IIT-JEE Engineering Entrance Exam
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"SOLID STATES" problem....

by R.S. Verma
Posted on 14 April, 2011

Please solve.....

The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is

(1) 144 pm 
(2) 288 pm
(3) 398 pm
(4) 618 pm

by sukrit dubey posted on 15 April, 2011
The lattice can either be ZnS or NaCl type.......as both exhibit fcc lattice
in NaCl the edge length = 2(r+) + 2 (r-)  [r+ is radius of the cation & r- is radius of anion ]
since in NaCl type Na is present on each edge and centre , Chlorine on each corner and face center

so we get r(-)=144 pm

in ZnS type then we get a diiferent answer not in the options by using root(3)a = 4[r(+) + r (-)]
by sukrit dubey posted on 15 April, 2011
we are getting an integral radii......all the other type of lattices have an irrational number in the relation
in anti fluorite structure the -ive ion forms ccp and the ive ion occupies tetrahedral voids (opposite for fluorite)

so in the relation we would be getting root(3)a = 2r( ) 2r(-)
by R.S. Verma posted on 15 April, 2011
ok.....
but why u haven't taken Na2O or CaF2........
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