Molecular Symmetry PROBLEMS

Size: px

Start display at page:

Download "Molecular Symmetry PROBLEMS"

Madison Fletcher
6 years ago
Views:

4 Molecular Symmetry PROBLEMS 4. A student sets up the coordinates for NH3 rather unconventionally, as shown on the right. In this coordinate system, the x-axis is the three-fold axis.

By considering the transformation of a general point in space, deduce the six 3 3 matrices which form a set of reducible representations for the C3v point group. 4.

1 4 Molecular Symmetry PROBLEMS 4. A student sets up the coordinates for NH3 rather unconventionally, as shown on the right. In this coordinate system, the x-axis is the three-fold axis. One of the σ -planes, σb, makes an angle of 45 with the y-axis. By considering the transformation of a general point in space, deduce the six 3 3 matrices which form a set of reducible representations for the C3v point group. 4. There are eight equivalent lobes for the fxyz orbital. They point toward the corners of a cube with the origin as its center and have alternating signs, as shown in Fig. A below. The cube is orientated in such a way that x-, y-, and z-axis pass through the centers of the six faces, as shown in Fig. B below. Rotation of fxyz orbital about z-axis by 45 results in fz(x y ) orbital. The top views of fxyz and fz(x y ) orbitals are shown below in Figs. C and D, respectively. Assign these two orbitals to their proper irreducible representation in point groups Cv and D4h. A 64 B C D

2 Problems 4.3 The square pyramidal [InCl5 ] anion has C4v symmetry. Of the four vertical mirror planes, one σv is in the xz-plane, the other σv in the yz-plane, and the two σd s lie in between them. (i) With reference to the coordinate system based on the orthonormal base vectors e, e, and e 3 (pointing in the x-, y-, and z-directions, respectively), write down all eight matrices that represent the symmetry operations of the C4v group. (ii) Deduce the character of Γe, the representation based on {e, e, e 3 }. Reduce Γe to the irreducible representations of C4v. REFERENCE: D. S. Brown, F. W. B. Einstein, and D. G. Tuck, Tetragonal-pyramidal indium(iii) species. Crystal structure of tetraethylammonium pentachloroindate(iii). Inorg. Chem. 8, 4 8 (969). 4.4 Point groups Sn, have only two symmetry elements: E and Sn. Explain why n must be an even integer and n Trans-N F has Ch symmetry and its coordinates are taken as those shown below. The character table of the Ch point group is shown below. Ch E C i σh Ag Bg Au Bu (i) To which representations do the Cartesian coordinates x, y, and z belong? (ii) To which representations do the binary products x, y, z, xy, xz, and yz belong? 65

3 Molecular Symmetry 4.6 In a reference book, the character table of the D3d point group has the following form: D3d Ag Ag Eg Au Au Eu E C 3 3C i S6 3σ d (i) To which representations do the Cartesian coordinates x, y, and z belong? (ii) Taking the z-axis as the C3 axis, to which representations do the binary products x, y, z, xy, xz, and yz belong? 4.7 (i) Borane dianions (BH) adopt structures with high symmetry. For instance, n closo-dodecaborane() dianion, (BH), has the structure of an icosahedron, a polyhedron with vertices and faces. The symmetry point group for this structure is Ih, which has symmetry operations: E, C5, C5, C3, C, i, S, S3, S6, 5σ. The structure of (BH) is shown below (left), where each vertex is occupied by a BH group. When one or more BH group(s) in (BH) is (are) substituted by C (which is isoelectronic to BH), carborane dianions Cm (BH) m are obtained. The only possible mono- and di-substituted closo-dodecaborane() dianions are displayed below [(a) to (d), where C atoms are represented by black dots]. Determine the symmetry point group of each carborane dianion shown. (ii) Fullerene C has the structure of a dodecahedron, a polyhedron with vertices and faces. The symmetry point group for this structure is also Ih. The structure of C is shown below (left), where each vertex is occupied by a C atom. When one or more C atom(s) of C is (are) replaced by Si, substituted fullerenes Sim C m are obtained. The only possible mono- and di-substituted C with the 66

