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A compound consisting of the monovalent ions A^+ ,B^- crystallizes in the body-centred cubic lattice. (i)What is the formula of the compound ? (ii)If one of A^+ ions from the corner is replaced by a monovalent ions C^+, what would be the simplest formula of the resulting compound ?

Answer» <html><body><p></p>Solution :(i)Contribution of `A^+` ions present at the <a href="https://interviewquestions.tuteehub.com/tag/corners-934262" style="font-weight:bold;" target="_blank" title="Click to know more about CORNERS">CORNERS</a> towards unit <a href="https://interviewquestions.tuteehub.com/tag/cell-25680" style="font-weight:bold;" target="_blank" title="Click to know more about CELL">CELL</a> =`8xx1/8=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>` <br/> Contribution of `B^-` ion present at the body-centre=1 <br/> `therefore` Ratio of `A^+ :B^-` =1:1 .Hence , the formula is AB. <br/> (ii)`A^+` ions at the corners =7. Hence, their contribution towards unit cell=`7/8` <br/> `C^+` ion at <a href="https://interviewquestions.tuteehub.com/tag/one-585732" style="font-weight:bold;" target="_blank" title="Click to know more about ONE">ONE</a> <a href="https://interviewquestions.tuteehub.com/tag/corner-409931" style="font-weight:bold;" target="_blank" title="Click to know more about CORNER">CORNER</a> contributes =`1/8`<br/> `B^-` ion at the body centre has contribution =1 <br/> Hence, ratio of A:B:C=`7/8:1/8:1=7:1:8 therefore ` Formula will be `A_7BC_8`</body></html>


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