Presentation on theme: "1 Vanessa N. Prasad-Permaul Valencia Community College CHM 1045."— Presentation transcript:
1 Vanessa N. Prasad-Permaul Valencia Community College CHM 1045
Molecular 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. 2
Molecular Shapes: VSEPR Two Electron Groups: Electron groups point in opposite directions. LINEAR ARRANGEMENT 3
Molecular Shapes: VSEPR Three Electron Groups: Electron groups lie in the same plane and point to the corners of an equilateral triangle. TRIGONAL PLANAR ARRANGEMENT Trigonal planar geometry or bent (angular) 4
Molecular Shapes: VSEPR Four Electron Groups: Electron groups point to the corners of a regular tetrahedron. TETRAHEDRAL ARRANGEMENT 5
Molecular 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 6
Molecular Shapes: VSEPR 7 Table of the Molecular Geometry and Examples of Certain Compounds
8 Molecular Shapes: VSEPR EXAMPLE 10.1 : Predict the geometry of the following Molecules or ions, using the VSEPR method: A.BeCl 2 B. NO 2 - C. SiCl 4 Be = 2 electrons Cl = 7 electrons each = 14 Total of 16 electrons :Cl:Be:Cl: 2 sets of electron pairs, 0 lone pairs linear ¨ ¨ ¨ ¨
9 Molecular Shapes: VSEPR B. NO 2 - N = 5 electrons, O = 6 electrons each = 12 Total 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 ¨ ¨ ¨ ¨
Molecular Shapes: VSEPR 10 C. SiCl 4 Si = 4 electrons, Cl = 7 electrons each = 28 electrons Total of 32 electrons 4 electron pairs, 0 lone pairs tetrahedral
11 Molecular Shapes: VSEPR EXERCISE 10.1 : Use the VSEPR method to predict the geometry of the following ion and molecules: A.ClO3- B.OF2 C.SiF4
Molecular Shapes: VSEPR Five Electron Groups: Electron groups point to the corners of a trigonal bipyramid. 12
Molecular Shapes: VSEPR Six Electron Groups: Electron groups point to the corners of a regular octahedron. 13
14 Molecular Shapes: VSEPR EXAMPLE 10.2 : What do you expect for the geometry of tellurium tetrachloride? Tellurium has 6 electron pairs in its valence shell Cl = 7 electrons each = 28 electrons Total of 34 electrons There are 4 bonding pairs and 1 lone pair of electrons Trigonal bipyramidal arrangement and a seesaw geometry ¨
15 Molecular Shapes: VSEPR EXERCISE 10.2 : According to the VSEPR model, what molecular geometry would you predict for iodine trichloride?
Dipole Moment & Molecular Geometry 16 Dipole Moment: a quantitative measure of the degree of charge separation in a molecule. + - H Cl If a molecule has a dipole moment it is polar A compound could contain polar bonds but the molecule could be non-polar because there is no dipole moment Bond dipoles can cancel each other out O C O
Dipole moment Based on electronegativity differences between atoms in a molecule The most electronegative atom is partially negative The less electronegative atom is partially positive The dipole moment is the average of all dipoles in the molecule Exception (C-H bonds do not have a dipole) 17
18 Molecular 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. SO 2 : linear or bent BENT 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.
19 Molecular Shapes: VSEPR PH 3 : trigonal planar or trigonal pyramidal In a trigonal planar geometry, the dipole moment is zero, therefore PH 3 has a trigonal pyramidal geometry. Tetrahedral arrangement; 3 electron pairs and one lone pair.
