solns21b-ps5-Feb11

# solns21b-ps5-Feb11 - CH21b Problem Set#5 Solution Set 2011...

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CH21b Problem Set #5 Solution Set 2011 1. a) The symmetry elements of diborane are: identity element (E), C 2 , 2 perpendicular C 2 , an inversion center( i ), 2 vertical mirror planes ( σ v ), and a horizontal mirror plane ( σ h ). Therefore, the point group is D 2h . Since there are 8 symmetry elements the group will be 8 × 8. b) Let Ξ correspond to the column matrix of x, y, z. Therefore: The matrices for the symmetry elements are constructed as follows. For E, we know that nothing is moved so all diagonal elements are +1. For any rotation axis, the matrix representing this has the following form: where θ = 2 π /n where n corresponds to the n-fold rotation axis (C n ). For i all points in x, y, z space are inverted so the sign changes. And for σ h the reflection occurs in the xy-plane so both x and y remain the same, while z is inverted. Note that he above matrix is for clockwise rotation. c) Determine Γ x , Γ y , and Γ z using the matrices calculated in part b. Get Γ xy by taking the direct product of Γ x and Γ y . Similar for Γ xz and Γ yz .

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D 2h E C 2x C 2y C 2z i σ xy σ xz σ yz Γ x 1 1 -1 -1 -1 1 1 -1 Γ y 1 -1 1 -1 -1 1 -1 1 Γ z 1 -1 -1 1 -1 -1 1 1 Γ xy 1 -1 -1 1 1 1 -1 -1 Γ xz 1 -1 1 -1 1 -1 1 -1 Γ yz 1 1 -1 -1 1 -1 -1 1 d) All properties of group representations and their characters can be derived from one basic theorem concerning the elements of the matrices that constitute the irreducible representations of the group. The order of a group is denoted as h and the dimension of the i th representation, which is the order of each of the matrices which constitute it, will be denoted by l i . The various operations in the group will be generically called R . The element in the m th row and n th column of the matrix corresponding to an operation R in the i th irreducible representation will be denoted by Γ i ( R ) mn . Given this, we can construct the great orthogonality theorem . In the D 2h point group, h = 8. Therefore, the sum of the squares of the elements in a given row must be 8. Now let’s construct the character table. From part c we already have 6 representations, so we must find two more. We know that every group has a totally symmetric representation so we can write that down immediately. The characters of the remaining representation are obtained using orthogonality requirements, namely that the vectors whose components are the characters of two different irreducible representations are
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## This note was uploaded on 01/03/2012 for the course CH 21b taught by Professor List during the Fall '10 term at Caltech.

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solns21b-ps5-Feb11 - CH21b Problem Set#5 Solution Set 2011...

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