A Foundation Tier Chemistry: Bonding and Structure - Assignment Example

OF TEESSIDE SCHOOL OF SCIENCE AND TECHNOLOGY CHEMICAL & BIOPROCESS ENGINEERING SECTION MODULAR SCIENCE STAGE Session 2008/2009 STRUCTUREAND BONDING (Resit home paper) ______________________________________________________________ DATE: August 2009 TIME: 2 hours ______________________________________________________________ INSTRUCTIONS TO CANDIDATES This paper contains SIX questions in three sections. Answer ALL questions. All questions carry equal marks. The pass mark is 60% Answer ALL PARTS of a question unless specifically instructed otherwise. The use of calculators is permitted provided they do not have a QWERTY keyboard. The following are provided: Chemistry data sheet SECTION A 1(a) This part of the question refers to the complexes A FeCl3(H2O)3 and B [Mn(CN)2(phen)2]+ For each of the above complexes: (i) State what types of isomerism are possible and illustrate each with an appropriate sketch. A FeCl3(H2O)3 Iron has a coordination number 6, so the complex has an octahedral structure. Geometric isomerism is possible, giving facial and meridional isomers as in the figure below:. Here, M is the metal (Fe), X the ligand Cl- and L the ligand H2O. When ligands of the same type are in the same plane, you have meridional isomer, and when they are adjacent, they are facial, forming the triangular face of the octahedron. B [Mn(CN)2(phen)2]+ Here Mn has a coordination number of 5. So, the complex has a trigonal bipyramidal or square pyramidal structure. Optical isomers exist. (ii) Determine the metal oxidation state and the number of d electrons. A FeCl3(H2O)3 Overall charge = 0 Removing 3 Cl- ions gives a charge of +3, while H2O ligands do not contribute to the charge. Therefore, oxidation state of Fe in the complex is +3. Therefore, the d electron configuration for Fe here is d5. B [Mn(CN)2(phen)2]+ Overall charge = +1 Removing 2(CN)- ions gives a total charge of +3, while phenyl ligands do not contribute to the charge. Therefore, oxidation state of Mn in the complex is +3. Therefore, the d electron configuration for Mn here is d4. (iii) State whether you would expect it to be high or low spin, giving a reason for your answer. A FeCl3(H2O)3 Chloride ions cause a small splitting energy of d-orbitals. Hence, the spin is high. B [Mn(CN)2(phen)2]+ CN- and phenyl ligands cause a huge splitting energy of d-orbitals. Hence, the spin is low. (iv) Give the electron configuration in terms of occupation of t2g and eg (or e and t2) orbitals, and calculate the magnetic moment in Bohr magnetons. A FeCl3(H2O)3 Here, the energy difference between t2g and eg orbitals is low. So, the 5 d electrons are split as t2g3 eg2. Magnetic moment is calculated as µ = sqrt(n(n+2)), where n is the number of unpaired electrons. Here, n=5. Therefore, µ = sqrt(n(n+2))=sqrt(35)= 5.92 Bohr magnetons. B [Mn(CN)2(phen)2]+ Here, the energy difference between t2g and eg orbitals is high. So all d electrons are in t2g level or t2g4. Here, n=2 Therefore, µ=sqrt(8)=2.83 Bohr magnetons. (b) Calculate the wavelength of (i) The photon absorbed by an electron undergoing the transition from the n = 3 to the n = 7 level in a hydrogen atom. For an electron undergoing transition between different series in the hydrogen atom, the wavenumber is calculated as: Wavenumber = RH(1/n12 – 1/ n22) where RH is the Rydberg constant for hydrogen and n1 and n2 are the level numbers 3 and 7, respectively. For an infinitely heavy nucleus, RH =10,973,731.57m-1. Wavenumber = 10,973,731.57m-1 (1/32 – 1/72)=995349.8m-1. Wavelength