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Without significant anomalous scattering, the electron density is effectively real and Friedel's law gives $F_0(-\mathbf{h})=F_0(\mathbf{h}{)}^{\ast }$, hence $I(\overline{222})=13130$.

With anomalous scattering the atomic imaginary term retains its sign when the reflection index reverses. Here $\Delta F(-222)=19.2i$, soThus the second expected intensity is approximately $1.157\times 10^4{\mathrm{electrons}}^2$. Conjugating the complete anomalous structure factor would incorrectly impose Friedel equality on the second experiment.

Use fractional coordinates and the structure factor conventionFor real atomic scattering factors, $F(-\mathbf{h})=F(\mathbf{h}{)}^{\ast }$, immediately proving Friedel's law. If an inversion centre is chosen as origin, each site at ${\mathbf{r}}_j$ has an identical partner at $-{\mathbf{r}}_j$, with the same possibly complex atomic scattering factor. Each pair contributes $2f_j\cos (2\pi \mathbf{h}\cdot {\mathbf{r}}_j)$; sites fixed by inversion modulo a lattice translation also have even phase factors at reciprocal-lattice vectors. Henceeven for scalar anomalous X-ray scattering in a centrosymmetric structure. This last equality is an equality of amplitudes, not generally complex conjugacy.

With $A=S_{x_{\mathrm{Si}}}(h,k,l)$ and $B=S_{x_{\mathrm{Mn}}}(h,k,l)$,For $l=0$, the transformation in part c maps $(h,k,0)$ onto its Friedel pair with only a phase factor. Cyclic permutations give the same result when either other index vanishes. Friedel's law therefore survives for every such reflection, including any systematic absences.

For the $(111)$ reflection set $t=2\pi x$. ThenSince the purely positional sums obey $S_x(-\mathbf{h})=S_x(\mathbf{h}{)}^{\ast }$, the scalar resonant Friedel-pair intensity difference isup to the same positive experimental scale factor. In particular,Different positional phases can therefore produce a nonzero resonant intensity difference at $(111)$, allowing the opposite structural handednesses to be distinguished. The claimed violation is generic, not an unconditional identity: the source supplies no numerical internal coordinates. For a concrete noncentrosymmetric counterexample, take $x_{\mathrm{Si}}=1/8$ and $x_{\mathrm{Mn}}=5/8$. Their orbit sums at $(111)$ are $\sqrt{2}(1-i)$ and $-\sqrt{2}(1-i)$, so the difference vanishes even if $f^{\prime \prime }\ne 0$; each generic chemical orbit is noncentrosymmetric, so this cancellation does not require inversion symmetry. Accidental cancellation, or $f^{\prime \prime }=0$, must be excluded before asserting a violation for a particular specimen.