Question 1

Copper crystallizes in a face-centered cubic lattice. The radius of a copper atom is 128 pm. What is the edge length of the unit cell?
A) 362 pm
B) 256 pm
C) 512 pm
D) 272 pm
E) 128 pm

Accepted Answer

In a face-centered cubic (FCC) lattice, the edge length (a) can be related to the atomic radius (r) by the formula a = 2√2r. Substituting the given radius of 128 pm into the formula gives a = 2√2(128 pm) = 2(1.414)(128 pm) ≈ 362 pm.

Question 2

Which one of the following statements is incorrect?
A) Polonium is the only metal that has a primitive cubic lattice.
B) The lattice structure of the alkali metals is body-centered cubic.
C) Many metals can crystallize in more than one type of crystal lattice.
D) Metals with a body-centered cubic lattice contain a net of four metal atoms per unit cell.
E) A hexagonal close packed structure is not an example of a cubic unit cell.

Accepted Answer

In a body-centered cubic (BCC) lattice, there is one atom at each of the eight corners of the cube and one atom in the center. However, each corner atom is shared among eight adjacent unit cells, contributing only 1/8 of an atom to the unit cell, resulting in a total of 2 atoms per unit cell (8 corners * 1/8 + 1 center = 2), not four.

Question 3

The metal vanadium crystallizes in a body-centered cubic lattice. If the density of vanadium is 6.11 g/cm³, what is the atomic radius of vanadium? A) 25.3 pm B) 104 pm C) 35.3 pm D) 31.9 pm E) 131 pm

Accepted Answer

E

Question 4

Which equation represents the number of atoms in a body-centered cubic unit cell?
A) $\# \text { atoms } = \frac { 1 } { 8 } ( 8 )$
B) $\# \text { atoms } = \frac { 1 } { 4 } ( 8 )$
C) $\# \text { atoms } = 1 + \frac { 1 } { 8 } ( 8 )$
D) $\# \text { atoms } = \frac { 1 } { 2 } ( 6 ) + \frac { 1 } { 8 } ( 8 )$
E) $\# \text { atoms } = 1 + \frac { 1 } { 2 } ( 6 ) + \frac { 1 } { 8 } ( 8 )$

Accepted Answer

A body-centered cubic unit cell has 1 atom at the center and 8 corner atoms, each contributing 1/8th of an atom, totaling 2 atoms per unit cell.

Question 5

What is the distance, in atomic radii, along any edge of a body-centered cubic unit cell?
A) $\frac { 4 \times r } { \sqrt { 3 } }$
B) 2 × r
C) 4 × r
D) $\frac { 2 \times r } { \sqrt { 4 } }$
E) r

Accepted Answer

In a body-centered cubic (BCC) unit cell, the body diagonal is equal to 4 times the atomic radius (4r), and the body diagonal is also equal to the square root of 3 times the edge length ($\sqrt{3} \times a$). Therefore, the edge length $a$ is $\frac{4r}{\sqrt{3}}$.

Question 6

A metal crystallizes in a face-centered cubic lattice. The atomic radius of the metal is 198 pm and the density of the metal is 6.57 g/cm³. What is the volume of the unit cell? A) 3.25 × 10⁷ pm³ B) 1.52 × 10⁹ pm³ C) 1.76 × 10⁸ pm³ D) 7.76 × 10⁶ pm³ E) 1.3 × 10⁸ pm³

Accepted Answer

First, we need to determine the edge length of the unit cell. In a face-centered cubic lattice, there are 4 atoms per unit cell. The radius of each atom (r) is half of the diagonal of the face-centered cubic unit cell (d). Using the Pythagorean theorem:

d^2 = 4r^2
d^2 = 4(198 pm)^2
d = 558.6 pm

The edge length of the unit cell (a) is related to the diagonal (d) by:

a = d/√2
a = 558.6 pm/√2
a = 395.3 pm

The volume of the unit cell is:

V = a^3
V = (395.3 pm)^3
V = 1.76 × 10^18 pm^3

Converting to cm^3:

V = 1.76 × 10^18 pm^3 × (1 cm/10^10 pm)^3
V = 1.76 × 10^-6 cm^3

Finally, the density of the metal is:

density = mass/volume
6.57 g/cm^3 = mass/(1.76 × 10^-6 cm^3)
mass = 1.16 × 10^-2 g

Therefore, the volume of the unit cell is 1.76 × 10^18 pm^3 or 1.76 × 10^-6 cm^3. The answer choice that matches this volume is C.

