School of Chemistry, University of KwaZulu-Natal, Westville Campus GENERAL PRINCIPLES CHEMISTRY - CHEM110 TEST 3 Date: WEDNESDAY, 5 May 2010 Total marks: 25 Time: 18h00 18h45 Examiners: Mrs H Govender and Dr B Moodley IMPRTANT: Complete this part immediately. Name: Student No: Tutorial Day: Tutorial Venue: Tutor s Name: INSTRUCTINS: 1. Answer ALL questions. 2. or Section A which contains the multiple choice questions, follow the instructions given in the question. 3. Calculators may be used but all working must be shown. 4. The pages of this test must not be unpinned. 5. Your answers for Section B must be written on the question paper in the spaces provided. The left-hand pages may be used for extra space or for rough work. 6. Marks will be deducted for the incorrect use of significant figures and the omission of units. 7. You must write legibly in black or blue ink. Pencils and Tipp-Ex are not allowed. 8. This test consists of 11 pages. Please check that you have them all.. A data sheet and a periodic table are provided. Question No. Mark Section A 1 2 3 4 5 6 Total
SECTIN A - Multiple Choice Questions or each of the following questions, select the correct answer from the list provided. There is only one correct answer for each question. There is N negative marking. 1. Which one of the following is not the electronic configuration of an atom of a noble gas? (1) A 1s 2 2s 2 2p 6 3s 2 3p 6 B 1s 2 2s 2 2p 6 C 1s 2 2s 2 D 1s 2 2. How many single covalent bonds must a silicon, Si, atom form to have a complete octet in its valence shell? (1) A 3 B 4 C 1 D 2 3. Which of the following molecules has a bond that is non-polar covalent? (1) A B C D H-Cl C- H-B C- 2
4. Which of the following molecules is expected to have the strongest intermolecular force of attraction? (1) A NH 3 B H 2 Se C SiH 4 D CH 4 5. If a chemical reaction is endothermic, the reactants have: (1) A B C D a lower potential energy than the products a higher potential energy than the products the same potential energy as the products None of the above [5] End of Section A 3
SECTIN B QUESTIN 1 a) List all the possible values for each of the four quantum numbers: n, l, m l and m s, for an electron in a 4f orbital. (1) n = 4 ( ¼) l = 3 ( ¼) m l = +3, +2, +1, 0, -1, -2, -3 ( ¼) m s = + ½, - ½ ( ¼) b) Write out the full electronic configuration for Ca 2+. (1) Z = 20 e - - 2 e - = 18 e - 1s 2 2s 2 2p 6 3s 2 3p 6 (1) (If answer is partly correct then a half mark should be given) c) Consider the valence electrons of lead and draw an orbital diagram showing the distribution of these valence electrons. State whether the atom is paramagnetic or diamagnetic. (1½) Pb has 4 valence e - - 6s 2 6p 2 6p 6s ( ) Paramagnetic (½) 4
QUESTIN 2 Draw THREE plausible Lewis structures for carbonic acid, H 2 C 3, and determine the formal charge on each carbon and oxygen atom. Show all bonds to atoms, any lone pairs of electrons and all formal charge calculations. The skeletal structure for carbonic acid is: (5) H C H Answer: H C H H C H (a) H C H H C H H C H Structure (a) (b) (c) C(C) = 4-(0+4) = 0 ( ) C(=) = 6-(4+2) = 0 ( ) C(-) = 6-(4+2) = 0 ( ) Structures (b) and (c) C(C) = 4-(0+4) = 0 ( ) C(=) = 6-(2+3) = +1 ( ) C(-) = 6-(6+1) = -1 ( ) C(-H) = 6-(4+2) = 0 ( ) (A ½ mark for each correctly drawn Lewis structure and a ½ mark for each calculated formal charge. If any part of the structure is incorrect, the student loses the ½ mark for the structure). 5
