Summer 2001 Notes June 13 June 15 June

![2+ [Co(H 2 O)6]2+ 2+ [Co(H 2 O)6]2+](https://slidetodoc.com/presentation_image_h2/73cf282b1e01b413c78a51ae070a0257/image-2.jpg)






![p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3 p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3](https://slidetodoc.com/presentation_image_h2/73cf282b1e01b413c78a51ae070a0257/image-9.jpg)
![p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3 p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3](https://slidetodoc.com/presentation_image_h2/73cf282b1e01b413c78a51ae070a0257/image-10.jpg)
![p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3 p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3](https://slidetodoc.com/presentation_image_h2/73cf282b1e01b413c78a51ae070a0257/image-11.jpg)
![p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3 p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3](https://slidetodoc.com/presentation_image_h2/73cf282b1e01b413c78a51ae070a0257/image-12.jpg)





















































































- Slides: 97
Summer 2001 Notes June 13 June 15 June 18 June 20 July 2 Fall 2001 Lectures 9/28 10/1 10/3 10/5 – 10/8
2+ [Co(H 2 O)6]2+
Hydrolysis by complex ions
+ H 2 O(l) + H 3 O+
Conjugate base acid + H 2 O(l) + H 3 O+
Fe(H 2 O)63+(aq) + H 2 O(l) H 3 O+(aq) + Fe(H 2 O)5 OH 2+(aq)
Fe(H 2 O)63+(aq) + H 2 O(l) H 3 O+(aq) + Fe(H 2 O)5 OH 2+(aq) Ka = [H 3 O+][Fe(H 2 O)5 OH 2+] [Fe(H 2 O)63+] = 7. 7 x 10 -3
Fe(H 2 O)63+(aq) + H 2 O(l) H 3 O+(aq) + Fe(H 2 O)5 OH 2+(aq) Ka = [H 3 O+][Fe(H 2 O)5 OH 2+] [Fe(H 2 O)63+] p. H of 0. 10 M Fe(H 2 O)63+ = 7. 7 x 10 -3
p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3 O+] [Fe(H 2 O)5 OH 2+] Start change equil. 0. 10 0 0
p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3 O+] [Fe(H 2 O)5 OH 2+] Start change equil. 0. 10 -x 0 +x
p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3 O+] [Fe(H 2 O)5 OH 2+] 0. 10 Start change -x equil. 0. 10 - x 0 +x x
p. H of 0. 10 M Fe(H 2 O)63+ [Fe(H 2 O)63+] [H 3 O+] [Fe(H 2 O)5 OH 2+] 0. 10 Start change -x equil. 0. 10 - x (x)(x) Ka = (0. 10 - x) 0 +x x = 7. 7 x 10 -3
p. H of 0. 10 M Fe(H 2 O)63+ (x)(x) Ka = (0. 10 - x) = 7. 7 x 10 -3
p. H of 0. 10 M Fe(H 2 O)63+ (x)(x) Ka = (0. 10 - x) = 7. 7 x 10 -3 x 2 = (7. 7 x 10 -3)(0. 10 - x)
p. H of 0. 10 M Fe(H 2 O)63+ (x)(x) Ka = (0. 10 - x) = 7. 7 x 10 -3 x 2 = (7. 7 x 10 -3)(0. 10 - x) x 2 + (7. 7 x 10 -3)x - 7. 7 x 10 -4 = 0
p. H of 0. 10 M Fe(H 2 O)63+ (x)(x) Ka = (0. 10 - x) = 7. 7 x 10 -3 x 2 = (7. 7 x 10 -3)(0. 10 - x) x 2 + (7. 7 x 10 -3)x - 7. 7 x 10 -4 = 0 x = 0. 024
p. H of 0. 10 M Fe(H 2 O)63+ (x)(x) Ka = (0. 10 - x) = 7. 7 x 10 -3 x 2 = (7. 7 x 10 -3)(0. 10 - x) x 2 + (7. 7 x 10 -3)x - 7. 7 x 10 -4 = 0 x = 0. 024 p. H = 1. 6
Symmetry
Symmetry Molecular symmetry BF 3
Symmetry Molecular symmetry F BF 3 B F F
Symmetry Molecular symmetry BF 3 F F F B B F F F
Symmetry Molecular symmetry F B BF 3 F F Rotate 120 o around an axis through B the plane of the screen. to
Symmetry Molecular symmetry BF 3 F F F B B F Rotate 120 o F F
Symmetry Since the fluorines are all identical, we cannot tell the two molecules apart. F BF 3 F B F F Rotate 120 o = B F F
Symmetry Since the fluorines are all identical, we cannot tell the two molecules apart. F F B F F = B F F Rotate 120 o
Symmetry Since the fluorines are all identical, we cannot tell the two molecules apart. F F F B F F = B F = F Rotate 120 o B F F
Symmetry A third 120 o rotation brings the molecule back to the starting position. This is a 3 -fold axis of symmetry. F F BF 3 B = B F F Rotate 120 o
Symmetry Rotate 180 o around the B - F axis. BF 3 F B F F
