Exam 2 - Version 220 Exam 2 mccord (50950) 1 This print-out...

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Unformatted text preview: Version 220 Exam 2 mccord (50950) 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 3.0 points Consider the radial distribution function (RDF) plot shown below. Which of the orbitals given as choices below would corre- spond to the RDF shown. r 4 r 2 R 2 1. 4 s 2. 4 d 3. 6 d 4. 4 f 5. 5 p 6. 5 d correct Explanation: The plot has 2 radial nodes (spherical nodes). The number of spherical nodes will al- ways be equal to n- - 1. Only the 5 d orbital fits this case. Other orbitals that WOULD fit this would have been the 3 s , 4 p , and 6 f . 002 3.0 points Which do you think would be larger and why? The first ionization energy, IE 1 , of Ne, The second ionization energy, IE 2 , of Na 1. IE 1 of Ne, because Ne has a smaller radius than Na. 2. IE 2 of Na, because Na + and Ne have the same electron configuration but Na has more protons than Ne. correct 3. IE 1 of Ne, because Ne is a noble gas and Na is an alkali metal. 4. IE 1 of Ne, because the electrons in Ne are not as well shielded from the nucleus as those in Na + . 5. IE 2 of Na, because Na + and Ne have the same number of protons, but Na + has fewer electrons than Ne. Explanation: The second ionization energy of Na is the ionization energy of Na + . Na + and Ne have identical electronic structures. They are both [He]2s 2 2p 6 . However, Na (and Na + has 11 protons and Ne only has 10. Therefore the electrons will be held more tightly onto Na + than Ne. 003 3.0 points Write the ground-state electron configuration of a lead atom. 1. [Xe] 4 f 14 5 d 10 6 s 2 6 p 2 correct 2. [Xe] 4 f 14 5 d 9 6 s 2 6 p 3 3. [Xe] 4 f 14 5 d 10 6 p 4 4. [Xe] 4 f 14 5 d 5 6 s 1 6 p 6 7 s 2 5. [Xe] 4 f 14 5 d 10 6 s 1 6 p 3 Explanation: The Aufbau order of electron filling is 1 s , 2 s , 2 p , 3 s , 3 p , 4 s , 3 d , 4 p , 5 s , 4 d , 5 p , 6 s , 4 f , 5 d , 6 p , etc . s orbitals can hold 2 electrons, p orbitals 6 electrons, and d orbitals 10 electrons. Note some exceptions do occur in the electron con- figuration of atoms because of the stability of either a full or half-full outermost d-orbital, so you may need to account for this by shuffling an electron from the ( n- 1) s orbital. Finally use noble gas shorthand to get the answer: [Xe] 4 f 14 5 d 10 6 s 2 6 p 2 . Version 220 Exam 2 mccord (50950) 2 004 3.0 points Rank the following diatomic molecules by bond energy, from greatest to least: F 2 , N 2 , O 2 . 1. N 2 > O 2 > F 2 correct 2. F 2 > O 2 > N 2 3. N 2 > F 2 > O 2 4. Not enough information. 5. F 2 > N 2 > O 2 Explanation: 005 3.0 points The atomic number ( Z ), electronic configu- ration, and number of unpaired electrons for an ion are 21, [Ar] 4 s 1 3 d 1 , and 2, respectively....
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This note was uploaded on 10/10/2011 for the course CHEM 301 taught by Professor Wandelt during the Fall '08 term at University of Texas at Austin.

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Exam 2 - Version 220 Exam 2 mccord (50950) 1 This print-out...

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