In a crystal, oxide ions are arranged in FCC and A2+ ions are at 1/8 of the terahydralhyderal void. ions of B3+ occupied 1/2 of octahyderal voids. calculate the packing fraction of the crystal, if o2- ions are removed from aletrnate corners and A2+ is being placed at two of the corners.
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Answer:
The answer is 3) 4.
In FCC crystal the number of octahedral voids = 4
50% of them are occupied by the metal ion so no of voids occupied =2, hence the number of metal ions =2
In FCC lattice the total number of O2- atoms are =4 ( 8 at corners and 6 at face centers)
Therefore the formula of the compound becomes M2O4 = MO2.
Now since the cahrge on a compond is zero so 1 x n +( 2 x -2) = 0
=> n=4
amritstar:
read question carefully
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here is your answer mate ⤵️
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