Foundations of Materials Science and Engineering 6th Edition By William Smith – Test Bank

Description

Foundations of Materials Science and Engineering 6th Edition By William Smith – Test Bank

 Sample Questions

Instant Download With Answers

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
12
CHAPTER 2
Knowledge and Comprehension Problems:
Problem 2.1 Describe the laws of (a) multiple proportions and (b) mass conservation as related to
atoms and their chemical properties.
Answer 2.1: (a) The law of multiple proportions states that atoms of one pure substance are
different from the atoms of other pure substances and, when combined in specific simple
fractions, form different compounds. (b) The law of mass conservation states that a chemical
reaction is explained by separation, combination, or rearrangement of atoms, and that a chemical
reaction does not lead to creation or destruction of matter.
Problem 2.2 How did scientists find out that atoms themselves are made up of smaller particles?
Answer 2.2: Henri Becquerel and Marie and Pierre Curie showed that some atoms
spontaneously emit rays and named this phenomenon radioactivity. The radiation was shown to
consist of a (alpha), b (Beta), and g (gamma) rays. It was also shown that a and b particles have
both charge and mass while g particles have no detectable mass or charge.
Problem 2.3 How was the existence of electrons first verified? Discuss the characteristics of
electrons.
Answer 2.3: Joseph J. Thompson, using cathode ray tube experiments, concluded that atoms in
all matter are made of smaller particles that are negatively charged. The negatively charged plate
(cathode) emits an invisible ray that is attracted by the positively charged plate (anode). The
invisible ray is called a cathode ray and is made up of electrons. He also calculated the ratio of
mass to charge of these electrons to be 5.60 X 10-19 g/C where Coulomb, C, is the unit of
electrical charge. Robert Millikan, in his oil-drop experiments, determined the fundamental
quantity of charge or the charge of an electron (regardless of the source) to be 1.60 X 10-19 C. For
an electron, this quantity of charge is represented by -1. Using the ratio of mass to charge of the
electron measured by Thompson and the charge of the electron measured by Millikan, the mass
of an electron was determined to be 8.96 X 10-28 g.
Problem 2.4 How was the existence of protons first verified? Discuss the characteristics of
protons.
Answer 2.4: Ernest Rutherford bombarded a very thin foil of gold with positively charged a
particles. He noticed that many of the a particles pass through the foil without deflection, some
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
13
are slightly deflected, and a few are either largely deflected or completely bounce back. He
concluded that 1) most of the atom must be made up of empty space (thus most particles pass
through without deflection) and 2) a small neighborhood at the center of the atom, the nucleus,
houses positively charged particles of its own. He suggested that those alpha particles that
deflected intensely or bounced back must have interacted closely with the positively charged
nucleus of the atom. The positively charged particles in the nucleus were called protons. It was
later determined that the proton carries the same quantity of charge as an electron but opposite
in sign and has a mass of 1.672 X 10-24 g (1840 times the mass of the electron). For a proton this
quantity of charge is represented by +1. Also, since atoms are electrically neutral, they must have
an equal number of electrons and protons.
Problem 2.5 What are the similarities and differences among protons, neutrons, and electrons?
Compare in detail.
Answer 2.5: Using the information in Table 2.1, one can summarize that protons and neutrons
have significantly higher mass, and basically constitute the total mass of the atom (the mass of
electron is minimal in comparison). On the other hand, the charge of the atom comes equally
from its electrons (negative) and protons (positive). Neutrons are not charged.
Particle Mass (g)
Charge
Coulomb (C) Charge Unit
Electron 9.10939 X 10-28 -1.06022 X 10-19 -1
Proton 1.67262 X 10-24 +1.06022 X 10-19 +1
Neutron 1.67493 X 10-24 0 0
Problem 2.6 One mole of iron atoms has a mass of 55.85 grams, without any calculations
determine the mass in amu of one iron atom.
Answer 2.6: One mole of iron corresponds to the number of atoms needed to create a mass in
units of grams (55.85 grams) numerically equal to the atomic mass in amu of the substance under
consideration. Thus one atom of iron has an atomic mass of 55.85 amu.
Problem 2.7 One atom of oxygen has a mass of 16.00 amu, without any calculations determine
the mass in grams of one mole of oxygen atoms.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
14
Answer 2.7: One mole of oxygen corresponds to the number of atoms needed to create a mass
in units of grams (16.00 grams) numerically equal to the atomic mass in amu of the substance
under consideration. Thus one mole of oxygen has a weight of 16 grams.
Problem 2.8 Define a) atomic number, b) atomic mass, c) atomic mass unit (amu), d) mass
number, e) isotopes, f) mole, g) relative atomic mass, h) average relative atomic mass, and
i) Avogadro’s number.
Answer 2.8:
a) atomic number – the number of protons in the nucleus of an atom is the atomic
number (Z).
b) atomic mass – the mass of one atom of a substance expressed in amu.
c) atomic mass unit (amu) – one amu is defined as exactly 1/12th the mass of a carbon atom
with 6 protons and 6 neutrons.
d) mass number – the sum of protons and neutrons in a nucleus of an atom.
e) isotopes – atoms with the same atomic number but different mass numbers.
f) mole – one mole or gram-mole (mol) of any element is defined as the amount of substance
that contains 6.02 x 1023 atoms.
g) relative atomic mass – the mass in grams of one mole of an element is called the relative
atomic mass, molar mass, or the atomic weight.
h) average relative atomic mass
i) Avogadro’s number – number of atoms in one mole of an element.
