Presentation on theme: "CH 10: Molecular Geometry & chemical bonding theory"— Presentation transcript:
1CH 10: Molecular Geometry & chemical bonding theory Vanessa N. Prasad-PermaulValencia Community CollegeCHM 1045
2Molecular Shapes: VSEPR Molecular Geometry: the general shape of a molecule, as determined by the relative positions of the atomic nuclei.Valence-Shell Electron-Pair Repulsion Theory: predicts the shapes of molecules and ions by assuming that the valence-shell electron pairs are arranged about each atom so that the electron pairs are kept as far away from one another as possible, thus minimizing electron-pair repulsions.
3Molecular Shapes: VSEPR Two Electron Groups: Electron groups point in opposite directions.LINEAR ARRANGEMENT
4Molecular Shapes: VSEPR Three Electron Groups: Electron groups lie in the same plane and point to the corners of an equilateral triangle.TRIGONAL PLANAR ARRANGEMENTTrigonal planar geometry or bent (angular)
5Molecular Shapes: VSEPR Four Electron Groups: Electron groups point to the corners of a regular tetrahedron.TETRAHEDRAL ARRANGEMENT
6Molecular Shapes: VSEPR The approximate shape of molecules is given by Valence-Shell Electron-Pair Repulsion (VSEPR).Step #1: Count the total electron groups.Step #2: Arrange electron groups to maximize separation.Groups are collections of bond pairs between two atoms or a lone pair.Groups do not compete equally for space:Lone Pair > Triple Bond > Double Bond > Single Bond
7Molecular Shapes: VSEPR Table of the Molecular Geometryand Examples of Certain Compounds
8¨ ¨ ¨ ¨ EXAMPLE 10.1 : Predict the geometry of the following Molecular Shapes: VSEPREXAMPLE 10.1 : Predict the geometry of the followingMolecules or ions, using the VSEPR method:BeCl B. NO C. SiCl4Be = 2 electrons Cl = 7 electrons each = 14Total of 16 electrons:Cl:Be:Cl:2 sets of electron pairs, 0 lone pairs linear
9¨ ¨ ¨ ¨ B. NO2- N = 5 electrons, O = 6 electrons each = 12 Molecular Shapes: VSEPRB. NO N = 5 electrons, O = 6 electrons each = 12Total of 17 electrons plus 1 electrons = 18 e-:O:N::O:3 sets of electron pairs, 1 lone pair trigonal planar arrangement and a bent geometry
10Molecular Shapes: VSEPR C. SiCl4Si = 4 electrons, Cl = 7 electrons each = 28 electronsTotal of 32 electrons4 electron pairs, 0 lone pairs tetrahedral
11EXERCISE 10.1 : Use the VSEPR method to predict the Molecular Shapes: VSEPREXERCISE 10.1 : Use the VSEPR method to predict thegeometry of the following ion and molecules:ClO3-OF2SiF4
12Molecular Shapes: VSEPR Five Electron Groups: Electron groups point to the corners of a trigonal bipyramid.
13Molecular Shapes: VSEPR Six Electron Groups: Electron groups point to the corners of a regular octahedron.
14Molecular Shapes: VSEPR EXAMPLE 10.2 : What do you expect for the geometry of tellurium tetrachloride?Tellurium has 6 electron pairs in its valence shellCl = 7 electrons each = 28 electronsTotal of 34 electronsThere are 4 bonding pairs and 1 lone pair of electrons Trigonal bipyramidal arrangement and a seesaw geometry
15Molecular Shapes: VSEPR EXERCISE 10.2 : According to the VSEPR model, what molecular geometry would you predict for iodine trichloride?
16Dipole Moment & Molecular Geometry Dipole Moment: a quantitative measure of the degree of charge separation in a molecule. -H ClIf a molecule has a dipole moment it is polarA compound could contain polar bonds but the molecule could be non-polar because there is no dipole momentBond dipoles can cancel each other outO C O
17Based on electronegativity differences between atoms in a molecule Dipole momentBased on electronegativity differences between atoms in a moleculeThe most electronegative atom is partially negativeThe less electronegative atom is partially positiveThe dipole moment is the average of all dipoles in the moleculeException (C-H bonds do not have a dipole)
18Molecular Shapes: VSEPR EXAMPLE 10.3 : Each of the following molecules has a non-zero dipole moment. Select the molecular geometry that is consistent with this information. Explain.SO2: linear or bentBENT because in linear geometry, the dipole moment would be zero. There are 2 electron pairs and 1 lone pair which is a trigonal planar arrangement giving a bent geometry.
19PH3: trigonal planar or trigonal pyramidal Molecular Shapes: VSEPRPH3: trigonal planar or trigonal pyramidalIn a trigonal planar geometry, the dipole moment is zero, therefore PH3 has a trigonal pyramidal geometry. Tetrahedral arrangement; 3 electron pairs and one lone pair.
20Molecular Shapes: VSEPR EXERCISE 10.3 : Bromine trifluoride BrF3, has a nonzero dipole moment. Indicate the correct arrangement and molecular geometry is consistent with this information.
21Molecular Shapes: VSEPR EXERCISE 10.4 : Which of the following would be expected to have a dipole moment of zero on the basis of symmetry. Explain.SOCl2SiF4OF2
223. The greater the amount of orbital overlap, the stronger the bond. Valence Bond Theory1. Covalent bonds are formed by overlapping of atomic orbitals, each of which contains one electron of opposite spin.2. Each of the bonded atoms maintains its own atomic orbitals, but the electron pair in the overlapping orbitals is shared by both atoms.3. The greater the amount of orbital overlap, the stronger the bond.
