An introduction to molecular symmetry

Transcription

1 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 configuration. The operator E leaves the molecule unchanged. All objects can be operated upon by the identity operator E. (b) A plane of symmetry (mirror plane) is denoted by G. (c) The symmetiy operation of rotation about anra-foldaxis (the symmetry element) is denoted by C,,, in which the angle of rotation is 360y«, where H - 1,2,3,4... (d) An S^ axis is an n-fold improper rotation axis: rotation through-360 M about the axis is followed by reflection through a plane perpendicular to the same axis. If the mirror plane lies perpendicular to the principal axis, it is denoted a^.. If the plane contains the principal axis, it is denoted o;. To see the difference between o^ and cr,', it is best to use an examplei'-'h^o is a simple example. igure 4.1 shows the principal (Cj) axis in a molecule of HjO^.and the two mirror planes, both of which contain the principal rotation axis. The plane which bisects the molecule is labelled cr,, and the plane in which the molecule lies is labelled o^'. A <J^ plane contains the principal rotation axis, and also bisects the angle between two adjacent 2-fold axes. 4.3 (a) An 8"Vertex star. The symmetry of the star is such that rotation thi'ough (360/8) - 43 gives another star superimposable on the first one; the * is used in the diagram to clarify the rotation that has occurred. This operation is repeated 7 more times to get back to the first orientation. The highestorder axis is an 8-fold axis (Cg) running through the centre of the star, perpendicular to the plane of the paper. (b) An ellipse. The symmetry is such that rotation thi'ough (360/2) ^ 180" as shown in the diagram gives another ellipse superimposable on the first one. This operation is repeated once more to get back to the first orientation. The highest-order rotation axis of the ellipse is a 2-fold axis (C2). There are 2 other Cj axes: both lie in the plane of the paper, one horizontal, and one vertical with respect to the ellipse drawn. (c) A pentagon. The symmetry is such that rotation through (360/5) = TZ'' gives another pentagon superimposable on the first one. This operation is repeated 4 more times to get back to thefirstorientation. The principal axis is a 5-fold axis (Cg). (d) The symmetry of this shape is such that rotation through (360/3) = 120 ^ves a shape superimposable on the first one. This operation is repeated twice more to get back to thefirstorientation. The principal axis is a 3-fold axis (C3). Rotate through 180 Rotate through

2 An introduction to molecular symmetry SOj is a bent molecule (see answer 2.4b). It must possess an E operator - all molecules do. The other symmetry operators are: A Cj axis: A o" plane A a' plane: The symmetry operators are E, C_, &' 'md a'. 4.5 igure 2.1 in H&S shows one view of the Hp^ molecule. Another 2 views are shown here in structures 4.8 and 4.9. In 4.9, the molecule is viewed along the 0-0 bond. The symmetry operator that Kp^ possesses is a C^ axis running through the midpoint of the 0-0 bond in the direction shown in the left-hand diagram below. Its operation is shown in the right-hand diagram below; look carefully at the perspective in the diagrams. (4.8) (4.9) H(l) H(2) HC2) ^H(l) 4.6 The diagrams below summarize the answer. ci (4.10) (4.11) EachC7^ plane contains one B- bond. Cl (4.12) Br 4.7 (a) On going from B3 to BClj (4.10 to 4.11): the C3 axis is lost, two C^ axes are lost, and two o; planes are lost. (Colour the left-hand atom in the diagrams above to signify a change to Cl, and confirm which of the symmetry operators are no longer valid). (b) On going from BCI2 to BBrCl (4.11 to 4.12): C^ axis is lost, and the a^ plane is lost. (Use the diagrams above again to help you confirm these). (c) Bach molecule has a c7^ plane ~ the plane containing the atoms.

3 46 An introduction to moleculac-^ymmetry 4.8 irst, draw out the structm-es of the molecules and ions: See Chapters 15 and 16 for representations of the bonding in SO3, [SO^]^- and [NO3]" so that each atom obeys the octet rule 4.9 As above, the structure drawn for [SOJ^" emphasizes its symmetry and does not illustrate a bonding picture Br' H H Br O'^ ^ O Cl Br C! A! "Cl 0 c) v:"o \ ^ 0 The structures of [SOJ^- and [NO3]- are drawn so as to emphasize the equivalence of the bonds (delocalized bonding), and therefore the symmetry; see margin-note. te; (i) each possesses a C3 axis;. j,--^ _ (ii) only the planar molecules possess a a^, plane. The answers are therefore: (a) C3 axis but no a^^ plane: NH3, PBr3, [SO^". (b) C3 axis and a ctj^ plane: SO3, AICI3, [NO3]-. irst, draw out the structures of the molecules and ions. Use the VSEPR model to show why [ICIJ- and Xe, are square planar (2 lone pairs and 4 bonding pairs): cr Cl C! Cl O 2 Si);/,, c) ^^0 '"'^Xe-'" ' The only two species with a Q axis and a <J^^ plane are the square planar ones: [ICy-andXe^. c.,o^>' (4.13) ^ (4.14) 4.10 or each, first draw out the structure (using VSEPR to help you) and then work out the number of mirror planes. (a) S^: the structure is disphenoidal (4.13). S^ contains 2 mirror planes as shown in 4.14: one plane contains the S and 2 (axial) atoms, and the other contains the S and 2 (equatorial) atoms. (b) HjS: the bent structure is analogous to that of HjO (although the bond angle is different) and so the 2 mirror planes are as shown in igure 4.1. (c) Sg: octahedral structure (6 bonding pairs, 0 lone pairs). There are 9 mirror planes. The 3 Oj, planes are shown in 4.15, and each contains the S and 4 atoms. The 6 (7d planes can be considered in 3 sets of 2., One set is shown in Each plane contains the S and two opposite atoms (i.e. it contains a Q axis), and bisects the other two -S- axes. The other 2 sets of a^ planes are similarly constructed, starting with a different Q axis (i.e. -S- axis). (4.15) (4.16) (d) SO^: the structure is disphenoidal (igure 2.2, p. 21). SO^ contains 2 mirror planes, analogous to those of S^ (4.14).

