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Exam 1-solutions.pdf

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Department
Chemistry
Course
CH 301
Professor
Sutcliffe
Semester
Spring

Description
Version 192 – Exam 1 – mccord – (51600) 1 This print-out should have 34 questions. 2. the electron can exist in any one of a set Multiple-choice questions may continue on of discrete states (energy levels) and can move the next column or page – find all choices from one to another by emitting or absorbing before answering. radiation. correct 3. because a hydrogen atom has only one electron, the emission spectrum of hydrogen c =3 .00 ▯ 10 m/s ▯34 should consist of only one line. h =6 .626 ▯ 10 J·s ▯31 m e9 .11 ▯ 10 kg 4. the electron in a hydrogen atom can jump 15 ▯1 from one energy level to another without gain R =3 .29 ▯ 10 s or loss of energy. ▯18 R =2 .178 ▯ 10 J N =6 .022 ▯ 10 23mol ▯1 5. energy, in the form of radiation, must A be continually supplied to keep the electron g =9 .81 m/s2 moving in its orbit. Explanation: E = h▯ c = ▯/▯ 002 10.0 points Which of the following subshells CANNOT ▯ = h = h exist in an atom? p mv 1 1. 3f correct mv = h▯ ▯ ▯ 2 2. 4f R E n ▯ 2 (hydrogen atom) 3. 5g n ▯ ▯ 1 1 4. 5f ▯ = R 2▯ 2 n 1 n2 2 2 5. 4d h d ▯ Explanation: ▯ 2m dx 2 + V (x)▯ = E▯ The notation is ▯ ,where ▯ ▯ 2 ▯ ▯ n =1 , 2, 3, 4, . , ▯ =0 , 1, 2,. (n ▯ 1) ▯ (x)= 2 sin n▯x represented as a letter: n L L n h 2 ▯ 012 3 45 etc. E n n =1 ,2,3,··· 8mL 2 orbital spd f gh 001 10.0 points and m ▯ = ▯▯, ▯(▯ ▯ 1), ▯(▯ ▯ 2),. 0,, +(▯ ▯ 2), +(▯ ▯ 1),▯ . Bohr’s theory of the hydrogen atom assumed Only 3f does NOT follow the rules given. that 003 10.0 points 1. electromagnetic radiation is given o▯ Which is NOT true about light? when the electrons move in an orbit around the nucleus. 1. As the frequency of light increases, the wavelength decreases. Version 192 – Exam 1 – mccord – (51600) 2 2. X-rays and visible light both travel at the 2. g same speed. 3. f 3. The energy of a photon is directly propor- tional to its wavelength. correct 4. s 4. It can behave like either particles or 5. p waves. Explanation: Explanation: Anodeisapot,aplaneorathreedimen- sional region of space where the sign of the 004 10.0 points wavefunction changes from positive to neg- How many d electrons does Cr have? ative (or negative to positive), so it equals zero. The lowest energy orbital of each type 1. 3 will have no three dimensional nodes but dif- fering numbers of nodal planes: 2. 4 1s –nonodalplanes 2p –onenodalplane(twoopposinglobesof 3. 5 correct di▯erent signs) 3d –twonoallesouropoglobs 4. 2 of two di▯erent signs) 4f –threenodalplanes(sixopposinglobes 5. 6 of two di▯erent signs) Explanation: (One orbital has only one nodal plane and one The electron configuration for Cr is conical node for both 3d and 4f.) 5 1 If we go to the next higher value of n for the [Ar]3d 4s which shows that it has 5 elec- trons in the d-orbital. same type of orbital we would add a spherical (radial) node but the number of nodal planes would remain the same. 005 10.0 points You shine white light through a sample of ar- gon atoms and then into a spectrophotometer. 007 10.0 points An element with 20 protons and 20 electrons From this experiment, you would expect gains one electron. The electron goes into 1. acotuousspectrum. what type of orbital? 2. mostly black space with thin lines of 1. 3p color. 2. 4d 3. anosmb nbyi black lines. correct 3. 3s Explanation: 4. 4s 006 10.0 points 5. 3d correct Agivenatomicorbitalhasatotaloftwonodal planes. What type of orbital is this? 6. 4p 1. d correct Explanation: Version 192 – Exam 1 – mccord – (51600) 3 This element has electron configuration 2 n [Ar]4ongexn d 9 +▯ orbital according to the Aufbau principle. =8 n + 8 ▯ 8s 008 10.0 points =7 n + 7 ▯ =6 7s 7p What is the correct order of decreasing fre- n + quency? 6 ▯ =5 6s 6p 6d n + 1. ultraviolet radiation, visible light, in- 5 ▯ =4 5s 5p 5d 5f frared radiation, radio waves correct value n n + 4 ▯ =3 4s 4p 4d 4f 2. visible light, ultraviolet radiation, in- n + frared radiation, radio waves 3 ▯ =2 3s 3p 3d n + 3. radio waves, visible light, ultraviolet radi- 2 ▯ =1 2s 2p ation, infrared radiation 1 1s 4. infrared radiation, radio waves, visible light, ultraviolet radiation ▯ value 0 1 2 3 until exponents sum to 16. 