23According to Valence Bond Theory: An orbital on one atom comes to occupy a portion of the same region of space as an orbital on the other atom. The two orbitals are said to overlap.The total number of electrons in both orbitals is no more than two.
24Combined hybrid orbitals Valence Bond TheoryCombined hybrid orbitalsHybrid orbitals# of orbitalsVSEPR geometry
31Steps to obtain the bonding description about an atom in a molecule: Valence Bond TheorySteps to obtain the bonding description about an atom in a molecule:Write the Lewis electron-dot formulaFrom the Lewis formula, use the VSEPR model to obtain arrangementFrom the geometry, s what type of hybrid orbitals are requiredAssign valence electrons to the hybrid oritals of this atom one at a time, pairing them only when necessaryForm the bonds by overlapping singly occupied orbitals of other atoms
32EXAMPLE 10.4: Describe the bonding in H2O according to Valence Bond TheoryEXAMPLE 10.4: Describe the bonding in H2O according tovalence bond theory.Using the VESPR theory, notice there are 3 sets of electron pairs, and two lone pairs giving a tetrahedral arrangement and a bent geometry
33Valence Bond TheoryEXERCISE 10.5: Using hybrid orbitals, describe the bonding in NH3 according to valence bond theory.
34EXAMPLE 10.5: Describe the bonding in XeF4 using hybrid orbitals. Valence Bond TheoryEXAMPLE 10.5: Describe the bonding in XeF4 using hybrid orbitals.4 single bonds, 2 lone pairs octahedral arrangement and square planar geometry.
35EXERCISE 10.6: Describe the bonding in PCl5 using hybrid orbitals. Valence Bond TheoryEXERCISE 10.6: Describe the bonding in PCl5 using hybrid orbitals.
36Molecular Orbital Theory Additive and subtractive combination of p orbitals leads to the formation of both sigma and pi orbitals.
37Valence Bond TheoryEXAMPLE 10.6: Describe the bonding on a given N atom in dinitrogen difluoride, N2F2, using valence bond theory.:F N N F:
38Valence Bond TheoryEXERCISE 10.7: Describe the bonding on the carbon atom in carbon dioxide using valence bond theory.
39Valence Bond TheoryEXERCISE 10.8: Dinitrogen difluoride exists as cis and trans isomers. Write structural formulas for these isomers and explain using valence bond theory why they exist.
40Molecular Orbital Theory The molecular orbital (MO) model provides a better explanation of chemical and physical properties than the valence bond (VB) model.Atomic Orbital: Probability of finding the electron within a given region of space in an atom.Molecular Orbital: Probability of finding the electron within a given region of space in a molecule.
41Molecular Orbital Theory Additive combination of orbitals (s) is lower in energy than two isolated 1s orbitals and is called a bonding molecular orbital.
42Molecular Orbital Theory Subtractive combination of orbitals (s*) is higher in energy than two isolated 1s orbitals and is called an antibonding molecular orbital.
43Molecular Orbital Theory Bond Order is the number of electron pairs shared between atoms.Bond Order is obtained by subtracting the number of antibonding electrons from the number of bonding electrons and dividing by 2.BO = Bonding electrons – antibonding electrons2
44*2p *2p 2p 2p *2s 2s *1s 1s B2*2p *2p 2p 2p *2s 2s *1s 1sB has 5 electronsSo B2 has 10 elecCore electrons don’t count toward BO
45*2p *2p 2p 2p *2s 2s *1s 1s C2*2p *2p 2p 2p *2s 2s *1s 1sCarbon has 6 electrons so C2 has 12 electrons
46Molecular Orbital Theory Molecular Orbital Diagram for H2:
47Molecular Orbital Theory Molecular Orbital Diagrams for H2– and He2:
49Molecular Orbital Theory Second-Row MO Energy Level Diagrams:
50Valence Bond TheoryEXAMPLE 10.7: Give the orbital diagram of the molecule. Is the molecular substance diamagnetic or paramagnetic? What is the electron configuration? What is the bond order?
51Valence Bond TheoryEXERCISE 10.9: The C2 molecule exists in the vapor phase over carbon at high temperature. Describe the molecular orbital structure of this molecule (orbital diagram and electron configuration). Is the substance paramagnetic or diamagnetic? What is the bond order?
52Valence Bond TheoryEXAMPLE 10.8: Write the orbital diagram for dinitrogen monoxide (nitric oxide). What is the bond order for NO?
53Valence Bond TheoryEXERCISE 10.10: Give the orbital diagram and electron configuration for the carbon monoxide molecule. What is the bond order and is this molecule paramagnetic or diamagnetic?
54Molecular Orbital Theory MO Diagrams Can Predict Magnetic Properties:
55Example 1: VSEPRDraw the Lewis electron-dot structure and predict the shapes of the following molecules or ions:O3 H3O+ XeF2PF6– XeOF4 AlH4–BF4– SiCl4 ICl4– AlCl3
56Example 2: Molecular Orbital Theory The B2 and C2 molecules have MO diagrams similar to N2. What MOs are occupied in B2 and C2, and what is the bond order in each? Would any of these be paramagnetic?
57Example 3: Dipole moment Draw the dipole moment for the following molecules, are they polar?HCl NH3 CHCl3H2O SF6 CCl4CO2 CH2Cl2