Chemistry, asked by amritstar, 1 year ago

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.

Answers

Answered by Anonymous
15

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
amritstar: what is question, and what are you answering
Anonymous: sorry yrr bro i will edit it
Anonymous: plzz report it
Answered by Anonymous
3

here is your answer mate ⤵️

Attachments:

12Ranvijay: mera account delete ker de yarr
Anonymous: U can also delete ur account bro
12Ranvijay: oye yarr pls mera account delete ker
Anonymous: I already done it
Anonymous: soon it will delete
12Ranvijay: fir mai ager apna password change ker loon toh v delete ho jayega na
Anonymous: Haan Haan
12Ranvijay: that's cool... milte hai kal..
Anonymous: hihihi get lost
12Ranvijay: Tera kuch ni ho skta...
Similar questions