Symmetry Rotate 180 o around the B - F axis. BF 3 F F F B B F F F
Symmetry Rotate 180 o around the B - F axis. BF 3 F = B F F F B F F A second 180 o rotation gives the original molecule.
Symmetry Rotate 180 o around the B - F axis. BF 3 F = B F F F B F This is a 2 -fold symmetry axis F
Symmetry BF 3 F B F F BF 3 has 3 2 -fold symmetry axes.
Symmetry BF 3 F F BB F mirror F
Symmetry BF 3 F F F = BB F F Mirror plane of symmetry
Symmetry BF 3 F F F = BB F F BF 3 has 3 mirror planes of symmetry along the B-F bonds.
There is a mirror plane in the plane of the molecule.
F B F F 1 3 -fold axis normal to plane 3 2 -fold axes along B - F bonds 3 mirror planes along bonds 1 mirror plane in molecular plane
2+
2+ 4 -fold rotation axis
2+ 4 -fold rotation axis = 4 90 o operations to get back to original configuration.
2+ The octahedral complex will have 3 4 -fold axes. 4 -fold rotation axis = 4 90 o operations to get back to original configuration.
Mirror planes? 2+
Mirror planes? 2+ O O Co O O
Mirror planes? 2+ O O Co O O 3 mirror planes with Co and 4 H 2 O’s.
Mirror planes? 2+ O O Co O O
Mirror planes? 2+ O O Co O O
2+ Any other rotation axes?
2+ Any other rotation axes?
2+ Any other rotation axes? Octahedral complexes have 3 -fold axes.
2+ Any other symmetry elements?
2+ Inversion center Any other symmetry elements?
2+ Any other symmetry elements? Inversion center The Co is the inversion center.
2+ Any other symmetry elements? Inversion center The Co is the inversion center. At any point where there is a ligand, there is a ligand the same distance in the opposite direction.
Tetrahedron
Tetrahedron Perchlorate Cl. O 4 -
Tetrahedron Perchlorate Cl. O 41 3 2 4 3 4 = 2 -fold 2 1
Tetrahedron Perchlorate Cl. O 41 1 2 4 3 4 = 2 Mirror plane 3
Tetrahedron Perchlorate Cl. O 41 4 2 4 3 1 = 3 -fold axis Cl-O 3 2 3
Tetrahedron Perchlorate Cl. O 41 4 3 -fold rotations 3 2 -fold rotations 2 4 3 3 mirror planes + others
octahemioctahedron
octahemioctahedron 4 -fold rotation axes
octahemioctahedron 4 -fold rotation axes This is not a 3 -fold
octahemioctahedron 4 -fold rotation axes This is not a 3 -fold
octahemioctahedron 4 -fold rotation axes This is not a 3 -fold
octahemioctahedron 4 -fold rotation axes This is not a 3 -fold a b The points a and b are related.
octahemioctahedron 4 -fold rotation axes a b The combination of 120 o rotation and a mirror leads to a new symmetry element
octahemioctahedron 4 -fold rotation axes a b The combination of 120 o rotation and a mirror leads to a new symmetry Element - S 3
Symmetry elements to look forrotations mirrors inversions
Crystals and solid-state structure
Crystals and solid-state structure octahedron
Crystals and solid-state structure
Crystals and solid-state structure Tetrahedral coordination
Crystals and solid-state structure Tetrahedral coordination C - C = 1. 544 Å
Å = ångström = 10 -10 m
Å = ångström = 10 -10 m The ångström is a useful unit when describing bonding distances.
Symmetry of a tetrahedron
Tetrahedrons and cubes have 3 -fold axes of symmetry
Graphite Crystal
Graphite Structure
Hexagonal bond array leads to hexagonal crystal Graphite Structure
Bonds - strong attraction Graphite Structure
Bonds - strong attraction Graphite Structure van der Waal’s forces- weak attraction
Hard structure - bonds are 3 -dimensional Soft structure bonds are in two dimensions
Hard structure - bonds are 3 -dimensional Soft structure bonds are in two dimensions van der Waal’s forces easy to break
Bucky. Ball a fullerene Individual molecule of carbon atoms OFB page 79 crystals. C 60
Bucky. Ball a fullerene C 60
Na. Cl
Na. Cl
Si. O 2
Si. O 2