Problem 2.9 Explain the law of chemical periodicity.
Answer 2.9: The law of chemical periodicity states that the properties of elements are functions
of their atomic number in a periodic manner.
Problem 2.10 What is the nature of visible light? How is the energy released and transmitted in
visible light?
Answer 2.10: Light is in the form of electromagnetic radiation. Energy is released and transmitted
in the form electromagnetic waves.
Problem 2.11 (a) Rank the following emissions in increasing magnitude of wavelength: microwave
oven emissions, radio waves, sun lamp emissions, x-ray emissions, and gamma ray emissions
from the sun. (b) Rank the same emissions in terms of frequency. Which emission has the highest
energy?
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
15
Answer 2.11:
(a) Increasing wave length : Gamma Ray – x-ray – sun lamp – microwave – radio wave
(b) Increasing frequency of emission: Gamma Ray– x-ray– sun lamp – microwave – radio
wave. The highest frequency emission, gamma ray, has the highest energy (E = hn).
Problem 2.12 Describe the Bohr model of the hydrogen atom. What are the shortcomings of the
Bohr model?
Answer 2.12: Bohr suggested that electrons travel in circular paths around the nucleus with
discrete angular momenta (a product of velocity and radius). Furthermore, he suggested that the
energy of the electron is restricted to a specific energy level that places the electron at that fixed
circular distance from the nucleus. He called this the orbit of the electron.
Bohr’s model worked very well for a simple atom such as hydrogen but it did not explain the
behavior of more complex (multi-electron) atoms and left many unanswered questions. Also,
Bohr’s model required that we know the position and speed (momentum) of a particle at a given
instant. However, Werner Heisenberg proposed the uncertainty principle stating that “it is
impossible to simultaneously determine the exact position and the exact momentum (product of
speed and mass) of a body, for instance an electron. Heisenberg also rejected Bohr’s concept of
an “orbit” of fixed radius for an electron; he asserted that the best we can do is to provide the
probability of finding an electron with a given energy within a given space.
Problem 2.13 Describe the Uncertainty Principle. How does this principle contradict Bohr’s
model of the atom?
Answer 2.13: The uncertainty principle states that it is impossible to simultaneously determine
the exact position and the exact momentum (product of speed and mass) of a body. Any attempt
at measurement of, for instance, position would alter the velocity and vice versa. For Bohr’s
theory to work, he needed the knowledge of the position and momentum of an electron
simultaneously.
Problem 2.14 Describe the following terms (give a diagram for each): a) electron density diagram,
b) orbital, c) boundary surface representation, and d) radial probability.
Answer 2.14:
a) Electron density diagram – An array of dots representing the probability of finding an electron
(electron density) of a given energy level in a given region of space (see Figure 2.7 a).
b) Orbital – not to be confused with Bohr’s “orbit” is a wave function that is the solution to the
wave equation. An orbital has a characteristic energy level as well as a characteristic distribution
of electron density (expressed geometrically in Figure 2.8).
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
16
c) Boundary surface representation – Another way to probabilistically represent the location of an
electron with a given energy level is by drawing the boundary inside which we have a 90%
chance of finding that electron (see Figure 2.7 b).
d) Radial probability – radial probability also called total probability, considers the probability of
the electron being at a spherical layer with respect to the volume of that layer (see Figure 2.7 c).
Problem 2.15 Name and describe all quantum numbers.
Answer 2.15:
1-The Principal Quantum Number, n – represents the principal energy level and only takes on
integer values of one or greater than one, n = 1, 2, 3, … Each principal energy level is also known
as a shell representing a collection of subshells and orbitals with the same principal number, n.
As n increases, so does the energy of the electron under consideration, indicating that the
electron is less tightly bonded to the nucleus (easier to ionize).
2-The Subsidiary Quantum Number, l – within each principal shell, n, there exists a subshell. The
shape of the electron cloud or the boundary space of the orbital is determined by this number.
The quantum number l may be represented by an integer ranging from 0 to n-1, or by letters.
3-The Magnetic Quantum Number, m l – represents the orientation of the orbitals within each
subshell. The quantum number, m l – will take on values ranging from + l to – l.
4-The Spin Quantum Number, ms – represents the direction of the spin of the electron. The spin
quantum number can take on either +1/2 or -1/2.
Problem 2.16 Explain the Pauli’s exclusion principle.
Answer 2.16: No more than two electrons can occupy the same orbital of an atom, and the two
electrons must have opposite spins. In other words, no two electrons can have the same set of
four quantum numbers.
Problem 2.17 Describe (a) the nucleus charge effect and (b) the shielding effect in multi-electron
atoms.