QUESTIN 3 a) Draw the Lewis structure for arsenic pentafluoride, As 5. (1) As (Give a ½ mark if no lone pairs of electrons are shown on As). b) Explain why the octet rule is not obeyed for the Lewis structure of arsenic pentafluoride, As 5. (1½) Arsenic is surrounded by 10 electrons ( ) which is 2 more than the octet accommodates. Arsenic therefore has an expanded octet ( ) which allows the 2 extra electrons to fill into the d orbitals ( ). (Each is a ½ mark) 6
QUESTIN 4 Use VSEPR theory to predict the electron group geometry, molecular geometry and bond angle for ICl 2. (2) Answer: Cl I Cl ( ) Electron group geometry is trigonal bipyramidal ( ) and molecular geometry is linear ( ). Bond angle is 180 ( ). (Each is a ½ mark. The central I atom must show 3 lone pairs of electrons to get the ½ mark for the structure if lone pairs of electrons on Cl are not shown, student can still get the ½ mark) 7
QUESTIN 5 The following fluorides and their boiling points are shown: H boiling point = + 1.5 C N 3 boiling point = - 12.1 C 2 boiling point = - 21.6 C Identify the type of intermolecular bonding found in each molecule and explain why H has the highest boiling point and 2 has the lowest boiling point. (3) Answer: H boiling point = + 1.5 C - hydrogen bonding ( ) N 3 boiling point = - 12.1 C - dipole-dipole interactions ( ) 2 boiling point = - 21.6 C - London or dispersion forces ( ) H has hydrogen bonding which is a stronger intermolecular force ( ) than dipoledipole interactions and London dispersion forces. The London or dispersion forces found in 2 are induced dipoles that are instantaneous and momentary ( ) and are weak forces of attraction ( ). (Each is a ½ mark) 8
QUESTIN 6 The total volume of the Pacific cean is 7.2 x 10 8 km 3. A large atomic bomb produces 1.0 x 10 24 J of energy upon explosion. Calculate the number of atomic bombs needed to release enough energy to raise the temperature of the water in the Pacific cean by 1 C. The specific heat of the water in the Pacific cean is 4.184 J g -1 C -1 and its density is 1.00 g ml -1. (4) Note: 1 km 3 is equal to 1 x 10 15 cm 3. Answer: 1 km 3 = 1 x 10 15 cm 3 7.2 x 10 8 km = x ( ) x = (7.2 x 10 8 km 3 x 1 x 10 15 cm 3 )/1 km 3 ( ) (correct substitution) = 7.2 x 10 23 cm 3 ( ) q = m x s x T = (7.2 x 10 23 ml x 1.00 g ml -1 ) ( ) x 4.184 J g -1 C -1 x (1 C) ( ) (conversion of volume to mass and realizing that T is 1 C) = 3.0 x 10 24 J ( ) No. of atomic bombs needed = 3.0 x 10 24 J/1.0 x 10 24 J ( ) = 3.0 ( ) (Each is a ½ mark. If calculation of volume in cm 3 is incorrect, but method of calculating the no. of atomic bombs is correct, then student must be given part marks for using the incorrect value in the correct method to calculate q and no. of bombs final answer will be incorrect.) Note: This test is not being marked for significant figures.
