Question

Oxygen atoms forms fcc unit cell with 'A' atoms occupying all tetrahedral voids and 'B' atoms occupying all octahedral voids. If atoms are removed from two of the body diagonals then determine the formula of resultant compound formed.

1. A4B4O7

2. A8B6O7

3. A8B8O7

4. A6B8O6
Ans is 2 nd option....
please explain with steps
I will mark u as brainlist............​

Answer 1

Answer:

See below.

Explanation:

First, we'll consider that all atoms are present in their respective positions ( no atoms are removed ).

So, atoms of O which are in FCC lattice,

$8\times \dfrac{1}{8}+6\times \dfrac{1}{2}= 4$

The number of tetrahedral voids is 2N, therefore no of atoms of A = 2 * 4  = 8

The number of octahedral voids is N, therefore no of atoms of B = 4

Now, we'll first see the atoms present on the body diagonal, see the 1st image attached.

On every body-diagonal, there are 2 atoms ( coloured blue ) present in T voids. There are also 1/8th parts of atom present at the vertices of the unit cell. So there are 2 parts of 1/8th atom,

$2\times \dfrac{1}{8}=\dfrac{1}{4}$

of an atom present on the vertice ( in our case it will be the atom of O as it makes an FCC lattice )

For 1 body diagonal we remove 1/4th of an O atom so for 2 body diagonals,

we have to remove 2 * 1/4 = 1/2th of O atoms.

Total O atoms remaining = 4 - 1/2 = 7/2

Also, each body diagonal had 2 atoms present in T voids, so for 2 body diagonals, we have 4 atoms.

In T voids, A atoms were present. So remaining A atoms,

8 - 4 = 4

Now, see image 2 attached. The body diagonal also passes through an octahedral void. As two diagonals intersect at the O void and an atom of B is present in that void, we remove 1 B atom.

Remaining B atoms = 4 - 1 = 3

So at last we have,

$\begin{aligned}A:B:O\\ 4:3:\dfrac{7}{2}\\ 8:6:7\end{aligned}$

Which gives the answer,

A8B6O7.