Chimica Inorganica 3
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1 A symmetry operation carries the system into an equivalent configuration, which is, by definition physically indistinguishable from the original configuration. Clearly then, the energy of the system must be the same before and after carrying out the symmetry operation! ˆR H ˆ = H ˆ ˆR [ ˆR, ˆ H ] = ˆR H ˆ ˆ -H ˆR = 0 ˆ H cψ = ch ˆ Ψ ˆ H ˆRΨ i = ˆR H ˆ Ψ i = ˆRE i Ψ i = E ˆRΨi i ˆRΨ i = ±Ψ i By applying each of the operations in the group to an eigenfunction Ψ i belonging to a non degenerate eigenvalue, we generate a representation of the group with each matrix, Γ i (R), equal to ±1. Since the representations are one dimensional, they are irreducible!
2 If E i is k-fold degenerate: ˆ H ˆRΨ i = E i ˆRΨi ˆRΨ i = k r j Ψ ij j=1 For some other operation ŜΨ ij = k m=1 s mj Ψ im k ˆT = Ŝ ˆR ˆTΨi = t m Ψ im Ŝ ˆRΨ i = Ŝ r j Ψ ij = m=1 k j=1 k k j=1 m=1 s mj r j Ψ im t m = k j=1 s mj r j
3 Matrices describing the transformation of a set of k eigenfunctions corresponding to a k- fold degenerate eigenvalue are a k-dimensional irreducible representation for the group Let us consider 2p x and 2p y nitrogen AOs of NH 3. p x = R N 2,1 p y = R N 2,1 ( r)sinθ cosϕ ( r)sinθ sinϕ None of the operation of the C 3v group will affect θ so that: sinθ 1 = sinθ 2 but ϕ 2 = ϕ 1 + 2π 3
4 cosϕ 2 = cos ϕ 1 + 2π 3 sinϕ 2 = sin ϕ 1 + 2π 3 = cosϕ 1 cos 2π 3 sinϕ sin 2π 1 3 = 1 2 cosϕ sinϕ 1 = sinϕ 1 cos 2π 3 + cosϕ sin 2π 1 3 = 1 2 sinϕ cosϕ 1 If we reflect in the xz plane, we have ϕ 2 = ϕ 1 ; cosϕ 2 = cosϕ 1 ; sinϕ 2 = sinϕ 1
5 Ê: ( ) = sinθ 2 cosϕ 2 = sinθ 1 cosϕ 1 = p x ( ) = sinθ 2 sinϕ 2 = sinθ 1 sinϕ 1 = p y Ê p x = Ê sinθ cosϕ 1 1 Ê p y = Ê sinθ sinϕ 1 1 Ĉ 3 : Ĉ 3 p x = Ĉ3 sinθ cosϕ 1 1 Ĉ 3 p y = Ĉ3 sinθ sinϕ 1 1 ˆσ v : ( ) = sinθ 2 cosϕ 2 = sinθ ( ) = sinθ 2 sinϕ 2 = sinθ ( ) = sinθ 2 cosϕ 2 = sinθ 1 cosϕ 1 = p x ( ) = sinθ 2 sinϕ 2 = sinθ 1 sinϕ 1 = p y ˆσ v p x = ˆσ v sinθ 1 cosϕ 1 ˆσ v p y = ˆσ v sinθ 1 sinϕ 1 cosϕ + 3 sinθ 1 1 ( ) = 1 2 p x 3 sinϕ 3 cosθ 1 1 ( ) = 3 2 p x 1 2 p y 2 p y
6 Ê: p x p y = Ê p x p y χ ( Ê ) = 2 Ĉ 3 : ˆσ v : p x p y = Ĉ3 p x p y χ ( ) Ĉ3 = p x p y = s v p x p y χ ( ˆσ v ) = 0 The characters are seen to be those of the E representation of C 3v. Thus we see that the p x and p y orbitals, as a pair, provide a basis for the E representation.
