Information sheet. and [X] is the concentration of X at time t = t. Joules n 2 c = m s 1 ε = hν c = νλ. Total Energy = K.E. + P.E.

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1 1 Dartmouth College Illustrative Chemistry 6 Credit test questions and their solutions As the following examples illustrate, the Chemistry 6 Credit test is, in the main, not a multiple-choice examination, but rather, it consists of a collection of short problems that provide the student an opportunity to display her or his problem-solving abilities. On such questions, major partial credit is awarded for the development of an approach that will lead to a successful answer. So, make sure that your answer shows clearly the approach you are using to solve the problem. The solutions to these questions are provided, not only for you to check your answers, but also to indicate the depth of answer that is expected. You should not be surprised to find that the depth of understanding expected in your answers to these questions sometimes goes beyond that expected in high school. To be awarded credit for Chemistry 6, a student must score at least 65% on the Chemistry 6 Credit test. You will be provided with an information sheet similar to the one included below. Information sheet dx x = ln x + C dx x 2 = 1 x + C ln [ X ] [ X ] 0 = k t 1 [ X ] 1 = k t [ X ] 0 k = A e E a RT where [X] 0 is the concentration of X at time t = 0 and [X] is the concentration of X at time t = t ln x = log 10 x R = J K 1 mol 1 = 2.0 cal K 1 mol 1 N A = atoms/mol K.E. = mv 2 p = 2 E 2 2 m n = Z 2 Joules n 2 c = m s 1 ε = hν c = νλ h = J s e = Coulombs 1nm = 10 9 m 1Å = 10 8 cm = m 1 J = ev 1 Volt 1 Coulomb = 1 Joule m(electron) = kg P.E. = Q 1 Q 2 α r Total Energy = K.E. + P.E. α = C 2 J 1 m 1 1 D = C m

2 2 H 2.1 Li Be B C N O F Na Mg Al Si P S Cl K Ca Ga Ge As Se Br Rb Sr In Sn Sb Te I Cs Ba Tl Pb Bi Po At Electronegativities of the representative elements

3 3 Question 1 Measurements of initial reaction rates were performed to determine the details of the differential rate law for the overall reaction (in aqueous solution): BrO 3 + 5Br + 6H + 3Br 2 + 3H 2 O with the following results: [BrO 3 ] [Br ] [H + ] Initial Reaxn Rate (mol L 1 ) (mol L 1 ) (mol L 1 ) (mol L 1 s 1 ) The differential rate law may be written in the form: d BrO 3 d t = k exp BrO 3 a Br b H + c (i). Determine the values of a, b and c. (ii). Using the data given in the above table and the result of part (i) of this question, calculate a value for the rate constant, k exp (state your units clearly). Question 2 The decomposition of species A was studied at 576K by recording the concentrations of A at known times with the results shown below: A Products (rate constant, k exp ) t (min) [A] mol L Calculate a value for the rate constant k exp. Show your method clearly and be sure to state the units of k exp.

4 4 Question 3 Rate constants k measured at different temperatures for the reaction CH 3 N 2 CH 3 2 CH 3 + N 2 (1) are given in the following table: T ( C) k (s 1 ) Calculate the activation energy, E a for the reaction described in equation (1). (Show your method clearly and state your units clearly). (Results which are inaccurate by more than 5% will not receive full credit). Question 4 The maximum wavelength (λ) of electromagnetic radiation required to eject electrons from the surface of tungsten metal is 272 nm. Calculate the maximum kinetic energy observed amongst the electrons ejected from tungsten metal by electromagnetic radiation with wavelength λ = 2000 Å. Express your answer in kj mol 1 and show your method clearly. Question 5 The energy required to remove the outermost electron from a ground state sodium (Na) atom is 496 kj mol 1. If a collection of Na atoms is heated in an electric discharge, two prominent emissions occur, one at a wavelength of 589 nm, and one at a wavelength of 820 nm. The former is emission resulting from an excited state with the electron configuration [Ne]3p 1, while the latter is emission resulting from an excited state with the electron configuration [Ne]3d 1. (i) (ii) Write down the electron configuration for the ground state of a Na atom. We take the zero of energy to correspond to Na + and a free electron, e, at infinite separation and both at rest. Calculate the energy (relative to the zero of energy defined above) of the 3p level for a Na atom. Express your answer in kj mol 1 and show your method clearly.

