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1 Announcements Final Exam TIME: October 8, 7:30-9:30AM VENUE: CTC Multiple Choice Questions 3 Questions Each Chapter Questions Each Chapter Questions From Chapter 9-11 Saturday Review C PM RESURRECTION: There is talk among the instructors to offer 45 bonus points to anyone who scores 55/65 on the Exam. Any F would automatically turn to a D.
2 Electronegativity is an element s inherent property to draw electrons to itself when chemically bonded to another atom in a molecule. The units are dimensionless (all relative measurements to Li). Rank F O N Cl Br
3 Differences in elements electronegativity between bonding atoms result in the formation of polar-covalent bonds and net dipole moments in molecules. Polar Bond Polar Bond Polar Bond Polar Bond No Net Dipole Moment Net Dipole Moment Think of the dipole moment as a molecule with separated charges + and -.
4 For a poly-atomic molecule we must consider the vector sum of polar bonds in the molecule to see if there is a net dipole moment. Dipole Moment Dipole Moment No Net Dipole Moment No Net Dipole Moment Dipole Moment
5 VSEPRT explains the geometry of molecules but NOT how covalent bonds are formed with that geometry. Molecular formula Lewis structure VSEPRT Geometry Hybrid orbitals Lewis Structure VSEPRT Valence Bond Theory
6 Valence Bond Theory explains covalent bonding by the spatial overlap of atomic orbitals on bonding atoms and the sharing of electron pairs. Bonding in H 2 1s 1 + 1s 1 Electrons that must have opposite spins. Bonding in HF 1s 1 + 2p 5 Bonding in F 2 2p 5 + 2p 5
7 Bonding in carbon presents a problem as combining pure atomics orbitals does not conform to experiment or explain bonding in C. VBT solves this by allowing the blending of pure atomic orbitals in a proces called hybridization. sp 3 hybridized orbitals hybridization Pure atomic orbitals (valence orbitals)
8 For example, if we look at the ground state of carbon we would expect 2 possible bonds in its 2p oribitals. This molecule is not observed, however, what we see is CH4 and tetrahedral geometry. We rationalize this by making an sp 3 hybrid as shown by the excited state carbon. It can form 4-bonds. sp 3 hybridized orbitals hybridization we expect 2 bonds: CH 2 but this does not exist. Pure atomic orbitals (valence orbitals)
9 By combing different numbers of atomic orbitals we can make different hybrids that match one of the VSEPRT geometries. For example one sp pure orbital + 1 p-orbital combine to give and two sp hybrids that are linear when superimposed s-orbital + p-orbital ----> Two sp hybrids = Linear s-orbital + Two p-orbital --> Three sp 2 hybrids = Trig Planar
10 The process of combining pure atomic orbitals to form hybrid orbitals on central bonding atoms in a molecule is called hybridization. sp 3 hybrid orbitals s-orbital + Three p-orbitals -> Four sp 3 hybrids = Tetrahedral
11 Some generalized rules and comments on VBT and the formation of hybridized orbitals. 1. The number of hybrid orbitals obtained equals the number of atomic orbitals mixed. 2. The type of and shape of a hybrid orbital varies with the types of atomic orbitals mixed. 3. Each hybrid orbital has a specific geometry that matches one of five VSEPRT shapes (show below). sp sp 2 sp 3 sp 3 d sp 3 d 2 Linear Trigonal Planar Tetrahedral Trigonal Bipyramidal Octahedral
12 The goal is to understand geometry (via VSEPRT) and to relate it to a picture of covalent bonding in molecules. Molecular formula Lewis structure VSEPRT Geometry Hybrid orbitals sp sp 2 sp 3 sp 3 d sp 3 d 2 Linear Trigonal Pyramidal Tetrahedral Trigonal Bipyramidal Octahedral
