2024 Prelims NJC H2 Chem P2 (Ans)

[Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on Question Paper. Additional Materials: Data Booklet 9729/02 23 August 2024 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use paper clips, highlighters, glue or correction fluid. Answers all questions. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 /9 2 /22 3 /6 4 /7 5 /14 6 /17 Paper 2 Total /75 Marks Weightings Paper 1 /30 15% Paper 2 /75 30% Overall Percentage Paper 3 /80 35% Paper 4 /55 20% Grade This document consists of 19 printed pages and 1 blank page.

NJC/H2 Chem Preliminary Examination/02/2024 2 Answer all the questions in the spaces provided. 1 (a) Element A is from Period 4 of the Periodic Table. The first eight ionisation energies of element A, in kJ mol-1, are 947 1798 2735 4837 6043 12310 14300 16800 (i) Identify element A and explain your answer. ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… [2] (ii) Explain the difference between the first ionisation energy of element A compared to the element to its right on the Periodic Table. ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… [2] Element A is Arsenic. There is a large difference between the 5th and 6th ionisation energies of A. The 6th electron is from an inner electronic shell. A has 5 valence electrons, hence it is from Group 15. The element to the right is Selenium. 1st ionisation energy of Se involves the removal of paired 4p electrons which experiences inter-electronic repulsion. Less energy is needed to remove the electron from Se than the unpaired 4p electron in A. Hence, the IE of Se is lower.

NJC/H2 Chem Preliminary Examination/02/2024 3 [Turn over (b) In organic chemistry, an aldol is a structure consisting of a hydroxy group ( -OH) two carbons away from either an aldehyde or a ketone. Aldols are the product of a carbon -carbon bond-formation reaction, giving them wide applicability as a precursor for a variety of other compounds. (i) Give the IUPAC name of the structure below: O HO Cl IUPAC name: ………………………………. [1] (ii) The compound above undergoes elimination to form the following product: CH(OH)CHCHO. Give the reagents and conditions for the reaction. ……………………………………………………………………………………… [1] (iii) State the isomerism which the product displays. ……………………………………………………………………………………… [1] (iv) With the use of a diagram, explain why the cis isomer is formed preferentially. ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… [2] [Total:9] 2-chloro-3-hydroxypropanal [1] Ethanolic KOH and heat [1] Cis-trans isomerism [1] [1] The cis isomer is favoured due to the intra-molecular hydrogen bonding that stabilises the cis isomer. [1]

NJC/H2 Chem Preliminary Examination/02/2024 4 2 (a) Some information about NO3− and NO2− are provided in Table 2.1 and Table 2.2 Table 2.1 Electron arrangement around N in NO3− 3 bond pairs & 0 lone pair Electron arrangement around N in NO2− 2 bond pairs & 1 lone pair Table 2.2 Nitrogen−oxygen bond in NO3− 0.124 nm Theoretical N−O bond length 0.136nm Theoretical N=O bond length 0.115nm Fig. 2.1 shows a possible structure of NO3− . N O O O Fig.2.1 (i) Use relevant information from the above tables to explain why Fig. 2.1 does not represent an accurate model for the bonding in NO3− . ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… [1] (ii) Nitrogen atoms undergo the same type of hybridisation as carbon atoms do. Suggest the mixing and overlap of atomic orbitals which accounts for the N−O bond length in NO3− . ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… [3] One s and 2p orbitals mixed to give 3 degenerate sp2 orbitals in N. The sp2 hybridised orbital of N undergo head on overlap with the p/sp2 orbitals of each O to form the sigma bond. The p orbitals of N undergo continuous p orbital sideway overlap with the the O atoms, resulting in delocalisation of pi electrons / a resonance structure with a partial N-O double bond. Mixing of correct orbitals [1] Correct orbitals identified for head on [½] Correct orbitals identified for side on [½] Correct overlap leading to sigma bond [½] Correct explanation on equal bond length [½] If the model is correct, there should be 1 shorter bond (N=O) and 2 longer bonds (N-O) in NO3− . However, there is only one N−O bond length in NO3− in table 2.2.

NJC/H2 Chem Preliminary Examination/02/2024 5 [Turn over (iii) The bond angle around nitrogen in NO3− and NO2− are different. Use VSEPR theory to explain how the bond angle in NO3− is different from that in NO2− . ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… [2] (b) When solid magnesium nitrate is heated, it decomposes to give magnesium oxide, oxygen and nitrogen dioxide gas. (i) Write an equation for the decomposition of solid magnesium nitrate. ……………………………………………………………………………………… [1] (ii) Zinc nitrate decomposes in the same way as magnesium nitrate when it is heated. Use data from the Data Booklet to explain the difference in decomposition temperature between zinc nitrate and magnesium nitrate. ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… [3] 2Mg(NO3)2(s)→ 2MgO(s) + O2(g) + 4NO2(g) Zn2+: 0.074nm Mg2+: 0.065nm Mg2+ has higher 𝑐ℎ𝑎𝑟𝑔𝑒 𝑠𝑖𝑧𝑒 ratio and has higher polarising power. It can polarise and weaken the N─O bond in NO3− to a greater extent. [1] As such, MgNO 3 is less thermally stable and require a lower decomposition temperature [1] as compared to ZnNO3. Quote data : [1] VSEPR states that electron pair arrange themselves as far apart as possible to minimise interelectronic repulsion. However, since lone- pair bond-pair repulsion is stronger than bond-pair bond-pair repulsion, the bond angle in NO2− is smaller than that in NO3−. (NO3− has an angle of 120o while NO2− has an angle of 118o)

NJC/H2 Chem Preliminary Examination/02/2024 6 (c) Magnesium and beryllium are Group 2 elements, but beryllium behaves differently from that of magnesium. There is said to be a ‘diagonal relationship’ between beryllium and aluminium as they show similar chemical behaviour due to their similarities in electronegativity and charge density. (i) Draw the dot-and-cross diagram to show the bonding in BeCl2.