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🧪 Chemistry  ·  Solid State  ·  NEET & JEE

If edge length a = 500 pm and the structure is FCC, the radius of the atom is:

Answer: 176.8 pm.

  • A 125 pm
  • B 176.8 pm
  • C 216.5 pm
  • D 250 pm

Correct answer: B. 176.8 pm

Explanation: For FCC: r = a√2/4 = 500 × 1.414 / 4 = 707.1 / 4 ≈ 176.8 pm.

Cubic Unit Cells: Atom PositionsSimple CubicCN=6, 52.4% packedBody-Centred (BCC)CN=8, 68% packedFace-Centred (FCC)CN=12, 74% packed (densest)Corner atoms (shared by 8 cells) shown lighter; body/face-centre atoms shown solid

The three cubic unit cells: Simple Cubic has atoms only at corners; Body-Centred adds one atom at the centre; Face-Centred adds one atom at the centre of each of the 6 faces, giving the highest packing efficiency.

Concept context

Explore the ordered world of crystalline solids: unit cells, packing, defects, and how structure determines electrical and magnetic properties.

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