Chemical Bonding [Autosaved].pptx

Chemical Bonding CHEM 1213 Fall 2023 Types of Chemical Bonds Ionic Compounds • Ionic Bonds NaCl MgO Ca(OH)2 Fe2Cl3 K2HPO4 NH4Br • Atoms lose or gain electrons to become ions and isoelectronic with a noble gas; ions of opposite charge associated through electrostatic interactions • Covalent Bonds • Atoms become isoelectronic with a noble gas through sharing of electrons • Metallic Bonds • Occurs in metals where electrons are free to move Molecular Compounds H2O CH4 CO2 N(CH3)3 C6H12O6 DNA Lewis Dot Structures for Ionic Compounds Energetics of Ionic Compound Formation 1. Vaporization of Na(s) 2. Ionization of Na(g) 3. Bond breakage in Cl2(g) 4. Ionization of Cl(g) 5. Condensation to form 3D lattice Net reaction: Net Reaction Revisited ______________________________________________ X Na(s) + Na(g) + X X 1/2Cl2(g) + Cl(g) + X + Na(g) + Na(g) + X - e + X X Cl(g) + - e X + + Na(g) X - Cl(g) + X - + Cl(g) NaCl(s) DE (kJ/mol) 108 495.6 120 -348.5 -781 ___________ -406 Experimental DE = 411 kJ/mol Chemical Formulas NaCl MgO Ca(OH)2 FeCl3 Chloride K2HPO4 NH4Br CuNO3 Nitrate Sodium Chloride Magnesium Oxide Calcium Hydroxide Iron (III) Chloride or Ferric Potassium Hydrogen Phosphate Ammonium Bromide Copper (I) Nitrate or Cuprous Monoatomic Ions Simple Cations: Simple Anions Na+ sodium K+ potassium Ca+2 calcium Ag+ silver Zn+2 zinc Sc+3 scandium F- fluoride Clchloride I- iodide O-2 oxide S-2 sulfide N-3 nitride Metal Cations with Different Charges Cr+2 Cr+3 Fe+2 Fe+3 Cu+1 Cu+2 chromous chromic ferrous ferric cuprous cupric Lower charge: ous Higher charge: ic Polyatomic Ions Cation NH4+ s: Anions: OH hydroxide NO3nitrate NO2nitrite HSO4- hydrogen sulfate SO4-2 sulfate Ammonium H2PO4- dihydrogen phosphate HPO4-2 hydrogen HCO phosphate 3 hydrogen -3 PO phosphate carbonate 4 CO3-2 carbonate C2H3O2- acetate Naming Acids HCl acid HBr acid HNO2 HI HNO3 hydrochloric hydrobromic nitrous acid hydroiodic acid nitric acid H2SO3 sulfurous acid H2SO4 sulfuric acid HClO hypochlorous acid HClO2 chlorous acid HClO3 chloric acid HClO “ous”4 perchloric acid givesacid “ite” polyatomic ion “ic” acid gives “ate” polyatomic ion Reaction of Acids with Water H3O+ is hydronium ion Percent Composition 1. Determine formula weight NaCl: Na = 22.99 Cl = 35.45 Total 58.44 2. Divide each elements contribution to the total by the total %Na = 22.99/58.44 x 100% = 39.34% %Cl = 35.45/58.44 x 100% = 60.66% K2SO4 K = 2 x 39.10 = 78.20 S = 1 x 32.07 = 32.07 O = 4 x 16.00 = 64.00 % K = 78.20/174.23 x 100% = 44.88% % S = 32.07/174.23 x 100% = 18.41% %O = 64.00/174.23 x 100% = 36.73% Total: Check 174.27 100.02% Empirical Formula from % Composition A compound was analyzed and found to be 74.83% C and 25.17% H. What is the empirical formula? Assume 100 g of “stuff”: Moles of C = 74.83 gC/12.01 gC/mol = 6.23 mol C Moles of H = 25.17gH/1.01 gH/mol = 24.92 Determine mole ratio (divide by the smallest mol H number of moles): 6.23 mol C/6.23 = 1 mol 24.92 mol H/6.23 = 4 CH 4 mol H A compound was found to be 40.00% C, 6.73% H and the rest was O. Determine the empirical formula. Moles of C = 40.00 gC/12.01 g C/mol = 3.33 mol C Moles of H = 6.73 gH/1.01 gH/mol = 6.66 mol H Determine molegO/16.00 ratio: Moles of O = 53.30 gO/mol = 3.33 mol C/3.33 = 1 mol 3.33 O CH2O mol 6.69 mol H/3.33 = 2 mol The molecular weight of that compound was determined to be 60.04. Determine the molecular formula. The empirical formula weight is: 1 C x 12.01 = 