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Chemistry 1 Chapter 6 (Part 1)

by: Courtney Beckwood

Chemistry 1 Chapter 6 (Part 1) CHEM 1110 - 02

Marketplace > University of Memphis > CHEM 1110 - 02 > Chemistry 1 Chapter 6 Part 1
Courtney Beckwood
University of Memphis

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This is the first couple of notes dealing with Chapter 6 of Chemistry 1. These notes cover Lewis structures and resonance structures.
General Chemistry I Lecture
Henry Kurtz
Class Notes
Chemistry, Courtney Beckwood, memphis
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This 5 page Class Notes was uploaded by Courtney Beckwood on Monday March 14, 2016. The Class Notes belongs to CHEM 1110 - 02 at University of Memphis taught by Henry Kurtz in Spring 2016. Since its upload, it has received 50 views.

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Date Created: 03/14/16
Chapter 6 Drawing Lewis Structures and Determining Molecular Shapes  Polar Covalent Bond (Polar Bond): Covalent bond with greater electron density around one of the two atoms  Electronegativity: The ability of an atom to attract toward itself the electrons in a chemical bond o The electronegativity increases as you move up and to the right of the periodic table  Difference in Electronegativity o o If the difference between a bond is:  0 – Covalent Bond  Between 0 and 2 – Polar Covalent Bond  Greater than or equal to 2 – Ionic Bond Writing Lewis Structures  1. Draw skeletal structures of compounds showing what atoms are bonded to each other. Put least electronegative element in the center.  2. Count total number of valence e . Add 1 for each negative charge. Subtract 1 for each positive charge.  3. Complete an octet for all atoms except hydrogen.  4. If structure contains too many electrons, form double and triple atoms on central atom as needed. Example 1: Write the Lewis Structure for Nitrogen Trifluoride (NF ) i3 which all three atoms are bonded to the N atom. o o You can determine step 2 by looking at the periodic table. Every row starts a new dot structure and every spot determines the number of electrons. th  Since N is the 5 element on the row, it will have five electrons. By this, F will have seven. In all, there should be 26 valance electrons. o Example 2: Write the Lewis structure for nitric acid (HNO ) in which 3he three O atoms are bonded to the central N atom and the ionizable H atom is bonded to one of the O atoms.  th  Step 2: By looking at the periodic table, N is at the 5 spot, so it will have five valence electrons. O is in the 6 spot, so each O atom will have six valence electrons, and H is in the 1 spot, so that will have one valence electrons. In all, there will be 24 valence electrons.  2- Example 3: Write the Lewis structure for the carbonate ion (CO ). 3  Step 1:  Step 2: Six valence electrons for each O and four valence electrons for C. 22 total valence electrons.  Step 3:  Formal Change  Formal charge: The difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.  Formal charge = total number of valence electrons – total number of nonbonding electrons - .5(total number of bonding electrons)  The sum of the charges of the atoms in a molecule or ion must equal the charge on the molecule or ion. Example 1: With formaldehyde (CH O): 2  C has four valence electrons, O has six valence electrons and H will have two in this case. That all adds up to have 12 electrons. o According to this picture of CH2O, there are two single bonds (Two single lines), one double bond (One set of parallel lines) and two lone pairs. These add up to 12 nonbonding and bonding electrons. o To find the formal charge on C, there are 4 valence electrons STRICTLY for C minus 2 nonbonding electrons (the dots on top of C) minus .5 times six bonding electrons (each line represents two electrons) and you should get -1. o To find the formal charge on O, there are 6 valence electrons STRICTLY for O minus 2 nonbonding electrons (the dots on top of C) minus .5 times six bonding electrons (each line represents two electrons) and you should get +1. o According to this picture of CH 2, there are still 12 valence electrons and 12 bonding and nonbonding electrons. o However, the formal charge on C has 4 valence electrons minus 0 nonbonding electrons (because the electrons surrounding C are all bonding) minus .5 times 8 bonding electrons (each line represents two electrons) and altogether you should get 0. o The formal charge on O has 6 valence electrons minus 4 nonbonding electrons (because the four dots that are near O) minus .5 times 4 bonding electrons (each line represents two electrons) and altogether you should get 0 as well. Example 2: With the carbonate ion (CO ) 3  Write the formal charges for the carbonate ion: o (You can ignore the red lines) o Formal charge = total number of valence electrons – total number of nonbonding electrons - .5(total number of bonding electrons)  C atom: 4 – 0 - .5(8) = 0  O atom in C=O: 6 – 4 - .5(4) = 0  Remaining O atoms: 6 – 6 - .5(2) = -1  Since the O atoms in C-O equal up to -1, the Lewis structure with formal charges will look like this:  Resonance Structure  Resonance Structure: One of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.  Example 1: Draw there resonance structures for the molecule nitrous oxide, N2O (the atomic arrangement is NNO). Indicate formal charges and rank the structures in their relative importance to the overall properties of the molecule. o o B is the most important because the negative charge is on the more electronegative oxygen atom. C is the least important because the positive charge is on the more electronegative oxygen atom.


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