5. radio waves, infrared radiation, visible 010 10.0 points light, ultraviolet radiation An electron is confined in a 1-dimensional box 3nmlon. Ccutehenegyeurdfr it to move from the n =8tothe n =9energy Explanation: level. ultraviolet radiation > visible light > infrared radiation > radio waves ▯19 1. 5.238 ▯ 10 J 009 10.0 points ▯19 2. 2.133 ▯ 10 J The ground state electron configuration of a sulfur atom is ▯21 3. 6.694 ▯ 10 J 1. 1s 2s 2p 3s . 4 ▯28 4. 3.414 ▯ 10 J 2. 1s 2s 2p 3s 3p . correct ▯14 5. 2.133 ▯ 10 J 3. 1s 2s 2p 3p . 6 ▯14 6. 1.717 ▯ 10 J 4. 1s 2s 2p 3s 3p 4p .2 2 ▯19 7. 1.138 ▯ 10 J correct 2 2 6 2 2 2 Explanation: 5. 1s 2s 2p 3s 4s 5s . n =8 n =9 1 2 h =6 .626 ▯ 10 ▯34 J · s L =13 ▯ 10 ▯9 m ▯31 m =9 .109 ▯ 10 kg Explanation: Sulfur has atomic number 16. Follow the pyramid ▯E = E n2 ▯ E n1 Version 192 – Exam 1 – mccord – (51600) 4 (n ▯ n )h 2 Explanation: = 2 1 8mL 2 Use the rules for the quantum numbers: (9 ▯ 8 )(6 .626 ▯ 10▯34J · s) If n =4hn ▯ =0 , 1, 2, 3; however, for = 8(9 .109 ▯ 1031 kg)(3 ▯ 10▯9 m) 2 m ▯ ▯1, ▯ =1 , 2, 3. Each of these permit- ▯19 ted sets of values of n, ▯ a▯d m specifies ONE =1 .13802 ▯ 10 J orbital: n =4, ▯ =1, m ▯ ▯1: 4p n =4, ▯ =2, m ▯ ▯1: 4d 011 10.0 points n =4, ▯ =3, m = ▯1: 4f Each solution to the Schrodinger wavefunc- ▯ and each orbital can have s = ± 1 ; i.en,c tion equation may be described in terms of 2 hold two electrons. 1. photoelectric e▯ect. 013 10.0 points 2. ionization energies. Assume n 1nd n ar2 two adjacent energy levels of an atom. The emission of radiation 3. photon energies. with the longest wavelength would occur for which two values of 1 and n2? 4. quantum numbers. correct 1. 7,6 5. velocity and distance. 2. 5,4 Explanation: 012 10.0 points 3. 2,1 How many electrons can possess this set 4. 3,2 of quantum numbers: principal quantum number n =4 ,atumn mr 5. 8,7 correct m ▯ ▯1? 6. 6,5 1. 12 2. 8 7. 4,3 Explanation: 3. 18 The frequency of a photon emitted when an electron moves between levels1n and 2 is 4. 14 given by the Rydberg equation: ▯ ▯ ▯ = R 1 ▯ 1 , 5. 4 n2 n2 1 2 6. 0 15 where R =3 .29 ▯ 10 Hz. The emission of radiation with the longest wavelength corre- 7. 10 sponds to that with the smallest frequency. From inspection of the formula above we see 8. 6 correct that ▯ is smallest when1n =8and n2=7. Conceptual Solution: E = h▯ = ▯ ▯ 9. 2 hR 1 ▯ 1 gives the energy of the pho- n 2 n2 10. 16 tons emitted. The emission of radiation with Version 192 – Exam 1 – mccord – (51600) 5 the longest wavelength corresponds to pho- tons with the smallest energy. From the Bohr 3. more like waves than particles when lo- frequency condition the energy of the emit- calized in an atom. correct ted photon must be equal to the di▯erence Explanation: in energy between the higher and lower lev- els. An energy level diagram for the H-atom The current atomic theory states that elec- trons have both particle- and wave-like char- shows that as the energy levels get higher, the acteristics. gaps between them converge; of the transi- tions listed, the two adjacent levels which are 016 10.0 points the closest together are 1 =8a d n2=7 . Thus a transition from n =8ot n =7 For the reaction 1 2 will result in the emission of a photon with ?C 6 6?O 2 ▯ ?CO 2 +?H 2O the smallest energy, hence the longest wave- 44.1rsfC 6H 6 are mixed with 153.1 length. grams of O 2nd allowed to react. How much CO 2ould be produced by this reaction? 014 10.0 points How many V(r)m tsrdnie 1. 198.3g + solution to the Schodinger equation for a Li ion? 2. 231.1g 3. 186.4g 1. 3V( r)terms correct 4. 281.8g 2. 2V( r)terms 3. 1V( r)terms 5. 106.5g 4. 4V( r)terms 6. 131.9g 5. 6V( r)terms 7. 149.1g correct 8. 313.1g 6. 5V( r)terms 9. 119.3g Explanation: Each of the two electrons has an attrac- 10. 168.5g tive potential energy term with the nucleus. And they mutually repel each other in a final Explanation: potential energy term. m C H =44 .1g m O =153 .1g 6 6 2
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