A Science of Aesthetics

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					A Science of Aesthetics

Transformations You Cannot See.
Topics
   Point Symmetry Operations, E C  i S
    • Classes
   Point Groups
    • Identification Scheme
   Uses and Consequences
Symmetry Operations on Points
   The nuclei of molecules are the points.
   Symmetry operations transform identical
    nuclei into themselves.
    • After such operations, the molecule looks
      absolutely unchanged.
    • Nature finds all these ways to fool you, so
      symmetry has Entropy consequences!
    • We must be able to identify the operations too.
The Identity Operation, E
   One way of finding a molecule is after
    nothing has been done to it! That counts.
   No matter how asymmetrical a molecule
    is, it must have an identity operation, E.
    • The symbol “E” comes from the German,
      “eigen,” meaning “the same.”
    • CBrClFI, bromochlorofluoroiodomethane, has
      E as its only symmetry operation, for example.
The Rotation Element, Cn
   Axes can often be found in molecules
    around which rotation leaves the atoms
    identical.
   An n-fold rotation, if present, is symbolized
    by the element Cn., and represents n–1
    rotational operations about the axis.
    • Each operation is a rotation through yet another
      360°/n, but the operation C1 is merely E.
Finding Rotations
   Symmetrical molecules often have rotation
    axes through their atoms.
    • SF6 is octahedral and has fourfold axes through
      the atoms F–S–F that invisibly cycle the
      remaining 4 fluorine atoms.
   But the axes need not pierce any atom!
    • The crown-shaped S8 molecule also has a C4,
      but it goes through the empty center.
Principal Axis

   Molecules may have            But the axis with the
    many rotation axes.            highest n is designated
    • E.g., S8 has four C2         as the principal axis.
      axes, one through            • It is used to find other
      each pair of bonds             operations, and often
      opposing across the            lends its name to the
      middle.                        symmetry of the
    • Only one of these is           molecule.
      shown in the molecular       • In S8, that would be the
      model above.                   C4 axis.
                               Inobvious Axes
 Remember the trick        That purple C3 axis
  for drawing tetrahedra     is one of 4 diagonals
  like that for CCl4?        of the cube on which
 The chlorines occupy       you could spin the
  opposite corners on        molecular top. 
  opposite faces of the     But what about the 3
  cube.                      C2 axes straight
 Like this …                through each face?
                               • Only 1 shown here.
Mirror Planes, 
   Reflection in a mirror leaves some
    molecules looking identical to themselves.
   What distinguishes these operations is the
    physical placement of the mirror so that the
    image coincides with the original molecule!
   There are 3 types of mirror planes:
    • Vertical, horizontal, and dihedral?!?
           Vertical Mirrors, v
   If the reflection plane contains the Principle
    Axis, it is called a “vertical mirror plane.”
   Just as rotation axes need not pierce atoms,
    neither do v, but they often do.
                   E.g., in SF4, the principle axis is C2.
                   The two reflection planes are both
                   “vertical” and happen to contain all
                   of the atoms in the molecule.
Horizontal Mirror, h
   Horizontal is  to vertical so you can infer
    that h is  to the Principle Axis.
   While a molecule might have several
    vertical mirrors, it can have only one h.
    In PCl5, the horizontal plane is
    obvious. It’s what we’ve been
    calling the equatorial plane that
    contains the three 120° separated
    chlorines. (The polar axis is C3.)
Dihedral Mirror, d
   Greek “two-sided” doesn’t help.
    • All planes are two-sided!
   In symmetry it means vertical planes that lie
    between the C2 axes  the Principal Axis.
    • E.g., in S8 above, one of S8’s four dihedral
      planes contains the C4 Principal Axis and
      bisects the adjacent C2 axes.
       • The plane contains opposite sulfur atoms,
The Magic of Mirrors
   Rotations and the identity operation do not
    rearrange a molecule in any way.
   But mirrors (and inverse and improper rotations) do.
   Mirror planes reflect a mirror image whose
    “handedness” has changed. Left  right.
    • “Chiral” molecules have mirror images that
      cannot be superimposed on the original! So
      they cannot have a . You’d see the change.
Chiral Molecules
   Caraway flavor agent                Spearmint flavor agent


                  These molecules cannot be aligned.




    • S-Carvone                           • R-Carvone
Inversion, i
   Mirrors merely transpose along one axis
    (their  axis), but inversion transposes
    atoms along all 3 axes at once.
    • i is like xyz so points (x,y,z)(–x,–y,–z)
    • Therefore, there must be identical atoms
      opposite one another through the center of the
      molecule.


