1. Joanna Sabey Chemistry 1411 1

2.  Lewis Dot Symbol: consists of the symbol of an element and one dot for each valence electron.  Valence Electron: the outer shell electrons of an atom. The valence electrons are the electrons that participate in chemical bonding. Based on the s and p orbitals. Group# of electrons 1A1 2A2 3A3 4A4 5A5 6A6 7A7 8A8 2

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4.  Ionic Bond: The electrostatic force that holds together an ionic compound, metal and nonmetal bond.   Use Lewis dot symbols to show the formation of aluminum oxide (Al 2 O 3 ). 4 Li + F Li + F - Li Li + + e - e - + FF -

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6.  Born-Haber cycle: relates lattice energies of ionic compounds to ionization energies, electron affinities and other atomic and molecular properties. 6

7.  Covalent bond: a bond in which two electrons are shared by two atoms, nonmetals.  In covalent bonds, each elements wants to have or share 8 electrons.  H + H  H H or H-H  For Water:  Single Bond: two atoms are held together by one electron pair. 7 H H O ++ O HH O HHor

8.  Double Bond: two atoms share two pairs of electrons. or O=C=O  Triple Bond: Two atoms share three pairs of electrons. or 8 O C O N N N N

9.  Electronegativity: The ability of an atom to attract toward itself the electrons in a chemical bond.  Polar Covalent Bond: The electrons spend more time in the vicinity of one atom than the other.  To determine the type of bond, must find the difference in the electronegativity values 9

10.  Increase in Electronegativity: Covalent < Polar Covalent < Ionic Bond shared e - partial transfer e - transfer of e -  Difference in value: 0- Covalent bond 0.1 to 2.0- polar covalent bond 2.0 higher-Ionic bond 10

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12.  Classify the following bonds as ionic, polar covalent, or covalent: H= 2.1, C=2.5, Cl =3.0, K = 0.8, F=4.0  HCl: ◦ Find the difference 3.0-2.1= 0.9 ◦ Between 0.1 and 2, this is a polar covalent bond  KF ◦ Find the difference 4.0-0.8=3.2 ◦ Greater than 2, ionic bond  the CC bond in H 3 CCH 3 ◦ Find the difference 2.5-2.5 =0 ◦ Equals 0, covalent bond 12

13.  Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center, except Hydrogen, it will always been on the outside.  Count total number of valence e -. Add 1 for each negative charge. Subtract 1 for each positive charge.  Complete an octet for all atoms except hydrogen.  If structure contains too many electrons, form double and triple bonds on central atom as needed. 13

14.  Write the Lewis Structure for the following compounds:  Nitrogen trifluoride (NF 3 )  Nitric Acid (HNO 3 )  Carbonate ion(CO 3 2- ) 14

15.  Formal Charge: The electrical charge difference between the valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. Formal Charge: Valence e - - e - assigned to atom  All nonbonding electrons belong to the atom  Any shared pairs are split, one goes with one atom and the other with the other atom. O=O-O 1 2 3  O #1 = 6e - - 6e - = 0  O #2 = 6e - - 5e - = +1  O #3 = 6e - - 7e - = -1 15

16.  What are the formal charges for the atoms in the carbonate ion (CO 3 2- )? ◦ Draw Lewis Structure ◦ C atom: 4e - - 4e - = 0 ◦ O #1 atom: 6e - - 7e - = -1 ◦ O #2 atom: 6e - - 6e - = 0 ◦ O #3 atom:6e - - 7e - = -1  What are the formal charges for the atoms in the nitrite ion (NO 2 - )? ◦ Draw Lewis Structure ◦ N atom: 5e - - 5e - = 0 ◦ O #1 atom: 6e - - 7e - = -1 ◦ O #2 atom: 6e - - 6e - = 0 16

17.  If a compound has more than one acceptable Lewis Structure, use the one that has the least or no formal charge on the atoms.  Formaldehyde ( CH 2 O) 17

18.  Resonance: the use of two or more Lewis structures to represent a particular molecule.  Resonance Structure: one of the two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. O=O-O - or - O-O=O 18

19.  Draw three resonance structures for the molecule nitrous oxide, N 2 O.  Draw the Skeletal structure: N N O  Arrange electrons: - N=N + =O N N + -O - 19

20.  The Incomplete Octet: The central atom does not contain eight electrons. ◦ In Group 2A, Be ◦ Group 3A, Particularly Boron and Aluminum.  Odd-Electron Molecules: The total number of valence electrons is odd. ◦ NO and NO 2 for example ◦ Generally known as radical compounds and easily react to form a more stable compound. 20

21.  The Expanded Octet: central elements can contain more than eight electrons around it, elements that can expand with d shells. ◦ Sulfur can do this in SF 6  Draw the Lewis structure for the following compounds: ◦ Aluminum iodide(AlI 3 ) ◦ Phosphorus pentafluoride (PF 5 ) ◦ Beryllium Fluoride(BeF 2 ) ◦ Arsenic pentafluoride (AsF 5 ) ◦ Xenon tetrafluoride (XeF 4 ) 21

22.  Bond Enthalpy: The enthalpy change required to break a particular bond in 1 mole of gaseous molecules.  In thermochemistry: BE= Bond energy ΔH 0 = ΣBE(reactants)- ΣBE(products) 22 EndothermicExothermic

23.  Calculate the Enthalpy of: H 2 (g) + Cl 2 (g)  2 HCl (g) ◦ Types of Bonds broken: 1 H-H: 436.4 kJ/mol 1 Cl-Cl: 242.7 kJ/mol ◦ Types of Bonds Formed: 2 H-Cl: 431.9 X 2= 863.8kJ/mol ◦ ΔH 0 = ΣBE(reactants)- ΣBE(products) ◦ ΔH 0 =(436.4 kJ/mol+242.7kJ/mol)- (863.8kJ/mol) ◦ ΔH 0 = -184.7 kJ/mol 23

24.  Calculate the Enthalpy of the following reactions:  H 2 (g) + F 2 (g)  2 HF(g) ◦ Types of Bonds Broken:  1 H-H: 436.4  1 F-F: 156.9 ◦ Types of Bonds Formed:  2 H-F: 2 X 568.2= 1136.4 ◦ Solve for Enthalpy:  ΔH 0 =(436.4 kJ/mol+156.9kJ/mol)-(1136.4 kJ/mol)  ΔH 0 =-543.1 kJ/mol  2H 2 (g) + O 2 (g)  2 H 2 O(g) ◦ Types of Bonds Broken:  2 H-H: 2X 436.4 = 872.8  1 O=O: 498.7 ◦ Types of Bonds Formed:  4 H-O: 4X460 = 1840 ◦ Solve for Enthalpy:  ΔH 0 =(872.8 kJ/mol+498.7kJ/mol)- (1840kJ/mol)  ΔH 0 =-468.5 kJ/mol 24