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July WebAssign Sunday, 17 2011 05:17 PM EDT Logged in as cdbadger@ncsu Log out Home | My Assignments | Grades | Communication | Calendar Notifications | Help | My Options CH_223, section 651, Summer 1 2011 My Assignments Prerequisite checkup (Daily) Crystal Badger CH_223, section 651, Summer 1 2011 Instructor: Kay Sandberg Current Score : 9.25 / 10 Due : Monday, May 23 2011 09:00 PM EDT Ask Your Teacher Extension Requests Print Assignment Question Points 1 2 3 4 5 6 7 8 9 1/1 0.25/1 0.5/0.5 0.5/0.5 0.6/0.6 3.2/3.2 1.4/1.4 0.2/0.2 1.6/1.6 Total 9.25/10 (92.5%) Description This assignment is to help you see if you have retained important knowledge from your general chemistry course. Instructions If you did not score 100% on this assignment (or if you scored 100% but you were guessing) you should view the following review presentations: electronic configuration review presentation (click here) formal charge review presentation (click here) hybridization review presentation (click here) Make sure you check the number of submissions for each question type. Click "points" link to see the number of submissions allowed for each question part. If the presentation does not work properly, see the "Troubleshooting" forum for help. Assignment Submission For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer. Assignment Scoring Your last submission is used for your score. The due date for this assignment is past. Your work can be viewed below, but no changes can be made. Important! Before you view the answer key, decide whether or not you plan to request an extension. Your Instructor may not grant you an extension if you have viewed the answer key. Automatic extensions are not granted if you have viewed the answer key. Request Extension 1. 1/1 points | Previous Answers My Notes Question Part Points Submissions Used 1 2 3 4 0.25/0.25 0.25/0.25 0.25/0.25 0.25/0.25 1/2 1/2 1/2 1/2 1/1 Click here for the review presentation and here for a periodic table. Give the number of: a) protons in an O atom 18 8 b) electrons in a Li atom 23 3 c) electrons in a Cl1- ion 318 18 d) electrons in a Ca2+ ion 418 18 Solution or Explanation Proton number is the atomic number. When the atom is neutral, the electron number equals the proton number. For the anion electron number, you add the charge to the atomic number to determine the electron number. For the cation electron number, you subtract the charge from the atomic number to determine the electron number. Viewing Saved Work Revert to Last Response 2. 0.25/1 points | Previous Answers My Notes Question Part Points Submissions Used 1 2 3 4 0/0.25 0.25/0.25 0/0.25 0/0.25 1/1 1/1 1/1 1/1 Total 0.25/1 Select the word(s) that makes the statement true. Click here for review presentation. a) An electron in a 3s-orbital is lower higher in energy than an electron in a 2s-orbital. b) An electron in a 2s-orbital is lower lower in energy than an electron in a 2p-orbital. c) A metal typically gains loses its valence electron(s) becoming a(n) anion cation . Solution or Explanation The lower the energy level the lower the energy of the electron. Viewing Saved Work Revert to Last Response 3. 0.5/0.5 points | Previous Answers My Notes Question Part Points Submissions Used 1 2 0.25/0.25 0.25/0.25 1/2 1/2 Give the symbol for: Click here for review presentation. Periodic Table a) the cation possessing a positive one charge that is isoelectronic with neon. (NOTE: input format for a charged species: symbol^#sign. Example, H^1+ represents H1+.) 1Na^1+ Na^1+ b) the anion possessing a negative two charge that is isoelectronic with argon. (NOTE: input format for a charged species: symbol^#sign. Example, H^1- represents H1-.) 2S^2S^2Viewing Saved Work Revert to Last Response 4. 0.5/0.5 points | Previous Answers My Notes Question Part Points Submissions Used 1 2 0.25/0.25 0.25/0.25 1/3 Total 1/3 0.5/0.5 Following the energy ordering dictated by the periodic table, write the indicated configuration. Input format: [Ne]3s23p1 would be typed in as [Ne]3s^2,3p^1 Note: a comma is used to separate sublevels in the valence shell as shown. Periodic Table Click here for review presentation. a) Using a noble gas core write the electron configuration for S. 1[Ne]3s^2,3p^4 [Ne]3s^2,3p^4 b) Using a noble gas core write the electron configuration for Li. 2[He]2s^1 [He]2s^1 Viewing Saved Work Revert to Last Response 5. 0.6/0.6 points | Previous Answers My Notes Question Part Points Submissions Used 1 2 3 0.2/0.2 0.2/0.2 0.2/0.2 1/2 1/2 1/2 Total 0.6/0.6 Give the formal charge of the indicated atom. All hydrogens and lone pairs are shown. Click here for review presentation. Input format for non-zero formal charge: sign and then number, something like: -1 10 0 -or- 0 -or- 0 atom 1 2+1 1+ -or- +1 -or- + atom 2 3-1 1- -or- -1 -or- atom 3 Solution or Explanation FC = family # - possession # Family # is off the periodic table Possession # is determined by drawing a circle around the atom (making sure to all cut bonds in half) and counting the # of electrons in the circle. Viewing Saved Work Revert to Last Response 6. 