MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 12, Problem 12.98P

(a)

Interpretation Introduction

Interpretation:

The number of Zn and Se ions in each unit cell is to be calculated.

Concept introduction:

Crystal structure or lattice is the three-dimensional representation of atoms and molecules arranged in a particular manner. The unit cell is the smallest part of the lattice that is repeated in all directions to yield the crystal lattice. There are 3 types of cubic unit cells as follows:

1. The simple cubic unit cell

2. Body-centered unit cell

3. Face-centered unit cell

In the cubic unit cell, atom at the corner is shared by 8 adjacent cells so the contribution of an atom at the corner is 18, atom at the face is shared by 2 adjacent cells so the contribution of an atom at the face is 12 and atom at the body center is shared by 1 cell so the contribution of an atom at the body center is 1. The number of atoms present in a simple cubic, a body-centered cubic, and a face-centered cubic unit cell are 1, 2 and 4 respectively.

(a)

Expert Solution
Check Mark

Answer to Problem 12.98P

The number of Zn ions is 4 and Se ions in each unit cell are 4.

Explanation of Solution

The structure of zinc selenide (ZnSe) is similar to the zinc blende structure. Therefore selenide ions are present at the corners of the cell and the faces of the cell. The contribution of an atom present at the corner is 18 and the contribution of an atom at the face of a cell is 12. Therefore the number of selenide ions is calculated as follows:

Number of selenide ions=(18ion per cell)(8ions)+(12ion per cell)(6ions)=1ion+3ions=4ions

The charge ratio between the ions in ZnSe is 2:2 and therefore, the number of zinc ions is 4. Zinc ions are present in half of the tetrahedral voids.

Conclusion

In the cubic unit cell, atom at the corner is shared by 8 adjacent cells so the contribution of an atom at the corner is 18, atom at the face is shared by 2 adjacent cells so the contribution of an atom at the face is 12 and atom at the body center is shared by 1 cell so the contribution of an atom at the body center is 1.

(b)

Interpretation Introduction

The mass of a unit cell is to be calculated.

Concept introduction:

Crystal structure or lattice is the three-dimensional representation of atoms and molecules arranged in a particular manner. The unit cell is the smallest part of the lattice that is repeated in all directions to yield the crystal lattice. There are 3 types of cubic unit cells as follows:

1. The simple cubic unit cell

2. Body-centered unit cell

3. Face-centered unit cell

In the cubic unit cell, atom at the corner is shared by 8 adjacent cells so the contribution of an atom at the corner is 18, atom at the face is shared by 2 adjacent cells so the contribution of an atom at the face is 12 and atom at the body center is shared by 1 cell so the contribution of an atom at the body center is 1. The number of atoms present in a simple cubic, a body-centered cubic, and a face-centered cubic unit cell are 1, 2 and 4 respectively.

(b)

Expert Solution
Check Mark

Answer to Problem 12.98P

The mass of a unit cell is 577.48amu.

Explanation of Solution

The number of Zn ions is 4 and Se ions in each unit cell are 4.

The formula to calculate the mass of the unit cell is as follows:

Massofunitcell=(4)(massofZnatom)+(4)(massofSeatom) (1)

Substitute 65.41 amu for the mass of Zn atom and 78.96 amu for the mass of Se atom in the equation (1).

Massofunitcell=(4)(65.41 amu)+(4)(78.96 amu)=577.48amu

Conclusion

The mass of a unit cell is 577.48amu.

(c)

Interpretation Introduction

The volume of a unit cell is to be calculated.

Concept introduction:

Crystal structure or lattice is the three-dimensional representation of atoms and molecules arranged in a particular manner. The unit cell is the smallest part of the lattice that is repeated in all directions to yield the crystal lattice. There are 3 types of cubic unit cells as follows:

1. The simple cubic unit cell

2. Body-centered unit cell

3. Face-centered unit cell

In the cubic unit cell, atom at the corner is shared by 8 adjacent cells so the contribution of an atom at the corner is 18, atom at the face is shared by 2 adjacent cells so the contribution of an atom at the face is 12 and atom at the body center is shared by 1 cell so the contribution of an atom at the body center is 1. The number of atoms present in a simple cubic, a body-centered cubic, and a face-centered cubic unit cell are 1, 2 and 4 respectively.

The conversion factor to convert amu to g is as follows:

1amu=1.66054×1024 g

(c)

Expert Solution
Check Mark

Answer to Problem 12.98P

The volume of a unit cell is 1.77×1022cm3.

