Molecular Symmetry and Group Theory - Fall 2003 | CHEM 3510, Study notes of Physical Chemistry

Download Molecular Symmetry and Group Theory - Fall 2003 | CHEM 3510 and more Study notes Physical Chemistry in PDF only on Docsity! CHEM 3510 Fall 2003 Molecular Symmetry and Group Theory – The symmetry properties of molecules can be used to: – Reduce the high-order secular determinants in Hückel method – Determine the IR or Raman activity of vibrational normal modes – Label and designate molecular orbitals – Derive selections rules for spectroscopic transitions – The symmetry of a molecule can be described in terms of its symmetry elements. Each symmetry element has symmetry operations (one or more) associated with them. Symmetry elements Symmetry operations Examples Description Symbol Symbol Description Identity E Ê No change n-Fold axis of symmetry nC nĈ Rotation about the axis by 360/n degrees 2C 3C 4C Plane of symmetry (mirror plane)      d h v σ σ σ σ      d h v σ σ σ σ ˆ ˆ ˆ ˆ Reflection through a plane Center of symmetry i î Reflection through the center of symmetry n-Fold rotation reflection axis of symmetry (improper rotation) nS nŜ Rotation about the axis by 360/n degrees followed by reflection through a plane perpendicular to the axis – The axis with the highest n is called principal axis. – The planes of symmetry can be: vσ : the plane of symmetry is parallel to a unique axis or to a principal axis hσ : the plane of symmetry is perpendicular to a unique axis or to a principal axis dσ : the plane of symmetry bisects the angle between C axes that are perpendicular to a principal axis ( 2 dσ is a special type of a vσ plane) CHEM 3510 Fall 2003 – A symmetry element may have more than one symmetry operation associated with it. – The 3-fold axis (C ) has two symmetry operations C (rotation with 3 3ˆ 1203 360 = degrees) and (rotation with 240 degrees). 33 2 3 ˆˆˆ CC C= – The 4-fold axis (C ) has three symmetry operations C (rotation with 4 4ˆ 90 4 360 = 4 3 4 ˆˆˆ CCC = degrees), (rotation with 180 degrees), and (rotation with 270 degrees). 44 2 4 ˆˆˆ CCC = 44Ĉ – A set of symmetry operations constitutes a point group. – Each point group consists of a number of symmetry elements. Examples of common point groups of interest to chemists. (The number in front of a symmetry element is the number of times such a symmetry element occurs.) Point group Symmetry elements Molecular examples C2v E, C2, 2σv C3v E, C3, 3σv NH3, C2h E, C2, i, σh D2h E, 3C2, i, 3σv D3h E, C3, 3C2, σh, S3, 3σv D4h E, C4, 4C2, i, S4, σh, 2σv, 2σd D6h E, C6, 3C2, i, S6, σh, 3σv, 3σd D2d E, S4, 3C2, 2σd Td E, 4C3, 3C2, 3S4, 6σd – Examples: Cnv ⇒ n-fold axis and n σv mirror planes Cnh ⇒ n-fold axis and a mirror plane perpendicular to the n-fold axis Dn ⇒ n-fold axis and n 2-fold axes perpendicular to n-fold axis Dnh ⇒ n-fold axis and n 2-fold axes perpendicular to n-fold axis plus a plane perpendicular to n-fold axis Dnd ⇒ same as Dn plus n-dihedral mirror planes – The total number of symmetry operations (which may be greater than the total number of symmetry elements) is called the order of the point group. 9 © eo eee 2:28 8e @ ra re ld 4d Ht /¢ PQ lne* 5 a wn CHEM 3510 Fall 2003 – The set of symmetry operations of a molecule form a point group. – A group is a set of entities (A, B, C…) that satisfy certain requirements: 1). Combining (multiplying) any 2 members of the group gives a member of the group. (“A group must be closed under multiplication.”) 