= Solution
With $A=S_{x_{\rm Si}}(h,k,l)$ and $B=S_{x_{\rm Mn}}(h,k,l)$,
$$
F=f_{\rm Si}A+(f+if'')B.
$$
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 $t=2\pi x$. Then
$$
S_x(111)=e^{3it}+3e^{-it}=4(\cos^3t-i\sin^3t).
$$
Since the purely positional sums obey $S_x(-\mathbf h)=S_x(\mathbf h)^*$, the <scalar resonant Friedel-pair intensity difference> is
$$
\boxed{I(\mathbf h)-I(-\mathbf h)=4f_{\rm Si}f''\operatorname{Im}(AB^*)}
$$
up to the same positive experimental scale factor. In particular,
$$
\operatorname{Im}(AB^*)=16\left(\cos^3t_{\rm Si}\sin^3t_{\rm Mn}-\sin^3t_{\rm Si}\cos^3t_{\rm Mn}\right).
$$
Different positional phases can therefore produce a nonzero resonant intensity difference at $(111)$, allowing the opposite structural handednesses to be distinguished. \b[The claimed violation is generic, not an unconditional identity:] the source supplies no numerical internal <coordinates>. For a concrete noncentrosymmetric counterexample, take $x_{\rm Si}=1/8$ and $x_{\rm Mn}=5/8$. Their orbit sums at $(111)$ are $\sqrt2(1-i)$ and $-\sqrt2(1-i)$, so the difference vanishes even if $f''\ne0$; each generic chemical orbit is noncentrosymmetric, so this cancellation does not require <inversion symmetry>. Accidental cancellation, or $f''=0$, must be excluded before asserting a violation for a particular specimen.
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