7 The Scalar Product Besides the scalar product between an m 1 n matrix A by an n m 2 matrix B (same n different m values) to give an m 1 m 2 matrix C AB =!! a 1n!!!!! a i1 a i2!! a in " " " " " a m1 1 a m1 2!! a m1 n b 11 b 12! b 1 j! b 1m2!!!!!! b n1 b n2! b nj! b nm2 =! " AB ( ) ij! "
8 A = B = A B = a 21 a 22 b 11 b 12 b 13 b 21 b 22 b 23 b 31 b 32 b 33 B a 21 B B a 22 B = The Direct Product There is a second type of matrix multiplication that we will encounter in applications. It is called the direct product of two matrices and is written A B. We shall define this operation by example b 11 b 12 b 13 b 11 b 12 b 13 b 21 b 22 b 23 b 21 b 22 b 23 b 31 b 32 b 33 b 31 b 32 b 33 a 21 b 11 a 21 b 12 a 21 b 13 a 22 b 11 a 22 b 12 a 22 b 13 a 21 b 21 a 21 b 22 a 21 b 23 a 22 b 21 a 22 b 22 a 22 b 23 a 21 b 21 a 21 b 32 a 21 b 33 a 22 b 31 a 22 b 32 a 22 b 33 In general, the direct product is not commutative. There are no restrictions on the number of rows or columns in either matrix.
9 The Direct Product When A and B are both square (but not necessarily of the same order) the following important result applies: tr( A B) = tr( B A) = tr( A)tr( B) Let us see which are the consequences
10 The Direct Product Let us suppose that R is an operation in the symmetry group of a molecule and X and Y are two sets of functions, which are bases for representations of the group. m ˆRΧ i = x ji Χ j j=1 n ˆRY k = y lk Y l l=1 m n ˆRΧ i Y k = x ji y lk Χ j Y l = z jl,ik Χ j Y l j=1 l=1 m j=1 n l=1 The set of functions X j Y l, the direct product of X j and Y l, also form a representation of the group. The z jl,ik are the elements of a matrix Z of order (mn) (mn). Moreover, it is possible to demonstrate that characters of the representation of a direct product are equal to the product of characters of the representations based on the individual sets of functions
11 χ Z ( R) = z j, j = x jj y = χ X R j m ˆRΧ i Y k = x ji y lk Χ j Y l = z jl,ik Χ j Y l j=1 n l=1 m j=1 n =1 m j=1 n l=1 ( ) χ Y ( R) A 1 A 1 = A 1 ; A 2 A 2 = A 1 ; B 1 B 1 = A 1 ; B 2 B 2 = A 1 A 1 A 2 = A 2 ; A 1 B 1 = B 1 ; A 1 B 2 = B 2 A 1 E = E; A 2 E = E; B 1 E = E; B 2 E = E E 2 = A 1 + A 2 + B 1 + B 2
12 The K h (Kugel group) point group contains an infinite number of C axes and a center of inversion i. It also contains elements generated from these. This is the point group to which a sphere belongs and therefore the point group to which all atoms belong. Although an atom with partially filled orbitals may not be spherically symmetrical, the electronic wave function is classified according to the K h point group The spherically symmetrical point group K h contains all symmetry operations. The pure rotational subgroup K contains all rotations. In either of these groups, a complete set of orbitals is a basis for an irreducible representation of the group. The s orbital is the totally symmetric one-dimensional representation, the p orbitals give a 3 dimensional representation, d orbitals are five-dimensional, etc.
