= Solution
The two sites give phases one and $\exp[i\pi(h+k+l)]=(-1)^{h+k+l}$. Thus
$$
F_{\rm CsCl}=f_{\rm Cs^+}+(-1)^{h+k+l}f_{\rm Cl^-},\qquad
\boxed{I_{hkl}=K_{hkl}\left|f_{\rm Cs^+}+(-1)^{h+k+l}f_{\rm Cl^-}\right|^2.}
$$
For real nonresonant <Atomic scattering factor>[form factors], even $h+k+l$ gives the square of the sum and odd $h+k+l$ the square of the difference. The factor $K_{hkl}$ represents the remaining experimental corrections, not another basis-interference term. Since the two <Ion>[ions] differ, a half-body-diagonal translation does not preserve the decorated <crystal>. Its <Bravais lattice> is therefore <Simple cubic lattice>[simple cubic], with a two-ion basis; <Caesium chloride structure>[CsCl] is not a monatomic <Body-centered cubic lattice>[body-centered cubic] <Bravais lattice>.
Replacing \b[both] sites by identical Fe <Chemical atom>[atoms] changes the amplitude to
$$
F_{\rm Fe}=f_{\rm Fe}[1+(-1)^{h+k+l}],\qquad
\boxed{I_{hkl}^{\rm Fe}=\begin{cases}4K_{hkl}|f_{\rm Fe}|^2,&h+k+l\text{ even},\\0,&h+k+l\text{ odd}.\end{cases}}
$$
Now the body-centering translation does preserve the structure. It is a <body-centered cubic lattice> with one Fe <Chemical atom>[atom] per <Primitive unit cell>[primitive cell], or two per conventional cubic cell.
For 111 the parity is odd: <Caesium chloride structure>[CsCl] generally has a nonzero amplitude $f_{\rm Cs^+}-f_{\rm Cl^-}$, whereas the identical-Fe 111 reflection is systematically absent. A numerical absolute <Caesium chloride structure>[CsCl] <intensity> requires the angle-dependent <Atomic scattering factor>[form factors] and experimental normalization; no such values are supplied. Forward <electron> counts, 54 for Cs+ and 18 for Cl-, explain why their amplitudes are substantially unequal, but are not substituted for the actual finite-angle factors without an additional approximation.
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