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CHEMISTRY 161 Chapter 10 Chemical Bonding & Molecular Structure PowerPoint Presentation
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CHEMISTRY 161 Chapter 10 Chemical Bonding & Molecular Structure - PowerPoint PPT Presentation


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CHEMISTRY 161 Chapter 10 Chemical Bonding & Molecular Structure. PREDICTING THE GEOMETRY OF MOLECULES. H. O. H. 1. derive Lewis structure of the molecule. 2. discriminate between bonding and non-bonding electron pairs. 3. VALENCE SHELL ELECTRON PAIR REPULSION.

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CHEMISTRY 161 Chapter 10 Chemical Bonding & Molecular Structure


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    1. CHEMISTRY 161 Chapter 10 Chemical Bonding & Molecular Structure

    2. PREDICTING THE GEOMETRY OF MOLECULES H O H 1. derive Lewis structure of the molecule 2. discriminate between bonding and non-bonding electron pairs 3. VALENCE SHELL ELECTRON PAIR REPULSION

    3. VALENCE SHELL ELECTRON PAIR REPULSION VSEPR 1. identify in a compound the central atom 2. electrons repel each other 3. valence electron pairs stay as far apart as possible 4. non-bonding electrons repel more than bonding electrons

    4. central atom non-bonding pairs no non-bonding pairs

    5. AB2 BeCl2 Be Cl Cl TWO ELECTRON PAIRS AROUND BERYLLIUM ATOM

    6. Cl Be Cl 180° 270° Be Be 90° 180° LINEAR ARRANGEMENT BEST IT PUTS ELECTRON PAIRS FURTHEST APART

    7. AB3 BF3 F F B F THREE ELECTRON PAIRS AROUND THE BORON ATOM

    8. F F B F 120° 120° B 120° THREE ELECTRON PAIRS AROUND THE BORON ATOM TRIGONAL PLANAR ARRANGEMENT BEST

    9. B MOLECULAR SHAPE F F F THE SHAPE OF BF3 IS TRIGONAL PLANAR.

    10. AB4 H H C H H 90° 90° C 90° 90° CH4 four electron pairs expect square planar

    11. C better arrangement for four electron pairs TETRAHEDRAL 109.5° bigger than 90 ° in square planar tetrahedral 4 electron pairs put on the H-atoms

    12. C C TETRAHEDRAL H 109.5° H H H shape of CH4 is tetrahedral

    13. AB5 F F F P F F P PF5 FIVE ELECTRON PAIRS AROUND PHOSPHORUS 5 electron pairs trigonal bipyramidal

    14. Bond angle F 900 F P F P F 1200 F shape of PF5 is trigonal bipyramidal two of the F atoms different from the others

    15. AXIAL Bond angle F 900 F F P F 1200 EQUATORIAL F

    16. AB6 SF6 six electron pairs around the sulfur atom F F F S S F F F octahedral 6 electron pairs

    17. 900 F F F S S F F F shape of SF6 is octahedral

    18. central atom non-bonding pairs no non-bonding pairs

    19. AB2E AB3 SeO2 O Se O

    20. O Se O Se VSEPR treats double bonds like a single bond THREE ELECTRON PAIRS AROUND SELENIUM ELECTRON PAIR GEOMETRY TRIGONAL PLANAR

    21. Se Se O O ADD OXYGENS SeO2 IS V-SHAPED (OR BENT) THE MOLECULAR SHAPE IS THE POSITION OF THE ATOMS

    22. AB3E AB4 H N H NH3 H electron pairs around the nitrogen atom

    23. N H N H NH3 H PUT ON THE 3 H ATOMS N H H H NH3 is trigonal pyramidal

    24. AB2E2 AB4 four electron pairs around the oxygen atom H O H PUT ON THE 2 H-ATOMS O O H H shape of H2O is V-shaped or bent

    25. S AB4E AB5 SF4 F F F S F TRIGONAL BIPYRAMID

    26. F F F S F S F F F F WHERE DOES LONE PAIR GO? OR lone pairs occupy the trigonal plane (the “equator”) to minimize the number of 90° repulsions