13 The character for rotation by the angle Φ is given by the following equation: ( ) = sin l χ Φ ( )Φ sin Φ 2
14 ( ) = 2l + 1 χ E ( ) = sin l χ Φ χ σ ( )Φ sin Φ 2 ( )π = 1 ( ) = ±sin l ( ) l χ ( C 2 ) ( ) = ± ( 2l + 1) = ±χ ( E) = ( 1) l χ ( E) χ i
15 lϕ m n,l,m l = R n,l P l l ( cosθ ) eim 2π If we carry a rotation of Φ around z R n,l P l m l ϕ ϕ + Φ lϕ ( cosθ ) eim 2π R P m l n,l l ( cosθ ) eim l ( ϕ+φ) 2π The rotation is thus given by the matrix e i Φ e i( 1)Φ e i Φ χ ( Φ) = e ilφ + e i l 1 2l ( )Φ + + e ilφ = e ilφ ( e ) iφ n = sin ( l )Φ n=0 sin Φ 2
16 χ ( Φ) = e i Φ + e i 1 ( 2 )Φ + + e i Φ = e i Φ ( e iφ ) n = sin ( + 1 2)Φ n=0 sin Φ 2 Geometrical series from e -ilφ to e ilφ having path e iφ a + ar + ar ar n 1 + ar n = a rn+1 1 r 1 a = e i Φ ; r = e iφ ; n = 2 e i Φ ( ) 2 +1 eiφ 1 e iφ 1 e i Φ e iφ e i Φ e iφ 1 sin( )Φ sin Φ 2 e i ( 2 +1)Φ = e i Φ = ei +1 1 = e iφ 1 ( )Φ e i Φ = e iφ 1 cosθ = eiθ + e iθ ; sinθ = eiθ e iθ 2 2i cos2θ = ei2θ + e i2θ ; sinθ = ei2θ e i2θ 2 2i ( )Φ e i Φ = e i +1 2 Φ e i 1 2 Φ e i +1 2 Φ i 1 e 2 Φ = e i +1 2ie i1 2 Φ sin Φ e iφ 1 = e i1 2 Φ e i1 2 Φ e i1 2 Φ 1 = 2iei 2 Φ sin Φ 2
17 ( ) = sin χ Φ Φ = π χ C 2 Φ = π 2 ( )Φ sin Φ 2 ( ) = sin χ ( C 4 ) = ( )Φ 1 1 = 0,1, 4, 5 1 = 2,3, 6, 7 = ( 1) The characters of the reducible representation Γ spanned by the five d orbitals (l = 2) in the point group D 4h (we limit our attention to the pure rotational group D 4 ) E 2C 4 C 2 2 C 2 2 C The irreducible components of Γ are ( ) = 1 n a 1 ; n a 2 d z 2 ( ) = 0; n( b 1 ) = 1 ; n b 2 ; n e d x 2 y 2 ( ) = 1 d xy ( ) = 1 d xz,d yz
18 All other groups are subgroups of K h, and all rotational groups are subgroups of K. If we lower the symmetry to O, we find that the representation for the s orbital is A 1 and the representation for the p orbitals in O is T 1. There is no five-dimensional representation, so the representation obtained for the d orbitals is a reducible representation in the O group. It reduces to E + T 2. The representation in K for the f orbitals reduces to A 2 + T 1 + T 2 in the O group.
19 In general Orbitals of even parity (l even) are g and those of odd parity (l odd) are u. Hence, in O h symmetry the representations are s, A 1g ; p, T 1u ; d, E g + T 2g ; and f, A 2u + T 1u + T 2u. We can find the representations for the rotational group using for χ(ω) and add the g or u subscripts for the corresponding centrosymmetric group
20 Why is the direct product so important? ( ) f x dx = 0 if f x ( ) is odd, f ( x) = f ( x) f a f b dτ 0 If the integrand is invariant under all operation of the symmetry group to which the molecule belongs or unless some term in it remains invariant. The representation of a direct product, Γ ab, will contain the totally symmetric representation iff the irreducible Γ a = the irreducible Γ b. a i = 1 h h R=1 χ ab ( R) χ i ( R) a 1 = 1 h h R=1 χ ab ( R) a 1 = 1 h h R=1 χ a ( R) χ b ( R) = δ ab
21 f a f b f c dτ 0 if the direct product of the representations of f a, f b, and f c is or contains the totally symmetric representation. ψ a ˆPψ b dτ ψ a ˆ Hψ b dτ ψ a ψ b dτ An energy integral may be nonzero only if ψ a and ψ b belong to the same irreducible representation of the molecular point group. I ψ a ˆµψ b dτ µ is a transition moment operator corresponding to changes in electric or magnetic dipoles, higher electric or magnetic multipoles, polarizability tensors. = E
22 The electric dipole operator ha the form µ = e i x i + e i y i + e i z i i i i I x I y I z ψ a ˆxψ b dτ ψ a ŷψ b dτ ψ a ẑψ b dτ An electric dipole transition will be allowed with x, y, or z polarization if the direct product of the representations of the two states is or contains the irreducible representation to which x, y, or z, respectively, belongs.
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