5 5 Question 6 (A) The ground state electron configuration of the gas phase O atom is [He]2s 2 2p 4. Using this notation write down (i) the ground state electron configuration and (ii) the number of unpaired electrons for each of the following gas phase atoms or ions. (a) Si (b) Al 3+ (c) As (B) Explain why the energy spacing between the 2s and the 2p levels in the gas phase nitrogen atom, N, is greater than the energy spacing between the 2s and the 2p levels in the gas phase boron atom, B. Question 7 (i) The H atom 3p y orbital is given by ψ 3py = constant r (6 r) exp ( r/3) sinθ sinφ where the distance r is expressed in atomic units. On the graph below, sketch a plot of the Radial Probability Distribution function ( r 2 R 2 (r)) vs. r for a H atom 3p orbital. Mark off the horizontal axis in atomic units. Your plot should identify the positions of nodes, relative positions of maxima, and which maximum is the largest. Give a brief explanation of your method. (ii) Using the plot you made in part (i), explain how you would calculate the probability of finding an electron in a 3p orbital within a distance of 3 Å from the nucleus.

6 6 Question 8 Predict the molecular geometries of each of the following molecules or ions. Show your method clearly and represent your prediction with a simple sketch and a descriptive name or phrase. Include a qualitative statement about deviations of bond angles from their idealized values (i.e. <180, <120, etc.). (a) GaI 3 (b) (BiCl 5 ) 2 Question 9 Consider the molecule HSCN (H S C N) and the ion SCN (S C N) (the skeletal structures in parentheses are only meant to indicate which atoms are connected). From a consideration of Lewis electron dot structures for such species, predict (i) which species has the shorter bond between C and N; and, (ii) which species has the shorter bond between C and S. (Full justification of your prediction will be required for full credit). Question 10 (i) Which of the following molecules possesses a permanent electric dipole moment? (a) CS 2 (b) BF 3 (c) PCl 3 (e) CCl 4 (ii) Arrange the following substances in order of increasing radius: Br, Kr, Sr 2+, K +. Explain your method. Question 11 (a) Identify the hybrid orbitals used on all the atoms in the most important resonance structure for the following molecules; AND (b) identify the idealized local geometry around all atoms except the terminal atoms. ( Skeletal molecular structures are given in parentheses) (i) H CNN 2 ( H C N N ) H (ii) (CH CO) O 3 2 H H ( H C C O C C H ) H O O H

7 Solutions to Sample Questions Question 1 (i) From data sets 1 and 2, doubling [BrO 3 ] increases the initial rate by a factor of 2. a = 1. From data sets 1 and 3, increasing [Br ] by a factor of 3 increases the initial rate by a factor of 3. b = 1. From data sets 2 and 4, increasing [H + ] by a factor of 3 increases the initial rate by a factor of 9. c = 2. (ii) From part(i), the differential rate law is given by Rate = k [BrO 3 ] [Br ] [H + ] 2 Use any data set to evaluate the rate constant k. For example, using data set 1, mol L 1 s 1 = k (0.10 mol L 1 ) (0.10 mol L 1 ) (0.10 mol L 1 ) 2 k = mol 3 L 3 s 1 Question 2 The first step is to determine the order of the reaction -- note that explicit determination of the reaction order was required for major credit. There are two ways to determine the order of the reaction in this question. (i) Plot ln [A] vs. t. The resulting linear plot indicates that the reaction is first-order, where the slope of this plot = k. Also, this could have been shown by substituting at least two different pairs of data in the [ A ] expression ln = k t (see the information sheet). [ A ] 0 For example, ln = k 40 min k = min = min 1 7 or ln = k 80 min k = min = min 1 The constancy of k confirms that the reaction order = 1. (ii) An alternate method is to note that [A] 80 = [A] 0 /2 and [A] 120 = [A] 40 /2. That is, whether we start with [A] 0 or [A] 40, it takes 80 minutes to reduce [A] to one-half of its initial value. Thus, the reaction half-life, t 1/2, is independent of the initial concentration indicating that the reaction is first-order. When t = t 1/2, [A] = [A] 0 /2, and ln [ A ] 0/2 [ A ] 0 = k t 1/2 Thus, t 1/2 = ln 2 k or k = ln 2 t 1/2 = min = min 1 (Note that using the differential rate law here is incorrect since the concentrations and hence the reaction rates vary very considerably over the time intervals given).

