CHE285~1 - 1 AUFBAU 55 Chem 161-2006 Final Exam Chapter 8...

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Unformatted text preview: 1 AUFBAU 55 Chem 161-2006 Final Exam Chapter 8 – Electronic Configurations, Atomic Properties, and Periodic Table Trends Polyelectronic atoms (Aufbau principle) Which is the correct box diagram for the electronic configuration 3d 3 4s 2 ? A. ↑↓ ↑ ↑↓ 3d 4s B. ↑ ↓ ↑ ↑↓ 3d 4s C . ↑ ↑ ↑ ↑↓ 3d 4s D. ↑↑ ↑ ↑↓ 3d 4s E. ↑ ↑ ↑ ↑↑ 3d 4s 23 V = [Ar]4s 2 3d 3 According to Aufbau, first the 4s orbital is filled with two electrons, and then the last three electrons are put into the 3d subshell. According to Hund’s law, they are put into the 3 d orbitals with one electron per orbital before pairing up, and that these single electrons per orbital have parallel spins. C 2 49 Chem 161-2006 Final Exam Chapter 8 – Electronic Configurations, Atomic Properties, and Periodic Table Trends Polyelectronic atoms (Aufbau principle) Which atom is paramagnetic? A. Zn B. Ba C. Ar D. Sr E . Mn 30 Zn = [Ar]4s 2 3d 10 56 Ba = [Xe]6s 2 18 Ar = [He]2s 2 2p 6 3s 2 3p 6 38 Sr = [Kr]5s 2 25 Mn = [Ar]4s 2 3d 5 A paramagnetic atom has unpaired electrons in its orbitals. Only Mn has unpaired electrons in its orbitals. Each of the five d orbitals contains only a single electron, making Mn strongly paramagnetic. E 42 Chem 161-2006 Final Exam Chapter 8 – Electronic Configurations, Atomic Properties, and Periodic Table Trends Polyelectronic atoms (Aufbau principle) Which species is correctly shown in its ground state? A. Cu: [Ar] 3d 4 4s 2 B. V 2+ : [Ar] 3d 1 4s 2 C . Fe 2+ : [Ar] 3d 6 D. Cl- : [Ne] 3s 2 3p 5 4s 1 E. Pb: [Xe] 4f 14 5d 10 6s 2 6p 4 29 Cu: [Ar]4s 2 3d 9 → [Ar]4s 1 3d 10 23 V 2+ : [Ar]4s 2 3d 3 → [Ar]3d 3 26 Fe 2+ : [Ar]4s 2 3d 6 → [Ar]3d 6 17 Cl- : [Ne]3s 2 3p 5 → [Ne]3s 2 3p 6 82 Pb: [Xe]6s 2 4f 14 5d 10 6p 2 C 3 CHEM 161-2006 EXAM III HILL & PETRUCCI CHAPTER 8 – ELECTRONIC CONFIUGATION, ATOMIC PROPERTIES, AND THE PERIODIC TABLE ATOMIC AND IONIC RADIUS ELECTRON CONFIGURATIONS: THE AUFBAU PRINCIPLE 19. Select the atom/ion that has the incorrect configuration listed next to it. A. V : [Ar] 3d 3 4s 2 B . Sn 2+ : [Kr] 4d 8 5s 2 5p 2 C. Br – : [Ar] 3d 10 4s 2 4p 6 D. Cu + : [Ar] 3d 10 E. N 3– : [He] 2s 2 2p 6 A. Correct: 23 V = [Ar]4s 2 3d 3 (It makes no difference whether 4s is written before 3d or 3d is written before 4s. B. Incorrect: 50 Sn 2+ = [Kr]5s 2 4d 10 5p 2 → [Kr]5s 2 4d 10 5p = [Kr]5s 2 4d 10 C. Correct: 35 Br- = [Ar]4s 2 3d 10 4p 6 D. Correct: 29 Cu + = [Ar]4s 2 3d 9 → [Ar]4s 1 3d 10 → [Ar]3d 10 (Note that for cations, the atom is first filled, and then an electron (or electrons) is removed from the valence orbitals (4s in this case), which might not have been the last orbitals filled.) E. Correct: 7 N 3- = [He]2s 2 2p 6 CHEM 161-2006 EXAM III + Answers HILL & PETRUCCI CHAPTER 8 – ELECTRONIC CONFIUGATION, ATOMIC PROPERTIES, AND THE PERIODIC TABLE PERIODIC TABLE AND PERIODIC PROPERTIES MAGNETIC PROPERTIES PARAMAGNETISM AND DIAMAGNETISM 1. Which of the following would you expect to exhibit paramagnetism...
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CHE285~1 - 1 AUFBAU 55 Chem 161-2006 Final Exam Chapter 8...

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