4 Problems general formula Si m C m (m = or ) are displayed below [(a) to (f), where Si atoms are represented by black dots]. Determine the symmetry point group of each substituted fullerene shown. 4.8 Determine the point group of each of the following silicate anions: (i) [SiO 4 ] 4 (ii) [O 3 Si O SiO 3 ] 6 with linear Si O Si linkage. Consider both staggered and eclipsed conformation. (iii) [Si 3 O 9 ] 6 (iv) [Si 4 O ] 8 (v) [Si 6 O 8 ] The structures for the last three species are shown below. The (SiO) n (n =3,4,6)ringsare planar. 67

5 Molecular Symmetry 4.9 Assign point group symbols for the following cyclic S n molecules (n = 6, 7, 8,,,, 3, 8 in α and β forms, ): (i) S 6 ;(ii)s 7 ; (iii) S 8 ;(iv)s ;(v)s ;(vi)s ;(vii)s 3 ; (viii) α S 8 ;(ix)β S 8 ;(x)s. 4. (i) Give the point groups and chiral nomenclature for the isomeric forms of tris(ethylenediamine)cobalt(iii) illustrated below: (ii) List the remaining possible configurations of [Co(en) 3 ]

6 Problems A note on the nomenclature of chiral metal complex: First we view the optically active tris(chelate) complexes down the three-fold rotational axis. If the helix viewed in this way is left-handed, it is called a -isomer. Its mirror image is the -isomer. The naming is illustrated in the figure on the right. Besides the asymmetry due to the tris(chelate) structure of octahedral complexes, asymmetry may also be generated in the ligands such as ethylenediamine. The enantiomeric gauche conformations may be left-handed (represented by λ ) or right-handed (represented by δ ), which are shown in the figure on the right. Thus, one of the possible complete notations showing the configuration of tris(ethylenediamine)cobalt(iii) is -Mλλλ, where M denotes metal. 4. There are five possible conformations for cyclohexane-,4-dione: a chair form, two boat forms, a symmetric twist form, and a twist form of lower symmetry generally referred to as the twist-boat form. Sketch all five distinct conformations and identify their point groups. REFERENCE: C.-S. Tse, D. Y. Chang, K.-Y. Law, and T. C. W. Mak, The crystal and molecular structure of trans-,5-di-p-bromobenzyl-,5-diethoxycarbonylcyclohexane,4-dione. Acta Cryst. B3, 6 9 (976). 4. The structure of ethane, C H6, is shown on the right. Consider the fluoro-substituted ethanes with the general formula C H6 n Fn, n =,,..., 6. Draw members of this series that have the following symmetries: (i) C ; (ii) C ; (iii) Cs ; (iv) Ch ; (v) C3v ; (vi) D3d. 4.3 Cyclohexane molecule (C6 H, shown on the right) has D3d symmetry. Consider the chloro-substituted cyclohexanes with the general formula C6 Hm Cln, with n =,,... and m + n =. Write the structural formulas for members from the C6 Hm Cln series that have the following symmetries: (i) C ; (ii) C ; (iii) Cs ; (iv) Ch ; (v) C3v. 69

7 Molecular Symmetry 4.4 Elemental sulfur, S 8, has the crown structure shown below (left). Viewed from the top, this structure becomes an octagon. This octagon is also shown below (right), where the + sign denotes atoms that are above the paper plane and the sign indicates atoms below. All eight atoms are labeled in the figures. (i) (ii) What is the symmetry point group of S 8? Also, if one of the sulfur atoms in S 8 is replaced by (the isoelectronic) oxygen, the hypothetical compound S 7 Oisformed. What is the point group of S 7 O? When four of the sulfur atoms in S 8 are replaced by oxygen atoms, another hypothetical compound, S 4 O 4, is formed. Clearly there are several isomers for S 4 O 4.Write down which sulfur atoms are to be replaced to obtain the isomers which have the following symmetries: (a) C ;(b)c ;(c)c s ;(d)c 4v ;(e)d d. 4.5 Cyclooctatetraene, C 8 H 8, has the tub structure shown below (left). This molecule belongs to the D d group, whose symmetry operations are easily visualized as in the top view displayed below (right). Here + signs indicate atoms that are above the plane of the paper and the signs indicate atoms below. Consider the halogen-substituted cyclooctatetraenes with general formula C 8 H 8 n X n, n =,,,...,8.WritedownwhichHatomsaretobereplacedtoobtaintheisomerswhich have the following symmetries: (i) C ;(ii)c s ; (iii) C ;(iv)c v ;(v)s 4 ;(vi)d. REFERENCE: W.-K. Li, Identification of molecular point groups. J. Chem. Educ. 7, (993). 7

