Xenon crystallizes in FCC lattice and the edge of the unit cell is 620 pm, then the radius of xenon atom is

Name: JEE Advanced Chemistry
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Question

Prashant Kotian · Accepted Answer

Correct answer: A. For FCC, atoms touch along the face diagonal: 4r = a*√(2). r = 620*1.414/4 = 219.20 pm.

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