The lattice parameter of an element stabilized in a fcc structure is 4.04 Å. The atomic radius of the element is:

(b) 1.43 Å. Explanation: fcc: 4r = a√2 → r = 4.04/2.828 = 1.43 Å.

Q90 · CSIR-NET Chemistry, June 2012

Paper: CSIR-NET June 2012 · Subject: Physical Chemistry · Chapter: Solid State Chemistry · Topic: Crystal Systems Lattices · Marks: 2 · Difficulty: Medium

The lattice parameter of an element stabilized in a fcc structure is $4.04 \AA$. The atomic radius of the element is:

(a)$2.86 \AA$

(b)$1.43 \AA$

(c)$4.29 \AA$

(d)$5.72 \AA$

Answer

Answer: B ✓ checked by 4AB · confidence high

Explanation

fcc: 4r = a√2 → r = 4.04/2.828 = 1.43 Å.

Study loop for Solid State Chemistry

1. Practise the PYQsPrevious-year questions, with answers

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