 Solution

The two body-centre/origin sites and the other six sites have different local environments. Their effective atomic scattering factors therefore need not be identical even in an elemental, chemically pure crystal. Write them as $f_A$ and $f_B$, respectively. The crystal basis structure factor becomesFor the low-angle families this givesThus $\{110\}$ can produce a weak extra powder diffraction ring below $\{200\}$. This is an accidental diffraction extinction from equal atomic scattering factors: its cancellation in part d is exact only in the identical-atom-scattering approximation. Small differences in the bonding electron density, or site-dependent Debye–Waller factors, can leave a small nonzero amplitude. For instance, taking $f_A=f+\delta f$ and $f_B=f$ gives $F_{110}=2\delta f$ while $F_{200}=4f+2\delta f$; the corresponding intensity suppression is quadratic in $|\delta f/f|$. No impurity is required. The $\{100\}$ and $\{111\}$ cancellations remain for arbitrary scalar $f_A,f_B$ in this two-site-class model.