Question

Gold crystallises in fcc structure with edge length 396 pm, find atomic radius of gold?

• A. 198 pm
• B. 714 pm
• C. 140 pm
• D. 162 pm

Hint:

In fcc structures, the atomic radius is related to the edge length by \( a = 2\sqrt{2} \times r \).

Answer 1

The Correct Option is C

Step 1: Relationship in fcc structure.
In an fcc (face-centered cubic) structure, the relationship between the edge length \( a \) and the atomic radius \( r \) is given by: \[ a = 2\sqrt{2} \times r \] Given that the edge length \( a = 396 \, \text{pm} \), we can calculate the atomic radius \( r \) as: \[ r = \frac{a}{2\sqrt{2}} = \frac{396}{2\sqrt{2}} = 140 \, \text{pm} \] Step 2: Analyzing the options.
(A) 198 pm: Incorrect. This is not the correct atomic radius.
(B) 714 pm: Incorrect. This value is too large for the atomic radius.
(C) 140 pm: Correct — This is the correct atomic radius based on the given edge length.
(D) 162 pm: Incorrect. This is not the correct radius.
Step 3: Conclusion.
The correct answer is (C) 140 pm, which is the atomic radius of gold in an fcc structure.