NaCl crystallizes in face-centered cubic cell. what is the total number of ions(Na+ ions and Cr- ions) that lie withing a unit cell of NaCl?

2
4
8
6
5

If I interpret the question correctly, there are 4 Na^+ in a unit cell of NaCl and 4 Cl^- in a unit cell of NaCl making, I suppose, 8 for the total number.

To determine the total number of ions within a unit cell of NaCl, we need to consider the ratio of the ions in the crystal lattice.

In NaCl, the ratio of Na+ ions to Cl- ions is 1:1. This means that for every Na+ ion, there is one Cl- ion.

In a face-centered cubic cell, there are atoms at each of the eight corners and one atom at the center of each face. Each corner atom is shared by eight adjacent unit cells, and each face-centered atom is shared by two adjacent unit cells.

Considering this, we can calculate the number of ions within a unit cell as follows:

Number of corner atoms = 8 corners × (1/8 Na+ ion + 1/8 Cl- ion) = 1 Na+ ion + 1 Cl- ion

Number of face-centered atoms = 6 faces × (1/2 Cl- ion) = 3 Cl- ions

Total number of ions within a unit cell = Number of corner atoms + Number of face-centered atoms = 1 Na+ ion + 1 Cl- ion + 3 Cl- ions

Simplifying, the total number of ions within a unit cell of NaCl is 2.

Therefore, the correct answer is 2.

To determine the total number of ions within a unit cell of NaCl, we first need to understand the structure of a face-centered cubic (FCC) cell.

In an FCC cell, there are ions at the corners of the unit cell and additional ions at the center of each face. Each corner ion is shared by eight unit cells, while each face-centered ion is shared by two unit cells.

For NaCl, each unit cell contains one Na+ ion and one Cl- ion. Since the FCC unit cell has eight corners and six face centers, the total number of ions within a unit cell of NaCl is:

8 corners * 1 ion/corner + 6 face centers * 1 ion/face center = 8 + 6 = 14 ions

So, the correct answer is 14 ions.

since NaCl crystallizes in face-centered cubic cell, this means that we can have 6 halves of Na+ ions at the phases which makes 3 Na+ ions and 8 of (1\8) ions of Cl-, which make 1 ion of Cl-. Now the total number of ions(Na+ ions and Cl- ions) that lie withing a unit cell of NaCl is given by sum of the ions of Na+ ions and Cl-

(3 Na+ and 1 Cl-= 4 ions )