CHAPTER 4: SYMMETRY AND GROUP THEORY

Size: px
Start display at page:

Download "CHAPTER 4: SYMMETRY AND GROUP THEORY"

Transcription

1 hapter 4 Symmetry and Group Theory 33 APTER 4: SYMMETRY AND GROUP TEORY 4. a. Ethane in the staggered conformation has 2 3 axes (the line), 3 perpendicular 2 axes bisecting the line, in the plane of the two s and the s on opposite sides of the two s. No h, 3 d, i, S 6. D 3d. b. Ethane in eclipsed conformation has two 3 axes (the line), three perpendicular 2 axes bisecting the line, in the plane of the two s and the s on the same side of the two s. Mirror planes include h and 3 d. D 3h. c. hloroethane in the staggered conformation has only one mirror plane, through both s, the l, and the opposite on the other. s. d.,2-dichloroethane in the trans conformation has a 2 axis perpendicular to the bond and perpendicular to the plane of both l s and both s, a h plane through both l s and both s, and an inversion center. 2h. 4.2 a. Ethylene is a planar molecule, with 2 axes through the s and perpendicular to the bond both in the plane of the molecule and perpendicular to it. It also has a h plane and two d planes (arbitrarily assigned). D 2h. b. hloroethylene is also a planar molecule, with the only symmetry element the mirror plane of the molecule. s. c.,-dichloroethylene has a 2 axis coincident with the bond, and two mirror planes, one the plane of the molecule and one perpendicular to the plane of the molecule through both s. 2v. cis-,2-dichloroethylene has a 2 axis perpendicular to the bond, and in the plane of the molecule, two mirror planes (one the plane of the molecule and one perpendicular to the plane of the molecule and perpendicular to the bond). 2v. trans-,2-dichloroethylene has a 2 axis perpendicular to the bond and perpendicular to the plane of the molecule, a mirror plane in the plane of the molecule, and an inversion center. 2h. 4.3 a. Acetylene has a axis through all four atoms, an infinite number of perpendicular 2 axes, a h plane, and an infinite number of d planes through all four atoms. D h. b. luoroacetylene has only the axis through all four atoms and an infinite number of mirror planes, also through all four atoms. v. c. Methylacetylene has a 3 axis through the carbons and three v planes, each including one hydrogen and all three s. 3v. d. 3-hloropropene (assuming a rigid molecule, no rotation around the bond) has no rotation axes and only one mirror plane through l and all three atoms. s. e. Phenylacetylene (again assuming no internal rotation) has a 2 axis down the long axis of the molecule and two mirror planes, one the plane of the benzene ring and the other perpendicular to it. 2v

2 34 hapter 4 Symmetry and Group Theory 4.4 a. Napthalene has three perpendicular 2 axes, and a horizontal mirror plane (regardless of which 2 is taken as the principal axis), making it a D 2h molecule. b.,8-dichloronaphthalene has only one 2 axis, the bond joining the two rings, and two mirror planes, making it a 2v molecule. c.,5-dichloronaphthalene has one 2 axis perpendicular to the plane of the molecule, a horizontal mirror plane, and an inversion center; overall, 2h. d.,2-dichloronaphthalene has only the mirror plane of the molecule, and is a s molecule. 4.5 a.,-dichloroferrocene has a 2 axis parallel to the rings, perpendicular to the l e l h mirror plane. It also has an inversion center. 2h. b. Dibenzenechromium has collinear 6, 3, and 2 axes perpendicular to the rings, six perpendicular 2 axes and a h plane, making it a D 6h molecule. It also has three v and three d planes, S 3 and S 6 axes, and an inversion center. c. Benzenebiphenylchromium has a mirror plane through the r and the biphenyl bridge bond and no other symmetry elements, so it is a s molecule. d. 3 O + has the same symmetry as N 3 : a 3 axis, and three v planes for a 3v molecule. e. O 2 2 has a 2 axis perpendicular to the O O bond and perpendicular to a line connecting the fluorines. With no other symmetry elements, it is a 2 molecule. f. ormaldehyde has a 2 axis collinear with the =O bond, a mirror plane including all the atoms, and another perpendicular to the first and including the and O atoms. 2v. g. S 8 has 4 and 2 axes perpendicular to the average plane of the ring, four 2 axes through opposite bonds, and four mirror planes perpendicular to the ring, each including two S atoms. D 4d. h. Borazine has a 3 axis perpendicular to the plane of the ring, three perpendicular 2 axes, and a horizontal mirror plane. D 3h. i. Tris(oxalato)chromate(III) has a 3 axis and three perpendicular 2 axes, each splitting a bond and passing through the r. D 3. j. A tennis ball has three perpendicular 2 axes (one through the narrow portions of each segment, the others through the seams) and two mirror planes including the first rotation axis. D 2d. 4.6 a. yclohexane in the chair conformation has a 3 axis perpendicular to the average plane of the ring, three perpendicular 2 axes between the carbons, and three v planes, each including the 3 axis and one of the 2 axes. D 3d. b. Tetrachloroallene has three perpendicular 2 axes, one collinear with the double bonds and the other two at 45 to the l l planes. It also has two v planes, one defined by each of the l l groups. Overall, D 2d. (Note that the ends of tetrachlorallene are staggered.)

3 hapter 4 Symmetry and Group Theory 35 c. The sulfate ion is tetrahedral. T d. d. Most snowflakes have hexagonal symmetry (igure 4.2), and have collinear 6, 3, and 2 axes, six perpendicular 2 axes, and a horizontal mirror plane. Overall, D 6h. (or high quality images of snowflakes, including some that have different shapes, see K. G. Libbrecht, Snowflakes, Voyageur Press, Minneapolis, MN, 2008.) e. Diborane has three perpendicular 2 axes and three perpendicular mirror planes. D 2h. f.,3,5-tribromobenzene has a 3 axis perpendicular to the plane of the ring, three perpendicular 2 axes, and a horizontal mirror plane. D 3h.,2,3-tribromobenzene has a 2 axis through the middle Br and two perpendicular mirror planes that include this axis. 2v,2,4-tribromobenzene has only the plane of the ring as a mirror plane. s. g. A tetrahedron inscribed in a cube has T d symmetry (see igure 4.6). h. The left and right ends of B 3 8 are staggered with respect to each other. There is a 2 axis through the borons. In addition, there are two planes of symmetry, each containing four atoms, and two 2 axes between these planes and perpendicular to the original 2. The point group is D 2d. i. A mountain swallowtail butterfly has only a mirror that cuts through the head, thorax, and abdomen. s j. The Golden Gate Bridge has a 2 axis and two perpendicular mirror planes that include this axis. 2v 4.7 a. A sheet of typing paper has three perpendicular 2 axes and three perpendicular mirror planes. D 2h. b. An Erlenmeyer flask has an infinite-fold rotation axis and an infinite number of v planes, v. c. A screw has no symmetry operations other than the identity, for a classification. d. The number 96 (with the correct type font) has a 2 axis perpendicular to the plane of the paper, making it 2h. e. Your choice the list is too long to attempt to answer it here. f. A pair of eyeglasses has only a vertical mirror plane. s. g. A five-pointed star has a 5 axis, five perpendicular 2 axes, one horizontal and five vertical mirror planes. D 5h. h. A fork has only a mirror plane. s. i. Wilkins Micawber has no symmetry operation other than the identity..

4 36 hapter 4 Symmetry and Group Theory j. A metal washer has a axis, an infinite number of perpendicular 2 axes, an infinite number of v mirror planes, and a horizontal mirror plane. D h. 4.8 a. D 2h f. 3 b. D 4h (note the four knobs) g. 2h c. s h. 8v d. 2v i. D h e. 6v j. 3v 4.9 a. D 3h f. s (note holes) b. D 4h g. c. s h. 3v d. 3 i. D h e. 2v j. 4.0 ands (of identical twins): 2 Baseball: D 2d Atomium: 3v Eiffel Tower: 4v Dominoes: 6 6: 2v 3 3: 2 5 4: s Bicycle wheel: The wheel shown has 32 spokes. The point group assignment depends on how the pairs of spokes (attached to both the front and back of the hub) connect with the rim. If the pairs alternate with respect to their side of attachment, the point group is D 8d. Other arrangements are possible, and different ways in which the spokes cross can affect the point group assignment; observing an actual bicycle wheel is recommended. (If the crooked valve is included, there is no symmetry, and the point group is a much less interesting.) 4. a. Problem 3.4 * : a. VOl 3 : 3v b. Pl 3 : 3v c. SO 4 : 2v d. SO 3 : D 3h e. Il 3 : 2v f. S 6 : O h g. I 7 : D 5h h. XeO 2 4 : D 4h i. 2 l 2 : 2v j. P 4 O 6 : T d * Incorrectly cited as problem 4.30 in first printing of text.

