Question

Name Report Sheet: The Structure of Crystals-An Experiment Using Models 9.11 Section 1. Coordination Numbers: Number of nearest neighbors on the same level SC ВСС FCC Number of nearest neighbors on the level above Number of nearest neighbors on the level below Coordination Number (-total # of nearest neighbors) 2. Total mumber of lons in the NaCI Unit Cell # of ions x fraction contributed to the unit cell- # of jons contributed Chloride ions on corners of the unit cell Chloride ions on faces of the unit cell Total number of chloride ions per unit cell Sodium ions inside the unit cell Sodium ions on edges of the unit cell Total number of sodium ions per unit cell- 3. Density of Molybdenum Molybdenum crystallizes with a body-centered cubie unit cell. The atomic radius of molybdenum is 136.3 pm and its atomic weight is 95.94 amu. Calculate the density of molybdenum What is the length of the diagonal through the center of the unit cell of a molybdenum crystal? a. pm What is the length (in em) of the unit cell edge, de, in a molybdenum crystal? b. ст What is the volume (in em) of the molybdenum unit cell? c. cm

Answer 1

1.

	SC	BCC	FCC
Number of nearest neighbors on the same level	4	0	4
Number of nearest neighbors on the level above	1	4	4
Number of nearest neighbors on the level below	1	4	4
Coordination Number (=total number of nearest neighbors)	6	8	12

SC BCC FCC 国母回 * 米米

From the above figure, you can easily understand the answer of question 1.

2.

Chloride ions on the corner of the unit cell = 8 x 1/8 = 1 (Note: In a cube, there are 8 corners which contribute only 1/8 of the atom)

Chloride ions on the faces of the unit cell = 6 x 1/2 = 3  (Note: In a cube, there are 6 faces which contribute only 1/2 of the atom)

Total number of chloride ions per unit cell = 1+3 = 4

Sodium ions inside the unit cell = 1 x 1 = 1 (Note: Inside the cube, only one atom will occupy which contribute 1 atom.)

Sodium ions on the edges of the unit cell = 12 x 1/4 = 3 (Note: Inside the cube, there are 12 edges which contribute only 1/4 of the atom)

Total number of sodium ions per unit cell = 1+3 = 4

3. Given, Molybdenum has BCC structure => # of atoms per unit cell (Z) = 2

atomic radius (r) = 136.3 pm

atomic weight (M) = 95.94 amu = 95.94 g/mol

(a) We know that the length of the diagonal through the center of the unit cell is given by

=> $\sqrt{3}$ a=4r ................(1)

Where a= edge length

r = radius of the atom

$\sqrt{3}$a = diagonal of the unit cell

So, the length of the diagonal of through the center of the unit cell = 4r = 4x136.3 pm = 545.2 pm

(b) From equation (1), the length of the unit cell edge is

a = 4r/$\sqrt{3}$ =  4x136.3 pm/$\sqrt{3}$ = 378.02 pm

edge length in cm, 1 pm = 10^-12 m = 10^-10 cm

a = 378.02 x 10^-10 cm = 3.78 x 10^-8 cm

(c) We know that the volume of a cube (V) = a³ = (3.78 x 10^-8 cm)³ = 54.02x10^-24 cm³ = 5.402x10^-23 cm³

Density (d) = ZxM/VxN_a = (2x95.94g/mol)/(5.402x10^-23 cm³x6.022x10²³) = 5.898 g/cm³

Hence the density of molybedenum crystal is 5.898 g/cm³