VSEPR Valence Shell Electron Pair Repulsions
Covalent Bond: A Model Chemical bonds can be viewed as forces that cause a group of atoms to act as a unit They result from the tendency of a system to seek its lowest possible energy Bonds occur when collections of atoms are more stable (lower in energy) than the separate atoms Note:  The next three slides will repeat at the end.  This is preliminary intro info that may make more sense at the end.
Example:  Methane 1652 kJ of energy are required to break a mole of methane into separate C and H atoms OR 1652 kJ of energy are released when one mole of methane is formed from one mole of C atoms and four moles of H atoms  Methane is therefore a stable molecule relative to its stable atoms Since there are four H atoms arranged around the central C, it is natural to envision four individual attractions between C and H (bonds) Each bond has an associated bond energy, found by dividing the total energy by four (1652/4 = 413 kJ) The positive Bond Energy value indicates the energy required to break the bond between C and H atoms
Bonding Model Models originate from our observations of the properties of nature Atoms can form stable groups by sharing electrons, shared electrons give a lower energy state because simultaneously attracted to two nuclei Remember:  Models are human inventions that allow us to explain and predict.  A model is a useful way of thinking; they include simplifications and assumptions.  A model does not equal reality.
Covalent Bonds Electron pair(s)  shared between two atoms attracted to both nuclei Location of a single shared pair Directly between two nuclei Maximizes attractions with shortest distance between two positive nuclei Minimizes repulsions with negative electrons between positive nuclei that would repel one another
Multiple covalent bonds around the same atom determine the shape Negative e- pairs with same charge repel each other Repulsions push the pairs as far apart as possible
Single bonds Sigma bond  ө Overlap of orbitals allow electron pair to be shared between the two atoms Electron pair shared directly between two nuclei Only one pair may be shared in this space - just as only one pair of electrons may occupy a single atomic orbital
Double and Triple Bonds Pi bonds  π Since the space between the nuclei is occupied, e- pair is shared above and below the plane or front and back Overlap of p-orbital lobes allow for this sharing above and below OR front and back
 
Lewis Structures Drawn to show the bonds between the atoms in the structure Only shows whether single, double or triple bonds Does not show the shape
Lewis Structure Represents the arrangement of valence electrons among atoms in the molecule Rules based upon observations of thousands of molecules, which show that in most stable compounds the atoms achieve noble gas configurations Duet Rule  hydrogen stable with only a pair of e- Octet Rule  other atoms stable with 4 pairs of e-
Rules for Drawing Lewis Structures Count the number of valence electrons Draw the skeleton structure- the central is generally listed first in formula Distribute electrons to give each atom a stable octet Reconcile # e- Do you have enough electrons? You may need to use double or triple bonds. Do have too many electrons?  You may need to explain the octet, but only if empty d-orbital available
Determine # Valence e-  from column #
 
 
 
