= Solution
For the four conventional <face-centered cubic lattice> sites $000$, $0\frac12\frac12$, $\frac120\frac12$ and $\frac12\frac120$, their centering factor is
$$
F_{\rm fcc}=1+(-1)^{k+l}+(-1)^{h+l}+(-1)^{h+k}
=\begin{cases}4,&h,k,l\text{ all even or all odd},\\0,&\text{mixed parity}.\end{cases}
$$
This is the <face-centred cubic selection rule>. Enumerating the nonzero representatives with $h^2+k^2+l^2\le16$ gives
$$
\begin{array}{c|rrrrrr}
(hkl)&111&200&220&311&222&400\\\hline
h^2+k^2+l^2&3&4&8&11&12&16
\end{array}
$$
Symmetry-related permutations and signs belong to the same powder ring, not additional rings. For Al, the basis contains one atom, $S=F_{\rm fcc}f_{\rm Al}$. Thus
$$
\boxed{\mathrm{Al}:\quad111,\ 200,\ 220,\ 311,\ 222,\ 400}.
$$
For the stated KCl basis,
$$
S_{\rm KCl}=F_{\rm fcc}[f_K+(-1)^hf_{\rm Cl}].
$$
The all-even reflections add the two atomic factors; all-odd reflections subtract them. Ionic $K^+$ and $Cl^-$ each have 18 <electrons>, giving similar X-ray amplitudes. Their difference is much smaller than their sum, so the 111 and 311 rings are weak. Consequently the four strong rings are
$$
\boxed{\mathrm{KCl}:\quad200,\ 220,\ 222,\ 400}.
$$
This is an <accidental diffraction extinction from equal atomic scattering factors> in the equal-factor limit, not an extra exact fcc selection rule. Isoelectronic ions can have different radial <electron> distributions, so real odd reflections need not vanish exactly. Even the cruder neutral-atom estimate $f_K\simeq19$, $f_{\rm Cl}\simeq17$ makes the odd/even basis-intensity ratio only $(2/36)^2\simeq0.0031$, before other ring-intensity factors.
For Si's displaced identical basis atoms,
$$
S_{\rm Si}=F_{\rm fcc}f_{\rm Si}\left[1+e^{i\pi(h+k+l)/2}\right].
$$
For all-odd indices, the bracket is $1\pm i$, with squared magnitude 2. For all-even indices it vanishes when $h+k+l=2\pmod4$, and has squared magnitude 4 when the sum is a multiple of four. Thus 200 and 222 vanish, whereas 111, 220, 311 and 400 survive:
$$
\boxed{\mathrm{Si}:\quad111,\ 220,\ 311,\ 400}.
$$
These are the basis extinctions of the ideal <diamond cubic structure>. Intensity strengths within each surviving set can still differ because of atomic factors, multiplicity and geometry.
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