Ch9_Bonding - 1 CHEMICAL CHEMICAL BONDING BONDING Cocaine...

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Unformatted text preview: 1 CHEMICAL CHEMICAL BONDING BONDING Cocaine Chemical Bonding Problems and questions — How is a molecule or polyatomic ion held together? Why are atoms distributed at strange angles? Why are molecules not flat? Can we predict the structure? How is structure related to chemical and physical properties? 2 Forms of Chemical Bonds • There are 2 extreme forms of connecting or bonding atoms: 3 • Ionic—complete transfer of 1 or more electrons from one atom to another • Covalent—some valence electrons shared between atoms • Most bonds are somewhere in between. 4 Essentially complete electron transfer from an element of low IE (metal) to an element of high affinity for electrons (nonmetal) 2 Na(s) + Cl2(g) ---> Na(s) 2 Na+ + 2 ClTherefore, ionic compds. compds. exist primarily between metals at left of periodic table (Grps 1A and 2A and (Grps transition metals) and nonmetals at right (O and Ionic Ionic Bonds Bonds Covalent Bonding The bond arises from the mutual attraction of 2 nuclei for the same electrons. Electron sharing results. (Screen 9.9) 5 Chemical Bonding Objectives Objectives are to understand: 1. valence e- distribution in molecules and ions. 2. molecular structures 3. bond properties and their effect on molecular properties. 6 HA + HB HA HB Bond is a balance of attractive and repulsive forces. Page 1 7 Electron Electron Distribution Distribution in Molecules in Molecules • Electron distribution is depicted with Bond and Lone Pairs • Valence electrons are distributed as shared or BOND PAIRS and BOND and unshared or LONE PAIRS. LONE 8 Bond Formation A bond can result from a “head-tohead” overlap of atomic orbitals overlap of on neighboring atoms. 9 Lewis electron dot structures • Valence electrons Valence are distributed as shared or BOND BOND PAIRS and and unshared or or H Cl •• •• • • H lone pair (LP) + Cl •• •• • • H Cl •• •• • • shared or bond pair G. N. Lewis 1875 - 1946 LONE PAIRS. This is called a LEWIS ELECTRON DOT structure. Overlap of H (1s) and Cl (2p) Note that each atom has a single, unpaired electron. Valence Electrons Valence Electrons Electrons are divided between 10 Rules of the Game No. of valence electrons of a main group atom = Group number •For Groups 1A-4A (14), no. of bond pairs = group number. •For Groups 5A (15)-7A (17), BP’s = 8 - Grp. No. Grp. 11 Rules of the Game •No. of valence electrons of an atom = Group number •For Groups 1A-4A (14), no. of bond pairs = group number • For Groups 5A (15)-7A (17), BP’s = 8 - Grp. Grp. No. •Except for H (and sometimes atoms of 3rd and higher periods), 12 core and core and valence electrons B 1s2 2s2 2p1 2s 2p Core = [He] and valence = 2s 2 2p1 2p BP’s + LP’s = 4 Br [Ar] 3d10 4s2 4p5 [Ar] 4s 4p Core = [Ar] 3d10 and valence = 4s 2 4p5 [Ar] and 4p This observation is called the OCTET RULE Page 2 Building a Dot Structure Ammonia, NH3 1. Decide on the central atom; never H. Central atom is atom of lowest affinity for electrons. Therefore, N is central 2. Count valence electrons H = 1 and N = 5 Total = (3 x 1) + 5 = 8 electrons / 4 pairs 13 Building a Dot Structure 3. Form a single bond between the central atom and each surrounding atom 14 15 Sulfite ion, SO32Sulfite ion, SO32Step 1. Central atom = S Step 2. Count valence electrons S= 6 3 x O = 3 x 6 = 18 Negative charge = 2 TOTAL = 26 e- or 13 pairs Step 3. Form bonds HNH H NH H •• 4. Remaining electrons form LONE PAIRS to complete octet asH needed. 3 BOND PAIRS and 1 LONE PAIR. Note that N has a share in 4 pairs (8 electrons), while H shares 1 pair. O S 10 pairs of electrons are 10 pairs of electrons are now left. now left. O O 16 17 18 Sulfite ion, SO32Sulfite ion, SO32Remaining pairs become lone pairs, first on outside atoms and then on central atom. • • Carbon Dioxide, CO2 Carbon Dioxide, CO2 1. 2. 3. Central atom = _______ Valence electrons = __ or __ pairs Form bonds. • • Carbon Dioxide, CO2 Carbon Dioxide, CO2 4. Place lone pairs on outer atoms. O •• •• C O •• •• • • O S •• •• • • O C O •• This leaves 6 pairs. • • O •• •• O •• •• 5. So that C has an octet, we shall form DOUBLE BONDS between C and O. • • • • 4. Place lone pairs on outer atoms. • • Each atom is surrounded by an octet of elect rons. electrons. O •• •• C O •• • • O •• •• C O •• •• • • • • O •• C O •• • • The second bonding pair forms a pi (π) bond. pi bond. Page 3 Double and even triple bonds are commonly observed for C, N, P, O, and S 19 20 H2CO Sulfur Dioxide, SO2 Sulfur Dioxide, SO2 1. Central atom = S 2. Valence electrons = 18 or 9 pairs • • •• •• •• Sulfur Dioxide, SO2 Sulfur Dioxide, SO2 bring in left pair • • 21 OR bring in right pair O •• S O •• • • O •• •• S O •• •• • • •• O •• •• • • •• •• • • This leads to the following structures. • • SO3 • • 3. Form double bond so that S has an octet — but note that there are two ways of doing this. bring in left pair • • O •• S •• O S •• •• O • • C2F4 O •• C O •• • • O •• •• S •• O •• •• OR bring in right pair • • These equivalent structures are called RESONANCE STRUCTURES. The RESONANCE true electronic structure is a HYBRID of the two. 22 23 24 Urea, (NH2)2CO Urea, (NH2)2CO 1. Number of valence electrons = 24 e2. Draw sigma bonds. Urea, (NH2)2CO 3. Place remaining electron pairs in the molecule. • • O C O C •• • • •• HN H N H H HN H •• N H H Page 4 25 26 Urea, (NH2)2CO 4. Complete C atom octet with double bond. Violations of the Octet Rule Usually occurs with B and elements of higher periods. Boron Trifluoride • Central atom = _____________ • Valence electrons = __________ or electron pairs = __________ • Assemble dot structure • • •• 27 O HN H •• •• • • •• C N H H BF3 3 SF4 4 • • F •• • • F •• B • • F •• • • The B atom has a The B atom has a share in only 6 pairs share in only 6 pairs of electrons (or 3 of electrons (or 3 pairs). B atom in pairs). B atom in many molecules is many molecules is electron deficient. electron deficient. 28 Sulfur Tetrafluoride , SF4 Tetrafluoride, • Central atom = • Valence electrons = ___ or ___ pairs. • Form sigma bonds and distribute electron pairs. • • F •• • • •• S •• •• F •• •• • • F •• F •• •• • • 5 pairs around the S 5 pairs around the S atom. A common atom. A common occurrence outside the occurrence outside the 2nd period. 2nd period. Page 5 ...
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This note was uploaded on 01/11/2011 for the course ENGINEERIN MAE 107 taught by Professor Pozikrizdis during the Fall '08 term at San Diego.

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