5 hapter 4 Symmetry and Group Theory 37 b. Problem 3.42 * : a. P 3 : 3v b. 2 Se: 2v c. Se 4 : 2v d. P 5 : D 3h e. I 5 : 4v f. XeO 3 : 3v g. B 2 l: 2v h. Snl 2 : 2v i. Kr 2 : D h j. IO : D 5h 4.2 a. igure 3.8: a. 2 : D h b. SO 3 : D 3h c. 4 : T d d. Pl 5 : D 3h e. S 6 : O h f. I 7 : D 5h g. Ta 8 3 : D 4d b. igure 3.5: a. 2 : D h b. 2 : 2v c. NO 2 : 2v d. SO 3 : D 3h e. SN 3 : 3v f. SO 2 l 2 : 2v g. XeO 3 : 3v h. SO 4 2 : T d i. SO 4 : 2v j. lo 2 3 : 2v k. XeO 3 2 : D 3h l. IO 5 : 4v 4.3 a. p x has v symmetry. (Ignoring the difference in sign between the two lobes, the point group would be D h.) b. d xy has D 2h symmetry. (Ignoring the signs, the point group would be D 4h.) c. d x 2 y 2 has D 2h symmetry. (Ignoring the signs, the point group would be D 4h.) d. d z 2 has D h symmetry. e. f xyz has T d symmetry a. The superimposed octahedron and cube show the matching symmetry elements. The descriptions below are for the elements of a cube; each element also applies to the octahedron. 2, 4 E Every object has an identity operation. Diagonals through opposite corners of the cube are 3 axes. Lines bisecting opposite edges are 2 axes. Lines through the centers of opposite faces are 4 axes. Although there are only three such lines, there are six axes, counting the 4 3 operations. (= 4 2 ) The lines through the centers of opposite faces are 4 axes as well as 2 axes. * Incorrectly cited as problem 3.4 in first printing of text.

6 38 hapter 4 Symmetry and Group Theory i 6S 4 8S 6 3 h 6 d The center of the cube is the inversion center. The 4 axes are also S 4 axes. The 3 axes are also S 6 axes. These mirror planes are parallel to the faces of the cube. These mirror planes are through two opposite edges. b. O h c. O 4.5 a. There are three possible orientations of the two blue faces. If the blue faces are opposite each other, a 3 axis connects the centers of the blue faces. This axis has 3 perpendicular 2 axes, and contains three vertical mirror places (D 3d ). If the blue faces share one vertex of the octahedron, a 2 axis includes this vertex, and this axis includes two vertical mirror planes ( 2v ). The third possibility is for the blue faces to share an edge of the octahedron. In this case, a 2 axis bisects this shared edge, and includes two vertical mirror planes ( 2v ). b. There are three unique orientations of the three blue faces. If one blue face is arranged to form edges with each of the two remaining blue faces, the only symmetry operations are identity and a single mirror plane ( s ). If the three blue faces are arranged such that a single blue face shares an edge with one blue face, but only a vertex with the other blue face, the only symmetry operation is a mirror plane that passes through the center of the blue faces, and the point group is s. If the three blue faces each share an edge with the same yellow face, a 3 axis emerges from the center of this yellow face, and this axis includes three vertical mirror planes ( 3v ). c. If there are four different colors, and each pair of opposite faces has the identical color, the only symmetry operations are identity and inversion ( i ). 4.6 our point groups are represented by the symbols of the chemical elements. Most symbols have a single mirror in the plane of the symbol ( s ), for example, s! Two symbols have D 2h symmetry (, I), and two more (, S) have 2h. Seven exhibit 2v symmetry (B,, K, V, Y, W, U). In some cases, the choice of font may affect the point group. or example, the symbol for nitrogen may have 2h in a sans serif font () but otherwise s (N). The symbol of oxygen has D h symmetry if shown as a circle but D 2h if oval.

7 hapter 4 Symmetry and Group Theory a. on-deck circle D h e. home plate 2v b. batter s box D 2h f. baseball D 2d (see igure 4.) c. cap s g. pitcher d. bat v 4.8 a. D 2h d. D 4h g. D 2h b. 2v e. 5h h. D 4h c. 2v f. 2v i. 2 y N 2 S 4.9 SN 3 3 N (top view) x Symmetry Operations: 2 3 N N 3 N 3 2 after E after 3 after v (xz) Matrix Representations (reducible): 2 E: : cos 2 sin sin 2 cos v (xz): haracters of Matrix Representations: 3 0 (continued on next page)

8 40 hapter 4 Symmetry and Group Theory Block Diagonalized Matrices: Irreducible Representations: E v oordinates Used 2 0 (x, y) z haracter Table: 3v E v Matching unctions A z x 2 + y 2, z 2 A 2 R z E 2 0 (x, y), (R x, R y ) (x 2 y 2, xy)(xz, yz) l 4.20 a. 2h molecules have E, 2, i, and h operations. l b. E: 2 : i: h : c. These matrices can be block diagonalized into three matrices, with the representations shown in the table. (E) ( 2 ) (i) ( h ) B u from the x and y coefficients A u from the z coefficients The total is = 2B u +A u. d. Multiplying B u and A u : + ( ) + ( ) ( ) + ( ) = 0, proving they are orthogonal.

9 hapter 4 Symmetry and Group Theory a. D 2h molecules have E, 2 (z), 2 (y), 2 (x), i, (xy), (xz), and (yz) operations. b. E: (z): (y): (x): i: (xy): (xz): (yz): c. E 2 (z) 2 (y) 2 (x) i (xy) (xz) (yz) 3 3 d. matching B 3u 2 matching B 2u 3 matching B u e. 2 = + ( ) ( ) + ( ) + ( ) + ( ) ( ) + + ( ) + ( ) = 0 3 = + ( ) + ( ) ( ) + ( ) + ( ) ( ) + ( ) + + ( ) = = + ( ) + ( ) + ( ) ( ) + ( ) ( ) + ( ) + ( ) + = a. h = 8 (the total number of symmetry operations) b. A E = ( 2) = 0 A 2 E = ( 2) + 2 ( ) ( ) 0 = 0 B E = ( ) 0 + ( 2) ( ) 0 = 0 B 2 E = ( ) 0 + ( 2) + 2 ( ) = 0 c. E: = 8 A : = 8 A 2 : = 8 B : = 8 B 2 : = 8

10 42 hapter 4 Symmetry and Group Theory d. = 2A + B + B 2 + E: A : /8[ ] = 2 A 2 : /8[ ( ) ( ) 2] = 0 B : /8[ ( ) ( ) 2] = B 2 : /8[ ( ) ( ) ] = E: /8[ ( 2) ] = 2 = 3 A + 2A 2 + B : A : /8[ ] = 3 A 2 : /8[ ( ) ( ) 0] = 2 B : /8[ ( ) ( ) 0] = B 2 : /8[ ( ) ( ) ] = 0 E: /8[ ( 2) ] = v = 3A + A 2 + E: A : /6[ ] = 3 A 2 : /6[ ( ) 2] = E: /6[ ( ) ] = 2 = A 2 + E: A : /6[ ( ) + 3 ( )] = 0 A 2 : /6[ ( ) + 3 ( ) ( )] = E: /6[ ( ) ( ) ( )] = 2 O h 3 = A g + E g + T u : A g : /48[ ] = A 2g : /48[ ] = 0 E g : /48[ ] = T g : /48[ ] = 0 T 2g : /48[ ] = 0 A u : /48[ ] = 0 A 2u : /48[ ] = 0 E u : /48[ ] = 0 T u : /48[ ] = T 2u : /48[ ] = 0

11 hapter 4 Symmetry and Group Theory The d xy characters match the characters The d x2-y2 characters match the characters of the B 2g representation: of the B g representation: d xy d x 2 y 2 E E 4 4 y 2 y x 2 x 2 d xy i S 4 d x2 y 2 i S 4 h h v v d d 4.25 hiral: 4.5: O 2 2, [r( 2 O 4 ) 3 ] 3 4.6: none 4.7: screw, Wilkins Micawber 4.8: recycle symbol 4.9: set of three wind turbine blades, lying Mercury sculpture, coiled spring 4.26 a. Point group: 4v O 4v E v 2 d A z A 2 R z B B 2 E (x, y), (R x, R y ) Xe b. = 4 A + A 2 + 2B + B 2 + 5E c. Translation: A + E (match x, y, and z) Rotation: A 2 + E (match R x, R y, and R z ) Vibration: all that remain: 3 A + 2B + B 2 + 3E O d. The character for each symmetry operation for the Xe O stretch is +. This corresponds to the A irreducible representation, which matches the function z and is therefore IR-active. Xe

