CHM 101 Chapters 9,10 PDF

CHAPTER NINE Chemical Bonding I Basic Concepts CHAPTER TEN Chemical Bonding II Molecular Geometry & Hybridization of Atomic Orbitals MOLECULAR STRUCTURE DETERMINATION “GENERAL RULES” 1. The most important requirement for the formation of a stable molecule is tha...

CHAPTER NINE Chemical Bonding I Basic Concepts CHAPTER TEN Chemical Bonding II Molecular Geometry & Hybridization of Atomic Orbitals MOLECULAR STRUCTURE DETERMINATION “GENERAL RULES” 1. The most important requirement for the formation of a stable molecule is that its atoms achieve noble gas electron configurations (obey the octet rule). 2. In writing Lewis structures, the rule is that only the valence electrons (V e’s) are included (Lewis structures show only V e’s). 3. In drawing a Lewis structure, the charges should be minimized, if possible. 4. The 2nd row elements C, N, O & F should always be assumed to obey the octet rule. These elements never exceed the octet rule, since their valence orbitals (2s & 2p) can accommodate only eight electrons. 5. The 3rd row elements P, S & Cl and heavier elements are often satisfy the octet rule but can exceed the octet rule by using their empty valence d-orbitals. 6. According to Valence Shell Electron Per Repulsion (VSEPR) theory, electron-electron (e-e) repulsion is maximum when the two pairs of electrons are in a position of 90o and its minimum when they are in a position of 120o or 180o. 7. Lone pairs require more room than bonding pairs and tend to compress the angles between the bonding pairs. X A X A X X X A X 8. Formal Charge (F.C.) = Group number - # of V e’s assigned to the atom (# of e’s on the atom) in the structure. 9. For any compound to be polar, its dipole moment (µ) is different from zero (µ # 0). X A X A X A X X X  C C C C C C Single bond Double bond Triple bond       Bond Order (B.O.) 1 2 3 As the Bond Order (B.O.) As the Bond Length (B.L.) As the Bond Strength (B.S.) As the E to break the bond In Drawing Lewis Structures, the Following Points Should be Noticed  All terminal atoms should be surrounded by 8 e’s (obey octet rule) except Hydrogen (should be surrounded only by 2 e’s). Central Atom X A X Terminal Atom  In general, the atom which has the highest electronegativity should carry the negative charge.  Hydrogen  Should be surrounded by 2 e’s  Can not be a central atom  Can hold only one bond ()  Can not have a  bond Fluorine  Should be surrounded by 8 e’s  Can not be a central atom  Can hold only one bond ()  Can not have a  bond  F  Should be surrounded by 8 e’s  Can not be a central atom  Can hold only one bond ()  Can not have a  bond Cl, Br & I  Can exceed octet  Can be central atoms  Can hold more than one bond  Rare to have a  bond  In general, the atom which has smaller group # should be a central atom.  If the atoms belong to the same group, the atom which has larger row # should be a central atom. Molecular Shapes Expected With & Without Lone Pairs (L.P.)  # of Molecular Name of Predicted Type of Effective # of Hybrid- Geometry Molecular Bond Examples Molecule BP (# of  LP ization (Shape) Geometry Angle bonds) AX2 2 0 sp X A X Linear 180o NO2+, CO2, C2H2 X Trigonal AX3 3 0 sp 2 A planar 120o NO3, BF3, N2O4 X X V-shaped or AX2E 2 1 sp2 A Bent ~120o NO2, SO2 X X X AX4 4 0 sp3 A Tetrahedral 109.5o SO42, NH4+, BF4 X X X Trigonal AX3E 3 1 sp3 A pyramidal ~109.5o NF3, NH3 X X X V-shaped or ~109.5o OCl2, H2O, ClO2 AX2E2 2 2 sp3 A Bent X X  # of Molecular Name of Predicted Type of Effective # of Hybrid- Geometry Molecular Bond Examples Molecule BP (# of  LP ization (Shape) Geometry Angle bonds) X X AX5 5 0 sp3d X A Trigonal 90o, 120o, PCl5 bipyramidal 180o X X X X 90o, 120o, AX4E 4 1 sp3d A Distorted SH4, SF4 tetrahedral or 180o X Sea saw X X 3 2 sp3d T-shaped 90o, 180o BrF3, XeCl3+ AX3E2 X A X X 2 3 sp3d Linear 180o XeF2, I3 AX2E3 A X X X X 90o, 180o AX6 6 0 sp3d2 A Octahedral SF6 X X X X X AX5E 5 1 sp3d2 A Square 90o, 180o BrF5, XeF5+ pyramidal X X X X X AX4E2 4 2 sp3d2 A Square 