    27. AB4E AB3E2 AB2E3 SF4 1 lone pair See-saw shaped ClF3 2 lone pairs T-shaped XeF2 3 lone pairs Linear F F F F Xe S Cl F F F F F F lone pairs occupy the trigonal plane (the “equator”) first to minimize the number of 90° repulsions

    28. AB5E F F F Br F F Br AB6 BrF5 Square pyramidal

    29. AB6 AB4E2 F F F Xe F F : F Xe Xe F : F F Xe F F F XeF4 lone pairs MUST BE AT 1800

    30. Summary of Molecular Shapes Total valence electron pairs Electron Pair Geometry Lone electron pairs Shape of Molecule 2 Linear 0 Linear 0 Trigonal planar Trigonal planar 3 1 V-shaped 0 Tetrahedral 4 Tetrahedral 1 Trigonal pyramid 2 V-shaped

    31. Total valence electron pairs Electron Pair Geometry Lone electron pairs Shape of Molecule 0 Trig. bipyramid. 1 See-saw Trigonal bipyramidal 5 2 T-shaped 3 Linear 0 Octahedral 6 Octahedral 1 Square pyramid 2 Square planar

    32. POLYATOMICS molecules with no single central atom we apply our VSEPR rules to each atom in the chain Example: ETHANOL

    33. H H H C C O H H H ETHANOL C2H5OH The atoms around the carbons form a. tetrahedral arrangement The atoms around the oxygen form a V-shaped structure.

    34. H O H H C C H H H

    35. EXAMPLES BF4- ICl4- Cl2O SO2Cl2 Cl2CO Cl2SO N2F2 NH4+ NH2OH

    36. 1. Lewis structures 2. VSEPR model WHY DO MOLECULES FORM?

    37. simplest molecule H2 two H-atoms 1s1 two H-atoms approach each other and the electron waves interact OVERLAP to form a region of increased electron density between the atoms

    38. chemical bond with electron density in between the nuclei is called  bond

    39. VALENCE BOND THEORY a covalent bond is formed by an overlap of two valence atomic orbitals that share an electron pair the better the overlap the stronger the bond the orbitals need to point along the bonds

    40. H C H H H CH4 What orbitals are used? hydrogen atoms bond using their 1s orbitals carbon needs four orbitals to bond with. [He] 2s22p2 2s, 2px , 2py, 2pz

    41. 1. The electronic configuration of carbon is [He] 2s22p2 The orbital diagram is:

    42. . . . C . 1. The electronic configuration of carbon is [He] 2s22p2 the orbital diagram is: [He] the Lewis dot structure is necessary to promote one 2s electron

    43. C PROMOTE AN ELECTRON [He] [He] [He] 2s22p2 [He] 2s12p3 excited state (valence state) Lewis dot structure four unpaired electrons we can use these to form chemical bonds

    44. 1. a covalent bond is formed by an overlap of two valence atomic orbitals that share an electron pair 2. bonds formed with s orbitals will be different to bonds formed with p orbitals Experiment shows that all four bonds are identical 3. three p orbitals are mutually perpendicular, suggesting 90° bond angles experiment shows that methane has 109.5° bond angles combining the orbitals

    45. H C H H H we need four orbitals pointing to the vertices of a tetrahedron orbitals are just mathematical functions we can combine them HYBRIDIZATION

    46. COMBINING ORBITALS TO FORM HYBRIDS HYBRIDIZATION number of atomic orbitals that are combined IS EQUAL TO the number of resulting hybrid orbitals

    47. HYBRIDIZATION Combine one s and one p a sp- hybrid + ADD the orbitals + 2s+ 2p

    48. HYBRIDIZATION + Combine one s and one p a sp- hybrid s + p 2s+ 2p What do we get? + The positive part cancels negative part DESTRUCTIVE INTERFERENCE The positive part adds to positive part CONSTRUCTIVE INTERFERENCE