= 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> becomes
$$
F_{hkl}=f_A[1+(-1)^{h+k+l}]+2f_B\left[(-1)^k\cos\frac{\pi h}{2}+(-1)^h\cos\frac{\pi l}{2}+(-1)^l\cos\frac{\pi k}{2}\right].
$$
For the low-angle families this gives
$$
F_{100}=0,\qquad \boxed{F_{110}=2(f_A-f_B),}\qquad F_{111}=0.
$$
Thus $\{110\}$ 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.
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