2). The multiplication must be associative: A (B C) = (A B) C 3). The set of entities (members of the group) contains an identity element E such that: E A = A E (E B = B E) 4). For every entity in the group (for example A) there is an inverse (A–1) that is also a member of the group so that: A A–1 = A–1 A = E (B B–1 = B–1 B = E) – As an example of a group of symmetry operations let’s consider the case of water that has four symmetry operations: Ê , , 2Ĉ vσ̂ , (C2v point group). 'ˆ vσ First Operation Second Operation Ê 2Ĉ vσ̂ 'ˆ vσ Ê Ê 2Ĉ vσ̂ 'ˆ vσ 2Ĉ 2Ĉ Ê 'ˆ vσ vσ̂ vσ̂ vσ̂ 'ˆ vσ Ê 2Ĉ 'ˆ vσ 'ˆ vσ vσ̂ 2Ĉ Ê O HH u – The table above is called the group multiplication table of the C2v point group. – Determine the group multiplication table by looking at a vector. – The four symmetry operations of C2v satisfy the conditions of being a group and are collectively referred to as the point group C2v. – Use NH3 as an example of a molecule whose symmetry properties are described by the C3v point group. The symmetry operations Ê , , , 3Ĉ 2 3Ĉ vσ̂ , , form the C3v point group. 'ˆ vσ ''ˆ vσ First Operation Second Operation Ê 2Ĉ 2 3Ĉ vσ̂ 'ˆ vσ ''ˆ vσ CHEM 3510 Fall 2003 – Symmetry operations can be represented by matrices. – Example for C2v point group (H2O). Consider a vector: u = uxi + uyj + uzk For the 180 degree rotation along z axis (C ): 2ˆ ux= –ux; uy= –uy; uz = uz 2Ĉ 2Ĉ 2Ĉ One can write: where C           =           z y x z y x u u u C u u u C 22ˆ           − − = 100 010 001 2 Similarly: ;           = 100 010 001 E vσ =  ; =          − 100 010 001 ' vσ          − 100 010 001 – A set of matrices that multiply together in the same manner as a group multiplication table is said to be a representation of that group. – These four matrices form a representation of the C2v point group or, more specific, a three-dimensional representation because it consists of 3×3 matrices. – There are an infinite number of such representations, but some of them, called irreducible representations, can be used to express all the others that are called reductible representations. Finding the irreducible representations has been done already for all point groups. – The results for C2v point group are denoted A1, A2, B1, and B2. – Use notation A if the representation is symmetric with respect to principal axis (C2) and B if is antisymmetric with respect to principal axis. – A1 is the totally symmetric one-dimensional irreducible representation. The irreducible representations of the C2v point group. Ê 2Ĉ vσ̂ 'ˆ vσ A1 (1) (1) (1) (1) A2 (1) (1) (–1) (–1) B1 (1) (–1) (1) (–1) B2 (1) (–1) (–1) (1) CHEM 3510 Fall 2003 Mathematical relations involving the characters of irreducible representatives – Define some notations: R̂ is a arbitrary symmetry operation is the character of the matrix representation of )ˆ(Rχ R̂ ) is the character of the jˆ(Rjχ th irreducible representation of R̂ – Relation between the order of the group (h) and the dimension of the irreducible representations ( is the dimension of the jjd th irreducible representation): ∑ = = N j jdh 1 2 – But the dimension of the jth irreducible representation equal to the character of the jth irreducible representation for the identity operation ( ) jj dE =ˆχ ⇒ [ ] hE N j j =∑ = 2 1 )ˆ(χ – The irreducible representations are orthogonal. 