 
Electron Clouds repel each other, thus structure around an atom is determined principally by minimizing repulsions
2 electron pairs (2 EP) around central atom Two clouds pushed as far apart as possible Greatest angle possible 180 º LINEAR shape
3 electron pairs (3 EP)  around central atom Three clouds pushed as far apart as possible Greatest angle possible 120 º TRIGONAL PLANAR shape   (3)   (flat)
4 electron pairs (4 EP)  around central atom Four clouds pushed as far apart as possible Greatest angle no longer possible  in two dimensions Requires three-dimensional TETRAHEDRAL shape
Orbital Hybridization #1 Atomic orbitals such as s and p are not well suited for overlapping and allowing two atoms to share a pair of electrons Remember:  best location of shared pair is directly between two atoms e- pair spends little time in best location With overlap of two s-orbital With overlap of two p-orbitals
Orbital Hybridization #2 Hybrid orbitals (cross of atomic orbitals) Remember: The pink flower hybrid cross of the red and white flower Hybrid orbitals Shape  more suitable for bonding One large lobe and one very small lobe Large lobe oriented towards other nucleus Angles  more suitable for bonding Angles predicted from VSEPR
Overlap of two s-orbitals NOT A GOOD LOCATION-  far from one nucleus Note:  shared in this overlap the e- pair would spend most of the time in an unfavorable location GOOD SPOT between both nuclei
Overlap of two p-orbitals One atom & its p-orbital The other atom & its p-orbital represents the nucleus BAD location far from other nucleus GOOD SPOT between  both nuclei
Hybrid Orbitals yield more favorable shape for overlap Atomic orbitals are not shaped to maximize attractions nor minimize repulsions Hybrid orbital shape  One large lobe oriented towards other atom Notice the difference in this shape compared to p-orbital shape
Angles  and  Shape Atomic orbitals are not shaped to maximize attractions nor minimize repulsions BUT the angles are also not favorable p-orbitals are oriented at 90 º to  each other Other angles are required 180º,  120º or 109.5º
Orbital Hybridization #3 Each e-pair requires a hybrid orbital If two hybrid orbitals required than two atomic orbitals must be hybridized, an s and a p orbital forming two sp orbitals at 180 º sp hybrids 2  EP 4  EP 3  EP sp 2  hybrids sp 3  hybrids
Electron-Pair Geometry vs Molecular Geometry Electron-pair geometry Where are the electron pairs Includes  bonding pairs (BP) – shared between 2 atoms nonbonding pairs (NBP) – lone pair Molecular geometry Where are the atoms Includes only the bonding pairs
Examples of 3 EP 3 BP + 0 NBP = 3 EP 3 EP = EP geom is trigonal planar All locations occupy by an atom,  so molecular geometry is also trigonal planar 2 BP + 1 NBP = 3 EP 3 EP = EP geom is trigonal planar Only two bonding pairs One of the locations is only lone pair of e- so molecular geometry is bent O O O N O O O
Carbonate Ion (CO 3 2- )
Nitrate Ion (NO 3 - )
Nitrite Ion (NO 2 - )
Examples of 4 EP 4 BP + 0 NBP = 4 EP Both EP geom and molecular geom tetrahedral 3 BP + 1 NBP = 4 EP 4 EP so EP geom is tetrahedral Molecular geom is TRIGONAL PYRAMIDAL  No atom at top location  2 BP + 2 NBP = 4 EP 4 EP so EP geom is tetrahedral Molecular geom is BENT  no atoms at two locations
4 BP + 0 NBP = 4 EP TETRAHEDRAL Cl Cl Cl Cl S
3 BP + 1 NBP = 4 EP TRIGONAL PYRAMIDAL N H H H ●● lone pair of e- NBP H H H N 107
2 BP + 2 NBP = 4 EP BENT O H H ●● lone pair of e- NBP H H O ●● lone pair of e- NBP 104.5 H
Hydronium Ion (H 3 O + )
Ammonia Molecule (NH 3 )
 
Summary of 4 EP
Exceptions to Octet Rule Reduced Octet  H  only forms one bond- only one pair of e- Be  tends to only form two bonds  only two pair of e- B  tends to only form three bonds only three pair of e- Expanded Octet Empty d-orbitals can be used to accommodate extra e- Elements in the third row and lower can expand Up to 6 pairs of e- are possible
Lewis Structures in Which the Central Atom Exceeds an Octet
5 EP Trigonal bipyramidal Orbital hybridization Requires 5 hybrid orbitals So, 5 atomic orbitals required sp 3 d
Trigonal planar shape
5 EP = trigonal pyramidal molecular geometry 5 BP + 0 NBP = 5 EP  4 BP + 1 NBP = 5 EP
5 EP = trigonal pyramidal molecular geometry 3 BP + 2 NBP = 5 EP  2 BP + 3 NBP = 5 EP
6 EP Octahedral Orbital hybridization Requires 5 hybrid orbitals So, 5 atomic orbitals required sp 3 d
6 EP = octahedral 6 BP +0 NBP 6 EP
6 EP = octahedral 5 BP +1 NBP 6 EP
6 EP = octahedral 4 BP +2 NBP 6 EP
 
 
Summary: Molecular Geometry of Expanded Octets
Summary of EP Geometry 2 EP 3 EP 4 EP  5 EP  6 EP
Summary of EP Geometry
Predict the geometry, angles  and orbital hybridization Predict the geometry, angles  and orbital hybridization
 
 
Covalent Bond: A Model Chemical bonds can be viewed as forces that cause a group of atoms to act as a unit They result from the tendency of a system to seek its lowest possible energy Bonds occur when collections of atoms are more stable (lower in energy) than the separate atoms
Example:  Methane 1652 kJ of energy are required to break a mole of methane into separate C and H atoms OR 1652 kJ of energy are released when one mole of methane is formed from one mole of C atoms and four moles of H atoms  Methane is therefore a stable molecule relative to its stable atoms Since there are four H atoms arranged around the central C, it is natural to envision four individual attractions between C and H (bonds) An average bond energy associated with each bond is found by dividing the total energy by four  (1652/4 = 413 kJ) The positive Bond Energy value indicates the energy required to break the bond between C and H atoms
Bonding Model Models originate from our observations of the properties of nature Atoms can form stable groups by sharing electrons, shared electrons give a lower energy state because simultaneously attracted to two nuclei Remember:  Models are human inventions that allow us to explain and predict.  A model is a useful way of thinking; they include simplifications and assumptions.  A model does not equal reality.
 