Answer 2.17:
(a) The higher the charge of the nucleus (more protons), the higher the attraction force on an
electron, and the lower the energy of the electron (a more stable system); this is called the
nucleus charge effect.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
17
(b) The shielding effect takes place when there is more than one outer electron. In this case,
the outer electrons repel each other because of their charge similarity. This repulsion energy
works against the attraction energy between the nucleus and the electrons. As a result, it is
easier to remove these electrons from the nucleus compared to a situation where only one
outer electron exists.
Problem 2.18 Describe the terms a) metallic radius, b) covalent radius, c) first ionization energy,
d) second ionization energy, e) oxidation number, f) electron affinity, g) metals, i) nonmetals,
k) metalloids, and l) electronegativity.
Answer 2.18:
a) Metallic radius- is a measure of the size of an atom equal to half the distance between the
nuclei of two adjacent atoms in a solid metallic element.
b) Covalent radius – is a measure of the size of an atom equal to half the distance between the
nuclei of the identical atoms within the covalent molecule.
c) First ionization energy (IE1)- the energy required for the removal of the outermost electron in an
atom.
d) Second ionization energy – the energy required for the removal of the second outermost
electron in an atom after the first outermost electron has already been removed.
e) Oxidation number – the number of outer electrons that an atom can give up or receive through
the ionization process.
f) Electron affinity – the tendency to easily accept an outermost electron.
g) Metals – those elements with atoms that have low ionization energies and little to no electron
affinity (Groups 1A and 2A are exclusively metallic).
i) Nonmetals – elements with atoms that have a high ionization energy and very high electron
affinity (Group 6A and 7A are exclusively nonmetallic).
k) Metalloids – elements that can behave either in a metallic or a nonmetallic manner (some
elements in group 3A, 4A, and 5A; see the periodic table).
l) Electronegativity – indicates the degree by which an atom attracts electrons to itself.
Problem 2.19 Compare and contrast the three primary bonds in detail (draw a schematics for
each). Explain what the driving force in the formation of such bonds or, in other words, why do
atoms want to bond at all?
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
18
Answer 2.19:
Ionic Bonds – metals and nonmetals bond through electron transfer and ionic bonding. Ionic
bonding is typically observed between atoms with large differences in their electronegativities;
for instance atoms of group 1A or 2A (reactive metals) with atoms of group 6A or 7A (reactive
nonmetals). In short, one atom loses an electron and forms a cation, another atom gains the
electron lost by the first atom and forms an anion. After the electron transfer process is
completed, both atoms will have completed their outer electronic structure and take on the
structure of a noble gas. The electrostatic attraction forces between the two ions will then hold
the ions together to form an ionic bond.
Covalent Bonds – are typically observed between atoms with small differences in their
electronegativities, and mostly between non-metals. At first, the nucleus of one atom attracts the
electron cloud of the other; the atoms get closer to each other. As they get close, the two
electron clouds interact, and both atoms start to take ownership of both electrons (share
electrons). The atoms keep getting closer until they reach the equilibrium point in which the two
atoms will form a bond by sharing their electrons, both completing their outer electronic
structure, and reaching the lowest state of energy.
Metallic Bonds – during solidification, from a molten state, the atoms of a metal pack tightly
together, in an organized and repeating manner. All the atoms contribute their valence electrons
to a “sea of electrons” or the “electron charge cloud”. These valence electrons (free electrons)
are delocalized, move freely in the sea of electrons, and do not belong to any specific atom. The
nuclei and the remaining core electrons of tightly packed atoms form a cationic or a positive
core. What keeps the atoms together, in solid metals, is the attraction force between the positive
ionic core (metal cations) and the negative electron cloud.
The driving for atoms to bond with other atoms through primary bonds is to lower their potential
energy levels and become more stable.
Problem 2.20 Describe the factors that control packing efficiency (number of neighbors) in ionic
and covalent solids. Give an example of each type of solid.
Answer 2.20: In ionic solids, the number of cations that can pack around an anion (packing
efficiency) is determined by two factors: 1) their relative sizes and 2) charge neutrality. Example
six Cl- anions can pack around one Na+ cation.
In covalent solids the number of neighbors (packing efficiency) around an atom will depend on
the bond order (the number of shared pairs). The number of neighbors cannot be greater than
four.
Problem 2.21 Describe the five stages leading to formation of an ionic solid. Explain which stages
require energy and which stages release energy.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
19
Answer 2.21:
Stage 1: Solid metal to gaseous metal (atomization) – requires energy – DH1
Stage 2: Nonmetal molecules to nonmetal atoms – requires energy – DH2
Stage 3: Metal atoms, removing outer electrons – requires energy – DH3
Stage 4: Nonmetal atoms, adding outer electrons – produces energy – DH4
Stage 5: Formation of ionic solid from gaseous ions – produces energy – DH5
Problem 2.22 Describe a) Hess law, b) lattice energy and b) heat of formation.
Answer 2.22:
(a) Hess Law states that the total heat produced during formation of an ionic solid is the sum
of the heats required in each five stage: DH0 = DH1 + DH2 + DH3 + DH4 + DH5.
(b) Lattice energy, DH5, is the energy released when gaseous ions form solid ions due to
electrostatic attraction forces.
(c) The total heat, DH0, is called the heat of formation.