0)ˆ()ˆ( ˆ =∑ RR j R i χχ (Mathematically: the product of two vectors is given by u ) ∑ = =⋅ n k kk vu 1 v – Example: A and B representations of group are orthogonal: 1 2 v2C 011)1(1)1(111 )ˆ()ˆ()ˆ()ˆ( )ˆ()ˆ()ˆ()ˆ( 2121 2121 BABA 2B2ABA =×+−×+−×+× =′′+ ++ vvvv CCEE σχσχσχσχ χχχχ ∑ = classes 0)ˆ()ˆ()ˆ( RRRn ji χχ where n is the number of symmetry operations in the class containing R̂ – If i from iχ above is the totally symmetric irreducible representation: 1)ˆ( =Riχ 0)ˆ()ˆ()ˆ( classesˆ == ∑∑ RRnR j R j χχ if 1Aj ≠ CHEM 3510 Fall 2003 – Looking at a n-dimensional irreducible representation like a vector for which one define the length of the vector as ∑ = ==⋅ n k kv 1 22)length(vv ⇒ [ ] [ ]∑ ∑ == R jj hRRnR ˆ classes 22 )ˆ()ˆ()ˆ( χχ – Example: A representation of point group: 2 v3C [ ] ( ) ( ) ( ) 6131211)ˆ()ˆ( 222 classes 2 =−×+×+×=∑ RRn jχ – Combining the two conditions from above: ijjij R i hRRRnRR δχχχχ == ∑∑ classesˆ )ˆ()ˆ()ˆ()ˆ()ˆ( – Reducing a reducible representation Γ as a sum of irreducible representations – Assume that ( )R̂χ is the character of the symmetry operator in reducible representation Γ. Write this ( )R̂χ as a combination of the characters of irreducible representations: ( ) ( )∑= j jj RaR ˆˆ χχ – How to find the coefficients : ja Multiply by (iχ and sum over )R̂ R̂ : 44 344 21 ji j R i j j R i RRaRR = ≠ ∑∑∑ = ifonly 0 ˆˆ )ˆ()ˆ()ˆ()ˆ( χχχχ ijj R i hRR δχχ =∑ )ˆ()ˆ( ˆ ⇒ ∑= R ii RRha ˆ )ˆ()ˆ( χχ ⇒ ∑∑ == classesˆ )ˆ()ˆ()ˆ(1)ˆ()ˆ(1 RRRn h RR h a i R ii χχχχ CHEM 3510 Fall 2003 – Example: Reduce the reducible representation Γ= 4 2 0 2 as a sum of irreducible representations belonging to C group. v2 4)ˆ( =Eχ 2)ˆ( 2 =Cχ 0)ˆ( =vσχ 2)ˆ( =′vσχ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 212101214 4 1 1A =×+×+×+×=a ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 112101214 4 1 2A =−×+−×+×+×=a ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 012101214 4 1 1B =−×+×+−×+×=a ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 112101214 4 1 2B =×+−×+−×+×=a ⇒ 2212 BAA ++=Γ – Verification: 2 A 2×1 2×1 2×1 2×1 1 1 1 –1 –1 2A 1 –1 –1 1 2B Γ 4 2 0 2 Hückel theory for benzene – When using the zp orbitals on various C, Hückel theory leads to: 0 10001 11000 01100 00110 00011 10001 = x x x x x x – Use a combination of following orbitals (linear combination of zp orbitals): )( 6 1 6543211 ψψψψψψφ +++++= )( 6 1 6543212 ψψψψψψφ −+−+−= )22( 12 1 6543213 ψψψψψψφ +−−−+= CHEM 3510 Fall 2003 The use of the symmetry of a molecule to simplify numerical calculations Example: Hückel molecular-orbital calculations for butadiene, trimethylenemethane, and benzene. 1). Write the chemical formula and draw the molecular structure including the 2pz atomic orbitals that are included in the calculation. Butadiene Trimethylenemethane Benzene A). Write the Hückel secular determinant, solve it, and determine the energies of the π-molecular orbitals. 0 100 110 011 001 = x x x x 0 010 010 111 001 = x x x x 0 10001 11000 01100 00110 00011 10001 = x x x x x x 013 24 =+− xx 03 24 =− xx 0496 246 =−+− xxx 2 53± ±=x 0,0,3±=x 2,1,1 ±±±=x βα 618.1−=E βα 618.0−=E βα 618.0+=E βα 618.1+=E βα 3−=E α=E , α=E βα 3+=E βα 2−=E βα −=E , βα −=E βα +=E , βα +=E βα 2+=E CHEM 3510 Fall 2003 Butadiene Trimethylenemethane Benzene B1). Identify the symmetry elements, symmetry operations, and the point group that the molecule belongs to. E , , , 2C i hσ E , , 3 , 3C 2C hσ , , 33S vσ E , , , , 36C 3C 2C 2C′ , 3 2C ′′ , , , , i 6S 3S hσ , 3 dσ , 3 vσ Ê , , , 2Ĉ î hσ̂ Ê , 2 , 3 , 3Ĉ 2Ĉ hσ̂ , 2 , 33Ŝ vσ̂ Ê , 2 , 2 , , 3 , 3 , , 2 , 2 , 