Question

Gold and Platinum both have FCC structures with unit cell dimensions of \(4.08 \, \text{\AA}\) and \(3.91 \, \text{\AA}\), respectively. The metallic radii of atoms are:

• A. \(R_{\text{Au}} = 1.44 \, \text{\AA}, \, R_{\text{Pt}} = 1.38 \, \text{\AA}\)
• B. \(R_{\text{Au}} = 2.44 \, \text{\AA}, \, R_{\text{Pt}} = 1.38 \, \text{\AA}\)
• C. \(R_{\text{Au}} = 1.44 \, \text{\AA}, \, R_{\text{Pt}} = 2.38 \, \text{\AA}\)
• D. \(R_{\text{Au}} = 4.44 \, \text{\AA}, \, R_{\text{Pt}} = 1.38 \, \text{\AA}\)

Hint:

The metallic radius in FCC structures depends directly on the lattice parameter a. Use R =a/2√2for quick calculations.

Answer 1

The Correct Option is A

The metallic radius \(R\) in an FCC unit cell is determined by:
\[R = \frac{a}{2\sqrt{2}}\]
For Gold, where \(a = 4.08 \, \text{\AA}\), the radius is calculated as: \[R_{\text{Au}} = \frac{4.08}{2\sqrt{2}} \approx 1.44 \, \text{\AA}\].
For Platinum, with \(a = 3.91 \, \text{\AA}\), the radius is: \[R_{\text{Pt}} = \frac{3.91}{2\sqrt{2}} \approx 1.38 \, \text{\AA}\].