I initially thought that it had something to do with the number of slip systems in FCC vs. BCC, but they're both the same.
[Physics] Why don’t FCC metals have a brittle-to-ductile temperature transition
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please learn a few definitions Slip plane – is the plane of greatest atomic density. Slip direction – is the close-packed direction WITHIN the slip plane Slip system = slip plane and slip direction TOGETHER
THEN; 5 independent slip systems are necessary to make a polycrystalline material ductile.
HCP - Has three slip systems (one plane and three directions, giving 3x1= 3 slip systems, we know that minimum 5 independent slip systems are necessary to make a polycrystalline material ductile.therefore HCP is brittle.
FCC - has 12 slip systems (three {111} family of planes and four <110> family of directions, giving 3x4 =12 slip systems, which is more than 5 independent slip systems therefore FCC is ductile.
BCC -has 48 slip systems and expecting better ductile but it is brittle (six {110} family of planes and two <111> family of directions =6x2 = 12 slip systems + six {211} family of planes and two <111> family of directions =6x2 = 12 slip systems + six {321} family of planes and four <111> family of directions =6x4 = 24 slip systems; grand total 12+12+24 = 48 slip systems)
BCC lattice structure has too much of slip systems(48), here slip systems are INTERFERE OR MUTUALLY OBSTRUCT each other therefore slip movement in BCC is made very difficult thus BCC is brittle.
You're missing the fact that you're really not making an apples-to-apples comparison when you categorize KCl as a (almost) simple cubic structure and KBr as an fcc structure. KBr has an fcc Bravais lattice with lattice constant $a$, whereas in order for for KCl to look like an (almost) simple cubic structure, not only do you have to consider the K and Cl atoms to be the same (not a bad assumption since they have similar atomic numbers) but - as you yourself noted - you also have to view the lattice as having a reduced lattice constant $a/2$. So you're not really comparing fcc to sc. You're comparing a fcc structure with lattice constant $a$ to a simple cubic (sc) structure with lattice constant $a/2$.
It's actually not surprising that KCl has fewer strong diffraction peaks than KBr. Both of these ionic crystals have the NaCl crystal structure, but KCl extinguishes more diffraction peaks because K and Cl have very similar atomic numbers of 19 and 17, respectively. Br has an atomic number of 35 which is quite a bit different from the atomic number of K (atomic number 19) so, if one considers both KCl and KBr to have the same NaCl structure with about the same lattice constant $a$, fewer diffraction peaks will be extinguished for KBr than for KCl.
Best Answer
You are mistaken to imply that FCC metals do not undergo a transition from brittle to ductile behavior. FCC metals can fracture at sufficiently high stress - at some temperature. Please provide more information in your question if you believe otherwise.