of a photon is the reciprocal of wave number. λ=(1/Wavenumber)=(1/995349.8m-1)=1.00467×10-6 m. (iii) A photon absorbed by a transition metal complex with o = 305 kJ mol-1 We assume energy absorbed by the complex corresponds to energy lost by the photon in emission. Therefore, E= o = 305 kJ mol-1. E=hc/λ, where h is the Planck’s constant, c the speed of light, and λ the wavelength of the photon for one mole of photons. Here, we substitute h by h per mole : h = h×NA. Therefore, wavelength is calculated as λ =hc/E=hNAc/E =6.626×10-37kJs×6.022×1023mol-1 ×3×108ms-1/305 kJ mol-1 =3.92×10-7m 2(a) For each of the following molecules, give the hybridisation of the central atom, the molecular shape, and the approximate bond angles, indicating how you arrive at your answers. (i)SbF6- Sb has an electronic configuration of 1s22s22p63s23p63d104s24p64d105s25p3. To accommodate the 6 F- ions, sp3d2 hybridization of the antimony atom takes place. This gives rise to an octahedral structure, with a bond angle of 90°. (ii)SOCl2 S has an electronic configuration of 1s22s22p63s23p4. To accommodate the lone pair, double bond with oxygen and bonds with chloride ions, sp3 hybridization takes place. The molecule is pyramidal in shape, with lone pair on top. Structure of SOCl2 is pyramidal. In thionyl chloride, the O-S-Cl bond angle is greater than the Cl-S-Cl bond. The O-S-Cl bond angle is approximately 108°, while Cl-S-Cl bond angle is 99°. (b)The electron configuration of Br is 1s22s22p63s23p63d104s24p5. (i) How many electrons have the quantum number  = 1 ? (ii) How many electrons have m = -1? Explain your answers briefly. (i) All the p electrons have Azimuthal quantum number,  = 1. Here, the number of electrons in p orbitals of Br atom is 17. (ii) Magnetic quantum number m = -1, 0, +1 for  = 1 and m = -2,-1, 0, +1, +2 for  = 2 (d orbital). Two 2p, two 3p, and two 3d and two 4p electrons have m = -1. Therefore, m = -1 for 8 electrons in boron. (c) Draw a Born-Haber cycle for the formation of potassium oxide K2O Born Haber cycle for K2O (d) Calculate the heat of formation of potassium oxide using the following data. heat of sublimation of potassium = 90 kJ mol-1 first ionisation energy of potassium = 419 kJ mol-1 bond dissociation energy of O2 = 498 kJ mol-1 Electron attachment enthalpies of oxygen: First = -141 kJ mol-1 Second = 861 kJ mol-1 Lattice energy of potassium oxide = -2238kJ mol-1 Heat of formation of K2O, ΔHf0 = 2(Heat of sublimation of potassium)+2(First ionization energy of potassium)+Bond dissociation energy of O2 + First ionization energy of oxygen+ Second ionization energy of oxygen + lattice energy of K2O = 2(90)+2(419)+498+(-141)+(861)+(-2238) KJmol-1 = -2KJmol-1. Heat of formation of potassium oxide is -2 KJmol-1. (e) The radius of the Tl+ ion is 159 pm and that of Br- is 196 pm. What is the likely structure of TlBr? Briefly explain how you arrive at your answer. Radius of cation, r+ = 159pm Radius of anion, r- = 196pm r+/r- = 159pm/196pm = 0.811 A radius ratio between 0.732 and 1 indicates a coordination number of 8, giving a cubic structure similar to CsCl. Structure of TlBr SECTION B 1Answer ALL parts (a)Determine which of the following alkenes exist as pairs of stereoisomers and give appropriate illustrations of both isomers for those that do: (i) CH2=C(Cl)CH3 Does not have stereoisomers. (ii) CH3CH2CH=CHCH2Cl (iii) No stereoisomers. (iv) (b)Using Fischer projections, represent each stereoisomer of 2,3-dihydroxy-1,4-butandial, OHCCH(OH)CH(OH)CHO, and use these representations to explain the meaning of the terms enantiomer, diastereoisomer