CHAPTERS 2-22 COVERED
PROBLEM SOLUTIONS
,
, CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
Fundamental Concepts
Electrons in Atoms
2.1 Cite the difference between atomic mass and atomic weight.
Solution
Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the
atomic masses of an atom's naturally occurring isotopes.
, 2.2 Silicon has three naturally occurring isotopes: 92.23% of 28Si, with an atomic weight of 27.9769 amu,
4.68% of 29Si, with an atomic weight of 28.9765 amu, and 3.09% of 30Si, with an atomic weight of 29.9738 amu. On
the basis of these data, confirm that the average atomic weight of Si is 28.0854 amu.
Solution
The average atomic weight of silicon is computed by adding fraction-of-occurrence/atomic weight
products for the three isotopes—i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the percent
of occurrence divided by 100.) Thus
, 64
2.3 Zinc has five naturally occurring isotopes: 48.63% of Zn with an atomic weight of 63.929 amu;
27.90% of 66Zn with an atomic weight of 65.926 amu; 4.10% of 67Zn with an atomic weight of 66.927 amu; 18.75%
of 68Zn with an atomic weight of 67.925 amu; and 0.62% of 70Zn with an atomic weight of 69.925 amu. Calculate
the average atomic weight of Zn.
Solution
The average atomic weight of zinc is computed by adding fraction-of-occurrence—atomic weight
products for the five isotopes—i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the percent
of occurrence divided by 100.) Thus
Including data provided in the problem statement we solve for as
= 65.400 amu
, 113 115
2.4 Indium has two naturally occurring isotopes: In with an atomic weight of 112.904 amu, and In
with an atomic weight of 114.904 amu. If the average atomic weight for In is 114.818 amu, calculate the fraction-
of-occurrences of these two isotopes.
Solution
The average atomic weight of indium is computed by adding fraction-of-occurrence—atomic weight
products for the two isotopes—i.e., using Equation 2.2, or
Because there are just two isotopes, the sum of the fracture-of-occurrences will be 1.000; or
which means that
Substituting into this expression the one noted above for , and incorporating the atomic weight values
provided in the problem statement yields
Solving this expression for yields . Furthermore, because
then
,
, 2.5 (a) How many grams are there in one amu of a material?
(b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are
there in a pound-mole of a substance?
Solution
(a) In order to determine the number of grams in one amu of material, appropriate manipulation of the
amu/atom, g/mol, and atom/mol relationships is all that is necessary, as
= 1.66 × 10−24 g/amu
(b) Since there are 453.6 g/lbm,
= 2.73 × 1026 atoms/lb-mol
, 2.6 (a) Cite two important quantum-mechanical concepts associated with the Bohr model of the atom.
(b) Cite two important additional refinements that resulted from the wave-mechanical atomic model.
Solution
(a) Two important quantum-mechanical concepts associated with the Bohr model of the atom are (1) that
electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells.
(b) Two important refinements resulting from the wave-mechanical atomic model are (1) that electron
position is described in terms of a probability distribution, and (2) electron energy is quantized into both shells and
subshells--each electron is characterized by four quantum numbers.
, 2.7 Relative to electrons and electron states, what does each of the four quantum numbers specify?
Solution
The n quantum number designates the electron shell.
The l quantum number designates the electron subshell.
The ml quantum number designates the number of electron states in each electron subshell.
The ms quantum number designates the spin moment on each electron.
PROBLEM SOLUTIONS
,
, CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
Fundamental Concepts
Electrons in Atoms
2.1 Cite the difference between atomic mass and atomic weight.
Solution
Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the
atomic masses of an atom's naturally occurring isotopes.
, 2.2 Silicon has three naturally occurring isotopes: 92.23% of 28Si, with an atomic weight of 27.9769 amu,
4.68% of 29Si, with an atomic weight of 28.9765 amu, and 3.09% of 30Si, with an atomic weight of 29.9738 amu. On
the basis of these data, confirm that the average atomic weight of Si is 28.0854 amu.
Solution
The average atomic weight of silicon is computed by adding fraction-of-occurrence/atomic weight
products for the three isotopes—i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the percent
of occurrence divided by 100.) Thus
, 64
2.3 Zinc has five naturally occurring isotopes: 48.63% of Zn with an atomic weight of 63.929 amu;
27.90% of 66Zn with an atomic weight of 65.926 amu; 4.10% of 67Zn with an atomic weight of 66.927 amu; 18.75%
of 68Zn with an atomic weight of 67.925 amu; and 0.62% of 70Zn with an atomic weight of 69.925 amu. Calculate
the average atomic weight of Zn.
Solution
The average atomic weight of zinc is computed by adding fraction-of-occurrence—atomic weight
products for the five isotopes—i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the percent
of occurrence divided by 100.) Thus
Including data provided in the problem statement we solve for as
= 65.400 amu
, 113 115
2.4 Indium has two naturally occurring isotopes: In with an atomic weight of 112.904 amu, and In
with an atomic weight of 114.904 amu. If the average atomic weight for In is 114.818 amu, calculate the fraction-
of-occurrences of these two isotopes.
Solution
The average atomic weight of indium is computed by adding fraction-of-occurrence—atomic weight
products for the two isotopes—i.e., using Equation 2.2, or
Because there are just two isotopes, the sum of the fracture-of-occurrences will be 1.000; or
which means that
Substituting into this expression the one noted above for , and incorporating the atomic weight values
provided in the problem statement yields
Solving this expression for yields . Furthermore, because
then
,
, 2.5 (a) How many grams are there in one amu of a material?
(b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are
there in a pound-mole of a substance?
Solution
(a) In order to determine the number of grams in one amu of material, appropriate manipulation of the
amu/atom, g/mol, and atom/mol relationships is all that is necessary, as
= 1.66 × 10−24 g/amu
(b) Since there are 453.6 g/lbm,
= 2.73 × 1026 atoms/lb-mol
, 2.6 (a) Cite two important quantum-mechanical concepts associated with the Bohr model of the atom.
(b) Cite two important additional refinements that resulted from the wave-mechanical atomic model.
Solution
(a) Two important quantum-mechanical concepts associated with the Bohr model of the atom are (1) that
electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells.
(b) Two important refinements resulting from the wave-mechanical atomic model are (1) that electron
position is described in terms of a probability distribution, and (2) electron energy is quantized into both shells and
subshells--each electron is characterized by four quantum numbers.
, 2.7 Relative to electrons and electron states, what does each of the four quantum numbers specify?
Solution
The n quantum number designates the electron shell.
The l quantum number designates the electron subshell.
The ml quantum number designates the number of electron states in each electron subshell.
The ms quantum number designates the spin moment on each electron.