10 1 18 1 H 1.008 2 13 14 15 16 17 2 He 4.003 3 Li 6.41 4 Be.012 Periodic Table 5 B 10.81 6 C 12.01 7 N 14.01 8 16.00 1.00 10 Ne 20.18 11 Na 22. 12 Mg 24.31 3 4 5 6 7 8 10 11 12 13 Al 26.8 14 Si 28.0 15 P 30.7 16 S 32.07 17 Cl 35.45 18 Ar 3.5 1 K 3.10 20 Ca 40.08 21 Sc 44.6 22 Ti 47.88 23 V 50.4 24 Cr 52.00 25 Mn 54.4 26 e 55.85 27 Co 58.3 28 Ni 58.6 2 Cu 63.55 30 Zn 65.3 31 Ga 6.72 32 Ge 72.61 33 As 74.2 34 Se 78.6 35 Br 7.0 36 Kr 83.80 37 Rb 85.47 38 Sr 87.62 3 Y 88.1 40 Zr 1.22 41 Nb 2.1 42 Mo 5.4 43 Tc 8.1 44 Ru 101.1 45 Rh 102. 46 Pd 106.4 47 Ag 107. 48 Cd 112.4 4 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126. 54 Xe 131.3 55 Cs 132. 56 Ba 137.3 57* La 138. 72 Hf 178.5 73 Ta 180. 74 W 183.8 75 Re 186.2 76 s 10.2 77 Ir 12.2 78 Pt 15.1 7 Au 17.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 20.0 84 Po (20) 85 At (210) 86 Rn (222) 87 r (223) 88 Ra (226) 8* * Ac (227) 104 Db (261) 105 Jl (262) 106 Rf (263) 107 Bh (262) 108 Hn (?) 10 Mt (?) * Lanthanide Series 58 Ce 140. 1 5 Pr 140. 60 Nd 144. 2 61 Pm (147) 62 Sm 150. 4 63 Eu 152. 0 64 Gd 157. 2 65 Tb 158. 66 Dy 162. 5 67 Ho 164. 68 Er 167. 3 6 Tm 168. 70 Yb 173. 0 71 Lu 175. 0 ** Actinide Series 0 Th (232) 1 Pa (231) 2 U (238) 3 Np (237) 4 Pu (23) 5 Am (243) 6 Cm (247) 7 Bk (247) 8 Cf (252) Es (252) 100 m (257) 101 Md (256) 102 No (25) 103 Lr (260)
DATA SHEET Physical Constants Boltzmann constant k = 1.381 x 10-23 J K -1 Planck constant h = 6.626 x 10-34 J s Elementary charge e = 1.602 x 10-1 C Speed of light in vacuum c = 2.8 x 10 8 m s -1 = 2.8 x 10 10 cm s -1 Avogadro constant L or N A = 6.022 x 10 23 mol -1 Gas constant R = kl = 8.315 J K -1 mol -1 = 8.315 L kpa K -1 mol -1 = 0.08206 L atm K -1 mol -1 Molar volume of an ideal gas V = 22.414 L mol -1 o m (at 1.000 atm and 273.2 K) V m = 24.78 L mol -1 (at 100.0 kpa and 28.2 K) araday constant = el =.6485 x 10 4 C mol -1 Atomic mass unit (amu) u = 1.661 x 10-27 kg Rest mass of electron m e =.10 x 10-31 kg Rest mass of proton m p = 1.673 x 10-27 kg Rest mass of neutron m n = 1.675 x 10-27 kg Vacuum permittivity Standard acceleration ε υ = 8.854 x 10-12 J -1 C 2 m - 1 of free fall g =.807 m s - 2 Rydberg constant for the H atom R H = 10677 cm -1 Conversion actors 1 micron (μ) = 10-6 m = 1 μm 1 Ångström (Å) = 1 x 10-10 m = 0.1 nm = 100 pm 1 L = 10-3 m 3 = 1 dm 3 1 atm = 1.013 x 10 5 N m -2 = 1.013 x 10 5 Pa = 760 mmhg = 760 Torr 1 bar = 1.000 x 10 5 Pa 1 J = 0.230 cal = 1 Pa m 3 = 1 m 2 kg s -2 1 cal = 4.184 J 1 ev = 1.602 x 10-1 J 1 L atm = 101.3 J 1 W = 1 J s -1 1 ppm = 1 μg g -1 = mg kg -1 = 1 mg L -1 (dilute aqueous solutions only) 1 tonne = 1000 kg Prefixes to Units P T G M k d c m μ n p f peta tera giga mega kilo deci centi milli micro nano pico femto 10 15 10 12 10 10 6 10 3 10-1 10-2 10-3 10-6 10-10 -12 10-15 11