90o, 180o XeF4, ICl4 planar X X  If C, N or O is a central atom & F or H is NOT a terminal atom, Octet e's - Valence e's # of Bonds = 2  # of  bonds = # of terminal atoms  # of  bonds = # of Bonds - # of  bonds  C, N & O could have easily  bonds  No  bonds for Cl, Br & I if they are terminal atoms  AX2 (0) (+) (0) o O N O Linear, 180 , sp In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). In general, the atom which has smaller group # should be a central atom. All terminal atoms should be surrounded by 8 e’s (obey octet rule) except Hydrogen (should be surrounded only by 2 e’s). Formal Charge (F.C.) = Group number - # of V e’s assigned to the atom (# of e’s on the atom) in the structure. In general, the atom which has the highest electronegativity should carry the negative charge.  AX3 O N O O In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). Formal Charge (F.C.) = Group number - # of V e’s assigned to the atom (# of e’s on the atom) in the structure. In general, the atom which has the highest electronegativity should carry the negative charge.  AX2E N O N O O O In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). Because of electron-electron (e-e) repulsion, lone pairs require more room than bonding pairs and tend to compress the angles between the bonding pairs. Formal Charge (F.C.) = Group number - # of V e’s assigned to the atom (# of e’s on the atom) in the structure In general, the atom which has the highest electronegativity should carry the negative charge. In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s).  Resonance Structures (R.S.) O O O N N N O O O O O O 1 2 3 4 = 1.33 NO3 B.O. (N-O bond) = 3  Resonance Structures (R.S.) N N O O O O 1 2 3 = 1.5 NO2 B.O. (N-O bond) = 2  Resonance Structures (R.S.) O N O O N O O N O 1 2 3 4 =2 NO2 B.O. (N-O bond) = 2 In drawing a Lewis structure, the charges should be minimized, if possible. As the Bond Order (B.O.) As the Bond Length (B.L.) As the Bond Strength (B.S.) As the E to break the bond  4 NO2 B.O. (N-O bond) = =2 2 4 NO3 B.O. (N-O bond) = = 1.33 3 3 NO2 B.O. (N-O bond) = = 1.5 2 NO3 has the weakest & the longest N---O bond  If P, S, Cl, or beyond 3rd row elements like Br & Xe is a central atom and at least one of the terminal atoms is oxygen:  # of  bonds = # of oxygen atoms + charge  The formal charge of the central atom is Zero.  No  bonds for Cl, Br & I if they are terminal atoms  AX4 O S O O O In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). If the atoms belong to the same group, the atom which has larger row # should be a central atom. # of  bonds = # of oxygen atoms + charge All terminal atoms should be surrounded by 8 e’s (obey octet rule) except Hydrogen (should be surrounded only by 2 e’s). The formal charge of the central atom is Zero. Formal Charge (F.C.) = Group number - # of V e’s assigned to the atom (# of e’s on the atom) in the structure  AX3E N F F F F F N F In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). No  bond for F atom. Because of electron-electron (e-e) repulsion, lone pairs require more room than bonding pairs and tend to compress the angles between the bonding pairs.  AX2E2 O Cl O Cl Cl Cl In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). No  bonds for Cl, Br & I if they are terminal atoms Because of electron-electron (e-e) repulsion, lone pairs require more room than bonding pairs and tend to compress the angles between the bonding pairs..  AX5 In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). The formal charge of the central atom is Zero.  AX4E H S H H H In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). The formal charge of the central atom is Zero.  AX3E2 F F Br F In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). e-e Repulsion is maximum when the two pairs of electrons are in a position of 90o and its minimum when they are in a position of 120o or 180o. The formal charge of the central atom is Zero.  