Lecture 12 120709
The corrected parameters (x ± Δx) will not be the “best” parameters because LS refinement is not linear (we have neglected all but the 1st order terms!!).
Therefore, need many cycles of refinement to get the best agreement between Fobs (the experiment) and Fcalc (our model). Also, there is NO guarantee that the refinement will converge; certainly a bad “trial” structure will yield NON-convergence.
How find A-1? There are several ways to invert matrices:
Method of co-factors (NOT used in crystallography):
Ex.: find inverse of 1 2 3
A = -1 1 2 which are the elements of A = aij
-2 4 1
Inverse of A = A-1 (elements = bij )
Can find bij = Aji /│A │ where A is the cofactor of Aji matrix and│A│is
the determinant of the matrix A.
This has to be a square matrix; the determinant cannot be singular, and cannot be = 0.
a11 a12 a13 · · · a1n A11/│A │ A12/│A │ A13/│A │ · · · A1n/│A │ b11 b12 b13 · · · b1n
a21 a22 a23 · · · a2n → A21/│A │ A22/│A │ A23/│A │ · · · A2n/│A │ = b21 b22 b23 · · · b2n
· · · · · · · · · · · ·
an1 an2 an3 · · · ann An1/│A │ An2/│A │ An3/│A │ · · · Ann/│A │ bn1 bn2 bn3 · · · bnn
A → A-1 = B
To solve, evaluate the determinants and cofactors:
Determinant │A │ = +1 │1 2│ -2 │-1 2│ +3 │-1 1│
│4 1│ │-2 1│ │-2 4│
│A │ = 1(1-8) –2(-1+4) +3(-4+2) = 1(-7) – 2(+3) + 3(-2) = -19
A11 = │1 2│ = -7 A12 = - │-1 2│ = -3 A13 = │-1 1│ = -2
│4 1│ │-2 1│ │-2 4│
A21 = - │2 3│ = 10 A22 = │1 3 │ = 7 A23 = - │1 2│ = -8
│4 1│ │-2 1│ │-2 4│
A31 = │2 3│ = 1 A32 = - │-1 3│ = -5 A33 = │1 2 │ = 3
│1 2│ │-1 2│ │-1 1│
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┌ ┐
A-1 = (1 /│A│) A11 A21 A31 · · · An1 = (1/-19) │ -7 10 1 │
A12 A22 A32 · · · An2 │ -3 7 -5 │
│ -2 -8 3 │
A1n A2n A3n · · · Ann └ ┘
Check that A-1 A = I3 = unit matrix of order 3:
(1 / -19) -7 10 1 1 2 3 = -(1 / 19) -19 0 0 1 0 0
-3 7 -5 -1 1 2 0 -19 0 = 0 1 0
-2 -8 3 -2 4 1 0 0 -19 0 0 1
where I11 = -7(1) + 10(-1) + 1(-2) = -19
I12 = -7(2) + 10(1) + 1(4) = 0, etc.
Therefore, I3 = unit matrix of order 3
Apply this to simultaneous equations: solve the following “normal equations” for our purposes:
x1 + 2x2 + 3x3 = 4 1 2 3 x1 4
-x1 + x2 + 2x3 = 2 OR: -1 1 2 x2 = 2
-2x1 + 4x2 + x3 = 1 -2 4 1 x3 1
(A) (x) = V
Solve for Δx: (A)(A-1) Δx = (A-1)(V) OR Δx = (A-1)(V)
Therefore, multiply A-1 by V :
x1 -7 10 1 4 -7 x1 -7 / -19 = 0.36842
x2 = (1/-19) -3 7 -5 2 = (1 / -19) -3 Therefore: x2 -3 / -19 = 0.15789
x3 -2 -8 3 1 -21 x3 -21 / -19 = 1.10526
Wherefrom? (-7x4) + (10x2) + (1x1) = -7
(-3x4) + (7x2) + (-5x1) = -3
(-2x4) + (-8x 2) + (3x1) = -21
Check: x1 + 2x2 + 3x3 = 4 ; 7/19 + 2(3/19) + 3(21/19) = 4
We essentially always use full matrix refinement: use all matrix elements: x,y,z,B (if isotropic) or x,y,z, β11, β12, β13, β22, β23, β33 (6 tensors / atom if anisotropic). So, for 30 independent non-H atoms in the asymmetric unit, there would be 30 x 4 = 120 “normal equations” for isotropic refinement, and 30 x 9 = 270 “normal equations” for anisotropic refinement.
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This means that, in this example, for each cycle of refinement, 270 “normal equations” are calculated, a 270 x 270 matrix is set up and diagonalized (only need ½ of the matrix), this diagonalized matrix is then inverted and multiplied by each of the 9 variables for each atom (x,y,z, β11, β12, β13, β22, β23, β33) to give Δx, Δy, Δz, Δβ11, etc.), and then the new “normal equations” are recalculated.
For a 120 non-H atom problem: there are 4 x 120 atoms = 480 “normal equations” for an isotropic refinement, and 9 x 120 atoms = 1080 “normal equations” for an anisotropic refinement: meaning that you have a calculation for each data point, then you have to invert ½ of the 1080 matrix, multiply this by the 9 parameters for each atom to get the “new” values for these 9 parameters, and then recalculate the equations for each of the data points.
