A compound AB crystallises in bcc lattice with the unit cell edge length of 380 pm. Calculate (i) the distance between oppositely charged ions in the lattice ,(ii) radius of B^- if the radius of A^+ is 190 pm

A compound AB crystallises in bcc lattice with the unit cell edge leng

1.	A compound AB crystallises in bcc lattice with the unit cell edge length of 380 pm. Calculate (i) the distance between oppositely charged ions in the lattice ,(ii) radius of B^- if the radius of A^+ is 190 pm
Answer» <html><body><p><br/></p>Solution :(i)For <a href="https://interviewquestions.tuteehub.com/tag/bcc-389643" style="font-weight:bold;" target="_blank" title="Click to know more about BCC">BCC</a>, distance between `A^+` and `B^- =sqrt3/2a` , i.e., it is half of <a href="https://interviewquestions.tuteehub.com/tag/body-900196" style="font-weight:bold;" target="_blank" title="Click to know more about BODY">BODY</a> diagonal <br/> (ii)As the <a href="https://interviewquestions.tuteehub.com/tag/cations-414251" style="font-weight:bold;" target="_blank" title="Click to know more about CATIONS">CATIONS</a> and the <a href="https://interviewquestions.tuteehub.com/tag/anions-876055" style="font-weight:bold;" target="_blank" title="Click to know more about ANIONS">ANIONS</a> touch each other , <br/> `r_(A^+)+r_(B^-)`329 pm `therefore r_(B^-)`=329-190=139 pm</body></html>