JPJC 2025 H2 Chemistry WA1 Answer

2026 J1 H2 Chemistry (9476) 1 WA1 Revision Paper 2 (Worked Solution) REVISION PAPER for JC1 H2 WEIGHTED ASSESSMENT 1 (REVISION PAPER 2) – 2025 Weighted Assessment 1 Suggested Solutions Section A (25 marks): Compulsory 1 (a) Explain the increasing trend in atomic radius of the elements down the group. Down the group, number of principal quantum shells increases (). The increase in shielding effect by inner shell electrons () outweighs the effect of increasing nuclear charge / number of protons (). Therefore, net nuclear attraction on outermost/valence electrons decreases (OR attraction between nucleus and outermost/valence electrons) (). Thus, atomic radius increases down the group. Note:  Effective nuclear charge is NOT a required marking point and it does not replace the 4th marking point.  Effective nuclear charge Zeff = (no. of protons – no. of inner shell electrons) Down the group, the number of protons and number of inner shell electrons increase by the same number. For example, compare the Z eff of Group 1 elements: 3Li: 1s22s1 Zeff of Li = 3 – 2 = 1 11Na: 1s22s22p63s1 Zeff of Na = 11 – 10 = 1 11K: 1s22s22p63s23p64s1 Zeff of K = 19 – 18 = 1 37Rb: 1s22s22p63s23p63d104s24p65s1 Zeff of Rb = 37 – 36 = 1 55Cs: 1s22s22p63s23p63d104s24p64d105s25p66s1 Zeff of Cs = 55 – 54 = 1 Hence, effective nuclear charge is constant down the group. [2] (b) The element selenium, Se, is in the same group as sulfur in the Periodic Table. (i) Write the electronic configuration of a Se atom. 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4 [1] 4p 4 1s 2 2s 2 2p6 3s 2 3p6 4s 2 3d10 4p1 4p2 4p3

2026 J1 H2 Chemistry (9476) 2 WA1 Revision Paper 2 (Worked Solution) (ii) Write an equation to show the first ionisation energy of Se. First ionisation energy (1 st IE) of an element is the energy required to remove one mole of electrons from one mole of GASEOUS atoms to form one mole of GASEOUS singly charged cations. Se(g)  Se+(g) + e [State symbols are required!] [1] (iii) Explain why the first ionisation energy of Se is smaller in magnitude than that of arsenic (As). As: 4p3 Se: 4p4 Inter-electronic repulsion between paired 4p electrons in Se makes it easier to remove a paired 4p electron as compared to removing an unpaired 4p electron in As. [1] (c) The third to eighth ionisation energies for element A are shown in Table 1.1. Table 1.1 3rd 4th 5th 6th 7th 8th ionisation energy / kJ mol1 5320 7450 11 000 13 300 71 000 84 100 Use Table 1.1 to deduce the group of the Periodic Table that A belongs to. Explain your answer. group number 16 () explanation Since there is a large increase from 6th IE to 7 th IE (), the 7th electron is removed from the inner quantum shell () closer to the nucleus so there are 6 valence/outermost electrons (). [2] [Total: 7] 4p 4p paired 4p electrons

2026 J1 H2 Chemistry (9476) 3 WA1 Revision Paper 2 (Worked Solution) 2 (a) Compound B is a hydrocarbon with the formula of C xHy. A gaseous sample of B occupies 20 cm3 at 37 C and 100 kPa. The 20 cm3 sample was completely burnt in 200 cm3 oxygen gas (an excess). The final volume, measured at the same conditions as the gaseous sample, was 150 cm3. Under these conditions, water produced is in liquid state. Treating the remaining gaseous mixture with NaOH(aq), decreased the volume by 100 cm3. Use the data given to calculate the values of x and y. Before combustion (1st contraction) after combusion (2nd contraction) After passing through NaOH Second contraction due to reaction of CO2 with NaOH = = 100 cm3 () First contraction due to combustion = 200 + 20 – 150    ) x y 2 2 2C H O initial O remaining CO(V V ) (V V = 70   x y 2 2 2C H O initial O remaining COV (V V ) V = 70 = 70 20 +  100 = 70 Volume of O2 reacted, = 150 cm 3 () OR Vremaining O2 = 150 – 100 = 50 cm3 Vreacted O2 = 200 – 50 = 150 cm3 () CxHy(g) + (x + y4 ) O2(g)  xCO2(g) + y2 H2O(l) Volume reacted or formed / cm3 20 : 150 : 100 gas volume ratio = mole ratio 1 : 7.5 : 5 mole ratio from eqn 1 : (x + y4 ) : x - x = 5 () x + y 4 = 7.5  y = 10 () [2] 2COV   x y 2 2C H O reacted COV V V 2O reactedV CO2 formed + O2 unreacted O2 unreacted excess NaOH(aq) Acidic CO2(g) reacted with NaOH combustion CxHy + O2 added (excess) H2O(g) is condensed to H2O(l). Volume of liquid H2O is negligible as compared to volume of the gases. same basis of 1 mol of CxHy(g) 20 cm3 200 cm3 150 cm3 VCO2 = 100 cm3 150 – 100 = 50 cm3