and meso isomer. Eight stereoisomers exist: Stereoisomers of 2,3-dihydroxy-1,4-butandial Enantiomers: F & G are mirror images and are enantiomers. Meso isomer: A & D are superimposable on their mirror images and are hence meso isomers. Diastereoisomers: B, C, E, & H are nonsuperimposable and are diastereoisomers. (c) For ONE of the stereoisomers of 2,3-dihydroxy-1,4-butandial, specify the configurations of the chiral atoms using the Cahn-Ingold-Prelog (R, S) convention giving full details about how the assignment is made. (d) A chiral organic acid (293 mg) was dissolved in water (25 mL) and placed into a cell of path length 5 cm at 25 oC before recording a value for  of +19.5 o. Calculate the specific optical rotation of the acid. Density of acid d = 29310-3g/25mL= 0.01172gmL-1 Length l = 5 cm = 0.5 dm =+19.5° Specific optical rotation of the acid is calculated as: []=/(ld)=19.5°/(0.5dm0.01172gmL-1) = 3327.65 This value is very high, so we subtract the nearest multiple of 360. [] = 3327.65 -360(9) = 87.65°. 2 Answer ALL parts: (a) Account for the order of reactivity of the carbonyl compounds below in nucleophilic addition reactions: HCHO > CH3CHO > CH3COCH3 Nucleophilic addition of aldehydes and ketones involves a tetrahedral transition state with partial negative charge on oxygen. The carbonyl group in HCHO is least crowded in the transition state and is more open to attack. So, the reactivity of HCHO is highest during nucleophilic addition. An extra methyl group in acetaldehyde increases crowding in the transition state. The carbonyl group here is not as susceptible to attack as in HCHO. So acetaldehyde is less reactive than formaldehyde during nucleophilic addition. In acetone, the two methyl groups increase crowding in comparison with 1 methyl group in acetaldehyde. Hence, acetaldehyde is least reactive to nucleophilic addition. (b) Different classes of alcohol can be oxidised to give a variety of carbonyl containing compounds, using examples of YOUR choosing discuss the types of alcohol, oxidant and product formed in these reactions. Primary alcohols can lose one or two -hydrogens. Loss of one -hydrogen results in formation of aldehyde (1), while loss of two -hydrogens results in formation of carboxylic acid (2). In reaction (1), KMnO4 is used as the oxidant and in reaction (2), pyridinium chlorochromate in CH2Cl2 is used as oxidant. Secondary alcohols have only one -hydrogen to lose. Oxidation results in formation of ketone, with loss of -hydrogen (3). Tertiary alcohols have no -hydrogen to lose. They cannot be oxidised. (c) Give products for the following addition and substitution reactions. Choose ONE of the reactions & provide a curly arrow mechanism for the reaction: SECTION C 1 Sodium azide is used in air bags and can decompose explosively when heated, according to the equation: 2 NaN3 (s)  3 N2(g) + 2 Na(s) (a) How many litres of N2 would be formed at 30 oC and 1.20 atm pressure by the explosion of 45 g of NaN3 ? According to the equation, 2 moles of NaN3 explode to form 3 moles of nitrogen. Therefore, 1 mole of NaN3 gives 1.5 moles of N2. Molar mass of NaN3=65gmol-1. Number of moles of NaN3 in 45g = Mass/molar mass = 45g/65gmol-1 = 0.69mol. No. of moles of N2 obtained (n) = 0.691.5 = 1.04 mol. From the ideal gas law, we have PV=nRT Where P = 1.2atm, R = 0.0821LatmK-1mol-1 and T =303K. Therefore, volume is calculated as: V=nRT/P = (1.04mol 0.0821LatmK-1mol-1303K)/1.2 atm = 21.56L. Therefore, 21.56L of N2 is formed when 45g of NaN3 explodes. (b) How many grams of sodium azide would have been used if 38 