13 Electron Geometry Molecular Geometry AX n E m Hybridization Linear Linear AX 2 sp Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral Trigonal planar V-shaped bent Tetrahedral Trigonal pyramidal V-shaped bent Trigonal bipyramidal Seesaw T-shaped Linear Octahedral Square pyramidal Square planar AX 3 AX 2 E 1 sp 2 AX 4 AX 3 E 1 sp 3 AX 2 E 2 AX 5 AX 4 E 1 AX 3 E 2 AX 2 E 3 AX 6 sp 3 d AX 5 E 1 sp 3 d 2 AX 4 E 2
14 Linking VSEPRT To Valence Bond Theory Hybrids Atomic Orbitals Mixed Linear Trig Planar Tetrahedral Trig Bypyr Octahedral AX2 AX3 AX4 AX5 AX6 s + p s + 2 p s + 3 p s + 3 p + d s + 3 p + 2d Hybrid Orbitals Formed Two sp Three Four sp 3 Five sp 3 d Six sp 3 d 2 sp 2 Hybrid Shape Unhybrid Orbitals Leftover Two p one p none Four d Three d
15 An sp hybrid is formed from the combination of a one pure 1s orbital and a one 2p orbital from a central bonding atom producing two new orbitals called sp orbitals. 2s Hybridization s-orbital p-orbital Two sp hybrid orbitals sp hybrid orbitals superimposed --The number of hybrid orbitals formed is equal to the number of pure orbitals combined! --When superimposed the sp-hybrid give us bonding orbitals for a linear molecules.
16 Show the bonding scheme and hybridized orbitals used in BeCl 2 2 unhybridized p-orbitals Hybridization So after hybridization we have on the central atom, 2 pure p-orbitals and two sp hybrids.
17 Show the bonding scheme and hybridized orbitals in BeCl 2 hybridization Hybridized Be Atom Isolated Be Atom two lone p-orbitals two sp hybrids on Be
18 sp hybrid:ethylyne: HC CH:Linear Sigma bonds (σ bonds) and Pi bonds (π bonds)are two different types of covalent chemical bonds that form as a result of end to end spatial overlap of atomic orbitals or hybridized orbitals (σ bonds) or side to side overlap on bonding atoms (π bonds) Lone p orbitals that are not hybridized sp hybrid orbitals
19 An sp 2 hybrid is formed from the combination of a one pure 1s orbital and a two 2p orbitals from a central bonding atom producing two new orbitals called sp 2 orbitals. 3-atomic orbitals, s and two p s combine to form 3- sp 2 hybrid orbitals sp 2 = Triginal planar geometry, 120 bond angle
20 Example 2: sp 2 hybridizaton scheme BF 3. Boron Hybrid Box Diagram Boron Orbital Box Diagram Bonding of pure p-orbital in F with sp 2 hybridized orbitals in BF 3
21 Tetrahedral geometry = sp 3 hybrid orbitals combine to generate four sp 3 orbitals Note the number of hybrids formed is the number of atomic orbitals combined! which are represented collectively as: sp 3 sp 3 = Tetrahedral geometry = bond angle
22 Example: sp 3 orbital hybridization: CH 4. sp 3 hybridization mixes one 2s orbital with three 2p orbitals to produce four sp 3 orbitals on each carbon atom. End to end overlap with a 1s orbital from H gives four sigma bond in CH 4. This is the ground state configuration of valence atomic orbitals the four sp 3 hybrid orbitals form a tetrahedral shape CH 4
23 Example 3: sp 3 hybrid orbitals in H 2 O. sp 3 hybridization mixes one 2s orbital with three 2p orbitals to produce four sp 3 orbitals. The e- are distributed throughout the hybrids ready for bonding. End to end overlap with a 1s orbital from H gives four sigma bond in CH 4. sp 3 is tetrahedral shape. In water we have AX 2 E 2 Note the lone pairs occupy 2-of the sp 3 orbitals What is the electronic geometry? What is the molecular geometry? What orbitals contribute to bonding?