Problem 2. 23 Describe the terms a) shared pair, b) bond order, c) bond energy, d) bond length,
e) polar and non-polar covalent bonds, and f) network covalent solid.
Answer 2.23:
a) shared pair – the pair of electrons active in the bond between two covalently bonded atoms.
b) bond order – the number of shared pairs between covalently bonded atoms (maximum of four)
c) bond energy – the energy required to break the bonds between covalently bonded atoms.
d) bond length – the distance between the nuclei of two covalently bonded atoms at the
equilibrium point (point of minimum energy).
e) polar and non-polar covalent bonds – if the difference between the electronegativities of the
covalently bonded atoms is zero, the bond is non-polar (no dipole). As the difference in
electronegativity between the atoms increases, the bonds become polar. This means the shared
electron will lean toward the more electronegative element, thus creating a dipole.
f) network covalent solid – when all atoms in a solid are bonded through covalent bonds in the
form of an ordered network (example diamond).
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
20
Problem 2.24 Explain the hybridization process in carbon. Use orbital diagrams.
Answer 2.24: The full 2S orbital is promoted to a 2p orbital to form 4 partially occupied hybrid
SP3 orbitals. Each hybrid orbital will be available for bonding with another atom for a
maximum of 4.
Problem 2.25 Describe the properties (electrical, mechanical, etc…) of materials that are
exclusively made up of a) ionic bonds, b) covalent bonds, and c) metallic bonds. Name a material
for each type.
Answer 2.25:
a) ionic solids – will be hard (difficult to indent), strong (difficult to deform permanently or fail), stiff
(difficult to deform elastically), brittle (deform little before they fail), electrically insulating (in the
solid state), and have a high melting temperature. Examples are MgO and CsCl.
b) network covalent solids – will be hard (difficult to indent), strong (difficult to deform
permanently or fail), stiff (difficult to deform elastically), brittle (deform little before they fail), have
a low thermal conductivity, and have a high melting temperature. Examples are quartz and
diamond.
c) metallic solids – in a pure state, they are generally more malleable (soft and deformable), and
are less stiff than ionic or covalent networked materials. Strength can be increased through
alloying. Highly conductive (both heat and electricity). Examples are copper and aluminum.
Problem 2.26 What are secondary bonds? What is the driving force for formation of such bonds?
Give example of materials in which such bonds exist.
Answer 2.26: The bonding formed between molecules or atoms of noble gasses. Significantly
weaker than primary bonds. The driving force is the electrostatic attraction between polar
molecules and atoms.
1s 2s Two half-filled 1s
2p orbitals
Ground-state orbital
arrangement
sp3 hybridized orbital
arrangement
Hybridization
Four equivalent half-filled
sp3 orbitals
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
21
Problem 2.27 Discuss various types of mixed bonding.
Answer 2.27: Although different solids may be more inclined to form predominantly by a certain
type of bond, other types of bonds will also be normally present. For instance, it is possible to
have i) ionic-covalent, ii) metallic-covalent, and iii) metallic-ionic combinations.
Problem 2.28 Define the following terms: a) dipole moment, b) fluctuating dipole, c) permanent
dipole, d) van der Waals bonds, and e) hydrogen bond.
Answer 2.28:
a) dipole moment – a moment that produces a temporary or permanent separation of positive
and negative charge centers in an atom or a molecule (moment = q*d).
b) fluctuating dipole – Fluctuating dipole bonding is a secondary type of bonding between atoms
which contain electric dipoles. These electric dipoles, formed due to the asymmetrical electron
charge distribution within the atoms, change in both direction and magnitude with time. This
type of bond is electrostatic in nature, and is very weak and nondirectional.
c) permanent dipole – Permanent dipole bonding is also a secondary type of bonding between
molecules possessing permanent electric dipoles. The bonds, formed by the electrostatic
attraction of the dipoles, are directional in nature. They are slightly stronger than the fluctuating
dipole.
d) van der Waals bonds – all bonds involving dipoles are collectively called van der Waals bonds
(forces).
e) hydrogen bond – a special class of permanent dipole bonds forming between polar molecules
containing hydrogen.
Problem 2.29 The diameter of a soccer ball is approximately 0.279 m (11 inches). The diameter
of the moon is 3.476 x 106 m. Give an “estimate” of how many soccer balls it will take to cover the
surface of the moon (assume moon is a sphere with a flat terrain). Compare this number to
Avogadro’s number. What is your conclusion?
Solution 2.29
Surface area of the moon = 4pR2 where R is the radius of the moon
2
6 13 2
s,moon
3.476
A 4 10 3.79 10 m
2
æç ÷ö = pçç ´ ÷÷÷ = ´ è ø
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
22
Cross-sectional area of soccer ball = pR2
2
2
cs,ball
0.279
A 0.0611m
2
æç ö÷ =pçç ÷÷÷ = è ø
Number of soccer balls required to cover the surface of the moon
13 2
14
2
3.79 10 m
6.22 10 balls
0.0611m
= ´ = ´
Avogadro’s number is ~ 1 billion times larger.
