OCR A Jun 2016 Paper 1 Q1 (with explained solutions) |

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OCR A Jun 2016 Paper 1 Q1

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Answer all the questions. 1 Nitrogen is the most common gas in the atmosphere. (a) Atoms of nitrogen consist of protons, neutrons and electrons. Complete the table below. Particle Relative mass Relative charge Position within the atom Proton Neutron Electron shell [2] (b) The electrons in the second shell of a nitrogen atom are found in an s-orbital and three p-orbitals. (i) State, in words, the 3D shape of an s-orbital and a p-orbital. s-orbitalp-orbital[1] (ii) Describe the relative energies of the 2s orbital and each of the three 2p orbitals in a nitrogen atom.[2] (c) Draw a dot-and-cross diagram to show the bonding in a nitrogen molecule. Show outer electrons only.OCR 2016 [1]<br />
(d) Calculate the amount, in mol, of nitrogen atoms in 5.1171020 nitrogen molecules. Give your answer in standard form. amount of nitrogen atoms =mol [2] (e) N2O3 is an unstable oxide of nitrogen that decomposes in a redox reaction. (i) State the oxidation number of nitrogen in each oxide in the table below. N2O3(g) NO(g) + NO2(g) Oxidation number of nitrogen Oxide N2O3 NO NO2 [1] (ii) Name this type of redox reaction. In your answer you should use appropriate technical terms spelled correctly.[1] (f) N2O3 reacts with water to form an acid as the only product. This reaction is not a redox reaction. The empirical formula of the acid formed is the same as the molecular formula. (i) State what is meant by the term molecular formula.[1] (ii) Suggest the empirical formula of the acid formed.OCR 2016 empirical formula of acid =[1] [Total: 12] Turn over<br />

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F321 Question 1 (a) Answer particle proton neutron electron relative mass relative charge 1 1 1/2000 + 1 nil/01 position within the atom nucleus nucleus shell Relative mass column (cid:51) Relative charge AND position columns (cid:51) Mark Scheme June 2016 Mark Guidance For relative masses ALLOW 1/1800 to 1/2000 for electron value (0.00050.00056) ALLOW negligible for electron value IGNORE + in front of correct values DO NOT ALLOWin front of 1/2000 DO NOT ALLOW nil OR zero for mass of electron For relative charges ALLOW 1+ and neutral and 1 IGNORE(ie a dash) for neutron DO NOT ALLOW + orwithout 1 DO NOT ALLOW 1 without charge For position within the atom IGNORE middle OR centre for nucleus For s-orbital IGNORE circular For p-orbital ALLOW other words indicating 3-D shape of p-orbital eg Peanut-shaped OR hour glass etc ALLOW figure of eight OR figure of 8 IGNORE diagrams ALLOW reverse argument ALLOW suitable energy diagram for either part (b) (i) s-orbital = spherical AND p-orbital = dumb-bell shape (cid:51) (ii) p-orbitals have greater energy than s-orbitals (cid:51) (three) p-orbitals have equal energy (cid:51)<br />
F321 Question 1 (c) (d) Mark Scheme June 2016 Mark ALLOW all dots or all crosses. Guidance ALLOW one mark for 0.17 x 102 OR 0.017 x 101 OR 0.0017 (not standard form) ALLOW one mark for 4.25 x 104 (dividing by 2 in M2 + standard form) ALLOW one mark for 6.16 x 1044 (multiplying by 6.02 x 1023 in M1 + standard form) Answer x xx N x x (cid:51) First check the answer line. If answer = 1.7(0) x 103 award 2 marks. ------------------------------------------------------------------------ M1 (Dividing by 6.02 x 1023) Number of N2 molecules = 5.117 x 1020 = 8.5. x 104 6.02 x 1023 OR 0.85 x 103 OR 0.085 x 102 OR 0.0085 x 101 OR 0.00085 (cid:51) M2 (Correct conversion of molecules to atoms + standard form) M1 x 2 and in standard form (cid:51) From 0.0085, answer = 2 x 0.00085 = 0.00170 = 1.7(0) x 103 Alternative method M1 (Correct conversion of molecules to atoms) = 5.117 x 1020 x 2 = 1.02(34) x 1021 OR 10.2(34) x 1020 OR 102.(34) x 1019 etc M2 (Correct use of 6.02 x 1023 + standard form) 1.02(34) x 1021 = 1.7(0) x 103 6.02 x 1023<br />
Mark Scheme June 2016 Mark Guidance ALLOW 3 OR 3+ etc ALLOW oxidation numbers written over the equation but IGNORE if oxidation numbers are given on the answer lines QWC disproportionation spelled correctly. ALLOW compound for molecule IGNORE simplest whole before number ALLOW actual ratio IGNORE ratio alone DO NOT ALLOW simplest ratio ALLOW O2HN etc F321 Question (i) 1 (e) Answer N2O3 = +3 NO = +2 NO2 = +4 (cid:51) (ii) Disproportionation (cid:51) (f) (i) (Actual) number of atoms of each element present in a molecule (cid:51) (ii) HNO2 (cid:51) Total 1 12<br />

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