Explanation of Solution

The formula to calculate the volume of the unit cell is as follows:

Volume of unit cell=(mass ofunit cellDensityof solid) (2)

Substitute 577.48amu for the mass of unit cell and 5.42g/cm3 for the density of solid in the equation (2).

Volume of unit cell=(577.48amu5.42g/cm3)(1.66054×1024 g1amu)=1.76924×1022cm31.77×1022cm3

Conclusion

The volume of a unit cell is 1.77×1022cm3.

(d)

Interpretation Introduction

The edge length of a unit cell is to be calculated.

Concept introduction:

Crystal structure or lattice is the three-dimensional representation of atoms and molecules arranged in a particular manner. The unit cell is the smallest part of the lattice that is repeated in all directions to yield the crystal lattice. There are 3 types of cubic unit cells as follows:

1. The simple cubic unit cell

2. Body-centered unit cell

3. Face-centered unit cell

In the cubic unit cell, atom at the corner is shared by 8 adjacent cells so the contribution of an atom at the corner is 18, atom at the face is shared by 2 adjacent cells so the contribution of an atom at the face is 12 and atom at the body center is shared by 1 cell so the contribution of an atom at the body center is 1. The number of atoms present in a simple cubic, a body-centered cubic, and a face-centered cubic unit cell are 1, 2 and 4 respectively.

(d)

Expert Solution
Check Mark

Answer to Problem 12.98P

The edge length of a unit cell is 5.61×108cm.

Explanation of Solution

The formula to calculate the volume of the unit cell is as follows:

Volumeofunitcell=(edge length of unit cell)3 (3)

Rearrange the equation (3) t calculate edge length as follows:

Edge length of unitcell=Volumeofunitcell3

Substitute 1.76924×1022cm3 for the volume of the unit cell in the equation (3).

Edge length of cube=1.76924×1022cm33=5.6139×108cm5.61×108cm

Conclusion

The edge length of a unit cell is 5.61×108cm.