8 Question 3 From the information sheet, k = A e E a RT. Taking natural logarithms of both sides of this equation gives: ln k = ln A E a R T, where both A and E a are independent of temperature. Using this equation at any two temperatures T 1 and T 2 yields: ln k 1 = ln A E a R T 1 (1) and k 2 = ln A Subtracting equation (2) from equation (1) gives: E a R T 2 (2) 8 ln k 1 k 2 = E a R 1 T 1 1 T2. We may now use any pair of data points to evaluate E a. For example: ln s 1 = E a s 1 R i.e. ln = E a (J mol 1 K 1 ) K 1 E a = J mol 1 K K 1 = J mol 1 = 220 kj mol 1. Question 4 Applying the conservation of energy to the photoelectric effect experiment h ν = Φ + Kinetic Energy of ejected electron Here, h ν is the energy of the incident photon (wavelength λ) and Φ is the binding energy of the electron to the metal. Thus, the first step in answering this question is to determine the binding energy Φ. The maximum wavelength photon has an energy that is just sufficient to overcome the binding energy Φ. That is, h c = h c λ max 272 nm = Φ and Φ = J With the value of Φ in hand, Kinetic Energy of ejected electron = h ν J = h c 200 nm J = ( ) = J/electron This is easily converted to units of kj/mol, by multiplying by Avogadro's number. Thus, Kinetic Energy = J/electron electrons/mol = 158 kj/mol

9 9 Question 5 (i) Na has 11 electrons; electron configuration is 1s 2 2s 2 2p 6 3s 1 or [Ne] 3s 1. (ii) The ionization energy of the ground state Na atom = E final E initial where E final is the energy of Na + and a free electron at infinite separation and both at rest, and E initial is the ground state energy of Na. From the information given: E final E initial = 0 E([Ne] 3s 1 ) = 496 kj mol 1 The emission line at 589 nm results from an electronic transition that originates at the Na excited state [Ne] 3p 1 and terminates at the ground state [Ne] 3s 1. This is summarised in the following figure: Energy E ([Ne] 3p 1 ) Thus, for the transition E = E ([Ne] 3s 1 ) E ([Ne] 3p 1 ) = E ([Ne] 3s 1) h c 589 nm = J/atom = 203 kj/mol That is, the excited state [Ne] 3p 1 is 203 kj mol 1 higher in energy than the ground state [Ne] 3s 1. Relative to the zero of energy given (i.e. the ionization limit), the Na ground state has an energy of 496 kj mol 1. Thus, relative to this same zero of energy the energy of the Na excited state [Ne] 3p 1 is ( ) = 293 kj mol 1. Question 6 (A) (a) Si: 1s 2 2s 2 2p 6 3s 2 3p 2 [Ne] 3s 2 3p 2 : there are 2 unpaired electrons. (b) Al 3+ : 1s 2 2s 2 2p 6 [Ne]: there are 0 unpaired electrons (c) As: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3 [Ar] 4s 2 4p 3 : there are 3 unpaired electrons (B) An examination of the radial distribution plots for 2s and 2p atomic orbitals shows that a 2s electron penetrates to the nucleus more strongly than does a 2p electron. Thus, in a many-electron atom, the effective nuclear charge Z eff experienced by a 2s electron is greater than that experienced by a 2p electron. That is, Z eff (2s) > Z eff (2p) and for the orbital energies E(2s) < E(2p) -- recall that the orbital energies are negative. In boron, when the 2s electron penetrates to the nucleus it "feels" a nuclear charge of +5e, whereas in nitrogen it will "feel" a nuclear charge of +7e. Since the 2s electron is screened predominantly by the 1s

10 electrons, increasing Z from 5 to 7 produces an increase in Z eff. Thus, Z eff (2s N ) is greater than Z eff (2s B ) and E(2s N ) < E(2s B ). Using similar arguments for the 2p electron, Z eff (2p N ) > Z eff (2p B ) and E(2p N ) < E(2p B ). However, since the 2p electron penetrates to the nucleus to a smaller extent that does a 2s electron, the effect of increasing Z from 5 to 7 produces a smaller increase in Z eff (2p) than in Z eff (2s), and E(2p) decreases to a smaller extent than E(2s). Thus, the 2s 2p orbital energy spacing will be greater in nitrogen than in boron. (Note that all of these points had to be considered for full credit) Question 7 (i) From the wavefunction given, the radial part of the wavefunction for the 3p y orbital, R 3p r (6 r) exp ( r/3). Thus, R 3p will vanish at r = 0 au, at r = 6 au, and as r. Thus, r 2 R 3p 2 will also vanish at r = 0, 6 au and as r. However, only values of r, other than r = 0 and r, which make R 3p vanish represent radial nodes. Thus, the 3p y orbital has one radial node at r = 6 au. The plot of r 2 R 3p 2 vs. r has two maxima, with the principal (i.e. the largest) maximum lying at a larger value of r r R (r) 3p r (atomic units) (ii) The importance of the r 2 R 3p 2 vs. r plot is that the area under the curve between two values of r, r 1 and r 2, for example, defines the probability of finding the 3p y electron at a distance between r 1 and r 2 from the nucleus. The first step in this part is to convert r = 3 Å to atomic units: r (atomic units) = m m = 5.67 au. Thus, the required probability is the area under the r 2 R 3p 2 vs. r plot between r = 0 and r = 5.67 au.