12.01 2 H x 1.01 = 2.02 1 O x 16.00 = 16.00 Total : 30.03 60.04/30.03 = 2 Thus, the molecular formula is: C2H4O2 1.50 g of copper is heated with sulfur to form a sulfide of copper. The mass of the product was 2.26 g. Determine the empirical formula. Mass of Cu: 1.50 g Mass of S: 2.26 g product – 1.50 g cu = 0.76 g S reacted Moles of Cu = 1.50 gCu/63.55 gCu/mol = 2.36 x 10-2 mol Cu 2.36 x 10-2 mol Cu/2.36 x 10-2 = Moles of S = 0.76 gS/32.07 gS/mol CuS =2.36 x 10-2 1 mol S 2.36 x 10-2 mol S/2.36 x 10-2 = Covalent Bonding Consider H2 1s1 + 1s1  s bond: molecular orbital Lennard-Jones Potential r What about F2 and HF? 2p + 1s bond 2p +  2p s bond  s Writing Lewis Dot Structures for Compounds • Add up valence shell electrons, add 1 for each – charge or subtract 1 for each positive charge • Arrange symbols in proper geometry • Draw in electrons to: • 1. Satisfy the octet rule • 2. Have the correct total number of electrons CH4 1. Carbon: 4 valence electrons Hydrogen: 1 Valence electron Total number of electrons: 8 2. 3. NBr3 1. Nitrogen: 5 valence electrons Bromine: 7 valence electrons each Total number of electrons: 26 2. 3. ClO3 - 1. Chlorine: 7 valence electrons Oxygen: 6 valence electrons each Negative charge: 1 electron Total: 26 electrons 2. 3. -1 CO2 1. Carbon:4 valence electrons Oxygen: 6 valence electrons Total: 16 electrons 2. 3. NO3 - 1. Nitrogen: 5 valence electrons Oxygen: 6 valence electrons - charge: 1 electron Total: 24 electrons 2. 3. Resonance Formal Charges Formal charges (FC) are charges individual atoms may have in a molecule or polyatomic ion. FC = Z – (NB + B/2) Where Z is number of normal valence electrons, NB is the number of non bonding electrons and B is the number of bonding electrons For the nitrogen: Z=5 NB = 0 B=8 FC = 5 – (0 + 8/2) = +1 Note that the sum of the formal charges is the charge of For this oxygen: Z=6 NB = 4 B=4 FC = 6 –(4+4/2) =0 For these oxygens: Z=6 NB = 6 B=2 Exceptions to the Octet Rule A. Fewer than 8 electrons: AlCl3 1. Aluminum: 3 valence electrons Chlorine: 7 valence electrons each Cl Al Total: 24 Al Cl 2. Cl 3. Cl Cl Cl B. More than 8 electrons: SO4-2 1. Sulfur: 6 valence electrons Oxygen: 6 valence electrons each -2 charge: 2 electrons O Total: 32 O 2. O S O 3. O S O O -2 O What about formal charges? O -1 O S 0 0 O O -1 O0 Double bonded oxygens: FC = 6 – (4 + 4/2) = 0 Single bonded oxygens: FC = 6 – (6 + 2/2) = - -1 O -1 +2 S O -1 O -1 Oxygens: FC = 6 – (6 +2/2) = -1 Sulfur: FC = 6 – (0 + 8/2) = +2 Bond Lengths X C 154 H C 74 – 92 pm for X = H, C, N, O, F 127 – 161 pm for X = Cl, 154 Br, I– 143 pm for X = C, N, O, F 199 – 266 pm for X = Cl, Br, I C 134 C 120 pm H-H 74 pm H-N 100 H-Br141 C-C 154 pm C-O 143 pm C-Cl 178 pm Bond Dissociation Energies The energy requires to break a covalent bond: BE = 155 kJ/mol BE = 435 kJ/mol BE = 464 kJ/mol BE = 422 kJ/mol BE = 339kJ/mol BEtotal = 1660 kJ/mol Net reaction: DHtotal = BEtotal/4 = 1660 kJ/mol/4 = 415 kJ/mol BE increases with bond multiplicity: C 348 kJ/mol C 614 kJ/mol 839 kJ/mol DHreaction = D H(bonds broken) - D H(bonds formed) Calculate DH for the following reaction: CH3OH + HCl CH3Cl + H2O Bonds broken: C-O BE = 358 kJ/mol DH = (789-791) H-Cl BE = 431 kJ/mol = -2 kJ/mol Total = 789 kJ/mol Bonds formed: C-Cl BE = 328 kJ/mol Bond Polarity Consider the bond between atoms A and B: A-B The difference in electronegativity between A and B determines whether the bond is ionic, pure covalent, covalent or polar DEN Bond