Which statement best describes slip planes in HCP metals relative to FCC/BCC metals?

• A. There are fewer slip planes
• B. HCPs have lower melting points
• C. There are more slip planes
• D. HCPs have higher density of dislocations

Answer: (A)

Understanding slip planes helps explain how metals deform. Slip planes are the crystal planes along which dislocations glide under applied shear. The number of easily activated slip planes (and slip systems) determines how readily a metal can plastically deform.

In hexagonal close-packed (HCP) metals, there is one primary easy glide plane—the basal plane—plus only a few other planes that can support slip, and those non-basal slip systems are much less readily activated. By contrast, face-centered cubic (FCC) and body-centered cubic (BCC) metals have multiple close-packed planes and multiple slip directions, giving many available slip systems. That means FCC and BCC can deform more easily through numerous slip paths.

So the statement that there are fewer slip planes in HCP metals relative to FCC/BCC metals is the best description.

Why the other ideas aren’t right: slip plane count doesn’t determine melting point, so it isn’t correct to say HCPs have lower melting points because of slip planes. The claim that there are more slip planes in HCPs contradicts the actual crystal symmetry and observed slip behavior. And dislocation density isn’t set by the slip planes themselves—it depends more on processing and deformation history, not the inherent number of slip planes.