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Question: The unit of silver iodide (AgI) has 4 iodine atoms in it. How many silver atoms must be there in the...

The unit of silver iodide (AgI) has 4 iodine atoms in it. How many silver atoms must be there in the unit cell?
A. 2
B. 8
C. 4
D. 1

Explanation

Solution

To solve this problem, we must firstly understand the type of unit cell present in the structure of AgI. Then we can calculate the total number of atoms present in this unit cell and by first, correspond the value of the number of atoms in AgI.

Complete step by step answer:
The smallest repeating array of atoms in a crystal is called a unit cell. A third common packing arrangement in metals, the body-centered cubic unit cell has atoms at each of eight corners of a cube plus one atom in the center of the cube.
A cubic unit cell is the smallest repeating portion of a crystal lattice. It has eight corners, twelve edges, and six faces. There are three different types of unit cells:
1.Simple cubic
2.Face-centered cubic
3.Body-centered cubic
In case of AgI, the ratio of numbers of cation and anion is 1:11:1 . Therefore in the unit shell of silver iodide 4 atoms iodine is present. then the total number of silver should be 4.
Therefore, the correct answer is C.

Additional information:
The density of a unit cell can determine the value of formula units as it is the ratio of mass and volume of the unit cell. The mass of the unit cell is equal to the mass of each atom in the unit cell and the number of atoms in the unit cell.
As the density is given in the question, we can use the below formula to find the number of ions that belong to each unit cell.
ρ=ZMNA(a)3\rho = \dfrac{{ZM}}{{{N_A}{{(a)}^3}}}
Where the Z is the number of atoms per unit cell. NA{N_A} is the Avogadro number, a is the edge length of the lattice. M is the molar mass of a solid.

Note: Particularly in FCC, the eight atoms at the corners contribute 8×18=18 \times \dfrac{1}{8} = 1 . There is one atom at each of the six faces which are shared by two unit cells each. Therefore, the contribution of six face-centered atoms 6×12=36 \times \dfrac{1}{2} = 3 . The total number of atoms be 1+3=4. The lattice of F.C.C is shown below.