type Exampl covalent. e 0 Pure covalent F2 Small (0.1 to 0.4) Covalent CH4 If DEN is large, the bond would be ionic: For example, for Na (EN = 0.9 D) and Cl (EN = 3.0), DEN = 2.1, which would be considered large: thus, NaCl is ionic. For F (EN = 4.0 D), DEN = 0 for F-F: thus, F2 has a pure covalent bond. This would also be true for any diatomic molecule where both atoms are the same (H2, O2, Cl2, etc.) What about the C-H bond in CH4? For C, EN = 2.5 D and for H, EN = 2.1: thus, DEN = 0.4. The C-H bond is covalent What about the bond in H-F? EN for H is 2.1 D and EN for F is 4.0 D so DEN is 1.9. The bond in HF is polar covalent. Dipole Moments (m) Consider HF d + m d- m = qr, where q = charge, r = distance between m is a vector with both direction and magnitude m For the above system, r = 130 pm and q = 1.6 x 10-19 C m = (130 x 10-12 m)(1.6 x 10-19 C) = 2.1 x 10-29 Cm = 6.2 D (Debye) -30 Molecu le Cl2 ClF HF LiF DEN m (D) 0 1.0 1.9 3.0 0 0.88 1.82 6.33 % Ionic Character % Ionic Character = m/mo x 100% Where m is the measured dipole moment and mo is the dipole moment for electron completely transferred Let’s say r = 130 pm and m is measured to be 3.5 D: % Ionic Character = 3.5 D/6.2 D x 100% = 56% Molecular Geometry: VSEPR VSEPR is Valence Shell Electron Pair Repulsion: The valence shell electron pairs surrounding an atom in a molecule assume a geometry to maximize the distance between electron pairs and thereby resulting in the lowest energy arrangement by minimizing e :e Thus, geometry depends upon the number of repulsion. electron pairs! 2 electron pairs (BeCl2): Linear a = 180 o a Is bond angle 3 electron pairs (BF3): a= 120o Trigonal Planar 4 electron pairs (CH4): Tetrahedra l a= o 109.5 Wedge and Stick Projection: Solid lines: Bonds in plane of paper Wedge: Bond coming out of plane 4 electron pairs (NH3) Trigonal Pyramidal (3 bonding, 1 nonbonding): Bond angle is compressed from normal tetrahedral value of 109.5o to 107o due to repulsion between the non a = 107o bonding pair of electrons and the 4 electron pairs (H2O) (2 bonding pairs, 2 nonbonding pairs): Bent a= o 5 electron pairs (PCl Trigonal Bipyramidal 5): axial a = 90o equatorial equatorial equatorial axial a = 120o 5 electron pairs (SF4) (4 bonding, 1 nonbonding): Seesaw or X Actually: a = 186o 5 electron pairs (ClF3) T Shaped (3 bonding pairs, 2 nonbonding pairs): Nonbodning electron pairs occupy equatorial positions 5 electron pairs (XeF2) 2 bonding pairs, 3 nonbonding Linear pairs): 6 electron pairs Octahedral (SF6): a = 90o for all F-SF bond angles 6 electron pairs (BrF5): Square Pyramidal (5 bonding pairs, 1 nonbonding pair): a < 90o a = 90o Since all positions in octahedron are equivalent, the nonbonding pair of electrons can go on any position 6 electron pairs (XeF4) (4 bonding pairs, 2 nonbondingSquare Planar pairs): Nonbonding electron pairs 180o apart All atoms are in the same plane All F-Xe-F bond angles are o Molecular Polarity Molecules which contain polar bonds may themselves be polar. Molecular polarity depends upon the atoms comprising the bonds and the relative geometry of the polar bonds. Polar molecules have very different physical properties than nonpolar molecules. Consider the following molecules: Thus, CH4 and CF4 are nonpolar molecules m=0 m > 0 < m > 0 < m >> 0 And CHF3 is more polar than CH2F2 which is more polar than CH3F m=0 What about the following? > > > Molecular polarity decreases as C-X bond polarity decreases: F > Cl > Br > I in electronegativity What about these? m= 0 m = 0 ish m>0 m >> 0 Valence Bond Theory 1. Atomic orbitals overlap to make bonds (molecular orbitals) 2. No more than 2 electrons in an orbital What about H2? Another way to look at it: Now F2: 1s22s22p5 Overlap the p orbitals containing 1 + 2p 2p  End on end: Side by side: More favorable: better overlap What about CH4? C: 1s22s22p2 Need to create hybridized atomic orbitals hybridizati on CH4 thus has 4 CH bonds formed from the overlap of the 1s orbital of H with an sp3 orbital of C to form s bonds Hybridized atomic orbitals Bring in 4H with 1 electron each An sp3 hybridized How about NH3? N: Bring in 3 H Thus, the lone pair with 1 occupies an sp3 orbital electron each and the remaining 3 sp3 orbitals overlap with the 1s orbitals of H to form s bonds Now H2O O: Thus, the 2 lone pairs occupies sp3 orbitals and the remaining 2 sp3 orbitals overlap with the 1s orbitals of H Bring in 2 H with 1 electron each BF3 B: Thus, we have 3 s bonds from the overlap of a p orbital of F with an sp2 orbital of B and 1 p orbital that is not used which is Bring in 3 F with 1 electron each in a p orbital BeF2 Be: Thus, we have 2 s Bring in 2F with 1 electron each bonds from the in a p orbital overlap of a p orbital of F with an sp orbital of Be and 2 p orbitals that are not used. PCl5 P: [Ne]3s23p3 Thus, we have 5 s bonds from the overlap of a p orbital of Cl with an sp3d orbital of P Bring in 5Cl with 1 electron each in a p orbital SF6 S: [Ne]3s23p4 Thus, we have 6 s bonds from the overlap of a p orbital of F with an sp3d2 orbital of S Bring in 6F with 1 electron each in a p orbital C2H4 The C-C double bond is comprised of a s bond and a p bond 1. Overlap the s orbital of each 2 H with an sp2 orbital of the C to make the C-H s bonds 2. Overlap an sp2 orbital of 1 C with the sp2 orbital of the other carbon to make the C-C s bond 3. Overlap the p orbital of one C with the p Thus, C2H4 has 4 C-H s bonds (s + sp2), 1 C-C s bond (sp2 + sp2) and 1 C-C p bond (p + p). Each C is trigonal planar so all 6 atoms must be in the same plane to allow the p orbitals to overlap maximally. staggered Rotate one end by 180o eclipsed These are referred to as different conformations cis X These are referred as geometrical tran s C2H2 The C-C triple bond is comprised of 1 s bond and 2 p bonds 1. Overlap the s orbital of a H with an sp orbital of the C to make a C-H s bond 2. Overlap an sp orbital of 1 C with the sp orbital of the other carbon to make the CC s bond 3. Overlap the p orbitals of one C with the p Thus, C2H2 has 2 C-H s bonds (s + sp), 1 C-C s bond (sp + sp) and 2 C-C p bonds (p + p) for a linear geometry. Molecular Orbital Theory A molecular orbital is simply a linear combination of atomic orbitals: MO = LCAO However, the number of MOs generated must equal the number of AOs used: Y1 = a1y1 + b1y2 Y2 = a2y1 – b2y2 For H2: 1s + 1s  s bonding MO 1s - 1s  s* antibonding MO Energetica lly: H2: (s1s)2 for the ground state; (s1s)1(s*1s)1 for the excited state hn Energy Correlation Diagram Bond Order: = ½(nb – na): So for H2, BO = ½(2 – 0) = 1 What about He2? BO = ½(2-2) = 0 EC =(s)2(s*)2 Thus, He2 is unstable!! N2 BO = ½(8 – 2) = 3 EC = KK(s2s)2(s2s*)2(p2p)4(s2p)2 Delocalization and Resonance: Ozone (O3) Each oxygen is sp2 hybridized thus each oxygen has a p orbital which overlaps with the Benzene: C6H6 Antibonding Bonding MO Occupancy

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