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In the fluorite structure if the radius ratio is `(sqrt(3)/(2)-1)` how many ions does each cation touch?A. `4` anionsB. `12` cationsC. `8` anionsD. No cations |
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Answer» Correct Answer - B::C `(r_(o+))/(r_(ɵ)) = sqrt(3/2) - 1 = 0.225` Hence, it is the limiting case where cation in the void of fcc structure is not distorted. So, number of cations surrounding the particular cation `= 12`. But at the same time `8` anions (present in `TVs`) touch the particul,ar cation. |
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