12 44 hapter 4 Symmetry and Group Theory 4.27 or S 6, the axes of the sulfur should point at three of the fluorines. The fluorine axes can be chosen in any way, as long as one from each atom is directed toward the sulfur atom. There are seven atoms with three axes each, for a total of 2. O h E i 6S 4 8S 6 3 h 6 d T u (x,y,z) T g (R x, R y, R z ) A g A 2u E g (2z 2 x 2 y 2, x 2 y 2 ) T 2u T 2g (xy, xz, yz) Reduction of gives = 3T u + T g + A g + E g + T 2g + T 2u. T u accounts for translation and also infrared active vibrational modes. T g is rotation. The remainder are infrared-inactive vibrations a. cis-e() 4 l 2 has 2v symmetry. The vectors for stretching have the representation : 2v E 2 v (xz) v (yz) A z A 2 B x B 2 y l l O e O O O n(a ) = /4[ ] = 2 n(a 2 ) = /4[ ( ) + 2 ( )] = 0 n(b ) = /4[4 + 0 ( ) ( )] = n(b 2 ) = /4[4 + 0 ( ) + 2 ( ) + 2 ] = = 2 A + B + B 2, all four IR active. l b. trans-e() 4 l 2 has D 4h symmetry. O O e O l D 4h E i 2S 4 h 2 v 2 d A 2u z E u (x,y)

13 hapter 4 Symmetry and Group Theory 45 Omitting the operations that have zeroes in : n(a 2u ) = /6[ ( ) + 4 ( ) ] = 0 n(e u ) = /6[ ] = (IR active) Note: In checking for IR-active bands, it is only necessary to check the irreducible representations having the same symmetry as x, y, or z, or a combination of them. c. e() 5 has D 3h symmetry. The vectors for O stretching have the following representation : D 3h E h 2S 3 3 v E (x, y) A 2 z O O O e O n(e) = /2 [ (5 2) + (2 2 ) + (3 2)] = n(a 2 ) = /2 [(5 ) + (2 2 ) + (3 ) + (3 ) + (3 3 )] = There are two bands, one matching Eand one matching A 2. These are the only irreducible representations that match the coordinates x, y, and z In 4.28a, the symmetries of the stretching vibrations of cis-e() 4 l 2 ( 2v symmetry) are determined as 2 A + B + B 2. Each of these representations matches Raman-active functions: A (x 2, y 2, z 2 ) ; A 2 (xy), B (xz); and B 2 (yz), so all are Raman-active. In 4.28b, the symmetries of the stretching vibrations of trans-e() 4 l 2 (D 4h symmetry) are A g + B g + E u. Only A g (x 2 + y 2, z 2 ) and B g (x 2 y 2 ) match Raman active functions; this complex exhibits two Raman-active stretching vibrations. In 4.28c, the symmetries of the stretching vibrations of e() 5 (D 3h symmetry) are 2 A + E + A 2. Only A (x 2 + y 2, z 2 ) and E(x 2 y 2, xy) match Raman-active functions; this complex exhibits four Raman-active stretching vibrations a. The point group is 2h. b. Using the Si I bond vectors as a basis generates the representation: 2h E 2 i h A g x 2 + y 2, z 2 B g xz, yz A u z B u x, y = A g + B g + A u + B u The A u and B u vibrations are infrared active. c. The A g and B g vibrations are Raman active.

14 46 hapter 4 Symmetry and Group Theory 4.3 trans isomer (D 4h ): cis isomer ( 2v ): The simplest approach is to consider if the number of infrared-active I O stretches is different for these structures. (Alternatively, one could also determine the number of IR-active I stretches, a slightly more complicated task.) trans: D 4h E i 2S 4 h 2 v 2 d A g A 2u z There is only a single IR-active I O stretch (the antisymmetric stretch), A 2u. cis: 2v E 2 (xz) (yz) A z B x There are two IR-active I O stretches, the A and B (symmetric and antisymmetric). Infrared spectra should therefore be able to distinguish between these isomers. (Reversing the x and y axes would give A + B 2. Because B 2 matches y, it would also represent an IR-active vibration.) Because these isomers would give different numbers of IR-active absorptions, infrared spectra should be able to distinguish between them. The reference provides detailed IR data a. One way to deduce the number of Raman-active vibrations of AsP 3 is to first P determine the symmetries of all the degrees of freedom. This complex exhibits P P 3 symmetry, with the 3 axis emerging from the As atom. The (E) is 2; the x, y, and z axes of the four atoms do not shift when the identity operation is carried out. Only the As atom contributes to the character of the 3 transformation matrix; the P atoms shift during rotation about the 3 axis. The general transformation matrix for rotation about the z axis (Section 4.3.3) affords 0 as ( 3 ) for The As atom and one P atom do not shift when a v reflection is carried out, and ( v ) = 2 (see the v(xz) transformation matrix in Section 4.33 for the nitrogen atom of N 3 as a model of how the two unshifted atoms of AsP 3 will contribute to the character of the v transformation matrix). As 3v E 3 v A z x 2 + y 2, z 2 A 2 R x E 2 0 (x, y), (R x, R y ) (x 2 y 2, xy), (xz, yz)

15 hapter 4 Symmetry and Group Theory 47 Reduction of the reducible representation affords 3 A + A E. On the basis of the character table, the translational modes of AsP 3 have the symmetries A + E, and the rotational modes have the symmetries A 2 + E. The six vibrational modes of AsP 3 subsequently have the symmetries 2 A + 2 E. These vibrational modes are Raman-active, and four absorptions are expected (and observed) since the sets of E modes are degenerate. B. M. ossairt, M.. Diawara,.. ummins, Science 2009, 323, 602 assigns the bands as: 33 (a ), 345 (e), 428 (a ), 557 (e) cm -. Alternatively, the set of six bonds may be selected as the basis for a representation focused specifically on stretches of these bonds. This approach generates the following representation: 3v E 3 v This representation reduces to 2 A + 2 E, the same result as obtained by first considering all degrees of freedom, then subtracting the translational and rotational modes. b. As 2 P 2 exhibits 2v symmetry, and (like AsP 3 ) will have six vibrational modes (3N 6). Inspection of the 2v character table indicates that all vibrational P modes will be Raman active. Since each irreducible representation has a dimension of, the number of Raman absorptions expected is 6 (that is, there will be no degenerate vibrational modes). This prediction can be confirmed via deduction of the symmetries of the vibrational modes. As in part a, we will first determine the symmetries of all the degrees of freedom. The (E) of the transformation matrix is 2. In As 2 P 2, the 2 axis does not pass through any atoms, and all four atoms shift upon rotation; ( 2 ) = 0. Two atoms do not shift upon reflection through each of the v planes. The contribution to the character of the transformation matrix for each of these unshifted atoms is, and ( v (xz)) = ( v (yz)) = 2. As P As 2v E 2 v (xz) v (yz) A z x 2, y 2, z 2 A 2 R z xy B x, R y xz B 2 y, R x yz Reduction of the reducible representation affords 4 A + 2 A B + 3 B 2. On the basis of the above character table, the translational modes of As 2 P 2 have the symmetries A + B + B 2, and the rotational modes have the symmetries A 2 + B + B 2. The six anticipated Raman-active vibrational modes of As 2 P 2 subsequently have the symmetries 3 A + A 2 + B + B 2. (continued on next page)

16 48 hapter 4 Symmetry and Group Theory As in part a, an alternative approach is to select the set of six bonds as the basis for a representation focused specifically on stretching vibrations. This approach generates the following representation: 2v E 2 v (xz) v (yz) This representation reduces to 3 A + A 2 + B + B 2, the same result as obtained by first considering all degrees of freedom, then subtracting the translational and rotational modes. c. The issue here is whether or not P 4 (T d ) exhibits 4 Raman-active vibrations as does AsP 3. We will first determine the symmetries of all the degrees of freedom. The (E) of the transformation matrix is 2. Only one P 4 atom is fixed upon rotation about each 3 axis; ( 3 ) = 0. All four atoms are shifted upon application of the 2 and S 4 axes; ( 2 ) = (S 4 ) = 0. Two atoms do not shift upon reflection through each of the d planes. The contribution to the character of the transformation matrix for each of these unshifted atoms is, and ( d ) = 2. P P P P T d E S 4 6 d A x 2 + y 2 + z 2 A 2 E (2z 2 x 2 y 2, x 2 y 2 ) T 3 0 (R x, R y, R z ) T (x, y, z) (xy, xz, yz) Reduction of the reducible representation affords A + E + T + 2 T 2. Since the symmetries of the translational modes and rotational modes are T 2 and T, respectively, the symmetries of the vibrational modes are A + E + T 2, all of these modes are Ramanactive so three Raman absorptions are expected for P 4, and P 4 could potentially be distinguished from AsP 3 solely on the basis of the number of Raman absorptions. The alternative approach, using the set of six P P bonds as the basis for a representation focused specifically on bond stretches, generates the following representation: T d E S 4 6 d This representation reduces to A + E + T 2, the same result as obtained by first considering all degrees of freedom, then subtracting the translational and rotational modes.

