A substitutional alloy is formed by mixing metal M and metal N, where N atoms are significantly larger than M atoms. The alloy remains metallic with delocalized electrons. Which property change is most likely compared with pure M?
- Decreased ductility because lattice distortion makes it harder for layers of atoms to slide. (correct answer)
- Increased ductility because larger atoms create stronger directional covalent bonds.
- Increased ductility because electrons are no longer delocalized and cannot resist deformation.
- No change in ductility because only the total mass of the solid affects layer slippage.
- Decreased ductility because the alloy becomes an ionic solid with fixed alternating charges.
Explanation: This question examines how atomic size differences in substitutional alloys affect ductility. Pure metal M has a regular lattice where layers slide easily due to delocalized electrons, allowing ductility. Introducing larger N atoms distorts the lattice, creating strain that resists the movement of dislocations and reduces the ability to deform without cracking, decreasing ductility. This effect is pronounced because the size mismatch hinders uniform bonding and layer slippage. Choice C is tempting but wrong, suggesting increased ductility from stronger covalent bonds, which misunderstands that alloys remain metallic without directional covalent bonding. Always consider atomic size and lattice strain when predicting mechanical changes in alloys.