VJC Prelim P2 Answers

 VJC 2017 9729/02/PRELIM/17 [Turn over 1 Victoria Junior College 2017 H2 Chemistry Prelim Exam 9729/2 Suggested Answers 1 This question concerns some unusual nitrogen compounds. (a) Heating monomer X to 150 C produces the trimer, melamine, C3H6N6. Melamine is a cyclic planar molecule and is symmetrical. (i) Suggest a structure for melamine. C N C N C N NH2 NH2 H2N C N C N C N NH2 NH2 H2N N N N NH2 NH2 H2N OR OR [1] (ii) The average C–N bond length is 0.145 nm and the average C=N bond is 0.125 nm. Suggest why all the carbon-nitrogen bonds in melamine are intermediate in length between the given C–N and C=N values. The structure of melamine is resonance–stabilised due to delocalisation of p orbital electrons between carbon and nitrogen atoms. Hence all carbon-nitrogen double bonds are partial double bonds. [1] (iii) Deduce the structure of X. Calculate the number of protons, neutrons and electrons in a monomer of X. 22p, 20n, 22e– [2] (b) Group 1 metal azides, MN3, can be formed by passing heated dinitrogen oxide, N 2O over their corresponding amines, MNH2. (i) Explain why lattice energies of Group 1 azides become less exothermic down the group. Lattice energy  |q+.q– / (r+ + r–)|. Down Group 1, cationic radius increases, thus lattice energy becomes less exothermic. [1] (ii) Suggest why the azides become thermally more stable down the group. Down Group 1, cationic radius increases while charge remains the same. The charge density and polarising po wer of cations decrease. Extent of polarisation of anionic charge cloud N 3– is less. More energy is required to decompose and break the N-N bonds within N 3– down Group 1. Hence azides become more stable to heat. [2]

 VJC 2017 9729/02/PRELIM/17 [Turn over 2 (c) Nitrogen has been used extensively in the research of compounds involving quadruple (bond order of 4) and quintuple (bond order of 5) bonds. These compounds typically involve transition metal atoms that are able to form bonds between themselves using their d orbitals. An example is given below: (i) Sketches of the shapes of the atomic orbitals from the d subshells are shown below, in random order. Name and label each orbital. dz2 dx2 – y2 dyz dxy

 VJC 2017 9729/02/PRELIM/17 [Turn over 3 dxz [2] (ii) The d orbitals of an atom can overlap with d orbitals of the same type to form pi () and delta () bonds. While a single sigma ( ) bond involves the overlap of two orbital lobes in total, and a single pi ( ) bond four lobes, a single delta ( ) bond involves the overlap of eight lobes in total. When two atoms overlap, the z-axis is used to define the internuclear axis. Suggest two different d orbitals that could be involved in delta bonds (). d x2 – y2 and dxy [1] [Total: 10]

 VJC 2017 9729/02/PRELIM/17 [Turn over 4 2 Use of Data Booklet is relevant to this question. Aerozine 50 is a 50/50 mix of UDMH, (CH3)2N2H2 and hydrazine, N2H4. It is used as a rocket fuel, and is typically mixed with dinitrogen tetroxide, N2O4, as the oxidising agent. The equation for the reaction between UDMH and dinitrogen tetroxide under standard conditions is given as follows: (CH 3)2N2H2(l) + 2N2O4(g)  2CO2(g) + 3N2(g) + 4H2O(l) ---------- reaction 1 (a) Suggest an equation, including state symbols, for the reaction between hydrazine and dinitrogen tetroxide under standard conditions. 2N2H4(l) + N2O4(g)  3N2(g) + 4H2O(l) [1] (b) An experiment was set up such that hydrazine in a spirit burner was combusted beneath a copper can filled with water. It was found that 0.50 g of hydrazine was required to raise the temperature of 100 cm3 of water in the can by 15 ºC. Using relevant data from the Data Booklet and the information given below, calculate the efficiency of the system, expressed as a percentage.  The heat capacity of the copper can be taken to be 96.0 J K –1  The density of water can be taken to be 1.00 g cm –3.  The standard enthalpy change of combustion of hydrazine, found using a bomb calorimeter, has a value of –628 kJ mol–1. Qaborbed = (100)(4.18)(15) + (96.0)(15) = 7710 J Qreleased = 628000 x (0.5/32.0) = 9813 J Efficiency = 7710/9813 x 100 % = 78.6 % [3]