Bond Energy and Enthalpy
 

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VSEPR

  • 1. VSEPR Valence Shell Electron Pair Repulsions
  • 2. Covalent Bond: A Model Chemical bonds can be viewed as forces that cause a group of atoms to act as a unit They result from the tendency of a system to seek its lowest possible energy Bonds occur when collections of atoms are more stable (lower in energy) than the separate atoms Note: The next three slides will repeat at the end. This is preliminary intro info that may make more sense at the end.
  • 3. Example: Methane 1652 kJ of energy are required to break a mole of methane into separate C and H atoms OR 1652 kJ of energy are released when one mole of methane is formed from one mole of C atoms and four moles of H atoms Methane is therefore a stable molecule relative to its stable atoms Since there are four H atoms arranged around the central C, it is natural to envision four individual attractions between C and H (bonds) Each bond has an associated bond energy, found by dividing the total energy by four (1652/4 = 413 kJ) The positive Bond Energy value indicates the energy required to break the bond between C and H atoms
  • 4. Bonding Model Models originate from our observations of the properties of nature Atoms can form stable groups by sharing electrons, shared electrons give a lower energy state because simultaneously attracted to two nuclei Remember: Models are human inventions that allow us to explain and predict. A model is a useful way of thinking; they include simplifications and assumptions. A model does not equal reality.
  • 5. Covalent Bonds Electron pair(s) shared between two atoms attracted to both nuclei Location of a single shared pair Directly between two nuclei Maximizes attractions with shortest distance between two positive nuclei Minimizes repulsions with negative electrons between positive nuclei that would repel one another
  • 6. Multiple covalent bonds around the same atom determine the shape Negative e- pairs with same charge repel each other Repulsions push the pairs as far apart as possible
  • 7. Single bonds Sigma bond ө Overlap of orbitals allow electron pair to be shared between the two atoms Electron pair shared directly between two nuclei Only one pair may be shared in this space - just as only one pair of electrons may occupy a single atomic orbital
  • 8. Double and Triple Bonds Pi bonds π Since the space between the nuclei is occupied, e- pair is shared above and below the plane or front and back Overlap of p-orbital lobes allow for this sharing above and below OR front and back
  • 9.  
  • 10. Lewis Structures Drawn to show the bonds between the atoms in the structure Only shows whether single, double or triple bonds Does not show the shape
  • 11. Lewis Structure Represents the arrangement of valence electrons among atoms in the molecule Rules based upon observations of thousands of molecules, which show that in most stable compounds the atoms achieve noble gas configurations Duet Rule hydrogen stable with only a pair of e- Octet Rule other atoms stable with 4 pairs of e-
  • 12. Rules for Drawing Lewis Structures Count the number of valence electrons Draw the skeleton structure- the central is generally listed first in formula Distribute electrons to give each atom a stable octet Reconcile # e- Do you have enough electrons? You may need to use double or triple bonds. Do have too many electrons? You may need to explain the octet, but only if empty d-orbital available
  • 13. Determine # Valence e- from column #
  • 14.  
  • 15.  
  • 16.  
  • 17.  
  • 18. Electron Clouds repel each other, thus structure around an atom is determined principally by minimizing repulsions
  • 19. 2 electron pairs (2 EP) around central atom Two clouds pushed as far apart as possible Greatest angle possible 180 º LINEAR shape
  • 20. 3 electron pairs (3 EP) around central atom Three clouds pushed as far apart as possible Greatest angle possible 120 º TRIGONAL PLANAR shape (3) (flat)
  • 21. 4 electron pairs (4 EP) around central atom Four clouds pushed as far apart as possible Greatest angle no longer possible in two dimensions Requires three-dimensional TETRAHEDRAL shape
  • 22. Orbital Hybridization #1 Atomic orbitals such as s and p are not well suited for overlapping and allowing two atoms to share a pair of electrons Remember: best location of shared pair is directly between two atoms e- pair spends little time in best location With overlap of two s-orbital With overlap of two p-orbitals
  • 23. Orbital Hybridization #2 Hybrid orbitals (cross of atomic orbitals) Remember: The pink flower hybrid cross of the red and white flower Hybrid orbitals Shape more suitable for bonding One large lobe and one very small lobe Large lobe oriented towards other nucleus Angles more suitable for bonding Angles predicted from VSEPR