17 hapter 4 Symmetry and Group Theory The possible isomers are as follows, with the triphenylphosphine ligand in either the axial (A) or equatorial (B) sites. Ph 3 P O e O e PPh 3 (A) 3v (B) 2v Note that the triphenylphosphine ligand is approximated as a simple L ligand for the sake of the point group determination. Rotation about the e P bond in solution is expected to render the arrangement of the phenyl rings unimportant in approximating the symmetry of these isomers in solution. The impact of the phenyl rings would likely be manifest in the IR ν() spectra of these isomers in the solid-state. or A, we consider each bond as a vector to deduce the expected number of carbonyl stretching modes. The irreducible representation is as follows: 3v E 3 v 4 2 A z x 2 + y 2, z 2 A 2 R x E 2 0 (x, y), (R x, R y ) (x 2 y 2, xy), (xz, yz) Reduction of the reducible representation affords 2 A + E. These stretching modes are IR-active and three ν() absorptions are expected for A. or B, a similar analysis affords the following irreducible representation: 2v E 2 v (xz) v (yz) A z x 2, y 2, z 2 A 2 R z xy B x, R y xz B 2 y, R x yz Reduction of the reducible representation affords 2 A + B + B 2. These stretching modes are IRactive, and four ν() absorptions are expected for A. The reported ν() IR spectrum is consistent with formation of isomer A, with the triphenylphosphine ligand in the axial site.

18 50 hapter 4 Symmetry and Group Theory 4.34 As in 4.33, we consider the triphenylphosphine ligand as a simple L group for point group determination. The point groups for isomers A, B, and are as follows: PPh 3 PPh 3 O e PPh 3 PPh 3 O e PPh 3 O e PPh 3 (A) 2v (B) D 3h () s or A, the set of irreducible representations for the three stretching vibrational modes is 2 A + B. These modes are all IR-active in the 2v character table, and three ν() IR absorptions are expected for isomer A. or B, the set of irreducible representations for the three stretching vibrational modes is A + E. Only the E mode is IR-active in the D 3h point group, and one ν() IR absorption is expected for isomer B. or, the set of irreducible representations for the three stretching vibrational modes is 2 A These modes are all IR-active in the s point group, and three ν() IR absorptions are expected for isomer. The single ν() IR absorption reported for e() 3 (PPh 3 ) 2 supports the presence of the D 3h isomer B. The trans isomer B is reported in R. L. Keiter, E. A. Keiter, K.. ecker,. A. Boecker, Organometallics, 988, 7, 2466, and the authors observe splitting of the absorption associated with the Emode in l 3. The forbidden A stretching mode was observed as a weak absorption The IR spectrum exhibits two ν() absorptions. The two proposed metal carbonyl fragments will be considered independently for analysis. O O Ti O O Ti D 4h 4v The reducible representation for the four vectors associated with the bonds in the 4v fragment is as follows: 4v E v 2 d Reduction of this representation affords A + B + E. The A and E modes are IR-active, and a titanium complex with a square pyramidal titanium tetracarbonyl fragment is supported by the IR spectral data.

19 hapter 4 Symmetry and Group Theory 5 or the D 4h fragment, the reducible representation for the set of vectors associated with the bonds is: D 4h E i 2S 4 h 2 v 2 d Reduction of the reducible representation affords A g + B g + E u. Only the E modes are IR-active, and a complex with a square planar titanium tetracarbonyl fragment is expected to exhibit a single IR ν() stretching absorption. This D 4h possibility can therefore be ruled out on the basis of the spectrum A reasonable product is the 4v molecule Mo() 5 (P(OPh) 3 ), with replacing a triphenylphosphite ligand. The reducible representation for the five vectors associated with the bonds in this molecule is: O O (OPh) 3 P Mo O 4v E v 2 d 5 3 Reduction of this representation affords 2 A + B + E. The A and E modes are IR-active, and three IR ν() stretching absorptions are expected. The reported IR spectrum features three strong ν() absorptions, and one very weak absorption attributed to the forbidden B mode in D. J. Darensbourg, T. L. Brown, Inorg. hem., 968, 7, I has 2 symmetry, with a 2 axis running right to left, perpendicular to the l, N, l, N and l, P, l, P faces. II also has 2 symmetry, with the same 2 axis as I. (Lower left corner occupied by l, not.) III has only an inversion center and i symmetry An example for each of the five possible point groups: T d : 3v : 2v : s : : Br Br Br l 4.39 a. The S portion is linear, so the molecule has a 3 axis along the line of these three atoms, three v planes through these atoms and an atom on each end, but no other symmetry elements. 3v b. The molecule has only an inversion center, so it is i. The inversion center is equivalent to an S 2 axis perpendicular to the average plane of the ring. c. M 2 l 6 Br 4 is i. d. This complex has a 3 axis, splitting the three N atoms and the three P atoms (almost as drawn), but no other symmetry elements. 3

20 52 hapter 4 Symmetry and Group Theory e. The most likely isomer has the less electronegative l atoms in equatorial positions. Point group: 2v 4.40 The structures on the top row are D 2d (left) and s (right). Those on the bottom row are 2h (left) and 4v (right). l l P 4.4 a. 3v b. D 5h c. Square structure: D 2d (bottom ligand on lower left Re should be instead of L); orner structure: s d. D 3d 4.42 Web of Science and Sciinder Scholar should be helpful, but simply searching for these symmetries using a general Internet search should provide examples of these point groups. Some examples: a. S 6 : Mo 2 (S 6 2 Me 3 ) 6 (M.. hisholm, J.. orning, and J.. uffman, J. Am. hem. Soc., 983, 05, 5924) Mo 2 (NMe 2 ) 6 (M.. hisholm, R. A. otton, B. A. Grenz, W. W. Reichert, L. W. Shive, and B. R. Stults, J. Am. hem. Soc., 976, 98, 4469) [Nae 6 (OMe) 2 (dbm) 6 ] + (dbm = dibenzoylmethane, ) (. L. Abbati, A. ornia, A.. abretti, A. aneschi, and D. Garreschi, Inorg. hem., 998, 37, 430) b. T Pt( 3 ) 4, 44 c. I h 20, 80 d. T h [o(no 2 ) 6 ] 3, Mo(NMe 2 ) 6 In addition to examples that can be found using Web of Science, Sciinder, and other Internet search tools, numerous examples of these and other point groups can be found in I. argittai and M. argittai, Symmetry Through the Eyes of a hemist, as listed in the General References section.

The Unshifted Atom-A Simpler Method of Deriving Vibrational Modes of Molecular Symmetries

The Unshifted Atom-A Simpler Method of Deriving Vibrational Modes of Molecular Symmetries Est. 1984 ORIENTAL JOURNAL OF CHEMISTRY An International Open Free Access, Peer Reviewed Research Journal www.orientjchem.org ISSN: 0970-020 X CODEN: OJCHEG 2012, Vol. 28, No. (1): Pg. 189-202 The Unshifted

More information

CHAPTER 5: MOLECULAR ORBITALS

CHAPTER 5: MOLECULAR ORBITALS Chapter 5 Molecular Orbitals 5 CHAPTER 5: MOLECULAR ORBITALS 5. There are three possible bonding interactions: p z d z p y d yz p x d xz 5. a. Li has a bond order of. (two electrons in a bonding orbital;

More information

Symmetry and group theory

Symmetry and group theory Symmetry and group theory or How to Describe the Shape of a Molecule with two or three letters Natural symmetry in plants Symmetry in animals 1 Symmetry in the human body The platonic solids Symmetry in

More information

Group Theory and Chemistry

Group Theory and Chemistry Group Theory and Chemistry Outline: Raman and infra-red spectroscopy Symmetry operations Point Groups and Schoenflies symbols Function space and matrix representation Reducible and irreducible representation

More information

Group Theory and Molecular Symmetry

Group Theory and Molecular Symmetry Group Theory and Molecular Symmetry Molecular Symmetry Symmetry Elements and perations Identity element E - Apply E to object and nothing happens. bject is unmoed. Rotation axis C n - Rotation of object

More information

Deriving character tables: Where do all the numbers come from?