2026 J1 H2 Chemistry (9476) 4 WA1 Revision Paper 2 (Worked Solution) (b) Another hydrocarbon, C2H4, reacts with MnO4, as shown in equation 1. equation 1 5C2H4 + 12MnO4 + 36H+  10CO2 + 12Mn2+ + 28H2O (i) Explain, with reference to oxidation numbers, why this reaction is a redox reaction. Oxidation number of C increases from –2 in C2H4 to +4 in CO 2 () so C2H4 is oxidised (). Oxidation number of Mn decreases from +7 in MnO 4– to +2 in Mn2+ () so MnO4– is reduced. () [1] (ii) Write a half-equation to represent the conversion of C2H4 to CO2. K Balance the element oxidised/reduced in the half equation. C2H4  2CO2 O Balance each oxygen atom with one molecule of water. C2H4 + 4H2O  2CO2 H Balance each hydrogen atom with one H+ (in acidic medium). C2H4 + 4H2O  2CO2 + 12H+ e Balance total charge with electrons C2H4 + 4H2O  2CO2 + 12H+ charge on LHS = 0 charge on RHS = 12+ Balanced half-eqn: C2H4 + 4H2O  2CO2 + 12H+ + 12e [1] (c) Sodium nitrite, NaNO 2, can be oxidised to sodium nitrate, NaNO 3, by reacting it with acidified potassium manganate(VII). The equation below shows the reaction between MnO4 and NO2. 2MnO4 + 5NO2 + 6H+ → 2Mn2+ + 5NO3 + 3H2O A 1.00 g sample of impure NaNO 2 was dissolved in water and the volume of the solution made up to 250 cm3. A 25.0 cm3 portion of this solution was found to react, in acidic medium, with 27.90 cm3 of 0.0140 mol dm–3 KMnO4. (i) Calculate the mass of NaNO2 present in the sample. MnO4– : KMnO4 1 : 1 n(MnO4–) reacted = 0.0140  27.901000 = 3.906  104 mol () MnO4– : NO2– 2 : 5 n(NO2–) in 25.0 cm3 solution = 52  (3.906 x 10-4) = 9.765  104 mol () [2] add 12e on RHS

2026 J1 H2 Chemistry (9476) 5 WA1 Revision Paper 2 (Worked Solution) n(NO2–) in 250 cm3 solution = (9.765  104)  25025.0 = 9.765  103 mol () NaNO2 : NO2– 1 : 1 Mass of NaNO2 in impure sample = (9.765 x 103)  69.0 = 0.674 g () 3sf + units (ii) Calculate the percentage purity of NaNO2 in the sample. % purity = 0.674 1.00  100% = 67.4 % [3sf + units] [1] (iii) Draw a dot-and-cross diagram of the NO3 ion in NaNO3. N: 5 valence electrons; N cannot expand octet! O: 6 valence electrons Overall charge of 1– due to one additional electron which goes to more electronegative O [1] additional electron to be shown as a different symbol from valence electron on O [Total: 8] OR

2026 J1 H2 Chemistry (9476) 6 WA1 Revision Paper 2 (Worked Solution) 3 (a) A sample of an alloy containing rhodium (Rh) and iridium ( Ir) was analysed and three different types of atoms were identified; 103 45Rh , I 191 77 r and I 193 77 r . Table 3.1 shows information about the three types of atom found in the sample. Table 3.1 atom relative % abundance 103Rh 50.00 191Ir 15.18 193Ir 34.82 Use the data in Table 3.1 t