AX2E3 F Xe F In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). e-e Repulsion is maximum when the two pairs of electrons are in a position of 90o and its minimum when they are in a position of 120o or 180o. The formal charge of the central atom is Zero.  AX6 In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). The formal charge of the central atom is Zero.  AX5E In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). The formal charge of the central atom is Zero.  AX4E2 In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). e-e Repulsion is maximum when the two pairs of electrons are in a position of 90o and its minimum when they are in a position of 120o or 180o. The formal charge of the central atom is Zero.  Dipole Moment () of Molecule = Zero > Zero (Non Polar) (Polar) If the central atom is If the central atom is 1. surrounded by same 1. surrounded by different atoms & atoms or 2. no L.P. on it 2. at least one L.P. on it except in case of AX2E3 & AX4E2 systems In General  Same # of Atoms  Same # of V e’s  Same Geometry Exceptions:  AX2 & AX2E3  Same # of Atoms  Different # of V e’s  Same Geometry  AX2E & AX2E2  Same # of Atoms  Different # of V e’s  Same Geometry Question: How many of the following species are linear? I3– NCl3 NO2 – CO2 SCl2 V e’s  22 18 16 20 V e’s  16 22 Linear  Should have 3 atoms; AX2 AX2E3 EXTRA QUESTIONS ON CHAPTERS 9 & 10  Question (a to e): Use the VSEPR theory to predict the shape, hybridization and the expected bond angle of each of the following species. AX3 In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). Dipole Moment (µ) = 0  Non Polar Formal Charge (F.C.) = Group number - # of V e’s assigned to the atom (# of e’s on the atom) in the structure  AX4 In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). Formal Charge (F.C.) = Group number - # of V e’s assigned to the atom (# of e’s on the atom) in the structure  In general, the atom which has smaller group # should be a central atom. Except for poly atomic ions (O with Cl, Br or I)  oxygen cannot be a central atom because it cannot exceed octet & should carry the negative charge.  l, Br & Cl could have a  bond if they are central atoms. Examples: chlorate, chlorite, bromate …etc  AX2E2 In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s). # of  bonds = # of oxygen atoms + charge The formal charge of the central atom is Zero. O Cl O Because of electron-electron (e-e) repulsion, lone pairs require more room than bonding pairs and tend to compress the angles between the bonding pairs.  AX2 H C C H Dipole Moment (µ) = 0  Non Polar In writing Lewis structures, the rule is that only the V e’s are included (Lewis structures show only V e’s).  AX3 Dipole Moment (µ) = 0  Non Polar MOLECULAR ORBITAL (MO) FOR HOMO & HETERO NULCEAR DIATOMIC SPECIES  Homo Nuclear Diatomic Species Examples: B2, C2, N2, O2 & F2 Hetero Nuclear Diatomic Species Examples: CO, NO, CN- & OF For B2, C2 & N2 the following MO Diagram should be used For O2, F2 & OF the following MO Diagram should be used  MO Diagram for MO Diagram for B2, C2, N2, CO & NO O2, F2 & OF  Bond Order (B.O.) equal to # of bonding e’s - # of antibonding e’s 2 As the Bond Order (B.O.) As the Bond Length (B.L.) As the Bond Strength (B.S.) As the E to break the bond p B.O. = 8-2 2 =3 p p 2+ O2 p V e’s = 10 s B.O. = 3 s p p p + O2 p V e’s = 11 s B.O. = 2.5 s p p p O2 p V e’s = 12 s B.O. = 2 s p p p _ O2 p V e’s = 13 s B.O. = 1.5 s p p p 2- O2 p V e’s = 14 s B.O. = 1 s a. O22 b. O2 B.O. = 1 B.O. = 2 c. O2 d. O2+ B.O. = 1.5 B.O. = 2.5 e. O22+ B.O. = 3 As the Bond Order (B.O.) As the Bond Strength (B.S.) p B.O. = 6-2 2 =2 p p 2+ CO p V e’s = 8 s B.O. = 2 s p p p + CO p V e’s = 9 s B.O. = 2.5 s p p p CO p V e’s = 10 s B.O. = 3 s p p p _ CO p V e’s = 11 s B.O. = 2.5 s p p p 2- CO p V e’s = 12 s B.O. = 2 s a. CO2 b. CO B.O. = 2 B.O. = 3 c. CO d. CO+ B.O. = 2.5 B.O. = 2.5 e. CO2+ B.O. = 2 As the Bond Order (B.O.) As the Bond Lenght (B.L.) MO Summary of 2nd Row Diatomic The higher the bond order, the stronger is the bond.  Good Luck Dr. Imad A. Abu-Yousef