Deriving character tables: Where do all the numbers come from? 3.4. Characters and Character Tables 3.4.1. Deriving character tables: Where do all the numbers come from? A general and rigorous method for deriving character tables is based on five theorems which in

More information

3. Assign these molecules to point groups a. (9) All of the dichlorobenzene isomers Cl

3. Assign these molecules to point groups a. (9) All of the dichlorobenzene isomers Cl CE 610 Spring 2007 Problem Set 1 key Total points = 200 1. (10 points) What distinguishes transition metals from the other elements in the periodic table? List five physical or chemical features that are

More information

All About the Chair Conformation

All About the Chair Conformation All About the Chair Conformation Background Before we begin, here are some terms to know: 1. Conformation: the shape that a molecule can adopt due to rotation around one or more single bonds 2. Angle strain:

More information

18 electron rule : How to count electrons

18 electron rule : How to count electrons 18 electron rule : How to count electrons The rule states that thermodynamically stable transition metal organometallic compounds are formed when the sum of the metal d electrons and the electrons conventionally

More information

Copyright Page 1

Copyright Page 1 1 Introduction to Bonding: Two-Dimensional Lewis tructures Key Terms: abbreviated electron configuration combines the inert, noble core electrons (abbreviated via a noble element) with the outermost or

More information

Polyatomic Molecular Orbital Theory

Polyatomic Molecular Orbital Theory Polyatomic Molecular Orbital Theory Transformational properties of atomic orbitals When bonds are formed, atomic orbitals combine according to their symmetry. Symmetry properties and degeneracy of orbitals

More information

Symmetric Stretch: allows molecule to move through space

Symmetric Stretch: allows molecule to move through space BACKGROUND INFORMATION Infrared Spectroscopy Before introducing the subject of IR spectroscopy, we must first review some aspects of the electromagnetic spectrum. The electromagnetic spectrum is composed

More information

PROTEINS THE PEPTIDE BOND. The peptide bond, shown above enclosed in the blue curves, generates the basic structural unit for proteins.

PROTEINS THE PEPTIDE BOND. The peptide bond, shown above enclosed in the blue curves, generates the basic structural unit for proteins. Ca 2+ The contents of this module were developed under grant award # P116B-001338 from the Fund for the Improvement of Postsecondary Education (FIPSE), United States Department of Education. However, those

More information

VSEPR Model. The Valence-Shell Electron Pair Repulsion Model. Predicting Molecular Geometry

VSEPR Model. The Valence-Shell Electron Pair Repulsion Model. Predicting Molecular Geometry VSEPR Model The structure around a given atom is determined principally by minimizing electron pair repulsions. The Valence-Shell Electron Pair Repulsion Model The valence-shell electron pair repulsion

More information

Adding vectors We can do arithmetic with vectors. We ll start with vector addition and related operations. Suppose you have two vectors

Adding vectors We can do arithmetic with vectors. We ll start with vector addition and related operations. Suppose you have two vectors 1 Chapter 13. VECTORS IN THREE DIMENSIONAL SPACE Let s begin with some names and notation for things: R is the set (collection) of real numbers. We write x R to mean that x is a real number. A real number

More information

In part I of this two-part series we present salient. Practical Group Theory and Raman Spectroscopy, Part I: Normal Vibrational Modes

In part I of this two-part series we present salient. Practical Group Theory and Raman Spectroscopy, Part I: Normal Vibrational Modes ELECTRONICALLY REPRINTED FROM FEBRUARY 2014 Molecular Spectroscopy Workbench Practical Group Theory and Raman Spectroscopy, Part I: Normal Vibrational Modes Group theory is an important component for understanding

More information

Chapter 9. Chemical reactivity of molecules depends on the nature of the bonds between the atoms as well on its 3D structure

Chapter 9. Chemical reactivity of molecules depends on the nature of the bonds between the atoms as well on its 3D structure Chapter 9 Molecular Geometry & Bonding Theories I) Molecular Geometry (Shapes) Chemical reactivity of molecules depends on the nature of the bonds between the atoms as well on its 3D structure Molecular

More information

The Lewis electron dot structures below indicate the valence electrons for elements in Groups 1-2 and Groups 13-18

The Lewis electron dot structures below indicate the valence electrons for elements in Groups 1-2 and Groups 13-18 AP EMISTRY APTER REVIEW APTER 7: VALENT BNDING You should understand the nature of the covalent bond. You should be able to draw the Lewis electron-dot structure for any atom, molecule, or polyatomic ion.

More information

Math 215 HW #6 Solutions

Math 215 HW #6 Solutions Math 5 HW #6 Solutions Problem 34 Show that x y is orthogonal to x + y if and only if x = y Proof First, suppose x y is orthogonal to x + y Then since x, y = y, x In other words, = x y, x + y = (x y) T

More information

AN INTRODUCTION TO SPARTAN

AN INTRODUCTION TO SPARTAN AN INTRODUCTION TO SPARTAN OBJECTIVE This tutorial introduces a number of basic operations in Spartan Student required for molecule manipulation, property query and spectra and graphics display. Specifically

More information

CHAPTER 9 COVALENT BONDING: ORBITALS. Questions

CHAPTER 9 COVALENT BONDING: ORBITALS. Questions APTER 9 VALET BDIG: RBITALS Questions 9. In hybrid orbital theory, some or all of the valence atomic orbitals of the central atom in a molecule are mixed together to form hybrid orbitals; these hybrid

More information

7.14 Linear triatomic: A-----B-----C. Bond angles = 180 degrees. Trigonal planar: Bond angles = 120 degrees. B < B A B = 120

7.14 Linear triatomic: A-----B-----C. Bond angles = 180 degrees. Trigonal planar: Bond angles = 120 degrees. B < B A B = 120 APTER SEVEN Molecular Geometry 7.13 Molecular geometry may be defined as the three-dimensional arrangement of atoms in a molecule. The study of molecular geometry is important in that a molecule s geometry

More information

Suggested solutions for Chapter 3

Suggested solutions for Chapter 3 s for Chapter PRBLEM Assuming that the molecular ion is the base peak (00% abundance) what peaks would appear in the mass spectrum of each of these molecules: (a) C5Br (b) C60 (c) C64Br In cases (a) and

More information

Inorganic Chemistry with Doc M. Day 3. Covalent bonding: Lewis dot structures and Molecular Shape.

Inorganic Chemistry with Doc M. Day 3. Covalent bonding: Lewis dot structures and Molecular Shape. Inorganic Chemistry with Doc M. Day 3. Covalent bonding: Lewis dot structures and Molecular Shape. Topics: 1. Covalent bonding, Lewis dot structures in review, formal charges 2. VSEPR 6. Resonance 3. Expanded

More information

ORGANIC COMPOUNDS IN THREE DIMENSIONS

ORGANIC COMPOUNDS IN THREE DIMENSIONS (adapted from Blackburn et al., Laboratory Manual to Accompany World of hemistry, 2 nd ed., (1996) Saunders ollege Publishing: Fort Worth) Purpose: To become familiar with organic molecules in three dimensions

More information

Bonding Models. Bonding Models (Lewis) Bonding Models (Lewis) Resonance Structures. Section 2 (Chapter 3, M&T) Chemical Bonding

Bonding Models. Bonding Models (Lewis) Bonding Models (Lewis) Resonance Structures. Section 2 (Chapter 3, M&T) Chemical Bonding Bonding Models Section (Chapter, M&T) Chemical Bonding We will look at three models of bonding: Lewis model Valence Bond model M theory Bonding Models (Lewis) Bonding Models (Lewis) Lewis model of bonding

More information

5.04 Principles of Inorganic Chemistry II

5.04 Principles of Inorganic Chemistry II MIT OpenourseWare http://ocw.mit.edu 5.4 Principles of Inorganic hemistry II Fall 8 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 5.4, Principles of

More information

C has 4 valence electrons, O has six electrons. The total number of electrons is 4 + 2(6) = 16.