4 An introduction to molecular symmetry 47 (e) SO : this has a bent structure and so has the same symmetry as H,S in part (b) it has 2 mirror planes. ^ ^ (f) SO3: the stmcture is trigonal planar (see answer 2.17i, p. 20). There are 4 mirror planes: one <y, and 3 a^ planes. Their positions are analogous to those in B answer 4,6. ^' Questions : general notes A centre of inversion: If reflection of all parts of a molecule through the centre of the rnolecule produces an mdistmguishable configuration, the centre is a centre of inversion (centre of symmetry), designated by the symboh-., j.;-,. (4.17) (4.18) (b)astaggeredconformer(4.17) will be the most favoured in terms of steric energy s^uitult^^^^^^^^ ; ''f 'f' ^^" ^"^^"^ "^^"^^- ^ p-^i^ ^^o-^n structu e To confirm that this is acentre of inversion, reflect each point of the moeculehrough cen^e.-andshowthatan identical pointis generated. or example, take H atom a... reflect through i... you end up at H atom a' which is indistinguishable from atom a. (d) The eclipsed confoit^er 4.18 is the least favoured in terms of steric energy (e) This confonner does not contain a centre of inversion. To confinn this try a ing the mi Joint of the Si-Si bond again as in part (c). Reflection of^l pss of the molecule through this pomt does not lead to an indistinguishable configuration. ^ (4.19) \ ' (4.20) 4.12 l"-chpartofthisanswer,ifyouareunsureofthereasonwhyamoleculedpesor andtlr'.' '""'''''f''^''^' ^^^^ '^' -«"tral atom as a trial inversion centre, and try reflecting parts of the molecule through this point. (a) E3: B, group 13; 3 valence electrons; trigonal planar; structure 4 19 inversion centre. (b) Si,: Si, group 14; 4 valence electrons; tetrahedral; stmcture inversion centre. (c)_ Xe,: Xe, group 18; 8 valence electrons; square planar (4 bonding and 2 lone pairs); stmcture The Xe centre lies on an inversion centre. i n v 5 o n c f "^ ^^' ^ "^^''''^ ^^e^trons; trigonal bipyramidal; structure (e) EXe ]-: Xe, group 18; 8 valence electrons; pentagonal planar (5 bonding and 2 lone pairs); stmcture inversion centre. (f) Sg: S, group 16; 6 valence electrons; octahedral; stmcture 4.24 The S centre lies on an inversion centre. (g)c3. planar smacture 4.25; this is related to ethene. There is an inversion centre at the midpoint of the C=C bond. ""::x^f'»>»'> / (4.21) (4.22) Xe / \ (4.23) """"s---^ (4.24) n p (4.25)

5 48 An introduction to molecular symmetry \ (h) The C=C=C unit is linear, but the presence of two adjacent 7S-honds places a ^^^^^,, H restriction on the orientation of the two CHj groups: they are orthogonal (4,26). ""^ H Therefore, there is no inversion centre. (4.26) 4.13 A linear molecule possesses an oo-fold axis of rotation. This applies to both symmetrical (e.g. j, 4.27) and asymmetrical (e.g. HCN, 4.28) molecules. (4.27) (4.28) Questions : general notes Before attempting questions , make sure that you'have studied worked examples in H&S, in which point group assignments are made with accompanying explanations. When reading through answers , ensure that you have igure 4.10 from H&S available for reference. This gives a flowchart for assigning point groups. -N,,,,,^ (4.29) 4.14 N3; trigonal pyramidal structure To determine the point group, apply the strategy shown in igure 4.10 in H&S: START C Is the moiecule linear? Does it have T^, O^ or 4 symmetry? Are there 3 C^ axes perpendicular Is there a cr^ plane? Are there n (i.e. 3) cr^ planes containing the C^ axis? Conclusion: the point group is Cg^. Yes: C3 axis Yes [=> STOP 4.15 A member of the D^^ point group must contain a C^ axis and, therefore, the species is linear. See answer ^'"..g 4.16 SjCl: structure This is an example of a molecule that we loosely call 'octahedral', but which does not possess octahedral symmetry, i.e. it does not belong to the O^ point group. ' ci To determine the point group, apply the strategy in igure 4.10 in H&S: (4.30) START I Is the molecule linear? c\ (4.31: C ' Does it have T^, 0,^ or 4 symmetry? Are there 4 Cj axes perpendicular Is there a a^ plane? Are there n (i.e. 4) o; planes containing the C^ axis? Conclusion: the point group is C^^. Yes: Q (see 4.31) Yes c:^ STOP mm 1