litres of N2 is produced at 36 0C and 1.8 atm pressure? Rearranging the ideal gas equation to calculate the number of moles n, we get: n=PV/RT. Here, P = 1.8atm, V = 38L, R = 0.0821LatmK-1mol-1 and T =309K Therefore, n = (1.8 atm  38L)/(0.0821LatmK-1mol-1309K) = 2.7 mol. 1.5 moles of N2 is obtained from 1 mole of NaN3. Number of moles of NaN3=2.7mol/1.5 = 1.8mol. Mass of NaN3=Number of molesmolecular mass = 1.8 mol65gmol-1 = 117g. Therefore, 117g of NaN3 is required to produce 38L of N2 at 36°C and 1.8atm. (c) What is the total mass (in grams) of oxygen in a room measuring 4.0 m by 5.0 m by 2.5 m? Assume that the gas is at STP and that air contains 20.95% oxygen by volume. Volume of room = 4.0m5.0m2.5m = 50m3. Percentage of oxygen by volume = 20.95% per unit volume. Volume of oxygen V = (20.95/100) 50m3 = 10.475m3.=1.0475104L. Standard Temperature and pressure refers to T= 273K and P= 1atm. From the ideal gas equation, we calculate number of moles of oxygen in the room as: n = PV/RT = (1atm1.0475104L)/( 0.0821LatmK-1mol-1273K)=467.35mol. Mass of oxygen = nmolecular mass = (467.35mol32gmol-1)= 14,955.41g. Therefore, 14,955.41 g of oxygen is present in 4.0 m by 5.0 m by 2.5 m room at STP. 2 For the reaction of CO and H2 to produce CH3OH CO(g) + 2 H2(g)  CH3OH(g) Hf (CO(g)) = -110.5 kJ mol-1 , Hf (CH3OH(g)) = -200.7 kJ mol-1 Sf (CO(g)) = +197.7 J mol-1K-1 , Sf (CH3OH(g)) = +239.8 J mol-1K-1 (a) Calculate the Gibbs free energy of the reaction Gr at 300K and use your value to calculate the equilibrium constant for the reaction at 300K. Gibbs energy of formation is calculated using: Gf = Hf - TSf Gf(CO)= Hf(CO)-TSf(CO) = -110.5 kJ mol-1 -(300K×197.7 ×10-3KJmol-1K-1) = -169.81KJmol-1. Gf(CH3OH)= Hf(CH3OH)-TSf(CH3OH) = -200.7kJmol-1 -(300K×239.8×10-3 KJmol-1K-1) = -272.64KJmol-1. Gf(H2) = 0 The Gibbs free energy of a reaction is calculated as: where, v represents stoichiometric coefficients and Gf the Gibbs energy of formation. For the given reaction: Gr =(1Gf(CH3OH)) - (1Gf(CO)+ 2Gf(H2)) = -272.64KJmol-1 - (-169.81KJmol-1) = -102.83 KJmol-1. Gibbs energy of a reaction is related to equilibrium constant K as: Gr = -RT(lnK) Here, T=300K, R = 8.314JK-1mol-1, and Gr = -102.83 KJmol-1. Therefore, lnK =-Gr/RT =(102.83103 Jmol-1)/( 8.314JK-1mol-1300K) =41.22 K=e41.22 = 7.971017. Gibbs energy is -102.83 KJmol-1 and equilibrium constant is 7.971017 for the reaction at 300K. (b) An unknown gas is found to diffuse through a porous membrane 2.92 times more slowly than H2. What is the molecular mass of the gas? Graham’s law of effusion/diffusion states that the rates of movement of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. This is written as For a gas with molar mass x that diffuses 2.92 times more slowly than H2 (molar mass = 2gmol-1): 2.92 = x/2 x=17.05gmol-1. A gas that diffuses 2.92 times more slowly than H2 has a molar mass of 17.05gmol-1. (c) Which has the higher average speed, N2 at 190 K or Cl2 at 600 K? For an ideal gas using the kinetic model of gases, Where, P is the pressure, V the volume, n the number of moles, M the molar mass and c the root mean square velocity. At constant temperature, using the ideal gas equation (PV=nRT), we get: For N2 at temperature T=190K, c=sqrt(38.314JK-1mol-1190K/28gmol-1) = sqrt(38.314103gm2s-2K-1mol-1190K/28gmol-1) =sqrt(169249.29m2s-2)=411.4 ms-1. For Cl2 at temperature T=600K, c=sqrt(38.314JK-1mol-1600K/70.9gmol-1) = sqrt(38.314103gm2s-2K-1mol-1600K/70.9gmol-1) =sqrt(211074.75m2s-2)=459.43 ms-1. Therefore, the average speed of Cl2 at 600K is greater than the average speed of N2 at 190K. Read More