2021 VJC Prelim P2 Ans

1  VJC 2021 8873/02/PRELIM/21 [Turn over Victoria Junior College 2021 H1 Chemistry Prelim Exam 8873/2 Suggested Answers Section A Answer all the questions in this section in the spaces provided. 1 (a ) The nth (n = 1, 2, 3, 4 or 5) ionisation energies of consecutive elements Q to Z, with atomic number below 20, are shown in Fig. 1.1. Fig. 1.1 (i) Define the nth ionisation energy of element T. The nth ionisation energy of T is the amount of energy required to remove one mole of electrons from one mole of T (n-1)+ gaseous ions, producing one mole of Tn+ gaseous ions. T(n-1)+(g)  Tn+(g) + e– [1] (ii) Element U is identified as Al. State the value of n. n = 3 or 3rd ionisation energy [Since U is a Group 13 element and T is a Group 2. There is a large decrease from T to U indicating that electrons is removed from a higher principle quantum shell in U2+.] [1] (b ) Aluminium chloride has a wide number of uses in the chemical industry. Anhydrous AlCl3 adopts different structures depending on the temperature and the state it is in. In the liquid state, it exists as the dimer Al2Cl6 while a mixture of monomer and dimer can be found in the gaseous phase.

2  VJC 2021 8873/02/PRELIM/21 [Turn over (i) State the type of interactions between two molecules of AlCl3. instantaneous dipole–induced dipole interactions [1] (ii) Draw a diagram to show the bonding in an Al2Cl6 dimer, and state the bond angle about the Al atom. Bond angle: 109.5o [2] (c ) Beryllium chloride, BeCl2, has properties similar to those of AlCl3. (i) Explain why BeCl2 behaves similarly to AlCl3. Both BeCl2 and AlCl3 are Lewis acids, i.e. they are electron deficient and act as an electron pair acceptor. [OR Both are covalent compounds due to similar charge to size ratio.] [1] (ii) BeCl2 and dimethylamine, (CH3)2NH reacts in the molar ratio 1:2. By considering the number of bond pairs and lone pairs around the central atoms of each molecule, explain the molar ratio and draw a diagram to show the bonding in the product. BeCl2: 2 bond pairs, 0 lone pair (CH3)2NH: 3 bond pairs, 1 lone pair BeCl2 can accept 1 lone pair of electrons from 2 molecules of (CH3)2NH to achieve octet configuration. [2] [Total: 8]

3  VJC 2021 8873/02/PRELIM/21 [Turn over 2 (a ) Fig. 2.1 shows the trend in melting points of some Period 3 elements: Fig. 2.1 (i) Complete the trend in melting point of the Period 3 elements for aluminium, silicon, chlorine and argon on Fig 2.1. [1] (ii) Explain, with reference to structure and bonding, why magnesium has a higher melting point than sodium. Both Na and Mg have giant metallic structures. Both Na and Mg have strong electrostatic forces of attraction between the cations and sea of delocalised electrons (i.e. metallic bonds). Mg has more delocalised electrons than Na, leading to stronger metallic bonds for Mg compared to Na. More energy is required to overcome the stronger metallic bonds of Mg. Thus, Mg has a higher melting point than Na. [2] (iii ) Explain, with reference to structure and bonding, why sulfur has a higher melting point than phosphorus. P4 and S8 have simple molecular structures. Both P4 and S8 have weak instantaneous dipole-induced dipole interactions (id-id) between their molecules. However, S8 has a larger and more polarisable electron cloud due to greater number of electrons present, leading to stronger id-id interactions between their molecules compared to P4. More energy is required to overcome the stronger id-id attractions in S8. Thus, S8 has a higher melting point than P4. [2] Na Mg Al Si P S Cl Ar melting point / K

4  VJC 2021 8873/02/PRELIM/21 [Turn over (b ) Serpentine has the formula Mg3Si2O5(OH)x. There is currently a lot of interest in the use of serpentine as a “green” cement as it chemically absorbs carbon dioxide, a major constituent of greenhouse gases. (i) Deduce the value of x in Mg3Si2O5(OH)x. 3(+2) + 2(+4) + 5(–2) + x(–1) = 0 x = 4 [1] (ii) Serpentine reacts with carbon dioxide to form water, silicon( IV) oxide and one other product. Write a balanced equation for this reaction. Mg3Si2O5(OH)4 + 3CO2 → 2H2O + 2SiO2 + 3MgCO3 [1] (c ) Dolomite has the formula MgY(CO3)2, where Y is a metal element. Dolomite is insoluble in water but dissolves in aqueous acid due to the reaction between the carbonate ions and hydrogen ions. CO32– + 2H+ → CO2 + H2O A 4.50 g sample of dolomite is dissolved in 25.0 cm3 of 5.00 mol dm–3 hydrochloric acid, in excess. The resulting solution is made up to 100 cm 3 in a volumetric flask using distilled water and labelled solution A. 10.0 cm3 of solution A is titrated against a 0.100 cm3 solution of sodium hydroxide. An average titre of 27.30 cm3 is obtained. (i) Calculate the amount of hydrochloric acid in 100 cm3 of solution A. Amount of NaOH in titration = 27.30 1000 x 0.100 = 0.00273 mol Amount of excess HCl (10.0 cm3) = 0.00273 mol Amount of excess HCl (100 cm3) = 100 10.0 x 0.00273 = 0.0273 mol [2] (ii) Hence, calculate the amount of hydrochloric acid that reacted with the 4.50 g sample of dolomite. Amount of added HCl = 25.0 1000 x 5.00 = 0.125 mol Amount of reacted HCl = 0.125 – 0.0273 = 0.0977 mol [1]

5  VJC 2021 8873/02/PRELIM/21 [Turn over (iii ) Determine the mole ratio of MgY(CO3)2 : CO32 – : H+. MgY(CO3)2 : CO32 – : H+ = 1 : 2 : 4 [1] (iv ) Hence, calculate the molar mass of Y and suggest its identity. Amount of dolomite in 4.50 g = 0.0977 4 = 0.0244 mol Molar mass of dolomite = 4.50 0.0244 = 184.4 g mol–1 Molar mass of Y ion = 184.4 – 24.3 – 2[12.0 + 3(16.0)] = 40.1 g mol–1 Y is Ca. [2] [Total: 13] 3 This question involves the study of the kinetics of decomposition of hydrogen peroxide, H2O2 in the presence of iron(III) nitrate, Fe(NO3)3. H2O2(aq) → H2O(l) + ½O2(g) A small amount of Fe(NO3)3 was added to 100 cm3 of 0.200 mol dm–3 H2O2 and the stopwatch was started. At various time intervals, 10 cm3 samples of the reaction mixture was extracted into a conical flask. 50 cm 3 of 0.200 mol dm –3 H2SO4 was added into a sample and titrated against 0.0200 mol dm–3 KMnO4. 2MnO4–(aq) + 5H2O2(aq) + 6H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g) Based on the titration results, the concentration of H 2O2 in each 10 cm 3 sample was calculated. A graph of concentration of H2O2 against time was obtained as show in Fig 3.1.

6  VJC 2021 8873/02/PRELIM/21 [Turn over Fig. 3.1 (a) (i) It was observed that the concentration of Fe(NO 3)3 in the reaction mixture remained constant throughout the whole experiment. Suggest and explain the role of Fe(NO3)3 in the decomposition of H2O2. FA 1 serves as a catalyst, that provides an alternative pathway with lower activation energy. [1] (ii) Hence, suggest and explain how the shape of the graph in Fig. 3.1 would change if Fe(NO3)3 was not added to the reaction mixture. The graph will have a more gen