NTA Abhyas JEE Main2020ChemistrySolid StatePractice
CsBr has bcc structure with edge length 4.3 Å . The shortest inter ionic distance in between Cs + and B r − is –
Options
- A4.3 Å
- B7.44 Å
- C1.86 Å
- 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