A solid AB has CsCl typestructure. The edge length of the unit cell is404 pm.Calculate the distance of closest approach between A^(+)and B^(-)ions ?

A solid AB has CsCl typestructure. The edge length of the unit cell is

1.	A solid AB has CsCl typestructure. The edge length of the unit cell is404 pm.Calculate the distance of closest approach between A^(+)and B^(-)ions ?
Answer» <html><body><p><br/></p>Solution :Distance of closest approach is <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> to the distance between the nearest <a href="https://interviewquestions.tuteehub.com/tag/neighbours-1113488" style="font-weight:bold;" target="_blank" title="Click to know more about NEIGHBOURS">NEIGHBOURS</a> (d)As CsCl has BCC latttic .,<br/>`d = sqrt3/ <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> a = 1.732/2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/404-1874913" style="font-weight:bold;" target="_blank" title="Click to know more about 404">404</a>"pm"= 349.9 ` pm</body></html>