24 What is the electron geometry, the molecular geometry at each carbon atom? Use that information to determine the hybridization around each carbon atom in nicotinic acid? How many sigma and pi bonds are in nicotinic acid?
25 sp 3 d hybridization in PCl 5. Trigonal Bipyramidal Electron Geometry AX 5 E 0 Trigonal BiPyramidal Molecular Geometry Isolated P atom
26 The sp 3 d 2 hybrid orbitals in SF 6 Octahedral Electron Geometry AX 6 E 0 Octahedral Molecular Geometry
27 Electron Geometry Molecular Geometry AX n E m Hybridizaton Linear Linear AX 2 sp Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral Trigonal planar V-shaped bent Tetrahedral Trigonal pyramidal V-shaped bent Trigonal bipyramidal Seesaw T-shaped Linear Octahedral Square pyramidal Square planar AX 3 AX 2 E 1 sp 2 AX 4 AX 3 E 1 sp 3 AX 2 E 2 AX 5 AX 4 E 1 AX 3 E 2 AX 2 E 3 AX 6 sp 3 d AX 5 E 1 sp 3 d 2 AX 4 E 2
28 Determine the VSEPRT geometry, the bond angles and the hybridization of each indicated atom in the following molecule? How many sigma and pi bonds are in the molecule?
29 Determine the electron domain, molecular geometry, the bond angles and the hybridization of each indicated atom in the following molecule? How many sigma and pi bonds are in the molecule? trig planar 120, sp 2 sp 2 sp 2 bent, <109.5, sp 3 sp 3 tetrahedral, 180, sp 3 linear 180, sp
30 Describe the types of bonds and orbitals in acetone, (CH 3 ) 2 CO and in CO2 and in HCN? Step 1 Step 2 Step 3 Molecular formula Lewis structure VSEPRT Geometry Hybrid orbitals
31 Describe the types of bonds and orbitals in acetone, (CH 3 ) 2 CO. PLAN: Draw the Lewis structures to ascertain the arrangement of groups and shape at each central atom. Postulate the hybrid orbitals taking note of geometries predicted from VSEPRT. Draw the orbitals and show overlap. SOLUTION: sp 3 hybridized sp 3 hybridized sp 2 hybridized σ bonds π bond
32 Postulating Hybrid Orbitals in a Molecule PROBLEM: Use partial orbital diagrams to describe mixing of the atomic orbitals of the central atom leads to hybrid orbitals in each of the following: (a) Methanol, CH 3 OH (b) Sulfur tetrafluoride, SF 4 PLAN: Use the Lewis structures to ascertain the arrangement of groups and shape of each molecule. Postulate the hybrid orbitals. Use partial orbital box diagrams to indicate the hybrid for the central atoms. SOLUTION: (a) CH 3 OH The groups around C are arranged as a tetrahedron. O also has a tetrahedral arrangement with 2 nonbonding e - pairs.
33 Postulating Hybrid Orbitals in a Molecule (a) Methanol, CH 3 OH SOLUTION: (a) CH 3 OH The groups around C are arranged as a tetrahedron. O also has a tetrahedral arrangement with 2 nonbonding e - pairs. single C atom hybridized C atom single O atom hybridized O atom
34 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Postulating Hybrid Orbitals in a Molecule (b) SF 4 has a seesaw shape with 4 bonding and 1 nonbonding e - pairs. S atom hybridized S atom 11-
35 Bond order is the number of bonds between two bonded atoms. Single bond between 2 atoms = order = 1 Double bond between 2 atoms = order = 2 Triple bond between 2 atoms = order = 3 S C N F F F N N C-N: Bond order = 2 F S F F Bond order = 3 S-C: Bond order = 2 S-F Bond order = 1
36 Higher bond orders give shorter bond lengths and require more energy to break a bond. Note how bond energies (energy required to break a bond) goes up as bond order increases. Bond Lengths Triple bond < Double Bond < Single Bond
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