Problem 2.30 Each quarter produced by the US mint is made up of a copper and nickel alloy. In
each coin, there is 0.00740 moles of Ni and 0.0886 moles of copper. (a) What is the total mass of
a quarter? (b) What percentage of the mass of a quarter is nickel and what percentage is copper?
Solution 2.30
In each coin:
0.00740 moles of Ni
0.0886 moles of Cu
The masses of Ni & Cu are
58.69 gr
0.0074 moles 0.434 gr Ni
mole
´ =
63.55 gr
0.0886 moles 5.63 gr Cu
mole
´ =
Total mass = 0.434 + 5.63 = 6.06 gr
0.434
% Ni 100 7.16%
6.06
= ´ =
5.63
% Cu 100 92.9%
6.06
= ´ =
Problem 2.31 Sterling silver contains 92.5 wt% silver and 7.5 wt% copper. Copper is added to
silver to make the metal stronger and more durable. A small sterling silver spoon has a mass of
100 grams. Calculate the number of copper and silver atoms in the spoon.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
23
Solution 2.31
Sterling silver: 92.5 wt % Ag + 7.5 wt % Cu
Mass of spoon = 100 gr
Mass of silver = mAg = 92.5 gr  number of silver atoms
6.02 1023 atomsmol 92.5 g
107.9 grmol
´
= ´
= 5.2 ´1023 atoms (Ag)
Mass of copper = mCu = 7.5 gr  number of copper atoms
6.02 1023 atomsmol 7.5 g
63.55 grmol
´
= ´
= 7.1 ´1022 atoms (Cu)
Total number of atoms in the spoon = 5.91 ´1023 atoms
Note: it is smaller than NA
Problem 2.32 There are two naturally occurring isotopes for boron with mass numbers 10
(10.0129 amu) and 11 (11.0093 amu); The %abundances are 19.91 and 80.09 respectively.
(a) Find the average atomic mass and (b) the relative atomic mass (or atomic weight) of boron.
(c) Compare your value with that presented in the periodic table.
Solution 2.32
Boron isotopes : 10 B – 10.0129 amu 19.91%
11 B – 11.0093 amu 80.09%
Average atomic mass = êëé(10.0129´0.1991)+(11.0093´0.8009)úûù
= éë1.993+8.817ùû = 10.81 amu
Relative atomic mass = 10.81
Comparing with the value in the periodic table for B, we have a match.
Problem 2.33 A monel alloy consists of 70 wt % Ni and 30 wt % Cu. What are the atom
percentages of Ni and Cu in this alloy?
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
24
Solution 2.33 Using a basis of 100 g of alloy, there are 70 g of Ni and 30 g of Cu. The number of
gram-moles of each element is thus,
30 g
No. of gram-moles of Cu 0.472mol
63.54 g/mol
70 g
No. of gram -moles of Ni 1.192 mol
58.71 g/mol
Total gram-moles
= =
= =
= 1.664 mol
The atomic percentages may then be calculated as,
( )
( )
0.472mol
Atomic % Cu 100%
1.664 mol
1.192mol
Atomic % Ni 100%
1.664 mol
é ù
= ê ú = êë úû
é ù
= ê ú = êë úû
28.4 at%
71.6 at%
Problem 2.34 What is the chemical formula of an intermetallic compound that consists of 15.68
wt % Mg and 84.32 wt % Al?
Solution 2.34 The chemical formula, MgxAly, may be determined based on the gram-mole
fractions of magnesium and aluminum. Using a basis of 100 g of intermetallic compound,
15.68 g
No. of gram-moles of Mg 0.645 mol
24.31 g/mol
84.32 g
No. of gram-moles of Al 3.125 mol
26.98 g/mol
Total gram-moles
= =
= =
= 3.770 mol
Thus we have Mg0.17Al0.83 or, multiplying by 6, MgAl5.
0.645mol
Gram-mole fraction of Mg
3.770mol
3.125mol
Gram-mole fraction of Al
3.770mol
x
y
é ù
= = ê ú = êë úû
é ù
= = ê ú = êë úû
0.17
0.83
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
25
Problem 2.35 In order to raise the temperature of 100 grams of water from room temperature
(20oC) to boiling temperature (100oC), an energy input of 33,440.0 J is required. If one uses a
microwave oven (I, of radiation of 1.20 cm) to achieve this, how many photons of the microwave
radiation are required?
Solution 2.35
33,440 joules of energy to raise water temp from 20°C to 100°C
l microwave = 1.20 cm = 0.012 m
Emw photon
( 34 )( 8 )
c
mw photon
6.63 10 J sec 3.00 10 m h sec
0.012 m
´ – ⋅ ´
= =
l
Emw photon = 1.66 23 10-
´ J
 Number of photons needed 27
23
33,440 J
2.0 10
1.66 10- J = = ´
´
Note: number is greater than NA by ~ 10,000 times.
Problem 2.36 For problem 2.35, determine the number of photons to achieve the same increase
in temperature if (a) ultraviolet (I = 1.0 x 10-8 m), visible (I = 5.0 x 10-7 m), and infrared (I = 1.0 x 10-4 m)
lights were used. What important conclusions can you draw from this exercise?