6Ĉ 3Ĉ 2Ĉ 2Ĉ′ 2Ĉ ′′ î 6Ŝ 3Ŝ hσ̂ , 3 dσ̂ , 3 vσ̂ h2C h3D h6D B2). Deduce the characters of the reducible representation for the 2pz atomic orbitals that are used in the Hückel molecular-orbital calculation. This is the sum of the diagonal elements of the matrix representing the symmetry operation. It is easier to consider a value of +1 for each 2pz orbital that do not change sign and place during transformation, a value of –1 for each 2pz orbital that do not change place but change sign during transformation, and a value of 0 for the 2pz orbital that change place during transformation. Ê 2Ĉ î hσ̂ Ê 2 3 3Ĉ 2Ĉ hσ̂ 2 33Ŝ vσ̂ Ê 2 2 3 3 2 2 6Ĉ 3Ĉ 2Ĉ 2Ĉ′ 2Ĉ ′′ î 6Ŝ 3Ŝ hσ̂ 3 dσ̂ 3 vσ̂ Γ 4 0 0 –4 Γ 6 0 0 0 –2 0 0 0 0 –6 0 2 CHEM 3510 Fall 2003 B3). Reduce the reducible representation Γ to a sum of irreducible representations. This is done using )ˆ(χ)ˆ(χ1 i ˆ RR h a R i ∑= . =′1Aa 24 1 1 = gA a [6×1 + 2×0×1 + 2×0×1 + 0×1 + 3×(–2)×1 + 3×0×1 + 0×1 + 2×0×1 + 2×0×1 + (–6)×1 + 3×0×1 + 3×2×1] = 0 4 1 = gA a [4×1 + 0×1 + 0×1 +(–4) ×1] = 0 =′2Aa 24 1 2 = uA a [6×1 + 2×0×1 + 2×0×1 + 0×1 + 3×(–2)× (–1) + 3×0×(–1) + 0×(–1) + 2×0×(–1) + 2×0×(–1) + (–6)× (–1) + 3×0×1 + 3×2×1] = 1 4 1 = uA a [4×1 + 0×1 + 0×(–1) +(–4) ×(–1)] = 2 =′Ea 24 1 2 = gB a [6×1 + 2×0×(–1) + 2×0×1 + 0×(–1) + 3×(–2)× (–1) + 3×0×1 + 0×1 + 2×0×(–1) + 2×0×1 + (–6)× (–1) + 3×0×(–1) + 3×2×1] = 1 4 1 = gB a [4×1 + 0×(–1) + 0×1 +(–4) ×(–1)] = 2 =′′1Aa 24 1 1 = gE a [6×2 + 2×0×1 + 2×0×(–1) + 0×(–2) + 3×(–2)×0 + 3×0×0 + 0×2 + 2×0×1 + 2×0×(–1) + (–6)× (–2) + 3×0×0 + 3×2×0] = 1 4 1 = uB a [4×1 + 0×(–1) + 0×(–1) +(–4) ×1] = 0 =′′2Aa 24 1 2 = uE a [6×2 + 2×0×(–1) + 2×0×(–1) + 0×2 + 3×(–2)×0 + 3×0×0 + 0×(–2) + 2×0×1 + 2×0×1 + (–6)× (–2) + 3×0×0 + 3×2×0] = 0 =′′Ea Similarly, 0 1211212 ======= uuuuggg EBBAEBA aaaaaaa Γ = ug AB 22 + Γ = EA ′′+′′22 Γ = ugug EEAB 2122 +++ CHEM 3510 Fall 2003 B6). Use the symmetry orbitals to determine the elements of a new, simpler, Hückel secular determinant. In determining the new secular determinant one should take advantage of the fact that non-diagonal elements are zero between the symmetry orbitals that are bases of different irreducible representations. ααα τψψ τψψ τψψ τφφ =+⋅−= + − == ∫ ∫ ∫ ∫ ∗ ∗ ∗ ∗ )02( 2 1 )ˆ ˆ2 ˆ( 2 1 ˆ 44 41 11 1111 dH dH dH dHH βαβαααααα τψψτψψτψψ τψψτψψτψψτψψ τφφ 2)012( 6 1 )zeroarethattermsother ˆtypeofterm12ˆˆ ˆˆˆˆ( 6 1 ˆ 216655 44332211 1111 +=++++++++= = +++ ++++ == ∫∫∫ ∫∫∫∫ ∫ ∗∗∗ ∗∗∗∗ ∗ L dHdHdH dHdHdHdH dHH 1)1021( 2 1 )2 ( 2 1 4441 11 1111 =+⋅−= + − == ∫∫ ∫ ∫ ∗∗ ∗ ∗ τψψτψψ τψψ τφφ dd d dS 1)00111111( 6 1 )zeroarethattermsother ( 6 1 665544 3322111111 =++++++++= = +++ +++== ∫∫∫ ∫∫∫∫ ∗∗∗ ∗∗∗∗ L τψψτψψτψψ τψψτψψτψψτφφ ddd ddddS And so on for , , H , S , H , , , , , . The other elements are zero. 12H 33S 12S 34H22 34H 22 44H 33 44S And so on for , , , , , , , , etc. Many other elements are zero. 22H 22S 33H 33S 34H 34H 44H 44S xEESH βα =−=− 1111 )2()2(21111 +=+ − =−+=− xEEESH β β αββα CHEM 3510 Fall 2003 B7). Write the Hückel secular determinant, solve it, and determine the energies of the π-molecular orbitals. 0 1100 100 0011 001 = − + x x x x 0 2 00 2 00 003 003 = − − xx xx x x 0 1 2 10000 2 110000 001 2 100 00 2 1100 000020 000002 = − − − − + + + + − + xx xx xx xx x x 0)1)(1( 22 =−−−+ xxxx 0)3( 22 =− xx 0)1()1)(2)(2( 22 =−+−+ xxxx 2 51, 2 51 ±±− =x 0,0,3±=x 2,1,1 ±±±=x βα 618.1−=E βα 618.0−=E βα 618.0+=E βα 618.1+=E βα 3−=E α=E , α=E βα 3+=E βα 2−=E βα −=E , βα −=E βα +=E , βα +=E βα 2+=E