Question 7

The metal vanadium crystallizes in a body-centered cubic lattice. If the density of vanadium is 6.11 g/cm³, what is the unit cell volume? A) 5.44 × 10⁵ pm³ B) 1.99 × 10⁵ pm³ C) 2.77 × 10⁷ pm³ D) 1.38 × 10⁷ pm³ E) 3.98 × 10⁵ pm³

Accepted Answer

First, we need to determine the atomic weight of vanadium (V) which is 50.94 g/mol.

To find the unit cell volume, we can use the formula:

density = (mass of 1 unit cell * Avogadro's number) / (volume of 1 unit cell * atomic weight)

We are looking for the volume of 1 unit cell, so we can rearrange the formula as:

volume of 1 unit cell = (mass of 1 unit cell * Avogadro's number) / (density * atomic weight)

The mass of 1 unit cell can be found by multiplying the atomic weight by the mass of one atom in grams and dividing by Avogadro's number:

mass of 1 unit cell = (50.94 g/mol * 1 atom/6.022 x 10^23 atoms/mol) = 8.461 x 10^-23 g

Plugging in the given values for density and atomic weight:

volume of 1 unit cell = (8.461 x 10^-23 g * 6.022 x 10^23) / (6.11 g/cm^3 * 50.94 g/mol)

Simplifying the expression:

volume of 1 unit cell = 2.767 x 10^-23 cm^3

Converting to picometers:

volume of 1 unit cell = 2.767 x 10^-13 pm^3

The closest answer choice is C, 2.77 × 10^17 pm^3.

Question 8

Arrange the three common cubic unit cells in order from least dense to most dense packing.
A) Face-centered cubic < body-centered cubic < primitive cubic
B) Primitive cubic < body-centered cubic < face-centered cubic
C) Primitive cubic < face-centered cubic < body-centered cubic
D) Body-centered cubic < primitive cubic < face-centered cubic
E) Body-centered cubic < face-centered cubic < primitive cubic

Accepted Answer

The primitive cubic unit cell has only one atom at each corner, giving it the lowest packing efficiency and density. The body-centered cubic unit cell has one atom at each corner and one in the center, giving it a higher packing efficiency and density than the primitive cubic. The face-centered cubic unit cell has one atom at each corner and one in the center of each face, giving it the highest packing efficiency and density among the three common cubic unit cells. Therefore, the order from least dense to most dense is primitive cubic < body-centered cubic < face-centered cubic.

Question 9

Palladium crystallizes in a face-centered cubic lattice with an edge length of 388.8 pm. What is the density of palladium? A) 0.752 g/cm³ B) 3.01 g/cm³ C) 1.50 g/cm³ D) 6.01 g/cm³ E) 12.0 g/cm³

Accepted Answer

In a face-centered cubic (fcc) lattice, there are 4 atoms per unit cell. The edge length (a) is 388.8 pm (or 388.8 x 10^-10 cm). The volume of the unit cell is a^3. The atomic mass of palladium (Pd) is approximately 106.42 g/mol. Using Avogadro's number (6.022 x 10^23 atoms/mol), the density (ρ) can be calculated using the formula ρ = (Z*M)/(a^3*N_A), where Z = 4 (number of atoms per unit cell), M = atomic mass, a^3 = volume of the unit cell, and N_A = Avogadro's number. Plugging in the values gives a density close to 12.0 g/cm^3, making E the correct choice.

Question 10

For a metal that crystallizes in a body-centered cubic unit cell, what percentage of the space in the cell is occupied by the metal atoms?
A) 47%
B) 62.4%
C) 68%
D) 77%
E) 87%

Accepted Answer

The answer of For a metal that crystallizes in a...

Question 11

Iron crystallizes in a body-centered cubic lattice. If the radius of iron is 126 pm, what is the unit cell edge length?
A) 291 pm
B) 252 pm
C) 168 pm
D) 126 pm
E) 54.6 pm

Accepted Answer

In a body-centered cubic (bcc) lattice, the diagonal across the cube is equal to 4 times the radius of the atom (4r). The diagonal can also be expressed in terms of the edge length (a) as √3a. Setting these equal and solving for a gives a = 4r/√3. Substituting the given radius of iron (126 pm) into this formula yields an edge length of approximately 291 pm.

