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

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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
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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? C . Fe 2+ : [Ar] 3d 6 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
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