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Chapter 12 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 12.6 - For each of the following crystalline solids,...Ch. 12.6 - Prob. 12.6BFPCh. 12.6 - Prob. 12.7AFPCh. 12.6 - Iron crystallizes in a body-centered cubic...Ch. 12.6 - Prob. 12.8AFPCh. 12.6 - Prob. 12.8BFPCh. 12.6 - Prob. B12.1PCh. 12.6 - Prob. B12.2PCh. 12 - Prob. 12.1PCh. 12 - Prob. 12.2PCh. 12 - Prob. 12.3PCh. 12 - Prob. 12.4PCh. 12 - Prob. 12.5PCh. 12 - Prob. 12.6PCh. 12 - Prob. 12.7PCh. 12 - Name the phase change in each of these events: (a)...Ch. 12 - Prob. 12.9PCh. 12 - Many heat-sensitive and oxygen-sensitive solids,...Ch. 12 - Prob. 12.11PCh. 12 - Prob. 12.12PCh. 12 - Prob. 12.13PCh. 12 - Prob. 12.14PCh. 12 - Prob. 12.15PCh. 12 - Prob. 12.16PCh. 12 - Prob. 12.17PCh. 12 - Prob. 12.18PCh. 12 - From the data below, calculate the total heat (in...Ch. 12 - Prob. 12.20PCh. 12 - Prob. 12.21PCh. 12 - Prob. 12.22PCh. 12 - Prob. 12.23PCh. 12 - Prob. 12.24PCh. 12 - Prob. 12.25PCh. 12 - Prob. 12.26PCh. 12 - Prob. 12.27PCh. 12 - Prob. 12.28PCh. 12 - Prob. 12.29PCh. 12 - Prob. 12.30PCh. 12 - Use Figure 12.10 to answer the following: Carbon...Ch. 12 - Prob. 12.32PCh. 12 - Prob. 12.33PCh. 12 - Prob. 12.34PCh. 12 - Prob. 12.35PCh. 12 - Prob. 12.36PCh. 12 - Distinguish between polarizability and polarity....Ch. 12 - Prob. 12.38PCh. 12 - Prob. 12.39PCh. 12 - Prob. 12.40PCh. 12 - Prob. 12.41PCh. 12 - Prob. 12.42PCh. 12 - Prob. 12.43PCh. 12 - Prob. 12.44PCh. 12 - Prob. 12.45PCh. 12 - Prob. 12.46PCh. 12 - Prob. 12.47PCh. 12 - Prob. 12.48PCh. 12 - Prob. 12.49PCh. 12 - Which liquid in each pair has the lower vapor...Ch. 12 - Which substance has the lower boiling point?...Ch. 12 - Which substance has the higher boiling point?...Ch. 12 - Prob. 12.53PCh. 12 - Prob. 12.54PCh. 12 - Prob. 12.55PCh. 12 - Prob. 12.56PCh. 12 - Why does the antifreeze ingredient ethylene glycol...Ch. 12 - Prob. 12.58PCh. 12 - Prob. 12.59PCh. 12 - Why does an aqueous solution of ethanol (CH3CH2OH)...Ch. 12 - Prob. 12.61PCh. 12 - Prob. 12.62PCh. 12 - Prob. 12.63PCh. 12 - Prob. 12.64PCh. 12 - Prob. 12.65PCh. 12 - Prob. 12.66PCh. 12 - Prob. 12.67PCh. 12 - Prob. 12.68PCh. 12 - Prob. 12.69PCh. 12 - Prob. 12.70PCh. 12 - Prob. 12.71PCh. 12 - Prob. 12.72PCh. 12 - Prob. 12.73PCh. 12 - Prob. 12.74PCh. 12 - Prob. 12.75PCh. 12 - Prob. 12.76PCh. 12 - Prob. 12.77PCh. 12 - Prob. 12.78PCh. 12 - Prob. 12.79PCh. 12 - Prob. 12.80PCh. 12 - Prob. 12.81PCh. 12 - Prob. 12.82PCh. 12 - Prob. 12.83PCh. 12 - Prob. 12.84PCh. 12 - Besides the type of unit cell, what information is...Ch. 12 - What type of unit cell does each metal use in its...Ch. 12 - What is the number of atoms per unit cell for each...Ch. 12 - Calcium crystallizes in a cubic closest packed...Ch. 12 - Chromium adopts the body-centered cubic unit cell...Ch. 12 - Prob. 12.90PCh. 12 - Prob. 12.91PCh. 12 - Prob. 12.92PCh. 12 - Prob. 12.93PCh. 12 - Prob. 12.94PCh. 12 - Prob. 12.95PCh. 12 - Prob. 12.96PCh. 12 - Prob. 12.97PCh. 12 - Prob. 12.98PCh. 12 - Prob. 12.99PCh. 12 - Prob. 12.100PCh. 12 - Prob. 12.101PCh. 12 - Prob. 12.102PCh. 12 - Prob. 12.103PCh. 12 - Polonium, the Period 6 member of Group 6A(16), is...Ch. 12 - Prob. 12.105PCh. 12 - Prob. 12.106PCh. 12 - Prob. 12.107PCh. 12 - Prob. 12.108PCh. 12 - Prob. 12.109PCh. 12 - Prob. 12.110PCh. 12 - Prob. 12.111PCh. 12 - Prob. 12.112PCh. 12 - Prob. 12.113PCh. 12 - Prob. 12.114PCh. 12 - Prob. 12.115PCh. 12 - Prob. 12.116PCh. 12 - Prob. 12.117PCh. 12 - Prob. 12.118PCh. 12 - Prob. 12.119PCh. 12 - Prob. 12.120PCh. 12 - Prob. 12.121PCh. 12 - Prob. 12.122PCh. 12 - Prob. 12.123PCh. 12 - Prob. 12.124PCh. 12 - Prob. 12.125PCh. 12 - Prob. 12.126PCh. 12 - Bismuth is used to calibrate instruments employed...Ch. 12 - Prob. 12.128PCh. 12 - Prob. 12.129PCh. 12 - Prob. 12.130PCh. 12 - Prob. 12.131PCh. 12 - Prob. 12.132PCh. 12 - Prob. 12.133PCh. 12 - Prob. 12.134PCh. 12 - Prob. 12.135PCh. 12 - Prob. 12.136PCh. 12 - Prob. 12.137PCh. 12 - Prob. 12.138PCh. 12 - Prob. 12.139PCh. 12 - Prob. 12.140PCh. 12 - Prob. 12.141PCh. 12 - Prob. 12.142PCh. 12 - Prob. 12.143PCh. 12 - Prob. 12.144PCh. 12 - Prob. 12.145PCh. 12 - The crystal structure of sodium is based on the...Ch. 12 - Prob. 12.147PCh. 12 - One way of purifying gaseous H2 is to pass it...
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