8 Problems 4.6 Shown on the right is the structure of metal cluster Mo 6 Cl 4+ 8, with the six Mo atoms forming an octahedron and the eight Cl atoms placed above the eight faces of the octahedron. This cluster has O h symmetry. Note that Cl atoms a, b, c, d, e, f, g,andhare above faces 35, 45, 46, 36, 35, 45, 46, and 36, respectively. (i) If the following Mo atom(s) is (are) replaced by the isoelectronic W and/or Cr atom(s), determine the symmetry point group of the substituted clusters: (a)=w;(b)==w;(c)3=5=w;(d)==3=w;(e)=3=5=w; (f)=wand=cr;(g)3=wand5=cr;(h)=5=wand=4=cr;(i)== Wand3=4=Cr. (ii) Replace one or more Cl atoms with F atoms so that the resulting cluster will belong to the following point groups: (a) C v ;(b)c s ;(c)c 4v ;(d)d h ;(e)c 3v ;(f)d 3d ; (g) T d ;(h)c. You only need to give the labels of the Cl atoms (a to h) replaced. 4.7 Shown on the right is the structure of metal cluster Ta 6 Cl +, with the six Ta atoms forming an octahedron and the twelve Cl atoms placed above the twelve edges of the octahedron. This cluster has O h symmetry. Replace one or more Cl atoms with F atoms so that the resulted cluster will belong to the following point groups: (i) C 4v ;(ii)d 4h ; (iii) C s ;(iv)c ;(v)c v ; (vi) C ;(vii)c h ; (viii) C 3v ;(ix)c 3 ;(x)d 3d ;(xi)d 3 ; (xii) D h ; (xiii) D d ;(xiv)d. You only need to give the labels of the Cl atoms (a to l) replaced. 4.8 The structure of P 4 O is shown on the right. In this highly symmetrical and pretty structure, the four P atoms form a perfect tetrahedron. Each P atom is linked to a terminal oxygen (with a double bond); these O atoms are labeled a to d in the figure. The remaining six oxygens, with labels e to j,arebridging atoms, located above the edges of the P 4 tetrahedron. So P 4 O has T d symmetry. 7

9 Molecular Symmetry Replace one or more O atoms in P 4 O with S atoms so that the resulting molecules will have the symmetries given below. Also, you need to observe the condition imposed. You only need to give the label(s) of the O atoms replaced. Do not draw the structures of your answers. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv) D d [replacing only bridging O atom(s)]; D d [replacing both bridging and terminal O atom(s)]. C 3v [replacing only terminal O atom(s)]; C 3v [replacing only bridging O atom(s)]; C 3v [replacing both terminal and bridging O atom(s), but each P atom is bonded to only either terminal S atom or bridging S atom(s), and not both]; C v [replacing only terminal O atom(s)]; C v [replacing only bridging O atom(s)]; C v [replacing both terminal and bridging O atom(s), but each P atom is bonded to only either terminal S atom or bridging S atom(s), and not both]; C s [replacing only bridging O atom(s)]; C s [replacing both bridging and terminal O atom(s)]; C [replacing only bridging O atom(s)]; C [replacing both bridging and terminal O atom(s)]; C [replacing ONE terminal O atom and bridging O atom(s)]; C [replacing TWO terminal O atoms and bridging O atom(s)]. 4.9 Adamantane, C H 6, has the structure shown below. This molecule has T d symmetry and two types of carbon atoms. There are four methine carbons (each of which is bonded to one hydrogen with label a, h, k, orn) which occupy the corners of a tetrahedron. In addition, there are six methylene carbons (each of which is bonded to two hydrogens) which are located above the edges of the aforementioned tetrahedron. All 6 hydrogens are labeled in the figure. 7