6 An introduction to molecular symmetry 49 a If Br3 were trigonal planar or trigonal pyramidal, the molecule would have a C axis. The fact that it belongs to the q, point group means that the principal axis is' a C2 axis. The other 3-coordinate structure possibility is T-shaped, so the next step in the answer is to work out the symmetry properties of a T-shaped BrR molecule Apart from the E operator, T-shaped Br3 contains a C^ axis, a a plane, and a a ' plane as shown in igure 4.2. These are consistent with the C^. point group. ' igure 4.2 or answer 4.17: the principal axis of rotation, and the two mirror planes in Brg. w see if this agrees with the VSEPR model: Central atom is Br [ Br is in group 17; number of valence electrons =^ 7 Number of bonding pairs (3 Br- bonds) = 3 Number of lone pairs = 2 Total number of electron pairs = 5 (4.32) 'Parent' shape = trigonal bipyramidal Molecular shape ^ T-shaped, see structure 4.32 a 4.18 The structure of [Xe^]- is shown in igure 4.3. To determine the point group, apply the strategy shown in igure 4.10 in H&S: igure4.3 oranswer4.18:the pentagonal planar structure of [Xeg]-. One of the 5 Coaxes is shown; each of the other 4 C^ axes coincides with a different Xe- bond vector. START! Is the molecule linear? Does it have 7^, O^ or 4 symmetry? Is there a C,^ axis? Are there 5 C^ axes perpendicular Is there a a^ plane perpendicular Conclusion: the point group is D... Yes: C5 (perpendicular to plane of paper) Yes (igure 4.3) Yes n:^ STOP 4.19 TMs question is best done by first grouping compounds as follows, taking into account the number of different substituents- (a)ccl,and(e)c, (b)ccl3and(d)ca3 (c)ccl22 Ci Take (a) and (e) together: each is regular tetrahedral and belongs to the r. point group Take (b) and (d) together: molecules 4.33 and 4.34 belong to the same point group X/ which is assigned as follows: ^ ^ i-. CI '"' ' cr ci START cnj) Is the molecule linear? (4.33) (4.34) Does it have 7^, O,, or 4 synametry? Yes: C3 axis Are there 3 C^ axes perpendicular Is there a o^ plane? Are there n (i.e. 3) o^ planes containing the C,, axis? Conclusion: the point group is C^^. Yes l=l> STOP

7 50 An introduction to molecular symmetry CI ^ (4.36) / (4.35) 4.20,,,»' (4.37) The a^ planes are shown in structure 4.14, p. 46 (c) CCl^2- ^^^ structure 4.35 in which the orientation has been chosen so as to help you in finding the symmetry elements. To assign the point group: START [I=> Is the naolecule linear? Does it have T^, O^ or 1^ symmetry? Yes: C^ axis (horizontal through C in 4.35) Are there 2 C^ axes perpendicular Is there a c^ plane? Are there n (i.e. 2) a^ planes containing ' " the Cj axis? Yes ST0P,, Conclusion: the point group is C^^. (a) To assign a point group to S^, first draw the structure (see answer 3.24, p. 36) whichis shown in START Is the molecule linear? Does it have T^, O^^ or 7,^ symmetry? Is there a C,, axis? Are there 2 Cj axes perpendicular Is there ac7^ plane? Are there n (i.e. 2) a^ planes containing the q axis? Yes: C^ axis (horizontal through S in 4.36) Yes (UNSTOP Conclusion: the point group is 3^. (b) The structures of SO^ and S^ are related ~ compare 4.37 with The presence of the O atom does not change the symmetry. The O atom lies on the Cj axis. The point group is still C^^ The tetrahedron, octahedron and icosahedron are drawn below: Tetrahedron Octahedroo Icosahedron The highest symmetry is possessed by the icosahedron; the /, point group possesses the highest number of symmetry elements. Questions : general notes or a molecule containing n atoms, the number of degrees of vibrational freedom can be determined by using the equations: Linear molecule: Number of degrees of vibrational freedorn ~3n~5 n-linear molecule: Number of degrees of vibrational freedom = 3n~6 or a vibrational mode to be IR active, the vibration must give rise to a change in the molecular dipole moment.