                                     SF6
Improper Rotation, Sn
   Proper rotations (Cn) do not rearrange the
    molecule, but improper rotations (Sn) do;
    they are rotations by 360°/n followed by
    reflection in a plane  to the Sn axis.
    • For Sn to be, neither Cn nor the  need exist!
    • To see that, consider the molecule S8 above;
      believe it or not, it has an S8 axis coincident
      with the C4 Principle Axis.
                      Clearly there are seven S8 operations.
Point Groups
   “Point” refers to atoms, and “Group” refers
    to the collection of symmetry operations a
    molecule obeys.
    • The group is complete because no sequences of
      operations ever generates one not in the group!
                                                     C2
   E.g., H2O, besides E,
                                                          v
       has v and v’ and C2.
         It’s clear that C2•C2 = E, but
         C2•v = v’ needs a little thought.   v’
Motivational Factors
   The reason we want to know the Point
    Group of a molecule is that all symmetry
    consequences are encoded in the Group.
    • The nature and degeneracies of vibrations.
    • The legitimate AO combinations for MOs.
    • The appearances and absences of lines in a
      molecule’s spectrum.
    • The polarity and chirality of a molecule.
Common Point Groups
   Cs molecules have only E and one .
       • flat and asymmetrical.
   Cnv besides E have Cn and n v planes.
       • NH3, for example.
   Dnh has E, Cn, n C2 axes lying in a h.
       • like BF3.
   Td, Oh, and I are the Groups for
       • tetrahedral, octahedral, and icosahedral molecules
       • like CH4, SF6, and C60, for example.
The Grand Scheme, Part I
   Is the molecule linear?
    • If so, does it have inversion, i?
       • If so, then it is in the group Dh.
       • If no i, then it’s Ch.
    • If not linear, has it 2 or more Cn with n>2?
       • If so, does it have inversion, i?
           – If no i, then it is Td.
           – If it has i, then is there a C5?
                • That C5 means it’s the icosahedral group, I.
                • But if C5 is absent, it’s the octahedral group, Oh.
       • If n<2 or there aren’t 2 or more Cn, go to Part II.
The Grand Scheme, Part II
   OK, does it have any Cn?
    • No? How about any ?
       • Has a , thank Lewis, it’s a Cs molecule.
       • If no , then has it an inversion, i?
          – If an i, then it’s Ci.
          – But without the i, it’s only C1 (and it has only E left)!
    • There is a Cn? OK, pick the highest n and
      proceed to Part III.
The Grand Scheme, Part III
    to the (highest n) Cn, are there n C2 axes?
    • If so, a h guarantees it’s Dnh.
    • Without h, are there n d planes?
       • The dihedrals identify Dnd, without them it’s Dn.
   If no  C2,
    • A h makes it Cnh, but without h,
       • n v would make it Cnv, but if n h are missing,
          – Is there a 2n-fold improper rotation, S2n?
               • If so, it’s S2n, but if not, it’s just Cn.

   In fewer than 8 questions, we have it!
What we’ve bought.
   Cn or Cnv (n>1) means no dipole  to axis.
    • If no polarity along axis, molecule isn’t polar!
   No Cnh or D group molecule can be polar.

   Molecule can’t be chiral with an Sn!
    • But inversion, i is S2; so i counts.
    • Also a Cn and its  h is also Sn; so they both
      count too. If present, they deny chirality.
 Anatomy of a Character Table
Group Name Pair of C3 operations Order of the Group

      Identity   Trio of   v
                                  (# of operations)

 C3v E 2C3 3v h=6
 A1    1    1     1     z       z2, x2+y2
 A2    1    1     –1                                   Rz
  E    2    –1    0     (x,y) (xy,x2–y2),(xz,yz) (Rx,Ry)

            Double
Symmetry   Degeneracy            Functions transforming as the
 Species                        symmetry species (e.g. orbitals)
Motion in NH3, a C3v Molecule
      y
          x
          y       y                         z

              x       x             z           x z
      y
                                        x             x
          x


What part of a coordinate survives each symmetry operation?
E leaves all 12 coordinates alone. Therefore 12 survive.
C3 leaves only Nz and –½ each of Nx and Ny. So 1–½–½=0.
v leaves only Nx, Nz, H1x, & H1z but makes –1 each for
   Ny and H1y. So 1+1+1+1–1–1=2.
ammonia motion = 12 0 2
C3v E 2C3 3v h=6
A1   1     1     1   z      z2, x2+y2
A2   1     1    –1                               Rz
E    2    –1     0   (x,y) (xy,x2–y2),(xz,yz) (Rx,Ry)
    12    0     2   x+y+z+Rx+Ry+Rz+vibrations

     A1 = 1(12) + 2(1)(0) + 3(1)(2) = 18/h = 3A1
     A2 = 1(12) + 2(1)(0) + 3(–1)(2) = 6/h = 1A2
     E = 2(12) + 2(–1)(0) + 3(0)(2) = 24/h = 4 E
Ammonia Motion = 3A1+A2+4E
   Since E is doubly degenerate, that means
    3+1+8 = 12 motions (3 coords for 4 atoms).
   Since x+y+z+Rx+Ry+Rz = A1+A2+2E, the
    six vibrations in NH3 must be 2A1+2E.
C3v E 2C3 3v h=6
A1     1   1    1    z    z2, x2+y2
A2     1   1    –1                            Rz
E      2   –1   0    (x,y) (xy,x2–y2),(xz,yz) (Rx,Ry)