---
title: "A metallurgical engineer analyzes the atomic radii of four elements shown in the table below.  | Element | Atomic Radius (\\(\\text{pm}\\)) | | :—: | :—: | | \\(\\text{Fe}\\) | \\(126\\) | | \\(\\text{C}\\) | \\(77\\) | | \\(\\text{Cu}\\) | \\(128\\) | | \\(\\text{Ni}\\) | \\(124\\) |  Two binary alloys are synthesized: Alloy 1 is composed of \\(\\text{Fe}\\) and \\(\\text{C}\\), while Alloy 2 is composed of \\(\\text{Cu}\\) and \\(\\text{Ni}\\). Which alloy is expected to be more rigid and less malleable, and what is the correct structural justification?"
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url: "https://nerd-notes.com/ubq/123446/"
date_modified: "2026-09-28T11:59:44+00:00"
---

# A metallurgical engineer analyzes the atomic radii of four elements shown in the table below.

| Element | Atomic Radius (\(\text{pm}\)) |
| :—: | :—: |
| \(\text{Fe}\) | \(126\) |
| \(\text{C}\) | \(77\) |
| \(\text{Cu}\) | \(128\) |
| \(\text{Ni}\) | \(124\) |

Two binary alloys are synthesized: Alloy 1 is composed of \(\text{Fe}\) and \(\text{C}\), while Alloy 2 is composed of \(\text{Cu}\) and \(\text{Ni}\). Which alloy is expected to be more rigid and less malleable, and what is the correct structural justification?

A metallurgical engineer analyzes the atomic radii of four elements shown in the table below.

| Element | Atomic Radius (\(\text{pm}\)) |
| :---: | :---: |
| \(\text{Fe}\) | \(126\) |
| \(\text{C}\) | \(77\) |
| \(\text{Cu}\) | \(128\) |
| \(\text{Ni}\) | \(124\) |

Two binary alloys are synthesized: Alloy 1 is composed of \(\text{Fe}\) and \(\text{C}\), while Alloy 2 is composed of \(\text{Cu}\) and \(\text{Ni}\). Which alloy is expected to be more rigid and less malleable, and what is the correct structural justification?

- **A.** Alloy 2 is more rigid and less malleable because the similar atomic radii of \(\text{Cu}\) and \(\text{Ni}\) allow the substituted atoms to form directional covalent bonds within the crystal lattice that resist deformation.
- **B.** Alloy 2 is more rigid and less malleable because \(\text{Ni}\) atoms substitute directly into the \(\text{Cu}\) lattice sites, increasing the effective nuclear charge across the lattice to hold delocalized electrons more tightly.
- **C.** Alloy 1 is more rigid and less malleable because the significantly smaller \(\text{C}\) atoms occupy the interstitial spaces, distorting the regular lattice and preventing the planes of metal atoms from sliding past one another.
- **D.** Alloy 1 is more rigid and less malleable because the nonmetal \(\text{C}\) atoms accept valence electrons from \(\text{Fe}\) to form an ionic network solid, resulting in brittle cleavage planes.

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