Solution 2.36
Repeating problem 2.35 for UV and visible red lights
EUV photon
( 34 )( 8 )
8
6.63 10 J sec 3.00 10 msec
1.0 10 m
–
–
´ ⋅ ´
=
´
a) Ultraviolet EUV photon = 1.989´10-17 J
l= 1´10-8m number of UV photons needed 21
17
33,440 J
1.68 10
1.98 10- J = = ´
´
Evis photon
( 34 )( 8 )
7
6.63 10 J sec 3.00 10 msec
5.0 10 m
–
–
´ ⋅ ´
=
´
b) Visible Evis photon = 4.0´10-19 J
l= 5.0´10-7m number of vis photons needed 22
19
33,440 J
8.36 10
4.0 10- J = = ´
´
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
26
EIR photon
( 34 )( 8 )
4
6.63 10 J sec 3.00 10 msec
1.0 10 m
–
–
´ ⋅ ´
=
´
c) Infrared EIR photon = 2.0´10-21J
l= 1´10-4m number of IR photons needed 25
21
33,440 J
1.6 10
2.0 10- J = = ´
´
Note: EUV > Evis > EIR > Emw
Problem 2.37 In order for the human eye to detect the visible light, its optical nerves must be
exposed to a minimum energy of 2.0 x 10-17 J. (a) Calculate the number of photons of red light
needed to achieve this (I = 700 nm). (b) Without any additional calculations, determine if you
would need more or less photons of blue light to excite the optical nerves?
Solution 2.37
a) Energy for detection of light by optic nerve = 2.0 ´ 10-17 J
number of photons of red light needed ( l = 700 nm)
Ered
( 34 )( 8 )
19
9
6.63 10 J sec 3.00 10 msec 2.84 10 J
700 10 m
–
–
–
´ ⋅ ´
= = ´
´
number of red photons =
17
19
2.0 10 J
70.4
2.84 10 J
–
–
´ »
´
photons
(very little energy is needed).
b) Blue light has a lower wavelength 450-495 nm. Thus, each photon of blue light will have
more energy. You would need less photons to detect blue light.
Problem 2.38 Represent the wave length of the following rays by comparing each to the length
of a physical object (example: a ray with a wavelength of 1m (100 cm) would be approximately
that of a baseball bat) : a) rays from a dental ray, b) rays in a microwave oven, c) rays in a sun
lamp, d) rays in a heat lamp, and e) an FM radio wave.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
27
Solution 2.38
Dental X-ray : l 10-1 nm ( 100 times the diameter of an atom)
Micro wave : l 1mm to 1m (thickness of 10 sheets of paper to the length of a baseball bat)
Heat lamp : l 750 nm to 1mm (a bacterial cell to thickness of 10 sheets of paper)
(Infrared)
Sun lamp : l400 nm to 10 nm (size of a virus)
(UV)
FM radio : l~ 1m to 10m (baseball bat to a flagpole)
(radio wave)
Problem 2.39 For the rays in problem 2.38, without any calculations, rank them in increasing
order of the energy of the radiation.
Solution 2.39
As wavelength increases, energy decreases
As wavelength decreases, energy increases
FM radio  microvave  heat lamp  sun lamp  X ray
(low energy) (high energy)
Problem 2.40 In a commercial x-ray generator, a stable metal such as copper (Cu) or tungsten
(W) is exposed to an intense beam of high-energy electrons. These electrons cause ionization
events in the metal atoms. When the metal atoms regain their ground state they emit x-rays of
characteristic energy and wavelength. For example, a “tungsten” atom struck by a high-energy
electron may lose one of its K shell electrons. When this happens, another electron, probably
from the tungsten L shell will “fall” into the vacant site in the K shell. If such a 2p  1s transition
occurs in tungsten, a tungsten Ka x-ray is emitted. A tungsten Ka x-ray has a wavelength l of
0.02138 nm. What is its energy? What is its frequency?
Solution 2.40
34 8
9
15
34
(6.63 10 J s)(3.00 10 m/s)
(0.02138 nm)(10 m/nm)
E 9.30 10 J
h 6.63 10 J s
hc
E
l
n
–
–
–
–
= = ´ ⋅ ´ =
= = ´ =
´ ⋅
-15
19
9.30× 10 J
1.40 × 10 Hz
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
28
Problem 2.41 A hydrogen atom exists with its electron in the n = 4 state. The electron undergoes
a transition to the n = 3 state. Calculate (a) the energy of the photon emitted, (b) its frequency,
and (c) its wavelength in nanometers (nm).
Solution 2.41
(a) Photon energy emitted is:
2 2 2
13.6 13.6 13.6
4 3
E
n
é- ù é- ù é- ù D = Dê ú = ê ú-ê ú = êë úû êë úû êë úû
0.66 eV = 1.06 × 10-19 J
(b) Photon frequency is found as:
19
-34
1.06 10 J
6.63 10 J s
E
h
n
D ´ – = = =
´ ⋅
1.6 × 1014 Hz
(c) The wavelength is given as:
-34 8
-19 -9
(6.63 10 J s)(3.00 10 m/s)
(1.06 10 J)(10 m/nm)
hc
E
l = = ´ ⋅ ´ =
D ´
1876 nm
Problem 2.42 A hydrogen atom exists with its electron in the n = 6 state. The electron undergoes
a transition to the n = 2 state. Calculate (a) the energy of the photon emitted, (b) its frequency,
and (c) its wavelength in nanometers.
