Lecture 9 - Topics for the day Administrative stuff The...

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Topics for the day • Administrative stuff • The periodic table • Effective nuclear charge • Relative orbital energies • Atomic radii • Ionization energy Administrative stuff ALEKS prep grades now posted on Blackboard (“My Grades”) 3% if you Fnished the assessment by 2/5/10 + 2% = 5% if you already had 70% by 2/5/10 You have till 3/29/10 to get up to 70% Worth doing this because it’s assumed material Email announcements ±riday email about how to know your ALEKS HW score. If you didn’t get it, check your “ofFcial” email address. Sunday email to all students with no registered iClicker. Administrative stuff Students with an exam makeup have been sent email today Quest Homework One Answer key now online 6% of class didn’t do the homework.Class average 84% Quest Sample Exam Answer key online at noon today. ALEKS homework Not due until midnight ±riday 2/12 1 2 3
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Help before exam Exam review with Susan: TODAY, 4 - 5 pm, WEL 3.502 OfFce hours with Susan: Mon, 12 - 1 pm; Wed 9 - 10 am OfFce hours with DrRuth: Tue 2 - 3 pm; Wed 3:30 - 4:30 pm Download HW and sample exam answer keys from Quest Exam Review in class on Wednesday Page showing info available on scantron on Blackboard Where were we? The periodic table can be used to quickly determine the electronic confguration. Knowing the electronic confguration, you can predict the number oF valence electrons, and the number oF unpaired electrons. Isoelectronic species have the same number oF electrons, and hence the same electronic confguration. Species with unpaired electrons are paramagnetic (weakly attracted to magnetic felds). The more unpaired electrons, the stronger the attraction. Species with no unpaired electrons are diamagnetic (weakly repelled by magnetic felds). 4 5 6
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The expected confguration would have been Cr: [Ar] 4s 2 3d 4 , but the extra stability gained For a half-Flled subshell overcomes the Fact that the 4s orbital is lower in energy than the 3d orbitals. Similarly Mo is [Kr]5s 1 4d 5 not [Kr]5s 2 4d 4 4s 3d 4p Cr: [Ar]4s 1 3d 5 Cu: [Ar]4s 1 3d 10 The expected confguration would have been Cu: [Ar] 4s 2 3d 9 , but the extra stability gained For a Flled subshell overcomes the Fact that the 4s orbital is lower in energy than the 3d orbitals. Similarly Ag is [Kr]5s 1 4d 10 not [Kr]5s 2 4d 9 Au is [Xe]6s 1 4F 14 5d 10 not [Xe]6s 2 4F 14 5d 9 A. Al B. C C. Mg D. Na iClicker Time Which oF the Following is diamagnetic? Al: [Ne]3s 2 3p 1 C: [He]2s 2 2p 2 Mg: [Ne]3s 2 Na: [Ne]3s 1 A. Kr B. Ca + S 2– Na + E. ± iClicker Time 1s 2 2s 2 2p 6 3s 2 3p 6 is the electron confguration For which one oF these species? Ca Ca + Kr S S 2– Na ± ± /Na + 7 8 9
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Summary • After trying to describe the electrons in an atom, we have arrived at a set of quantum numbers which describe the relative energy, size, shape, orientation in space and spin for each electron.
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Lecture 9 - Topics for the day Administrative stuff The...

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