10 Problems Consider the chloro-substituted adamantanes with the general formula C H 6 n Cl n, n =,,,...,6.Drawmembersofthisseriesthathavethefollowingsymmetries:(i)C ; (ii) C ; (iii) C 3 ;(iv)c s ;(v)c v; (vi) C 3v ;(vii)d d. Provide only one answer for each part. In the answers, only list the label(s) of the hydrogen atom(s) to be replaced. REFERENCE: W.-K. Li, Identification of molecular point groups. J. Chem. Educ. 7, (993). 4. In a classic paper on optically active coordination compounds, Mills and Quibell reported the synthesis of a Pt complex containing one molecule of meso-stilbenediamine (H N CHPh CHPh NH ) and one molecule of isobutylenediamine (H N CH CMe NH ). Also, the complex was successfully resolved into its enantiomers. Determine whether the coordination around the metal ion is tetrahedral or square planar. Also illustrate these modes pictorially. REFERENCE: W. H. Mills and T. H. H. Quibell, The configuration of the valencies of 4-covalent platinum: the optical resolution of meso-stilbenediaminoisobutylenediaminoplatinous salts. J. Chem. Soc (935). 4. Consider the series of octahedral complexes with the general formula [MA m B n (XX) p (XY) q ], where m, n, p, q =,,,..., m + n +p +q =6,AandB are monodentate ligands such as Cl and F, XX is a bidentate ligand with two identical donating groups, and XY is a bidentate ligand with two different donating groups. An example of XX is H NCH CH NH, and an example of XY is (CH 3 ) NCH CH NH. Write the structural formulas for members from this series that have the symmetry requested below and that meet the following requirements. (i) C, with at least one monodentate ligand; (ii) C, with no monodentate ligand; (iii) C, with at least one monodentate ligand; (iv) C, with no monodentate ligand; (v) C 3 ;(vi)c s ;(vii)c v ; (viii) C h ;(ix)d h ;(x)c 3v ;(xi)d 3 ;(xii)c 4v ; (xiii) D 4h. REFERENCE: W.-K. Li, Identification of molecular point groups. J. Chem. Educ. 7, (993). 4. Draw all geometrical isomers of [Co(trien)Cl ] +, where trien is the tetradentate ligand triethylenetetramine which may be conveniently represented by N N N N in your drawings. Also, ignoring the conformation of chelate rings, identify the symmetry point group of each isomer and predict whether it is optically active or not. 73

11 Molecular Symmetry SOLUTIONS A4. E =, 3, C3 = 3 3 C, σb =, 3 = 3 3 σc = σb C3 = σa = C 3 σb =, 3 E C (z) σ v (xz) σ v (yz) Γ (fxyz ) = A Γ (fz(x y ) ) = A A4. Cv D4h E C4 (z) C C C i Γ (fxyz ) Γ (fz(x y ) ). 3 3 S4 σh σv σd Thus, Γ (fxyz ) transforms as Bu ; Γ (fz(x y ) ) transforms as Bu. A4.3 (i) For the symmetry operations σ va, σ vb, σ da and σ db, β =, 9, 45, and 45, respectively. E: C34 : σ va : 74 C4 : C : σ vb :

12 Solutions σ da : (ii) C 4v Γe σ db : E C4 C4 σv σd 3 = A + E. A4.4 If n is an odd number, Snn generates an additional operation σ perpendicular to the Sn axis and hence group Sn becomes Cnh. So n cannot be odd. In addition, S = i and hence the point groups S and Ci are equivalent. To sum up: S = Cs, S = Ci, and S3 = C3h. Hence n must be equal to or larger than 4. A4.5 Ch σh E C i Ag Bg Au Bu x, y, z, xy xz, yz z x, y Reasoning: (i) Functions x, y, z must have u symmetry. It is easily seen that z is symmetric with respect to C, while both x and y are antisymmetric with respect to C. (ii) All the binary products must have g symmetry. Also: Γ (z ) = Γ (z) Γ (z) = Au Au = Ag ; Γ (x ) = Γ (y ) = Γ (xy) = Bu Bu = Ag ; Γ (xz) = Γ (yz) = Au Bu = Bg. A4.6 D 3d E C3 3C Ag Ag Eg Au S6 3σd Au z Eu (x, y) i x + y, z (x y, xy); (xz, yz) 75