20 Molecular 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.
21 Molecular 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. A.SOCl 2 B.SiF 4 C.OF 2
Valence Bond Theory 1. 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. 22
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. 23 According to Valence Bond Theory:
24 Hybrid orbitals Combined hybrid orbitals # of orbitals VSEPR geometry Valence Bond Theory
sp hybrid: 25 Valence Bond Theory
sp 2 hybrid: 26 Valence Bond Theory
sp 2 hybrid (π bond): 27 Valence Bond Theory
sp 3 hybrid: 28 Valence Bond Theory
sp 3 d hybrid: 29 Valence Bond Theory
sp 3 d 2 hybrid: 30 Valence Bond Theory
31 Valence Bond Theory Steps to obtain the bonding description about an atom in a molecule: 1.Write the Lewis electron-dot formula 2.From the Lewis formula, use the VSEPR model to obtain arrangement 3.From the geometry, s what type of hybrid orbitals are required 4.Assign valence electrons to the hybrid oritals of this atom one at a time, pairing them only when necessary 5.Form the bonds by overlapping singly occupied orbitals of other atoms
32 Valence Bond Theory EXAMPLE 10.4: Describe the bonding in H 2 O according to valence 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
33 Valence Bond Theory EXERCISE 10.5: Using hybrid orbitals, describe the bonding in NH 3 according to valence bond theory.
34 Valence Bond Theory EXAMPLE 10.5: Describe the bonding in XeF 4 using hybrid orbitals. 4 single bonds, 2 lone pairs octahedral arrangement and square planar geometry.
35 Valence Bond Theory EXERCISE 10.6: Describe the bonding in PCl 5 using hybrid orbitals.
Molecular Orbital Theory Additive and subtractive combination of p orbitals leads to the formation of both sigma and pi orbitals. 36
37 Valence Bond Theory EXAMPLE 10.6: Describe the bonding on a given N atom in dinitrogen difluoride, N 2 F 2, using valence bond theory. :F N N F: ¨ ¨¨ ¨¨¨
38 Valence Bond Theory EXERCISE 10.7: Describe the bonding on the carbon atom in carbon dioxide using valence bond theory.
39 Valence Bond Theory EXERCISE 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.
Molecular 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. 40
Molecular Orbital Theory Additive combination of orbitals ( ) is lower in energy than two isolated 1s orbitals and is called a bonding molecular orbital. 41
Molecular Orbital Theory Subtractive combination of orbitals ( ) is higher in energy than two isolated 1s orbitals and is called an antibonding molecular orbital. 42
Molecular 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 electrons 2 43
B2B2 * 2p * 2p 2p 2p * 2s 2s * 1s 1s 44 B has 5 electrons So B 2 has 10 elec Core electrons don’t count toward BO
C2C2 * 2p * 2p 2p 2p * 2s 2s * 1s 1s 45 Carbon has 6 electrons so C 2 has 12 electrons
Molecular Orbital Theory Molecular Orbital Diagram for H 2 : 46
Molecular Orbital Theory Molecular Orbital Diagrams for H 2 – and He 2 : 47
Molecular Orbital Theory Second-Row MO Energy Level Diagrams: 49
50 Valence Bond Theory EXAMPLE 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?
51 Valence Bond Theory EXERCISE 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?
52 Valence Bond Theory EXAMPLE 10.8: Write the orbital diagram for dinitrogen monoxide (nitric oxide). What is the bond order for NO?
53 Valence Bond Theory EXERCISE 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?
Molecular Orbital Theory MO Diagrams Can Predict Magnetic Properties: 54
Example 1: VSEPR Draw the Lewis electron-dot structure and predict the shapes of the following molecules or ions: O 3 H 3 O + XeF 2 PF 6 – XeOF 4 AlH 4 – BF 4 – SiCl 4 ICl 4 – AlCl 3 55
Example 2: Molecular Orbital Theory The B 2 and C 2 molecules have MO diagrams similar to N 2. What MOs are occupied in B 2 and C 2, and what is the bond order in each? Would any of these be paramagnetic? 56
Example 3: Dipole moment Draw the dipole moment for the following molecules, are they polar? HClNH 3 CHCl 3 H 2 OSF 6 CCl 4 CO 2 CH 2 Cl 2 57