Solution 2.42
(a) Photon energy emitted is:
2 2 2
13.6 13.6 13.6
6 2
E
n
é- ù é- ù é- ù D = Dê ú = ê ú-ê ú = êë úû êë úû êë úû
3.02 eV = 4.84 × 10-19 J
(b) Photon frequency is found as:
19
-34
4.84 10 J
6.63 10 J s
E
h
n
D ´ – = = =
´ ⋅
7.3 × 1014 Hz
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
29
(c) The wavelength is given as:
-34 8
-19 -9
(6.63 10 J s)(3.00 10 m/s)
(4.84 10 J)(10 m/nm)
hc
E
l = = ´ ⋅ ´ =
D ´
410 nm
Problem 2.43 Using the information given in Examples 2.4 and 2.5 determine the uncertainty
associated with the electron’s position if the uncertainty in determining its velocity is 1%. Compare
the calculated uncertainty in the position with the estimated diameter of the atom. What is your
conclusion?
Solution 2.43
Speed of electron = 16.67 % of speed of light
= 0.166 × 3 × 108 = 50.0 × 106 m/sec
1% uncertainty in measurement of speed = 0.01 × 50.0 × 106 = 500 × 103
Uncertainty in position ( )
2
34
31 3
6.62 10 kg.m h s Δx
4π m Δ u 4 π 9.11 10 kgr (500 10 kgr)
–
–
´
³ =
´ ´ ´
Δx ≥ 1.15 × 10-10 m » 0.115 nm
Note: diameter of atom is ~ 0.1 nm. Uncertainty in the position of electron will be close to the size
of the atom.
Problem 2.44 Repeat problem 2.43 to determine the uncertainty associated with the electron’s
position if the uncertainty in determining its velocity is 2%. Compare the calculated uncertainty in
the position with that of problem 2.37. What is your conclusion?
Solution 2.44
2% uncertainty associated with speed
= 0.02 × 50.0 × 106 m/s = 1000 × 103 m/s
 The corresponding uncertainty in position will be ( )( )
2
34
31 3
6.62 10 kg.m Δx s
4 π 9.11 10 kgr 1000 10 m/s
–
–
´
³
´ ´
Δx ³ 5.268 ´ 10-11 m » 0.0526 nm
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
30
Note: the uncertainty in position decreases to the radius of the atom as the uncertainty in the
speed of electron increases to 2%.
Problem 2.45 For the principal quantum number, n, of value 4, determine all other possible
quantum numbers for l and ml.
Solution 2.45
Problem 2.46 For each pair of n and l given below, give the sublevel name, possible values of
ml, and the corresponding # of orbitals.
(a) n=1, l=0
(b) n=2, l=1
(c) n=3, l=2
(d) n=4, l=3
s mL
mL
mL
mL
−1
−2
−1
−3
−2
−1
0
0
0
1
2
3
1
2
1
0
P 1
d 2
f 3
n=4 => L=
0
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
31
Solution 2.46
Sublevel name ml number of orbitals
a) n=1, l=0  1s 0 1
b) n=2, l=1  2p -1,0,+1 3
c) n=3, l=2  3d -2,-1,0,+1,+2 5
d) n=4, l=3  4f -3,-2,-1,0,+1,+2,+3 7
Problem 2.47 Determine if the following combinations of quantum numbers are acceptable.
(a) n=3, l=0, ml=+1
(b) n=6, l=2, ml=-3
(c) n=3, l=3, ml=-1
(d) n=2, l=1, ml=+1
Solution 2.47
a) n=3, l=0 ml = +1 : not possible, ml = 0
b) n=6, l=2 ml = -3 : not possible, -2 < ml < +2
c) n=3, l=3 ml = -1: not possible, l = 0,1,2 for n=3
d) n=2 l=1 ml = +1 : possible
Problem 2.48 In each row (a through d) there is only one piece of information that is wrong,
highlight the information that is wrong (explain why).
n l m l Name
(a) 3 0 1 3s
(b) 2 1 -1 2s
(c) 3 1 +2 3d
(d) 3 3 0 4f
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
32
Solution 2.48
(Note: only one entry is wrong)
n l ml name
a) 3 0 1 3s : ml  0 since l=0
b) 2 1 -1 2s : name 2p since l=1
c) 3 1 +2 3d : l  2 since ml > +2 and name is 3d
d) 3 3 0 4f : n 4 since l=3 and name is 4f
Problem 2.49 Determine the four quantum numbers for the 3rd, 15th, and 17th electrons of the
Cl atom.
Solution 2.49
Cl has 17 electrons
3rd electron of Cl atom: n = 2, l = 0, s, ml = 0 ms = +1/2
15th electron of Cl atom: n = 3, l = 1, p, ml =+1 ms = +1/2
17th electron of Cl atom: n = 3, l = 1,p, ml = 0 ms = -1/2
Problem 2.50 Determine the electron configuration and group number of the atom in the ground
state based on the given partial (valence level) orbital diagram. Identify the element.