C has 4 valence electrons, O has six electrons. The total number of electrons is 4 + 2(6) = 16. 129 Lewis Structures G. N. Lewis hypothesized that electron pair bonds between unlike elements in the second (and sometimes the third) row occurred in a way that electrons were shared such that each element

More information

Solving Spectroscopy Problems

Solving Spectroscopy Problems Solving Spectroscopy Problems The following is a detailed summary on how to solve spectroscopy problems, key terms are highlighted in bold and the definitions are from the illustrated glossary on Dr. Hardinger

More information

TRANSITION METALS AND COORDINATION CHEMISTRY

TRANSITION METALS AND COORDINATION CHEMISTRY CHAPTER TWENTY-ONE TRANSITION METALS AND COORDINATION CHEMISTRY For Review 1. Chromium ([Ar]:4s 0 3d 5 ) and copper [Ar]:4s 1 3d 10 ) have electron configurations which are different from that predicted

More information

Laboratory 11: Molecular Compounds and Lewis Structures

Laboratory 11: Molecular Compounds and Lewis Structures Introduction Laboratory 11: Molecular Compounds and Lewis Structures Molecular compounds are formed by sharing electrons between non-metal atoms. A useful theory for understanding the formation of molecular

More information

What Are the Shapes of Molecules?

What Are the Shapes of Molecules? Lab 7 Name What Are the Shapes of Molecules? Pre-Lab Assignment Read the entire lab handout. There is no written pre-lab assignment for this lab. Learning Goals Derive the Lewis structure of a covalent

More information

Communicated March 31, 1951 CHR CHR CHR. *H* zz '" *H _ 0.-...H / k C,.. CHR CNR CHR CHR CHR *HN/' N 'H_N/' H_./ - H-(H.

Communicated March 31, 1951 CHR CHR CHR. *H* zz ' *H _ 0.-...H / k C,.. CHR CNR CHR CHR CHR *HN/' N 'H_N/' H_./ - H-(H. VOL. 37, 1951 CHEMISTR Y: PA ULING AND COREY 251 THE PLEATED SHEET, A NEW LAYER CONFIGURATION OF POL YPEPTIDE CHAINS BY LINUS PAULING AND ROBERT B. COREY GATES AND CRELLIN LABORATORIES OF CHEMISTRY,* CALIFORNIA

More information

2. If C is the midpoint of AB and B is the midpoint of AE, can you say that the measure of AC is 1/4 the measure of AE?

2. If C is the midpoint of AB and B is the midpoint of AE, can you say that the measure of AC is 1/4 the measure of AE? MATH 206 - Midterm Exam 2 Practice Exam Solutions 1. Show two rays in the same plane that intersect at more than one point. Rays AB and BA intersect at all points from A to B. 2. If C is the midpoint of

More information

MODERN ATOMIC THEORY AND THE PERIODIC TABLE

MODERN ATOMIC THEORY AND THE PERIODIC TABLE CHAPTER 10 MODERN ATOMIC THEORY AND THE PERIODIC TABLE SOLUTIONS TO REVIEW QUESTIONS 1. Wavelength is defined as the distance between consecutive peaks in a wave. It is generally symbolized by the Greek

More information

EXPERIMENT ONE COVALENT BONDING AND MOLECULAR MODELS. In order to draw proper Lewis structures chemists use two rules,

EXPERIMENT ONE COVALENT BONDING AND MOLECULAR MODELS. In order to draw proper Lewis structures chemists use two rules, EXPERIMENT ONE COVALENT BONDING AND MOLECULAR MODELS Today you will use ball-and-stick molecular model kits to better understand covalent bonding. You will figure out the structures of several different

More information

Molecular Geometry and VSEPR We gratefully acknowledge Portland Community College for the use of this experiment.

Molecular Geometry and VSEPR We gratefully acknowledge Portland Community College for the use of this experiment. Molecular and VSEPR We gratefully acknowledge Portland ommunity ollege for the use of this experiment. Objectives To construct molecular models for covalently bonded atoms in molecules and polyatomic ions

More information

Visualizing Molecular Orbitals: A MacSpartan Pro Experience

Visualizing Molecular Orbitals: A MacSpartan Pro Experience Introduction Name(s) Visualizing Molecular Orbitals: A MacSpartan Pro Experience In class we have discussed Lewis structures, resonance, VSEPR, hybridization and molecular orbitals. These concepts are

More information

The Clar Structure of Fullerenes

The Clar Structure of Fullerenes The Clar Structure of Fullerenes Liz Hartung Massachusetts College of Liberal Arts June 12, 2013 Liz Hartung (Massachusetts College of Liberal Arts) The Clar Structure of Fullerenes June 12, 2013 1 / 25

More information

Molecular Structures. Chapter 9 Molecular Structures. Using Molecular Models. Using Molecular Models. C 2 H 6 O structural isomers: .. H C C O..

Molecular Structures. Chapter 9 Molecular Structures. Using Molecular Models. Using Molecular Models. C 2 H 6 O structural isomers: .. H C C O.. John W. Moore onrad L. Stanitski Peter. Jurs http://academic.cengage.com/chemistry/moore hapter 9 Molecular Structures Stephen. oster Mississippi State University Molecular Structures 2 6 structural isomers:

More information

SHAPE, SPACE AND MEASURES

SHAPE, SPACE AND MEASURES SHPE, SPCE ND MESURES Pupils should be taught to: Understand and use the language and notation associated with reflections, translations and rotations s outcomes, Year 7 pupils should, for example: Use,

More information

5. Structure, Geometry, and Polarity of Molecules

5. Structure, Geometry, and Polarity of Molecules 5. Structure, Geometry, and Polarity of Molecules What you will accomplish in this experiment This experiment will give you an opportunity to draw Lewis structures of covalent compounds, then use those

More information

Molecular Models Experiment #1

Molecular Models Experiment #1 Molecular Models Experiment #1 Objective: To become familiar with the 3-dimensional structure of organic molecules, especially the tetrahedral structure of alkyl carbon atoms and the planar structure of

More information

10. Geometry Hybridization Unhybridized p atomic orbitals

10. Geometry Hybridization Unhybridized p atomic orbitals APTER 14 VALET BDIG: RBITALS The Localized Electron Model and ybrid rbitals 9. The valence orbitals of the nonmetals are the s and p orbitals. The lobes of the p orbitals are 90E and 180E apart from each

More information

B) atomic number C) both the solid and the liquid phase D) Au C) Sn, Si, C A) metal C) O, S, Se C) In D) tin D) methane D) bismuth B) Group 2 metal

B) atomic number C) both the solid and the liquid phase D) Au C) Sn, Si, C A) metal C) O, S, Se C) In D) tin D) methane D) bismuth B) Group 2 metal 1. The elements on the Periodic Table are arranged in order of increasing A) atomic mass B) atomic number C) molar mass D) oxidation number 2. Which list of elements consists of a metal, a metalloid, and

More information

Organic Chemistry Tenth Edition

Organic Chemistry Tenth Edition Organic Chemistry Tenth Edition T. W. Graham Solomons Craig B. Fryhle Welcome to CHM 22 Organic Chemisty II Chapters 2 (IR), 9, 3-20. Chapter 2 and Chapter 9 Spectroscopy (interaction of molecule with

More information

THREE DIMENSIONAL GEOMETRY

THREE DIMENSIONAL GEOMETRY Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,

More information

ABERLINK 3D MKIII MEASUREMENT SOFTWARE

ABERLINK 3D MKIII MEASUREMENT SOFTWARE ABERLINK 3D MKIII MEASUREMENT SOFTWARE PART 1 (MANUAL VERSION) COURSE TRAINING NOTES ABERLINK LTD. EASTCOMBE GLOS. GL6 7DY UK INDEX 1.0 Introduction to CMM measurement...4 2.0 Preparation and general hints

More information

Ch a p t e r s 8 a n d 9

Ch a p t e r s 8 a n d 9 Ch a p t e r s 8 a n d 9 Covalent Bonding and Molecular Structures Objectives You will be able to: 1. Write a description of the formation of the covalent bond between two hydrogen atoms to form a hydrogen

More information

SYMMETRY MATHEMATICAL CONCEPTS AND APPLICATIONS IN TECHNOLOGY AND ENGINEERING

SYMMETRY MATHEMATICAL CONCEPTS AND APPLICATIONS IN TECHNOLOGY AND ENGINEERING Daniel DOBRE, Ionel SIMION SYMMETRY MATHEMATICAL CONCEPTS AND APPLICATIONS IN TECHNOLOGY AND ENGINEERING Abstract: This article discusses the relation between the concept of symmetry and its applications

More information

Angles that are between parallel lines, but on opposite sides of a transversal.

