JEE Advanced levelSingle correct optionDifficulty 3 of 3ss050
A salt MX has r(M+)=100pm and r(X−)=181pm. Using the radius-ratio rule (and 2=1.414, 3=1.732), the coordination number of M and the unit cell edge length for the corresponding cubic structure (ions touching along the edge for CN 6; along the body diagonal for CN 8) are
ACN 6 and 397 pm
BCN 6 and 562 pm
CCN 4 and 562 pm
DCN 8 and 325 pm
Answer: B
r+/r−=100/181=0.552, between 0.414 and 0.732, so CN 6 (rock salt, fcc). Ions touch along the edge: a=2(r++r−)=2×281=562pm. The value 397 pm comes from wrongly dividing by 2, and 325 pm is the CsCl-type value.