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Correlation matrix: Can all parameters always be refined? No. For sure, we can only refine the independent parameters from 1 asymmetric unit. Besides this, we have to worry about symmetry.
Ex.: in a cubic structure, we wish to refine x,y,z for an atom. However, in cubic space groups, x,y,z; y,z,x; z,x,y are ALL equivalent!!! Therefore, these coordinates are correlated strongly, because in fact, they are equivalent!!!
The correlation matrix tells of the interdependence of the parameters. For large correlation coefficients (> 0.85), then it is best to refine these parameters separately. This is an indication of either pseudosymmetry or of a poor trial structure.
Estimated standard deviations (esd) of the parameters: (at convergence):
How can we estimate the precision of the corrected parameters?
Use the elements of A-1 to lead to estimated standard deviations, σ, of the parameters
that we have refined:
σ = { bij Σ whkl [│Fhkl(obs) - Fhkl(calc│2]1/2}/ m - n
hkl
where bij = diagonal elements of the inverse matrix A-1
whkl = weight assigned to each of the hkl reflections
m = number of data points = reflections
n = number of parameters varied in the refinement
For very precise parameters, need: 1) large number of reflections = m
2) small number of parameters = n
3) small values of ΔF = │Fhkl(obs) - Fhkl(calc)│
Typically, want 5/1 < m/n ~ 10/1 reflections-to-variable parameter ratio.
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If we have small ΔF, then this tells us that the refinement has converged and we have good data.
The best refinements give precision (esd) on the x,y,z parameters of 0.00001-0.00002
and are reported as: x = 0.23591(19), y = 0.40266(13), z = 0.70125(12)
Note: precision of the positional parameters leads to precision of the bond lengths & angles. However, need precise cell dimensions as well since they enter into the calculations of bond distances.
How do we know when the refinement converges? Since the LS is NOT linear, when it is done? After each cycle of refinement, we calculate σi for all the variable parameters. We assume that the refinement has converged when:
1) Δxi < σx, etc.
2) Δxi < 0.10 σx, etc.
3) Δxi < 0.010 σx, etc., depending on how much confidence we have in the data.
Weighting functions:
In crystallographic LS, minimize Σ rhkl2 = Σ [Fhkl(obs) – Fhkl(calc) – ΔF]2
hkl hkl
But, [Fhkl(calc) + ΔF] = Fhkl(calc) of the next cycle of refinement.
So, in effect, we are minimizing: Σ rhkl2 = Σ [Fhkl(obs) – Fhkl(calc)]2
hkl hkl
Assume that F(obs) has been corrected for all systematic errors: i.e., L, p, absorption,
extinction, etc.):
Are all F(obs) known experimentally with equal precision?
No, typically, some F(obs) are more precisely known than others.
Assign a σ to each of the F(obs):
Ex.: hkl │F(obs)│ σ [F(obs)]
111 232 25
240 480 100
Suppose that we have these errors (σ), then we could weight the LS refinement.
From statistics, the proper weight to assign is the reciprocal of the variance [σ2(Fhkl)].
Therefore, weights can be assigned to each F(obs) as 1 / [σ2(Fhkl)] = whkl.
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Now, minimize Σ whkl [Fhkl(obs) – Fhkl(calc)]2
hkl
How do we find σ(Fhkl)? Best way is to measure F(obs) many times and calculate
σ(Fhkl):
――
σ(Fhkl) = Σ [{[Fhkl(m) - Fhkl]2 } / m-1]1/2
This would mean that you would need many data sets to average; this is impractical.
Crystallographers use various approximations (more or less justified) to get at whkl.
Weighting scheme: from counting statistics, σ(Fhkl) = N1/2, where N = total # counts accumulated.
Following the LS refinement, how do you know if refinement is converging?
Look at 3 functions, which all should decrease:
1) wRF (weighted R based on F):
{ Σ whkl [Fhkl(obs) - Fhkl](calc)]2 / Σ whkl [F2hkl(obs)}1/2
Generally find 0.03 < wRF < 0.09 for a structure with < 50 atoms per asymmetric unit.
2) RF (R based on F) = {Σ [││Fhkl(obs) │ - │Fhkl](calc)] │ / Σl │F2hkl(obs) │
Typically find 0.03 < wRF < 0.07 for a structure with < 50 atoms per asymmetric unit.
3) GOOF = goodness of fit function: the error in an observation of unit weight. This
determines the quality of the weighting scheme and the model.
Anisotropic Thermal Parameters:
1) much more reasonably approximates the thermal motion of the atoms.
2) will lead to lower R factors.
3) requires more data.
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4) And of course, the end results of structural analysis:::::::>
Bond lengths:
Bond Length= [(Dxa)2 + (Dyb)2 + (Dzc)2 + 2abDxDycosg + 2acDxDzcosb + 2bcDyDzcosa]1/2
Bond Angles: B /
(AB)2 + (AC)2 - (BC)2 /
Bond Angle(BAC) = θ = cos-1 ______where A θ
2(AB)(AC) \
\
C
where AB = distance between atoms A and B (in Å), AC = distance between atoms
A and C (in Å), BC = distance between atoms B and C (in Å), and θ is in degrees.
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