Angles that are between parallel lines, but on opposite sides of a transversal. GLOSSARY Appendix A Appendix A: Glossary Acute Angle An angle that measures less than 90. Acute Triangle Alternate Angles A triangle that has three acute angles. Angles that are between parallel lines,

More information

12B The Periodic Table

12B The Periodic Table The Periodic Table Investigation 12B 12B The Periodic Table How is the periodic table organized? Virtually all the matter you see is made up of combinations of elements. Scientists know of 118 different

More information

Solving Simultaneous Equations and Matrices

Solving Simultaneous Equations and Matrices Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering

More information

1 Symmetries of regular polyhedra

1 Symmetries of regular polyhedra 1230, notes 5 1 Symmetries of regular polyhedra Symmetry groups Recall: Group axioms: Suppose that (G, ) is a group and a, b, c are elements of G. Then (i) a b G (ii) (a b) c = a (b c) (iii) There is an

More information

Digital Image Processing. Prof. P.K. Biswas. Department of Electronics & Electrical Communication Engineering

Digital Image Processing. Prof. P.K. Biswas. Department of Electronics & Electrical Communication Engineering Digital Image Processing Prof. P.K. Biswas Department of Electronics & Electrical Communication Engineering Indian Institute of Technology, Kharagpur Lecture - 27 Colour Image Processing II Hello, welcome

More information

Covalent Bonding & Molecular Orbital Theory

Covalent Bonding & Molecular Orbital Theory Covalent Bonding & Molecular Orbital Theory Chemistry 754 Solid State Chemistry Dr. Patrick Woodward Lecture #16 References - MO Theory Molecular orbital theory is covered in many places including most

More information

Computer Graphics Prof. Sukhendu Das Dept. of Computer Science and Engineering Indian Institute of Technology, Madras Lecture 7 Transformations in 2-D

Computer Graphics Prof. Sukhendu Das Dept. of Computer Science and Engineering Indian Institute of Technology, Madras Lecture 7 Transformations in 2-D Computer Graphics Prof. Sukhendu Das Dept. of Computer Science and Engineering Indian Institute of Technology, Madras Lecture 7 Transformations in 2-D Welcome everybody. We continue the discussion on 2D

More information

2.1. The Formation of Ionic and Covalent Bonds. Clues in Naturally Occurring Compounds SECTION. Key Terms

2.1. The Formation of Ionic and Covalent Bonds. Clues in Naturally Occurring Compounds SECTION. Key Terms SETI 2.1 The Formation of Ionic and ovalent Bonds Key Terms octet rule ionic bond ionic compound covalent bond molecular compound single bond double bond triple bond bonding pair lone pair Lewis structure

More information

Please note: use the ionic model unless asked otherwise and comment on any complexes that

Please note: use the ionic model unless asked otherwise and comment on any complexes that 1. For all complexes listed below, determine a) metal oxidation state b) total number of electrons contributed from metal c) total number of electrons contributed from the ligand set d) total electron

More information

Molecular Symmetry 1

Molecular Symmetry 1 Molecular Symmetry 1 I. WHAT IS SYMMETRY AND WHY IT IS IMPORTANT? Some object are more symmetrical than others. A sphere is more symmetrical than a cube because it looks the same after rotation through

More information

Chapter 10 Molecular Geometry and Chemical Bonding Theory

Chapter 10 Molecular Geometry and Chemical Bonding Theory Chem 1: Chapter 10 Page 1 Chapter 10 Molecular Geometry and Chemical Bonding Theory I) VSEPR Model Valence-Shell Electron-Pair Repulsion Model A) Model predicts Predicts electron arrangement and molecular

More information

Drawing, Representation, and Projection of Chemical Formulae.

Drawing, Representation, and Projection of Chemical Formulae. Drawing, Representation, and Projection of hemical Formulae. In order to learn and communicate organic chemistry you must learn the language of organic chemistry. Pictures play an important role in this

More information

Introduction to Group Theory with Applications in Molecular and Solid State Physics

Introduction to Group Theory with Applications in Molecular and Solid State Physics Introduction to Group Theory with Applications in Molecular and Solid State Physics Karsten Horn Fritz-Haber-Institut der Max-Planck-Gesellschaft 3 84 3, e-mail horn@fhi-berlin.mpg.de Symmetry - old concept,

More information

Section 1.1. Introduction to R n

Section 1.1. Introduction to R n The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to

More information

AP CHEMISTRY CHAPTER REVIEW CHAPTER 6: ELECTRONIC STRUCTURE AND THE PERIODIC TABLE

AP CHEMISTRY CHAPTER REVIEW CHAPTER 6: ELECTRONIC STRUCTURE AND THE PERIODIC TABLE AP CHEMISTRY CHAPTER REVIEW CHAPTER 6: ELECTRONIC STRUCTURE AND THE PERIODIC TABLE You should be familiar with the wavelike properties of light: frequency ( ), wavelength ( ), and energy (E) as well as

More information

CHEM1002 Worksheet 4: Spectroscopy Workshop (1)

CHEM1002 Worksheet 4: Spectroscopy Workshop (1) CHEM1002 Worksheet 4: Spectroscopy Workshop (1) This worksheet forms part of the Spectroscopy Problem Solving Assignment which represents 10% of the assessment of this unit. You should use the support

More information

An introduction to molecular symmetry

An introduction to molecular symmetry 44 An introduction to molecular symmetry 4.2 igure 4.1 or answer 4.2: the principal axis of rotation, and the two mirror pianes in H^O. (a) E is the identity operator. It effectively identifies the molecular

More information

Star and convex regular polyhedra by Origami.

Star and convex regular polyhedra by Origami. Star and convex regular polyhedra by Origami. Build polyhedra by Origami.] Marcel Morales Alice Morales 2009 E D I T I O N M O R A L E S Polyhedron by Origami I) Table of convex regular Polyhedra... 4

More information

Chapter 3 Applying Your Knowledge- Even Numbered

Chapter 3 Applying Your Knowledge- Even Numbered Chapter 3 Applying Your Knowledge- Even Numbered 2. Elements in a specific compound are always present in a definite proportion by mass; for example, in methane, CH 4, 12 g of carbon are combined with

More information

3.4 Covalent Bonds and Lewis Structures

3.4 Covalent Bonds and Lewis Structures 3.4 Covalent Bonds and Lewis Structures The Lewis Model of Chemical Bonding In 1916 G. N. Lewis proposed that atoms combine in order to achieve a more stable electron configuration. Maximum stability results

More information

EXPERIMENT 14: COMPARISONS OF THE SHAPES OF MOLECULES AND IONS USING MODELS

EXPERIMENT 14: COMPARISONS OF THE SHAPES OF MOLECULES AND IONS USING MODELS EXPERIMENT 14: CMPARISNS F TE SAPES F MLECULES AND INS USING MDELS PURPSE Models of various molecules and ions will be constructed and their shapes and geometries will be compared. BACKGRUND LEWIS STRUCTURES

More information

The Lewis structure is a model that gives a description of where the atoms, charges, bonds, and lone pairs of electrons, may be found.

