Exams › JEE Advanced › Chemistry
Correct answer: 2
In CaF2 structure: Ca²+ ions form FCC sublattice with lattice parameter a. F⁻ ions occupy ALL tetrahedral voids. Ca²+ positions: corners and face centres of the cube. F⁻ positions: all 8 tetrahedral voids at (a/4, a/4, a/4) and similar positions (body divided into 8 small cubes, F⁻ at centres of each). x (Ca-F distance): Ca²+ at (0,0,0), nearest F⁻ at (a/4, a/4, a/4). x = sqrt[(a/4)² + (a/4)² + (a/4)²] = (a/4)sqrt(3) = a*sqrt(3)/4 y (F-F distance): Two adjacent F⁻ sites, e.g., (a/4,a/4,a/4) and (3a/4,a/4,a/4) — but these differ by a/2 in x only... wait: adjacent tetrahedral void positions are (a/4,a/4,a/4) and (3a/4,3a/4,a/4): distance = sqrt[(a/2)²+(a/2)²+0] = a/sqrt(2). But nearest F-F: positions like (a/4,a/4,a/4) and (3a/4,a/4,a/4) differ by a/2, distance = a/2. Actually adjacent tetrahedral voids: (a/4,a/4,a/4) and (a/4,3a/4,a/4) differ by a/2 in y. But (a/4,a/4,a/4) and (3a/4,3a/4,a/4) differ by a/2 in both x and y: distance = a/sqrt(2). The shortest is between (a/4,a/4,a/4) and (3a/4,a/4,a/4): distance = a/2. So y = a/2. z (Ca-Ca distance): FCC, nearest neighbor Ca-Ca = a/sqrt(2). Compute: sqrt(6) * x*z / y² = sqrt(6) * [a*sqrt(3)/4] * [a/sqrt(2)] / (a/2)² = sqrt(6) * a²*sqrt(3)/(4*sqrt(2)) / (a²/4) = sqrt(6) * sqrt(3)/sqrt(2) = sqrt(6) * sqrt(3/2) = sqrt(6 * 3/2) = sqrt(9) = 3 Hmm, getting 3. Let me recheck y. In fluorite, F⁻ are at (1/4,1/4,1/4)a and all permutations with odd/odd/odd: (1/4,1/4,1/4), (3/4,3/4,1/4), (3/4,1/4,3/4), (1/4,3/4,3/4), (3/4,1/4,1/4)... wait no. ALL 8 tetrahedral voids: (1/4,1/4,1/4), (3/4,3/4,1/4), (3/4,1/4,3/4), (1/4,3/4,3/4), (3/4,3/4,3/4), (1/4,1/4,3/4), (1/4,3/4,1/4), (3/4,1/4,1/4). Nearest pair among these: e.g., (1/4,1/4,1/4) and (3/4,1/4,1/4) differ by a/2 in x only: distance = a/2. So y = a/2. With y = a/2: the answer is 3.