 VJC 2017 9729/02/PRELIM/17 [Turn over 5 Pure UDMH, (CH3)2N2H2, can be used as an alternative to Aerozine 50 in thruster rockets. (c) Using relevant data from the Data Booklet and the information given below, construct an energy cycle to calculate the enthalpy change for reaction 1. (CH3)2N2H2(l) + 4O2(g)  2CO2(g) + N2(g) + 4H2O(l) –1980 kJ mol–1 N2O4(g)  2N(g) + 4O(g) +1930 kJ mol–1 ∆Hr 4O2(g) + (CH3)2N2H2(l) + 2N2O4(g) 2CO 2(g) + 3N2(g) + 4H2O(l) + 4O2(g) –1980 2CO 2(g) + N2(g) + 4H2O(l) + 2N2O4(g) 4BE(O=O) + 2BE(N ≡N) = 4(496) + 2(944) 2(+1930) 2CO 2(g) + N2(g) + 4H2O(l) + 4N(g) + 8O(g) By Hess Law ∆Hr + 4(496) + 2(944) = –1980 + 2(+1930) ∆Hr = –1990 kJ mol–1 [3] (d) The total mass of propellant ( UDMH and dinitrogen tetroxide, N 2O4) used in the thruster rockets in the ascent stage of a lunar module was 366 kg. Assuming that UDMH ( M r = 60.0) and dinitrogen tetroxide ( Mr = 92.0) were mixed according to the stoichiometric ratio, calc ulate the mass of UDMH in the propellant mixture. Let amount of UDMH be x mol and amount of N2O4 be 2x mol x(60.0) + 2x(92.0) = 366000 x = 1500 mol mUDMH = 1500 x 60.0 = 90000 g = 90 kg [1] [Total: 8]

 VJC 2017 9729/02/PRELIM/17 [Turn over 6 3 Use of the Data Booklet is relevant to this question. Phosphorus and sulfur are elements in Period 3 in the Periodic Table and each can exist in various allotropic forms. Phosphorus can exist as white phosphorus and red phosphorus. White phosphorus exists as a tetrahedron with bond angle 60° while red phosphorus exists as a polymeric chain of regular tetrahedrons. Sulfur, on the other hand is thermodynamically most stable at room temperature as rhombic sulfur, which consists of puckered S 8 rings. Both elements form a wide range of compounds with the halogens. (a) (i) By considering the bond angles involved, suggest why white phosphorus is less stable than red phosphorus. The bond angle of 60° in white phosphorus, P 4 is smaller than 107° in red phosphorus. Hence there will be high angular strain in the bonding / strong electronic repulsion between the bond pairs. Thus, white phosphorus is less stable than red phosphorus. [1] (ii) The phosphorus halides fume in air because of reaction with water vapour. Explain briefly why phosphorus( V) chloride can react with water. Write a balanced equation for its reaction. PCl5 undergoes hydrolysis in water due to the presence of energetically accessible vacant 3d orbitals on phosphorus which can accommodate lone pair of electrons from oxygen atoms of water molecules. PCl5 + 4H2O  H3PO4 + 5HCl [2]

 VJC 2017 9729/02/PRELIM/17 [Turn over 7 (b) When sulfur is heated under pressure with chlorine, the major product is SCl2. S8(g) + 8Cl2(g)  8SCl2(g) (i) Using data from the Data Booklet, calculate the enthalpy change, ∆H, for this reaction. ∆H = Bonds broken – Bonds formed = 8 x BE(S-S) + 8 x BE (Cl-Cl) – 16 x BE(S-Cl) = (8 × 264) + (8 × 244) – (16 × 250) = +64 kJ mol –1 [2] Under suitable conditions, SCl2 reacts with water to produce a yellow precipitate and a solution X. Solution X contains a mixture of SO2(aq) and compound Y. (ii) By constructing an equation for the hydrolysis of SC l2, work out how the oxidation number of s