11 12 Question 8 (a) GaI 3 has (7) = 24 valence electrons and the Lewis electron dot structure shown: Because there are three electron groups around the central Ga atom the idealized electron group geometry is trigonal planar. All the electron groups are bonding electron groups and the idealized molecular geometry is also trigonal planar. I I Ga I Trigonal Planar Because all the bonding pairs around Ga are equivalent, there are no deviations from ideality and I Ga I = 120. (b) (BiCl 5 ) 2 has 5 + 5(7) + 2 = 42 valence electrons and the Lewis electron dot structure shown: There are six electron groups around the central Bi atom and the idealized electron group geometry is octahedral. Of the six electron groups, five are bonding, each representing a BiCl bonding electron pair, and the sixth is a nonbonding electron pair on Bi. Thus, the idealized molecular geometry is square-based pyramidal. Cl a Cl Clb Bi Cl Cl c Square-based Pyramidal Because nonbonding pair bonding pair repulsion > bonding pair bonding pair repulsion there will be deviations from this idealized structure. Thus, Cl b Bi Cl a and analogous angles will be less than 90 and Cl b Bi Cl c and analogous angles will be less than 90.

12 13 Question 9 The number of valence electrons in HSCN = = 16. There are three Lewis electron dot structures ((a), (b) and (c) shown below) that satisfy the octet rule: [ H S C N : ] [ H S C N: ] [ H S C N ] (a) (b) (c) A consideration of the formal charges shown in the above structures, indicates that the relative importance of these structures is (b) >> (c) >> (a). In structure (b) there is no separation of formal charge, while structure (c) separates smaller opposite formal charges than does structure (a). In view of the relative importance of these three structures, the CN bond in HSCN is almost a pure triple bond, and the CS bond in HSCN is almost a pure single bond. The number of valence electrons in SCN is = 16. There are three Lewis electron dot structures ((a), (b) and (c) shown below) that satisfy the octet rule: [ : S C N : ] [ : S C N: ] [ S C N ] (a) (b) (c) Neither (b) nor (c) involves a separation of formal charge. However, since nitrogen is more electronegative than S, structure (c) is of greater relative importance than structure (b). The overall order of relative importance is (c) > (b) >> (a). Thus, the CN bond in SCN, although predominantly a double bond has some triple bond character and the CS bond in SCN, although predominantly a double bond has some single bond character. From this analysis, we conclude that HSCN has the shorter CN bondlength and that SCN has the shorter CS bondlength.

13 14 Question 10 (i) From Lewis structure considerations, the most important resonance structure is that shown for each molecule: In CS 2 there are two electron groups around C and the molecule is linear; thus, the two C S bond dipole moment vectors cancel -- i.e. S C S and CS 2 does not possess a permanent dipole moment. The molecular geometry of BF 3 is trigonal planar, the three B F bond dipole moment vectors cancel and BF 3 does not possess a permanent dipole moment. The molecular geometry of PCl 3 is pyramidal. In this case, the three P Cl bond dipole moment vectors will not cancel and PCl 3 possesses a permanent dipole moment. The molecular geometry of CCl 4 is tetrahedral, the C Cl bond dipole moment vectors cancel and CCl 4 does not possess a permanent dipole moment. (ii) K + < Sr 2+ < Kr < Br In K +, the outermost electrons are in the n = 3 shell, whereas for all the other species the outermost electrons are in the n = 4 shell. Because the species Kr, Br and Sr 2+ are isoelectronic with the configuration [Ar] 3d 10 4s 2 4p 6, and because Z(Sr 2+ ) > Z(Kr) > Z(Br ), the effective nuclear charges have the same relative order -- i.e. Z eff (Sr 2+ ) > Z eff (Kr) > Z eff (Br ). Thus, the outermost electrons in Sr 2+ are held more tightly than those in Kr and Sr 2+ has a smaller radius than Kr. Similarly, the outermost electrons in Kr are held more tightly than those in Br and Kr has a smaller radius than Br.

14 15 Question 11 (i) The geometry around the C atom is trigonal planar and the geometry around the central N is linear. (ii) The geometry around the terminal C atoms is tetrahedral, that around the C = O carbon atoms is trigonal planar, and the geometry around the central O atom is bent.