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CsBr has bcc structure with edge length 4.3 Å . The shortest inter ionic distance in between Cs + and B r − is –

Options

  1. A4.3 Å
  2. B7.44 Å
  3. C1.86 Å
  4. D3.72 Å

Correct answer

D. 3.72 Å

Step-by-step solution

In BCC structure the cations and anions touch along the body diagonal 2 r cs + + r Br - = 3 × 4 · 3 (Body Diagonal = 3 a ) r cs + + r Br - = 1 · 7 3 2 × 4 · 3 2 = 3 · 7 2 Å

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