Question 12

Palladium crystallizes in a face-centered cubic unit cell. If the edge length of the unit cell is 387 pm, what is the radius of a palladium atom in picometers?
A) 137 pm
B) 548 pm
C) 1100 pm
D) 316 pm
E) 48.4 pm

Accepted Answer

In a face-centered cubic (fcc) unit cell, the diagonal across the face of the cube is equal to four times the radius of the atom (4r), and it also equals the square root of 2 times the edge length (a√2). Therefore, 4r = a√2. Given that the edge length (a) is 387 pm, we can solve for r: r = (387 pm * √2) / 4 = 137 pm.

Question 13

What is the length of the diagonal (in atomic radii) along the face of a face-centered cubic unit cell?
A) $\frac { 4 \times r } { \sqrt { 3 } }$
B) 2 × r
C) 4 × r
D) $2 \sqrt { 2 } \times r$
E) r

Accepted Answer

The answer of What is the length of the diagonal...

Question 14

Cesium crystallizes in the body-centered cubic system. If the edge of the unit cell is 612 pm, what is the radius of a cesium atom in picometers?
A) 265 pm
B) 1060 pm
C) 1410 pm
D) 1220 pm
E) 115 pm

Accepted Answer

In a body-centered cubic structure, the diagonal of the unit cell can be calculated as:
$$
\mathrm{diagonal} = \sqrt{3} 	imes \mathrm{edge}
$$
So, in this case:
$$
\mathrm{diagonal} = \sqrt{3} 	imes 612 \, \mathrm{pm} = 1056.57 \, \mathrm{pm}
$$
The diagonal of the unit cell can be expressed as the sum of the diameter of an atom and the space diagonal of a cube with edge length equal to the diameter of an atom (twice the atomic radius):
$$
\mathrm{diagonal} = 2r + \sqrt{3} 	imes 2r = 2r(1 + \sqrt{3})
$$
Equating these two expressions for the diagonal of the unit cell, we get:
$$
2r(1+\sqrt{3}) = 1056.57 \, \mathrm{pm}
$$
Solving for r, we obtain:
$$
r = \frac{1056.57 \, \mathrm{pm}}{2(1 + \sqrt{3})} = 265.26 \, \mathrm{pm} \approx \boxed{\mathrm{(A)}\ 265 \, \mathrm{pm}}
$$

Question 15

If a metal crystallizes in a body-centered cubic lattice, each metal atom has _____ "nearest neighbors."
A) 3
B) 4
C) 6
D) 8
E) 12

Accepted Answer

In a body-centered cubic lattice, each atom is surrounded by 8 corner atoms and 1 atom in the center of the cube. Therefore, each metal atom has 8 nearest neighbors.

Question 16

Gold (atomic mass 197 g/mol), with an atomic radius of 144.2 pm, crystallizes in a face-centered cubic lattice. What is the density of gold? A) 19.3 g/cm³ B) 1.21 g/cm³ C) 4.82 g/cm³ D) 2.41 g/cm³ E) 9.65 g/cm³

Accepted Answer

The density of gold can be calculated using the formula for the density of a crystal structure, $\rho = \frac{Z \times M}{a^3 \times N_A}$, where $Z$ is the number of atoms per unit cell (which is 4 for a face-centered cubic lattice), $M$ is the molar mass (197 g/mol for gold), $a$ is the lattice parameter (which can be calculated from the atomic radius, for FCC $a = 2\sqrt{2}r$), and $N_A$ is Avogadro's number ($6.022 \times 10^{23} \text{mol}^{-1}$). Given the atomic radius of 144.2 pm (or 144.2 x 10^-10 cm), the lattice parameter $a$ can be calculated and then used to find the density. The calculation confirms that the density of gold is approximately 19.3 g/cm^3, which matches choice A.

Question 17

The space-filling representation provided below is an example of a _____ unit cell, which contains _____ atom(s). 
A) face centered cubic, 4 atoms
B) primitive cubic, 1 atom
C) body centered cubic, 2 atoms
D) face centered cubic, 13 atoms
E) primitive cubic, 7 atoms

Accepted Answer

The answer of The space-filling representation provided below is an...

