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Published byEvan Stark Modified about 1 year ago

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A Science of Aesthetics Transformations You Cannot See.

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Topics w Point Symmetry Operations, E C i S Classes w Point Groups Identification Scheme w Uses and Consequences

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Symmetry Operations on Points w The nuclei of molecules are the points. w 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.

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The Identity Operation, E w 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.

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The Rotation Element, C n w 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 C n., and represents n–1 rotational operations about the axis. Each operation is a rotation through yet another 360°/n, but the operation C 1 is merely E.

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Finding Rotations w Symmetrical molecules often have rotation axes through their atoms. SF 6 is octahedral and has fourfold axes through the atoms F–S–F that invisibly cycle the remaining 4 fluorine atoms. w But the axes need not pierce any atom! The crown-shaped S 8 molecule also has a C 4, but it goes through the empty center.

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Principal Axis w Molecules may have many rotation axes. E.g., S 8 has four C 2 axes, one through each pair of bonds opposing across the middle. Only one of these is shown in the molecular model above. But the axis with the highest n is designated as the principal axis. It is used to find other operations, and often lends its name to the symmetry of the molecule. In S 8, that would be the C 4 axis.

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Inobvious Axes w Remember the trick for drawing tetrahedra like that for CCl 4 ? w The chlorines occupy opposite corners on opposite faces of the cube. w Like this … That purple C 3 axis is one of 4 diagonals of the cube on which you could spin the molecular top. But what about the 3 C 2 axes straight through each face? Only 1 shown here.

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Mirror Planes, Reflection in a mirror leaves some molecules looking identical to themselves. w What distinguishes these operations is the physical placement of the mirror so that the image coincides with the original molecule! w There are 3 types of mirror planes: Vertical, horizontal, and dihedral?!?

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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 SF 4, the principle axis is C 2. The two reflection planes are both “vertical” and happen to contain all of the atoms in the molecule.

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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 PCl 5, 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 C 3.)

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Dihedral Mirror, d w Greek “two-sided” doesn’t help. All planes are two-sided! w In symmetry it means vertical planes that lie between the C 2 axes the Principal Axis. E.g., in S 8 above, one of S 8 ’s four dihedral planes contains the C 4 Principal Axis and bisects the adjacent C 2 axes. The plane contains opposite sulfur atoms,

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The Magic of Mirrors w Rotations and the identity operation do not rearrange a molecule in any way. w 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.

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Chiral Molecules w Caraway flavor agent S-Carvone w Spearmint flavor agent R-Carvone These molecules cannot be aligned.

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SF 6 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.

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Improper Rotation, S n Proper rotations (C n ) do not rearrange the molecule, but improper rotations (S n ) do; they are rotations by 360°/n followed by reflection in a plane to the S n axis. For S n to be, neither C n nor the need exist! To see that, consider the molecule S 8 above; believe it or not, it has an S 8 axis coincident with the C 4 Principle Axis. Clearly there are seven S 8 operations.

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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! w E.g., H 2 O, besides E, has v vv and v ’ v’v’ and C 2. C2C2 It’s clear that C 2 C 2 = E, but C 2 v = v ’ needs a little thought.

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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.

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Common Point Groups C s molecules have only E and one . flat and asymmetrical. C nv besides E have C n and n v planes. NH 3, for example. D nh has E, C n, n C 2 axes lying in a h. like BF 3. T d, O h, and I are the Groups for tetrahedral, octahedral, and icosahedral molecules like CH 4, SF 6, and C 60, for example.

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The Grand Scheme, Part I w 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 C n with n>2? If so, does it have inversion, i? –If no i, then it is T d. –If it has i, then is there a C 5 ? That C 5 means it’s the icosahedral group, I. But if C 5 is absent, it’s the octahedral group, O h. If n<2 or there aren’t 2 or more C n, go to Part II.

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The Grand Scheme, Part II w OK, does it have any C n ? No? How about any ? Has a , thank Lewis, it’s a C s molecule. If no , then has it an inversion, i? –If an i, then it’s C i. –But without the i, it’s only C 1 (and it has only E left)! There is a C n ? OK, pick the highest n and proceed to Part III.

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The Grand Scheme, Part III w to the (highest n) C n, are there n C 2 axes? If so, a h guarantees it’s D nh. Without h, are there n d planes? The dihedrals identify D nd, without them it’s D n. w If no C 2, A h makes it C nh, but without h, n v would make it C nv, but if n h are missing, –Is there a 2n-fold improper rotation, S 2n ? If so, it’s S 2n, but if not, it’s just C n. w In fewer than 8 questions, we have it!

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What we’ve bought. C n or C nv (n>1) means no dipole to axis. If no polarity along axis, molecule isn’t polar! No C nh or D group molecule can be polar. Molecule can’t be chiral with an S n ! But inversion, i is S 2 ; so i counts. Also a C n and its h is also S n ; so they both count too. If present, they deny chirality.

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Anatomy of a Character Table C 3v E2C32C3 3v3v h=6 A1A1 111z z 2, x 2 +y 2 A2A2 11–1 R z E2–10(x,y) (xy,x 2 –y 2 ),(xz,yz) (R x,R y ) Group NameIdentityPair of C 3 operations Trio of v Order of the Group (# of operations) Symmetry Species Double Degeneracy Functions transforming as the symmetry species (e.g. orbitals)

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Motion in NH 3, a C 3v Molecule x y x z x z x z x y x y x y What part of a coordinate survives each symmetry operation? E leaves all 12 coordinates alone. Therefore 12 survive. C 3 leaves only N z and –½ each of N x and N y. So 1–½–½= 0. v leaves only N x, N z, H 1x, & H 1z but makes –1 each for N y and H 1y. So 1+1+1+1–1–1= 2.

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ammonia motion = 12 0 2 C 3v E2C32C3 3v3v h=6 A1A1 111z z 2, x 2 +y 2 A2A2 11–1 R z E2–10(x,y) (xy,x 2 –y 2 ),(xz,yz) (R x,R y ) 1202x+y+z+R x +R y +R z +vibrations A 1 = 1(12) + 2(1)(0) + 3(1)(2) = 18/h = 3A 1 A 2 = 1(12) + 2(1)(0) + 3(–1)(2) = 6/h = 1A 2 E = 2(12) + 2(–1)(0) + 3(0)(2) = 24/h = 4 E

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Ammonia Motion = 3A 1 +A 2 +4E w Since E is doubly degenerate, that means 3+1+8 = 12 motions (3 coords for 4 atoms). Since x+y+z+R x +R y +R z = A 1 +A 2 +2E, the six vibrations in NH 3 must be 2A 1 +2E. C 3v E2C32C3 3v3v h=6 A1A1 111z z 2, x 2 +y 2 A2A2 11–1 R z E2–10(x,y) (xy,x 2 –y 2 ),(xz,yz) (R x,R y )

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