3.2/3.2 points | Previous Answers My Notes Question Part Points Submissions Used 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2/2 2/2 1/2 1/2 1/2 1/2 1/2 2/2 1/2 2/2 1/2 1/2 2/2 1/2 2/2 2/2 Total 3.2/3.2 Each atom below has access to an octet of electrons (except for hydrogen which has access to a duet) and all atoms are shown explicitly. Not shown are the lone pairs of electrons and the formal charges. You will need to determine how many lone pairs, if any, are needed to give the indicated atom access to 8 valence electrons and the formal charge of the indicated "octeted" atom Click here for review presentation. (a) Select the number that corresponds to the number of lone pairs possessed by the indicated atom. (b) Select the formal charge of the indicated atom. atom a lone pair(s) atom a formal charge atom b lone pair(s) atom b formal charge 2 -1 3 -1 2 -1 3 -1 atom c lone pair(s) atom c formal charge atom d lone pair(s) atom d formal charge 0 +1 1 -1 0 +1 1 -1 atom e lone pair(s) atom e formal charge atom f lone pair(s) atom f formal charge 1 +1 4 -1 1 +1 4 -1 atom g lone pair(s) atom g formal charge atom h lone pair(s) atom h formal charge 2 0 1 0 2 0 1 0 Solution or Explanation For access to octet determination, count all of the electrons in the bond and then add as many lone pairs as needed to equal 8 electrons. The formal charge is determined by the formula: Family # minus possession #. The family # is from the periodic table. For example, oxygen is in the 6th family. The possession number is determined by circling the atom for which you are determining the formal charge making sure you cut all of the bonds in half. The electrons in the circle will include all of the lone pairs and half of the bonding electrons. Viewing Saved Work Revert to Last Response 7. 1.4/1.4 points | Previous Answers My Notes Question Part Points Submissions Used 1 2 3 4 5 6 7 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 1/2 1/2 2/2 1/2 2/2 2/2 1/2 You may want to draw the Lewis structure for the HCN molecule before attempting this question. Click here for review presentation. b a) Select the correct description for each region. Region V pi bonding region pi bonding region Region U sigma bonding region sigma bonding region Region X lone pair region lone pair region Region Y sigma bonding region sigma bonding region b) Write the capital letter (U, V, W, X or Y) of the region that results from the overlap of 2 hybrid orbitals. 5Y Y Only input a single capital letter c) Write the capital letter (U, V, W, X or Y) of one of the regions that results from the overlap of 2 pure p orbitals. 6V V -or- W Only input a single capital letter d) Type in the total number of electrons contained in all of the electron regions between the C and N. 76 6 Viewing Saved Work Revert to Last Response 8. 0.2/0.2 points | Previous Answers My Notes Question Part Points Submissions Used 1 0.2/0.2 1/2 Total 0.2/0.2 Imagine the following orbitals gradually approaching each other as indicated by the arrows so that they begin to overlap. Click here for review presentation. If two electrons with opposite spins reside in the overlap region, then which of the following is the true statement? 1 A & C result in a pi bond and B results in a sigma bond A & B result in a sigma bond and C results in a pi bond B & C result in a sigma bond and A results in a pi bond A & C result in a sigma bond and B results in a pi bond A & B result in a pi bond and C results in a sigma bond & C result in a pi bond and A results in a sigma bond Viewing Saved Work Revert to Last Response 9. 1.6/1.6 points | Previous Answers My Notes Question Part Points Submissions Used 1 2 3 4 5 6 7 8 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 2/2 2/2 1/2 1/2 1/2 1/2 1/2 2/2 Total 1.6/1.6 Click here for review presentation. The red labels a, b and c are to be used in part c only. Use the structure below to answer the following questions. Note: the structure below does not reflect its true 3-D shape. You should add lone pairs as needed so that each non-hydrogen atom obeys the octet rule. a) number of pi bonds in the entire molecule 13 3 b) number of sigma bonds in the entire molecule 211 11 B c) Select the hybridization of the indicated atom. a sp^2 b sp c sp^3 sp^2 sp sp^3 d) Give the approximate values of the indicated angles. Choose from the following values: 60, 90, 109, 120, 180. Do not include the degree symbol - I have provided that for you. 6120 7180 8109 120 180 109 o o o Viewing Saved Work Revert to Last Response Home My Assignments Ask Your Teacher Extension Request WebAssign 4.0 1997-2003 by North Carolina State University. Portions 2003-2011 by Advanced Instructional Systems, Inc. All rights reserved. ... View Full Document

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