Question 18

Chromium (atomic mass 52.00 g/mol) crystallizes in a body-centered cubic unit cell. If the length of an edge of the unit cell is 289 pm, what is the density (in g/cm³) of chromium? A) 3.58 g/cm³ B) 7.15 g/cm³ C) 13.7 g/cm³ D) 14.3 g/cm³ E) 21.3 g/cm³

Accepted Answer

In a body-centered cubic (bcc) unit cell, there are 2 atoms per unit cell. The density can be calculated using the formula: density = (mass of atoms in the cell / volume of the cell). The mass of the atoms in the cell is (2 atoms * 52.00 g/mol) / Avogadro's number (6.022 x 10^23 atoms/mol). The volume of the cell is (289 pm)^3, but first, convert pm to cm (1 pm = 1 x 10^-12 m = 1 x 10^-10 cm), so the volume in cm^3 is (289 x 10^-10 cm)^3. Plugging these values into the density formula and converting to the appropriate units gives a density of approximately 7.15 g/cm^3.

Question 19

Nickel has a face-centered cubic cell, and its density is 8.90 g/cm³. What is the radius (in pm) of a nickel atom? (The molar mass of nickel is 58.69 g/mol) A) 62.3 pm B) 88.1 pm C) 125 pm D) 249 pm E) 535 pm

Accepted Answer

The answer of Nickel has a face-centered cubic cell, and...

Question 20

Polonium (atomic mass 209.0 g/mol) crystallizes in a primitive cubic unit cell. If the density of polonium is 9.15 g/cm³, what is the radius of a polonium atom (in pm)? A) 168 pm B) 238 pm C) 336 pm D) 475 pm E) 672 pm

Accepted Answer

The density (ρ) of a substance is related to its molar mass (M), the volume of its unit cell (V), and the number of atoms per unit cell (Z) by the formula ρ = (Z*M)/(N_A*V), where N_A is Avogadro's number. For a primitive cubic unit cell, Z = 1. The volume of the unit cell can also be expressed in terms of the edge length (a) as V = a^3. The edge length is related to the radius (r) of the atom by a = 2r for a primitive cubic structure. Given the density and molar mass of polonium, and knowing it crystallizes in a primitive cubic structure, we can solve for the radius of a polonium atom. Rearranging the formula and solving for r gives a radius of approximately 168 pm.

Quiz 13: The Solid State

Copper crystallizes in a face-centered cubic lattice. The radius of a copper atom is 128 pm. What is the edge length of the unit cell?

Which one of the following statements is incorrect?

The metal vanadium crystallizes in a body-centered cubic lattice. If the density of vanadium is 6.11 g/cm³, what is the atomic radius of vanadium?

Which equation represents the number of atoms in a body-centered cubic unit cell?

What is the distance, in atomic radii, along any edge of a body-centered cubic unit cell?

A metal crystallizes in a face-centered cubic lattice. The atomic radius of the metal is 198 pm and the density of the metal is 6.57 g/cm³. What is the volume of the unit cell?

The metal vanadium crystallizes in a body-centered cubic lattice. If the density of vanadium is 6.11 g/cm³, what is the unit cell volume?

Arrange the three common cubic unit cells in order from least dense to most dense packing.

Palladium crystallizes in a face-centered cubic lattice with an edge length of 388.8 pm. What is the density of palladium?

For a metal that crystallizes in a body-centered cubic unit cell, what percentage of the space in the cell is occupied by the metal atoms?

Iron crystallizes in a body-centered cubic lattice. If the radius of iron is 126 pm, what is the unit cell edge length?

Palladium crystallizes in a face-centered cubic unit cell. If the edge length of the unit cell is 387 pm, what is the radius of a palladium atom in picometers?

What is the length of the diagonal (in atomic radii) along the face of a face-centered cubic unit cell?

Cesium crystallizes in the body-centered cubic system. If the edge of the unit cell is 612 pm, what is the radius of a cesium atom in picometers?

If a metal crystallizes in a body-centered cubic lattice, each metal atom has _____ "nearest neighbors."

Gold (atomic mass 197 g/mol) , with an atomic radius of 144.2 pm, crystallizes in a face-centered cubic lattice. What is the density of gold?

The space-filling representation provided below is an example of a _ unit cell, which contains _ atom(s) .

Chromium (atomic mass 52.00 g/mol) crystallizes in a body-centered cubic unit cell. If the length of an edge of the unit cell is 289 pm, what is the density (in g/cm³) of chromium?

Nickel has a face-centered cubic cell, and its density is 8.90 g/cm³. What is the radius (in pm) of a nickel atom? (The molar mass of nickel is 58.69 g/mol)

Polonium (atomic mass 209.0 g/mol) crystallizes in a primitive cubic unit cell. If the density of polonium is 9.15 g/cm³, what is the radius of a polonium atom (in pm) ?