The Lewis structure is a model that gives a description of where the atoms, charges, bonds, and lone pairs of electrons, may be found. CEM110 Week 12 Notes (Chemical Bonding) Page 1 of 8 To help understand molecules (or radicals or ions), VSEPR shapes, and properties (such as polarity and bond length), we will draw the Lewis (or electron

More information

Lecture 34: Symmetry Elements

Lecture 34: Symmetry Elements Lecture 34: Symmetry Elements The material in this lecture covers the following in Atkins. 15 Molecular Symmetry The symmetry elements of objects 15.1 Operations and symmetry elements 15.2 Symmetry classification

More information

CHAPTER 6: THE PERIODIC TABLE

CHAPTER 6: THE PERIODIC TABLE CHAPTER 6: THE PERIODIC TABLE Problems to try in the textbook. Answers in Appendix I: 5,9,13,15,17,19,21,25,27,29,31,33,35,41,43,45,47,49,55abcde,57,59,61,63,65,67,69,71,73,75,89,91 6.1 CLASSIFICATION

More information

Role of Hydrogen Bonding on Protein Secondary Structure Introduction

Role of Hydrogen Bonding on Protein Secondary Structure Introduction Role of Hydrogen Bonding on Protein Secondary Structure Introduction The function and chemical properties of proteins are determined by its three-dimensional structure. The final architecture of the protein

More information

Theoretical Methods Laboratory (Second Year) Molecular Symmetry

Theoretical Methods Laboratory (Second Year) Molecular Symmetry Theoretical Methods Laboratory (Second Year) DEPARTMENT OF CHEMISTRY Molecular Symmetry (Staff contacts: M A Robb, M J Bearpark) A template/guide to the write-up required and points for discussion are

More information

Chapter 18 Symmetry. Symmetry of Shapes in a Plane 18.1. then unfold

Chapter 18 Symmetry. Symmetry of Shapes in a Plane 18.1. then unfold Chapter 18 Symmetry Symmetry is of interest in many areas, for example, art, design in general, and even the study of molecules. This chapter begins with a look at two types of symmetry of two-dimensional

More information

CHAPTER 10 THE SHAPES OF MOLECULES

CHAPTER 10 THE SHAPES OF MOLECULES ATER 10 TE AE MLEULE 10.1 To be the central atom in a compound, the atom must be able to simultaneously bond to at least two other atoms. e,, and cannot serve as central atoms in a Lewis structure. elium

More information

discuss how to describe points, lines and planes in 3 space.

discuss how to describe points, lines and planes in 3 space. Chapter 2 3 Space: lines and planes In this chapter we discuss how to describe points, lines and planes in 3 space. introduce the language of vectors. discuss various matters concerning the relative position

More information

Solution Set To draw the correct Lewis diagrams, we use the rules on page

Solution Set To draw the correct Lewis diagrams, we use the rules on page Solution Set 1 9.21 To draw the correct Lewis diagrams, we use the rules on page 272... (a) ICl: Both atoms are halogens (Group 7A) with seven valence electrons, and are inclined to complete their octet

More information

2. Atoms with very similar electronegativity values are expected to form

2. Atoms with very similar electronegativity values are expected to form AP hemistry Practice Test #6 hapter 8 and 9 1. Which of the following statements is incorrect? a. Ionic bonding results from the transfer of electrons from one atom to another. b. Dipole moments result

More information

Geometry Chapter 1 Vocabulary. coordinate - The real number that corresponds to a point on a line.

Geometry Chapter 1 Vocabulary. coordinate - The real number that corresponds to a point on a line. Chapter 1 Vocabulary coordinate - The real number that corresponds to a point on a line. point - Has no dimension. It is usually represented by a small dot. bisect - To divide into two congruent parts.

More information

CHAPTER 10: COORDINATION CHEMISTRY II: BONDING

CHAPTER 10: COORDINATION CHEMISTRY II: BONDING 138 Chapter 10 Coordination Chemistry II: Bonding CHAPTER 10: COORDIATIO CHEMISTRY II: BODIG 10.1 a. Tetrahedral d 6, 4 unpaired electrons b. [Co(H 2 O) 6 ] 2+, high spin octahedral d 7, 3 unpaired electrons

More information

Figure 1.1 Vector A and Vector F

Figure 1.1 Vector A and Vector F CHAPTER I VECTOR QUANTITIES Quantities are anything which can be measured, and stated with number. Quantities in physics are divided into two types; scalar and vector quantities. Scalar quantities have

More information

Health Science Chemistry I CHEM-1180 Experiment No. 15 Molecular Models (Revised 05/22/2015)

Health Science Chemistry I CHEM-1180 Experiment No. 15 Molecular Models (Revised 05/22/2015) (Revised 05/22/2015) Introduction In the early 1900s, the chemist G. N. Lewis proposed that bonds between atoms consist of two electrons apiece and that most atoms are able to accommodate eight electrons

More information

Chapter 2 The Chemical Context of Life

Chapter 2 The Chemical Context of Life Chapter 2 The Chemical Context of Life Multiple-Choice Questions 1) About 25 of the 92 natural elements are known to be essential to life. Which four of these 25 elements make up approximately 96% of living

More information

Hybrid Atomic Orbitals

Hybrid Atomic Orbitals Hybrid Atomic Orbitals These materials were adapted from Prof. George Bodner, Purdue University (http:// chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/hybrid.html#geom; excerpted 08/25/2011). This

More information

Periodic Table of the Elements

Periodic Table of the Elements Periodic Table of the Elements Where did it come from? 1869 Demitri Mendeleev Russian chemist who discovered a pattern to the elements Wrote properties on cards Arranged cards according to properties

More information

12-1 Representations of Three-Dimensional Figures

12-1 Representations of Three-Dimensional Figures Connect the dots on the isometric dot paper to represent the edges of the solid. Shade the tops of 12-1 Representations of Three-Dimensional Figures Use isometric dot paper to sketch each prism. 1. triangular

More information

3. What would you predict for the intensity and binding energy for the 3p orbital for that of sulfur?

3. What would you predict for the intensity and binding energy for the 3p orbital for that of sulfur? PSI AP Chemistry Periodic Trends MC Review Name Periodic Law and the Quantum Model Use the PES spectrum of Phosphorus below to answer questions 1-3. 1. Which peak corresponds to the 1s orbital? (A) 1.06

More information

Chapter 1 Benzene Blues 27

Chapter 1 Benzene Blues 27 hapter 1 Benzene Blues 27 The ybridization Model of Atoms in Molecules An important question facing chemists about 80 years ago, was, ow does one go from recently invented atomic orbitals to rationalizing

More information

Assignment 7 and Practice Third Exam Solutions

Assignment 7 and Practice Third Exam Solutions Assignment 7 and Practice Third Exam Solutions 1. Draw the structures of the following molecules, and don t forget their lone pairs! (a) thiocyanide anion, SC Isoelectronic analogies: S is like, and is

More information

Two hydrogen atoms join together to attain the helium Noble gas configuration by sharing electrons and form a molecule.

Two hydrogen atoms join together to attain the helium Noble gas configuration by sharing electrons and form a molecule. Lecture Simple Molecular Orbitals - Sigma and Pi Bonds in Molecules 1 An atomic orbital is located on a single atom. When two (or more) atomic orbitals overlap to make a bond we can change our perspective

More information

Solid State Theory Physics 545

Solid State Theory Physics 545 Solid State Theory Physics 545 CRYSTAL STRUCTURES Describing periodic structures Terminology Basic Structures Symmetry Operations Ionic crystals often have a definite habit which gives rise to particular

More information

Infrared Spectroscopy 紅 外 線 光 譜 儀

Infrared Spectroscopy 紅 外 線 光 譜 儀 Infrared Spectroscopy 紅 外 線 光 譜 儀 Introduction Spectroscopy is an analytical technique which helps determine structure. It destroys little or no sample (nondestructive method). The amount of light absorbed

More information

We know a formula for and some properties of the determinant. Now we see how the determinant can be used.

We know a formula for and some properties of the determinant. Now we see how the determinant can be used. Cramer s rule, inverse matrix, and volume We know a formula for and some properties of the determinant. Now we see how the determinant can be used. Formula for A We know: a b d b =. c d ad bc c a Can we

More information

The Fundamentals of Infrared Spectroscopy. Joe Van Gompel, PhD

The Fundamentals of Infrared Spectroscopy. Joe Van Gompel, PhD TN-100 The Fundamentals of Infrared Spectroscopy The Principles of Infrared Spectroscopy Joe Van Gompel, PhD Spectroscopy is the study of the interaction of electromagnetic radiation with matter. The electromagnetic

More information

1.3 STRUCTURES OF COVALENT COMPOUNDS

1.3 STRUCTURES OF COVALENT COMPOUNDS 1.3 STRUTURES OF OVALENT OMPOUNDS 13 1.9 Draw an appropriate bond dipole for the carbon magnesium bond of dimethylmagnesium. Explain your reasoning. 3 Mg 3 dimethylmagnesium 1.3 STRUTURES OF OVALENT OMPOUNDS

More information

ATOMIC THEORY. Name Symbol Mass (approx.; kg) Charge

ATOMIC THEORY. Name Symbol Mass (approx.; kg) Charge ATOMIC THEORY The smallest component of an element that uniquely defines the identity of that element is called an atom. Individual atoms are extremely small. It would take about fifty million atoms in

More information

521493S Computer Graphics. Exercise 2 & course schedule change

521493S Computer Graphics. Exercise 2 & course schedule change 521493S Computer Graphics Exercise 2 & course schedule change Course Schedule Change Lecture from Wednesday 31th of March is moved to Tuesday 30th of March at 16-18 in TS128 Question 2.1 Given two nonparallel,

More information