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Lecture Presentation Unit 6: Chemical Bonding Day 2: Formal Charge, Resonance Structures, and Bond Enthalpy
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Warm Up TAKE OUT: Unit 6 Notes ( ) AND ChemActivity or ChemQuest to have me stamp them for full credit COME: Grab TODAY’S NOTES, Monday’s assignments, and Answers Keys (if wanted) THEN ANSWER: What is the formula for calculating the formal charge of an atom? What about a molecule? TIME: 6 MINUTES
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Agenda Lecture Formal Charge Resonance Structures
Bond Enthalpy and Bond Length Work Time
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Formal Charge Take Notes: on the following video for the following molecules CH4 SO2
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Writing Lewis Structures
The dominant Lewis structure is the one in which atoms have formal charges closest to zero. puts a negative formal charge on the most electronegative atom. Which one is it?
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The Best Lewis Structure?
Following our rules, this is the Lewis structure we would draw for ozone, O3. However, it doesn’t agree with what is observed in nature: Both O to O connections are the same. Start here 6/11/10
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Resonance One Lewis structure cannot accurately depict a molecule like ozone. We use multiple structures, resonance structures, to describe the molecule.
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Resonance The organic compound benzene, C6H6, has two resonance structures. It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring. Localized electrons are specifically on one atom or shared between two atoms; Delocalized electrons are shared by multiple atoms.
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Resonance Draw two equivalent resonance structures for the formate ion, HCO2-
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Resonance Draw two equivalent resonance structures for the formate ion, HCO2- (-) O O (-) H C O H C O Would you be able to show resonance with a molecule with only single bonds?
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Exceptions to the Octet Rule
There are three types of ions or molecules that do not follow the octet rule: ions or molecules with an odd number of electrons, ions or molecules with less than an octet, ions or molecules with more than eight valence electrons (an expanded octet).
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Odd Number of Electrons
Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons.
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Fewer Than Eight Electrons
Elements in the second period before carbon can make stable compounds with fewer than eight electrons. Consider BF3: Giving boron a filled octet places a negative charge on the boron and a positive charge on fluorine. This would not be an accurate picture of the distribution of electrons in BF3.
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Fewer Than Eight Electrons
The lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom.
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More Than Eight Electrons
When an element is in period 3 or below in the periodic table (e.g., periods 3, 4, 5, etc.), it can use d-orbitals to make more than four bonds. Examples: PF5 and phosphate below (Note: Phosphate will actually have four resonance structures with five bonds on the P atom!)
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More Than Eight Electrons
Which of the following atoms is never found with more than an octet of valence electrons around it? S, C, P, Br, or I? Carbon
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More Than Eight Electrons
In which of the following is there only one lone pair of electrons on the central sulfur atom? SF4 SF6 SOF4 SF2 SO42- A
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Covalent Bond Strength
Most simply, the strength of a bond is measured by determining how much energy is required to break the bond. This is called the bond enthalpy. The bond enthalpy for a Cl—Cl bond, 1(Cl— Cl), is measured to be 242 kJ/mol.
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Average Bond Enthalpies
Average bond enthalpies are positive, because bond breaking is an endothermic process. Note that these are averages over many different compounds; not every bond in nature for a pair of atoms has exactly the same bond energy.
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Using Bond Enthalpies to Estimate Enthalpy of Reaction
One way to estimate H for a reaction is to use the bond enthalpies of bonds broken and the new bonds formed. Energy is added to break bonds and released when making bonds. In other words, Hrxn = (bond enthalpies of all bonds broken) − (bond enthalpies of all bonds formed).
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Example From the figure on the last slide
CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) In this example, 4 C—H bonds and one Cl—Cl bond are broken; one C—Cl and 3 C—H and one H—Cl bond are formed.
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Answer = [4(413 kJ) + 1(242 kJ)] − [3(328 kJ) + 1(413) + (431 kJ)]
H = [4(C—H) + 1(Cl—Cl)] − [3(C—Cl) + 1(C—H) + 1(H—Cl)] = [4(413 kJ) + 1(242 kJ)] − [3(328 kJ) + 1(413) + (431 kJ)] = (1652 kJ kJ) − (984 kJ kJ kJ) = +66 kJ
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Using Bond Enthalpies to Estimate Enthalpy of Reaction
Hrxn = (bond enthalpies broken) − (bond enthalpies formed) Using table 8.4, estimate the Hrxn for the reaction: H2O (g) H2 (g) + ½ O2 (g) 242 kJ 417 kJ 5 kJ -46 kJ b
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Bond Enthalpy and Bond Length
We can also measure an average bond length for different bond types. As the number of bonds between two atoms increases, the bond length decreases. Why?
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The formal charge on the nitrogen atom in the nitrate ion (NO31–) is
+2. +1. 0. –1. Answer: b
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Which molecule below violates the octet rule?
PF5 CH4 NBr3 OF2 Answer: a
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Which molecule below has an unpaired electron?
NO2 NH3 BF3 PF5 Answer: a
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greater than less than equal to variable to (depending on X and Y)
For atoms X and Y, the bond enthalpy of an X—Y bond is _______ the bond enthalpy of an X=Y bond. greater than less than equal to variable to (depending on X and Y) Answer: b
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greater than less than equal to variable to (depending on X and Y)
For atoms X and Y, the bond length of an X—Y bond is _______ the bond length of an X=Y bond. greater than less than equal to variable to (depending on X and Y) Answer: a
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WORK TIME USE: This time to work on ChemQuest 25 WORK TOGETHER: With your table partners TIME: Until End of Class WHEN DONE: Begin Homework Due on Monday
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