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Friedif Calculator

https://xraytools.com/friedif?formula=C6H5SeCH3&energy=12.7&Moradiation=true

Single-crystal data

The Friedif value of Flack and Shmueli: how much resonant-scattering contrast a composition offers at a given radiation — the signal an absolute structure determination would have to work with, computed before the crystal is on the diffractometer.

Before this
Friedel’s law says hkl and its opposite have equal intensity, which is why a diffraction pattern normally cannot tell a structure from its mirror image. This page is about the effect that breaks it.
You supply
A chemical formula — brackets and charges included, as on the CHN page — and one or more radiations. It is a property of the composition, so no cell or density is required.
Reading it
A Friedif of 200 corresponds to a standard uncertainty of about 0.04 on the Flack parameter, the general requirement; 80 corresponds to 0.1, which is enough only for a compound known to be enantiopure. The notes below give the thresholds in full.

Worked examples: a nickel complex · an organic cation · C6H5SeCH3 just above the selenium K edge

Earlier on the path: Absorption Coefficient Calculator On Choosing the radiation, step 4 of 4

Notation here: x, u(x) · f0, f′, f″ — what each one means here

See also: Absorption Coefficient Calculator · Space Group Reflection Conditions · Refinement Statistics and R Factors

Teaching with this page
Objective
After this page a learner can choose between two anodes for a particular formula from its own numbers rather than from a rule of thumb.
Start from
this worked example
Ask first
The heaviest atom in your compound is bromine. Is Cu Kα still the better radiation for the absolute structure?
Watch for
“Yes — copper is the radiation for absolute structure”
Check yourself: The heaviest atom in your compound is bromine. Is Cu Kα still the better radiation for the absolute structure?

No — for bromine molybdenum gives about twice the signal Yes — copper is the radiation for absolute structure

f″ is not a property of the element alone: it depends on where the photon energy sits relative to that element’s absorption edges. Bromine’s K edge is at 13.47 keV, between Cu Kα (8.04) and Mo Kα (17.44), so molybdenum excites it and copper does not — f″ is 2.45 e at Mo Kα against 1.28 at Cu. The copper rule is real and it comes from light-atom compounds, where the opposite holds: for oxygen f″ is 0.032 at Cu and 0.006 at Mo, five times better with copper. This page computes the comparison for your formula instead of applying either rule.

Input

X-ray sources
Å
keV

Either one, not both, and it is added to whatever is ticked above.

Results

Formula as entered C6H5SeCH3
Read as C7H8Se
Friedif(Mo): 711 at 17.445 keV
Friedif(0.97625 Å): 1206 at 12.7 keV

The real part of each scattering factor is taken as the atomic number Z — its value at sin θ/λ = 0 — and f′ is left out of the sum; the table of factors below prints it for reference. For light atoms far from an edge that changes nothing. Just below an edge of a heavy atom, where f′ reaches several electrons, it changes the value by tens of per cent: with Z + f′ this formula would give 712 in place of 711 for Mo, 1533 in place of 1206 for 0.97625 Å.

0.97625 Å is 42 eV above the K edge of Se (12.658 keV). This close to an edge the tabulated f′ and f″ are those of an isolated atom: in a compound the edge shifts by several eV with the oxidation state and carries fine structure the tabulation does not hold, so the value for this radiation is an indication and not a figure to plan an experiment on.

The scattering factors behind it

Mo — 17.445 keV, 0.71073 Å

The anomalous scattering factors of each element at Mo. CSV
elementZ f′f″ nearest edge below
C 6 0.0051 0.0016 —
H 1 0.0000 0.0000 —
Se 34 −0.0246 2.2229 K at 12.658 keV

0.97625 Å — 12.7 keV

The anomalous scattering factors of each element at 0.97625 Å. CSV
elementZ f′f″ nearest edge below
C 6 0.0089 0.0033 —
H 1 0.0000 0.0000 —
Se 34 −4.8005 3.7999 K at 12.658 keV

f′ and f″ are Chantler’s relativistic tabulation (NIST FFAST, J. Phys. Chem. Ref. Data 24 (1995) 71 and 29 (2000) 597); the absorption coefficient is f″ through the optical theorem plus the tabulated scattering term. Compilations of these quantities differ from one another by a few per cent, and that is the accuracy to read them at.

Notes and references
  • What the number is about. Far from an absorption edge an atom’s scattering factor is real, and I(hkl) = I(−h−k−l) for every structure whether or not it has a centre of symmetry — that is Friedel’s law, and while it holds a diffraction pattern cannot tell a structure from its mirror image. Near an edge the factor becomes complex, f = f0 + f′ + if″, and the imaginary part f″ breaks it: the two members of a Friedel pair come out with measurably different intensities, and that difference — the Bijvoet difference — is the whole of the information an absolute structure is determined from. Friedif scores how much of it a given composition offers at a given wavelength, which is why the answer changes so much between Cu, Mo and Cr for the same formula, and why it can be computed before the crystal is on the diffractometer. The structure factor page builds F without any of this, so everything there obeys Friedel’s law exactly.
  • It is a property of the composition and the radiation, and so an estimate of the signal available rather than a prediction of the outcome. What is actually achieved also depends on resolution, redundancy, the absorption correction, crystal quality, inversion twinning and where the resonant scatterers sit in the cell.
  • This is an implementation of an Excel spreadsheet calculation which was created by Flack and Shmueli; their paper is linked at the foot of this page.
  • Flack found the standard uncertainty u of the Flack parameter to be inversely proportional to Friedif, u = m/Friedif, with u·Friedif normally between 8 and 12 for an ordinary structure determination. Taking m = 8, the two limits on u in the next note become limits on Friedif: 200 corresponds to u = 0.04, the general case, and 80 to u = 0.1, the limit for a compound established to be enantiopure. A Friedif below the limit that applies means an absolute-structure determination should not be expected routinely; a different radiation, a derivative or a cocrystal with a stronger resonant scatterer are the usual remedies. D-glucose, C6H12O6, is Flack’s example: Friedif 7 with Mo Kα and 36 with Cu Kα, both under its limit of 80, though with Cu careful measurement of selected reflections might still succeed. After the experiment, a u·Friedif much larger than 10 is a reason to examine the analysis very carefully. (See H. D. Flack, Acta Chim. Slov., 2008, 55, 689–691.)
  • The quality of an absolute structure determination is often derived from the Flack parameter x along with its associated standard uncertainty u, thus x(u). u < 0.04 is associated with a strong inversion-distinguishing power, 0.04 < u < 0.1 means enantiopure-sufficient inversion-distinguishing power. A reliable absolute structure determination requires u < 0.04 and |x| < 2u. If there is a priori biological, chemical, or physical evidence for true enantiopurity of the compound of interest, 0.04 < u < 0.1 and |x| < 2u is also acceptable. The classification and the limit of 0.04 are Flack and Bernardinelli’s, whose paper is linked at the foot of this page; the limit of 0.1 for an enantiopure compound is the one Flack uses in the 2008 paper cited in the note above.

Where this comes from