2023 CJC H2 Chem Prelims P2 Examiners comments

[Turn over 9729/2 CJC JC2 Preliminary Examination 2023 CANDIDATE NAME CLASS 2T CHEMISTRY 9729/02 Paper 2 Structured Questions 28 August 2023 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculation is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 20 printed pages. For Examiner’s Use Paper 1 30 Paper 2 Q1 /13 Q2 /12 Q3 /14 Q4 /14 Q5 /10 Q6 /12 75 Paper 3 80 Paper 4 55 OVERALL (100%) GRADE Catholic Junior College JC2 Preliminary Examination Higher 2 MARK SCHEME

2 9729/2 CJC JC2 Preliminary Examination 2023 1 (a) Nitrogen and phosphorus are elements of Group 15 in the Periodic Table. (i) Explain why phosphorus has a larger atomic radius than nitrogen. ……………………………………………………………………………………….....……… ……………………………………………………………………………………….....……… ……………………………………………………………………………………….....……… ……………………………………………………………………………………….....……… …………………………..………………………………………………………………......[2] (ii) Nitrogen exists naturally as gaseous diatomic N≡N molecules whereas phosphorus is a solid and exists as P4 molecules comprising P-P single bonds. Suggest why phosphorus does not occur naturally as P≡P molecules. …….…………………………………………………………………………………………… ….……………………………………………………………………………………………... ………………………………………………………………………………………………[1] (b) Nitrate, NO 3–, and phosphate, PO 43–, are oxoanions of nitrogen and phosphorous respectively. (i) Draw dot–and–cross diagrams to show the bonding in NO 3– and PO43– ions. Hence, suggest the shape of each of these ions. [4] ion NO3– PO43– dot-and-cross diagram shape trigonal planar tetrahedral O N O x xx xx x x • • x xx • x x x O x • x • x x x o ¯ O P x x ox xx x x • • • xx x x x 3– x O O • x • x x x o x O x x ox x x x Phosphorus is a relatively big atom with diffused (3p) orbitals, side-on overlap of its p orbitals to form π bonds is much less effective than head-on overlap to form sigma bond. P has a larger atomic radius than N as although P has a greater nuclear charge, it has one more principal quantum shell and hence the outermost (valence) electrons are further from the nucleus and experience greater shielding effect due to inner electron shells. Nuclear attraction on the valence electrons is lower, hence the atomic radius of P is larger. Most candidates were able to identify that increase in shielding effect was the dominant factor. Common mistakes were omitting the discussion on nuclear charge, failure to conclude the effect on attraction of the valence electrons to the nucleus or omitting key words (like nuclear charge, shielding effect). Few candidates managed to recognise that side -on overlap of the diffused p orbitals would be less effective. Many candidates skipped this part.

3 9729/02/CJC JC2 Preliminary Examination 2023 [Turn over (ii) The 2s and 2p orbitals of nitrogen atoms can hybridise in the same way as the 2s and 2p orbitals of carbon atoms. With reference to your answer in (b)(i), state the hybridisation state of nitrogen in NO3–. ……………………………………………………………………………………………...[1] (iii) Table 1.1 shows the bond lengths of two nitrogen–oxygen bonds. Table 1.1 bond N–O N=O bond length (pm) 136 115 The experimental bond length of each nitrogen–oxygen bond in the nitrate ion, NO3–, is 128 pm. With reference to your answers in (b)(i) and (b)(ii), explain this observed bond length in NO3–. ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………[1] sp2 The continuous overlap of p orbitals between the N and O atoms allow the lone pair / negative charge on the O atom to be delocalised into the N=O bond. As a result, all the N–O bonds have partial double bond character. Many candidates were able to apply hybridisation of C to N here, where 3  bonds and 1  bond indicates sp 2 hybridisation (based on the correct dot - and-cross diagram). Candidates should ensure they write the number 2 as a superscript. Few candidates managed to get the correct dot -and-cross diagram for nitrate. It is important to note that N is in period 2 and hence cannot expand its octet. Other common mistakes include not differentiating the extra electrons with a different symbol, omitting the overall charge, omitting lone pairs of electrons on surrounding atoms, drawing numerous dative bonds from the same atom. Most candidates managed to identify the correct shape. Candidates are reminded on the need to ensure their diagrams are well-spaced. Few candidates provided a good explanation. Candidates should deduce from the dot-and-cross diagram that the O atoms that are single bonded with N have a lone pair which can overlap with the unhybridised p orbital in N/ p orbital of the N=O bond, hence all the nitrogen -oxygen bonds have partial double bond character/ resonance structure. Common mistakes: -Some candidates misread the question and simply explained the difference in N–O and N=O bond lengths -Some stated that 128 pm is the average of single and double bond lengths but did not explain why the average could be taken -Some stated that hybrid orbitals with partial s or p character will form bonds with partial double bond character, which is completely incorrect

4 9729/2 CJC JC2 Preliminary Examination 2023 (c) A nuclear reaction is a reaction in which there is a change to an atomic nucleus. A scientist attempts to produce P15 32 synthetically using a nuclear reaction where a neutron collides with an isotope E of another element as shown in the equation below (n represents a neutron). isotope E + n0 1 → P15 32 + H1 3 Identify isotope E, showing clearly its mass number. …………………………………………………………………………………………......[1] (d) Another type of nuclear reaction is radioactive decay, which occurs spontaneously in elements with an unstable atomic nucleus. Uranium–234 is used in nuclear power generation, and emit s a constant stream of  particles, which are equivalent to Helium–4 nuclei. Iodine–131 is used in treatment for thyroid cancer. On decaying, it emits  particles, which can be considered as electrons. A small amount of Uranium –234 and Iodine–131 are separately placed in an ionisation chamber to emit a constant stream of radiation, and the emitted particles are passed through an electric field. (i) On the diagram below, sket