Mark Total (a) (1s2)2s22p6 (b) (i) The amount of energy required/energy change when one electron is removed from each atom in one mol of gaseous atoms (ii) Greater nuclear charge/number of protons Same shielding/number of shells/energy level (c) (i) mean/average mass of the isotopes/an atom(s) relative to 1/12 of the mass of an atom of 12C/on a scale where an atom of 12C is (exactly) 12 (ii) 20.2 ( 20 90.48 ( 21 0.27 ( 9.25 100 22.133 2020 1815.27 9.25 y = 22 (d) (i) pV mRT mRT pV 0.275 8.31 298 100 10 200 10 Mr = 34.05/34.1 (ii) (Let % Ne = x so % Ar = 100-x) 20.2x + x) 39.9(100 100 34.05 % Ne = 29.7 (e) (i) Van der Waals/London/dispersion Uneven electron distribution/temporary dipole Induced dipole-dipole attraction (ii) more electrons more polarisable/greater attraction/stronger IMFs [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [3] [2] [2] [1] [2] [1] [1] [2] [1] [1] [1] [1] [1] [1] [1] [3] [2] [18]Cambridge International Examinations 2015<br />](https://learnah.org/wp-content/uploads/2018/08/June_2015_v3_MS_-_Paper_2_CIE_Chemistry_A-level_1.jpg)