13 Molecular Symmetry Reasoning: (i) It is straightforward to obtain Γ (z) by applying all the operations on z. On the other hand, each of x and y is neither symmetric nor antisymmetric with respect to C3, so they must form a degenerate set. Since they have u symmetry, they form an Eu set. (ii) Γ (z ) = Γ (z) Γ (z) = Au Au = Ag. Since x + y = r z and Γ (r ) = Γ (z ) = Ag, Γ (x + y ) is also Ag. Γ (xz, yz) = Γ (x, y) Γ (z) = Eu Au = Eg. None of the remaining x y and xy is symmetric or antisymmetric with respect to C3, so they also form a degenerate set. Since they have g symmetry, they form another Eg set. A4.7 (i) (a) C5v ; (b) Cv ; (c) Cv ; (d) D5d. (ii) (a) C3v ; (b) Cv ; (c) Cs ; (d) C ; (e) Cv ; (f) D3d. A4.8 (i) Td. (ii) D3d for staggered conformation and D3h for eclipsed. (iii) D3h. (iv) D4h. (v) D6h. A4.9 (i) D3d ; (ii) Cs ; (iii) D4d ; (iv) D ; (v) C ; (vi) D3d ; (vii) C ; (viii) Ch ; (ix) Ch ; (x) D4. A4. (i) (a) -Mδδδ, D3 ; (b) -Mλλλ (more stable than -Mδδδ form by about 7.5 kj mol ), D3 ; (c) -Mδδλ, C ; (d) -Mλδδ, C. (ii) -Mλλλ, -Mδδδ, -Mλλδ, and -Mδλλ, which are enantiomers of (a), (b), (c), and (d), respectively. A4. 76

14 Solutions A4. A4.3 A4.4 (i) S 8 : D 4d ;S 7 O: C s. (ii) (a) abdf or abce;(b) abcd or abdg;(c)abde or abcf ;(d) aceg;(e) abef. A4.5 (i) a;(ii) ab; (iii) ah;(iv) abef ;(v) aceg;(vi)adeh. A4.6 (i) (a) C 4v ;(b)d 4h ;(c)c v ;(d)c v ;(e)c 3v ;(f)c 4v ;(g)c s ;(h)c ;(i)d h. (ii) (a) ab, ac;(b) abc, abg;(c) abcd;(d) abgh;(e)a;(f) ag;(g) acfh;(h) abdf. 77

15 Molecular Symmetry A4.7 (i) abcd; (ii) efgh; (iii) ab; (iv) ace; (v) e; (vi) aj; (vii) abkl; (viii) abe; (ix) abcehj; (x) abegkl;(xi) chj;(xii) ak; (xiii) acjl;(xiv) acegjl. A4.8 (i) ej; (ii) abcdej; (iii) a; (iv) efg; (v) ahij; (vi) ab; (vii) e; (viii) abj; (ix) ef ;(x) aef ; (xi) efi;(xii) abfi; (xiii) aeh;(xiv)abfj. A4.9 (i) af;(ii) bm; (iii) bdf;(iv)b;(v)de;(vi)a;(vii)gfij. A4. Square planar (optically active, as found by Mills and Quibell) Tetrahedral (may show slight optical activity since the chelate rings are non-planar) A4. There are other possible answers, in addition to the ones given below. 78

16 Solutions A4. 79

Electronic Spectra and Magnetic Properties of Transition Metal Complexes)

Subject Chemistry Paper No and Title Module No and Title Module Tag Paper 7: Inorganic Chemistry-II (Metal-Ligand Bonding, Electronic Spectra and Magnetic Properties of Transition Metal 22: Isomerism part