Question

In an AB unit cell (Rock salt type) assuming cations forms FCC lattice points, the:
A. Nearest neighbours of ${A^ + }$ are $6{B^ - }$ions
B. Nearest neighbours of ${B^ - }$ are $6{A^ + }$
C. Second nearest neighbours of ${A^ + }$ are $12{A^ + }$ ions
D. Packing fraction of AB crystals is $\dfrac{{\sqrt 3 \pi }}{8}$

Answer 1

We know that unit cell is the smallest repeating unit of the crystal lattice. We can call the building block of a crystal a unit cell. We have to know that a unit cell can be a face-centered cubic cell (FCC), body-centered cubic cell (BCC), and primitive cubic cell.

Complete step by step answer:
We know that an FCC unit cell comprises atoms at all the corners of the crystal lattice and atoms present at the centre of all the faces of the cube. The atom seen at the face-centered is shared between 2 adjacent unit cells and only ½ of each atom belongs to an individual cell.
In a face centered cubic unit cell, atoms are found in all corners of the lattice.
An atom is also present at the center of every face.
Two adjacent unit-cells are shared by a face-center atom.
Twelve atoms per unit cell.
The coordination number of FCC is twelve and the total number of atoms per unit cell is four.
For rock salt type, cations form 4 unit cells for FCC type $4{A^ + }$and ${B^ - }$ occupies octahedral voids $4{B^ - }$. The formula unit is $4AB$.
Since, the ${B^ - }$ occupies octahedral voids, the coordination number is six and in Binary compound AB, the ratio of coordination number will be equal to the charge number ratio.
The ratio of charge number is,
$A:B$
$1:1$
The ratio of coordination number is,
$A:B$
$6:6$
The nearest neighbor of $A +$ is $6{B^ - }$ ion and the nearest neighbor of ${B^ - }$ is $6{A^ + }$ ion. Therefore, Options (A) and (B) are correct.
In FCC, edge centers are also present and its coordination number is 12. Second nearest neighbours of ${A^ + }$ are $12{A^ + }$ ions.
$\therefore$ Option (C) is correct.
We have to know that the packing fraction FCC is $74\%$.
Let us solve the equation of Packing fraction of AB crystals $\dfrac{{\sqrt 3 \pi }}{8}$ by substituting the values of $\pi$. The value of $\pi$ is 3.14.
$\dfrac{{\sqrt 3 }}{8} \times 3.14 = 66.18\%$
We can see that the packing fraction AB crystals are not equal to the actual packing fraction. So, Option (D) is incorrect.

So, the correct answer is Option A,B,C..

Note:
An example of a compound that has face-centered cubic lattice is sodium chloride. Some other compounds that have face-centered cubic structures are lithium fluoride, lithium chloride, Sodium fluoride, potassium fluoride, potassium chloride etc. Some of the other crystal structures are Simple cubic cell, and body centered cubic cell.