  • 24. Overlap of two s-orbitals NOT A GOOD LOCATION- far from one nucleus Note: shared in this overlap the e- pair would spend most of the time in an unfavorable location GOOD SPOT between both nuclei
  • 25. Overlap of two p-orbitals One atom & its p-orbital The other atom & its p-orbital represents the nucleus BAD location far from other nucleus GOOD SPOT between both nuclei
  • 26. Hybrid Orbitals yield more favorable shape for overlap Atomic orbitals are not shaped to maximize attractions nor minimize repulsions Hybrid orbital shape One large lobe oriented towards other atom Notice the difference in this shape compared to p-orbital shape
  • 27. Angles and Shape Atomic orbitals are not shaped to maximize attractions nor minimize repulsions BUT the angles are also not favorable p-orbitals are oriented at 90 º to each other Other angles are required 180º, 120º or 109.5º
  • 28. Orbital Hybridization #3 Each e-pair requires a hybrid orbital If two hybrid orbitals required than two atomic orbitals must be hybridized, an s and a p orbital forming two sp orbitals at 180 º sp hybrids 2 EP 4 EP 3 EP sp 2 hybrids sp 3 hybrids
  • 29. Electron-Pair Geometry vs Molecular Geometry Electron-pair geometry Where are the electron pairs Includes bonding pairs (BP) – shared between 2 atoms nonbonding pairs (NBP) – lone pair Molecular geometry Where are the atoms Includes only the bonding pairs
  • 30. Examples of 3 EP 3 BP + 0 NBP = 3 EP 3 EP = EP geom is trigonal planar All locations occupy by an atom, so molecular geometry is also trigonal planar 2 BP + 1 NBP = 3 EP 3 EP = EP geom is trigonal planar Only two bonding pairs One of the locations is only lone pair of e- so molecular geometry is bent O O O N O O O
  • 34. Examples of 4 EP 4 BP + 0 NBP = 4 EP Both EP geom and molecular geom tetrahedral 3 BP + 1 NBP = 4 EP 4 EP so EP geom is tetrahedral Molecular geom is TRIGONAL PYRAMIDAL No atom at top location 2 BP + 2 NBP = 4 EP 4 EP so EP geom is tetrahedral Molecular geom is BENT no atoms at two locations
  • 35. 4 BP + 0 NBP = 4 EP TETRAHEDRAL Cl Cl Cl Cl S
  • 36. 3 BP + 1 NBP = 4 EP TRIGONAL PYRAMIDAL N H H H ●● lone pair of e- NBP H H H N 107
  • 37. 2 BP + 2 NBP = 4 EP BENT O H H ●● lone pair of e- NBP H H O ●● lone pair of e- NBP 104.5 H
  • 38. Hydronium Ion (H 3 O + )
  • 40.  
  • 42. Exceptions to Octet Rule Reduced Octet H only forms one bond- only one pair of e- Be tends to only form two bonds only two pair of e- B tends to only form three bonds only three pair of e- Expanded Octet Empty d-orbitals can be used to accommodate extra e- Elements in the third row and lower can expand Up to 6 pairs of e- are possible
  • 43. Lewis Structures in Which the Central Atom Exceeds an Octet
  • 44. 5 EP Trigonal bipyramidal Orbital hybridization Requires 5 hybrid orbitals So, 5 atomic orbitals required sp 3 d
  • 46. 5 EP = trigonal pyramidal molecular geometry 5 BP + 0 NBP = 5 EP 4 BP + 1 NBP = 5 EP
  • 47. 5 EP = trigonal pyramidal molecular geometry 3 BP + 2 NBP = 5 EP 2 BP + 3 NBP = 5 EP
  • 48. 6 EP Octahedral Orbital hybridization Requires 5 hybrid orbitals So, 5 atomic orbitals required sp 3 d
  • 49. 6 EP = octahedral 6 BP +0 NBP 6 EP
  • 50. 6 EP = octahedral 5 BP +1 NBP 6 EP
  • 51. 6 EP = octahedral 4 BP +2 NBP 6 EP
  • 52.  
  • 53.  
  • 54. Summary: Molecular Geometry of Expanded Octets
  • 55. Summary of EP Geometry 2 EP 3 EP 4 EP 5 EP 6 EP
  • 56. Summary of EP Geometry
  • 57. Predict the geometry, angles and orbital hybridization Predict the geometry, angles and orbital hybridization
  • 58.  
  • 59.  
  • 60. Covalent Bond: A Model Chemical bonds can be viewed as forces that cause a group of atoms to act as a unit They result from the tendency of a system to seek its lowest possible energy Bonds occur when collections of atoms are more stable (lower in energy) than the separate atoms
  • 61. Example: Methane 1652 kJ of energy are required to break a mole of methane into separate C and H atoms OR 1652 kJ of energy are released when one mole of methane is formed from one mole of C atoms and four moles of H atoms Methane is therefore a stable molecule relative to its stable atoms Since there are four H atoms arranged around the central C, it is natural to envision four individual attractions between C and H (bonds) An average bond energy associated with each bond is found by dividing the total energy by four (1652/4 = 413 kJ) The positive Bond Energy value indicates the energy required to break the bond between C and H atoms
  • 62. Bonding Model Models originate from our observations of the properties of nature Atoms can form stable groups by sharing electrons, shared electrons give a lower energy state because simultaneously attracted to two nuclei Remember: Models are human inventions that allow us to explain and predict. A model is a useful way of thinking; they include simplifications and assumptions. A model does not equal reality.
  • 63.  
  • 64. Bond Energy and Enthalpy
  • 65.