Solution 2.50
The outer electron structure is 4s2p1.
This means 28 subvalent electrons.
Total number of electrons will be 31.
The element is Ga (Gallium)
(Alternatively, the element is in the 3rd column since it has 3 outer electrons and fourth period
since 4s24p7  Ga)
Problem 2.51 Write the electron configurations of the following elements by using spdf notion:
(a) yttrium, (b) hafnium, (c) samarium, (d) rhenium.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
33
Solution 2.51
(a) Y (Z = 39): [Kr] 4d 15s (c) Sm (Z = 62): [Xe] 4f 66s 2
(b) Hf (Z = 72): [Xe] 4f 145d 26s 2 (d) Re (Z = 75): [Xe] 4f 145d 56s 2
Problem 2.52 Write the electron configuration of the following ions by using spdf notation:
(a) Cr2+, Cr3+, Cr6+; (b) Mo3+, Mo4+, Mo6+; (c) Se4+, Se6+, Se2-.
Solution 2.52
(a) Cr [Ar] 3d 5 4s 1 (b) Mo [Kr] 4d 5 5s 1 (c) Se [Ar] 3d 10 4s 24p 4
Cr2+ [Ar] 3d 4 Mo3+ [Kr] 4d 3 Se4+ [Ar] 3d 10 4s 2
Cr3+ [Ar] 3d 3 Mo4+ [Kr] 4d 2 Se6+ [Ar] 3d 10
Cr6+ [Ar] Mo6+ [Kr] 4d 3 Se2 – [Ar] 3d 10 4s 24p 6
Problem 2.53 Rank the following atoms in (a) increasing atomic size and (b) decreasing first
ionization energy, IE1. Use only the periodic table to answer the questions. Check your answer
using Figures 2.10 and 2.11.
i) K, Ca, Ga
ii) Ca, Sr, Ba
iii) I, Xe, Cs
Solution 2.53
i)
a) Atomic size (increasing)
As we move to the left in a period, the atomic size increases
Ga : period 4 group 3
Ca: period 4 group 2  Ga, Ca, K (increasing order)
K: period 4 group 1
b) Ionization energy (decreasing)
As we move to the left in a period, the ionization energy increases.
Ga, Ca, K (decreasing order)
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
34
ii)
a) Atomic size (increasing)
Ca : period 4 group 2
Sr: period 5 group 2
Ba: period 6 group 2
As we move down in a group, atomic size increases
 Ca, Sr, Ba (increasing size)
b) Ionization energy (decreasing)
Ionization energy will have the opposite trend as that of atomic size.
(as we move down in a group, ionization energy decreases)
Thus, Ca, Sr, Ba (decreasing)
iii)
a) Atomic size (increasing)
I : period 5 group 7
Xe: period 5 group 8
Cs: period 6 group 1
In period 5, Xe (group 8) will have a smaller radius than I (group 7).
Cs in period 6 will be larger than both.
Cs, I, Xe (increasing size)
b) Ionization energy (decreasing)
In period 5, I will have a lower ionization energy than Xe.
Cs, in period 6, will have the lowest ionization energy.
Xe, I, Cs (decreasing)
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
35
Problem 2.54 Rank the following atoms in (a) increasing atomic size and (b) decreasing first
ionization energy, IE1. Use only the periodic table to answer the questions. Check your answer
using Figures 2.10 and 2.11.
i) Ar, Li, F, O, Cs C
ii) Sr, H, Ba, He, Mg, Cs
Solution 2.54
a) Increasing atomic size
i)
Ar : period 3, group 8
Li : period 2, group 1
F : period 2, group 7
O : period 2, group 6
Cs : period 6, group 1
C : period 2, group 4
Period 2 elements will be smaller than period 3, and both will be smaller than period 6.
period 2 Li, F, O, C
1st organization period 3 Ar Increasing size
Period 6 Cs
Within each period, the size decreases as you move to the right
2nd organization Li F
C, O
Increasing size O, C
F, Li
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
36
Argon (Ar), the last element in period 3, will be smaller than Li but larger than F
F
O
C
Ar
Li
Cs
Cs, will be the largest since it is in period 6.
Problem 2.55 The first ionization energies of two atoms with electronic configurations
(a) 1S22s22p6 and (b) 1s22s22p63s1 are given to be 2080 kJ/mol and 496 kJ/mol. Determine which
IE1 belongs to which electronic structure and justify your answer.
Solution 2.55
1s22s22p6 This atom belongs to group VIII. Its outer electron structure is complete.
It will be difficult to ionize this atom. Thus, it must have a high IE1.
1s22s22p63s1 This atom belongs to group I, period III. There is a single electron in n=3.
It will be easier to ionize, so it must have a lower IE1.
Problem 2.56 The first ionization energies of three atoms with electronic configurations
(a) [He]2s2, (b) [Ne]3s1, and (c) [Ar]4s1 and (d) [He]2S1 are given to be 496 kJ/mol, 419 kJ/mol,
520 kJ/mol, and 899 kJ/mol. Determine which IE1 belongs to which electronic structure and
explain your answer.
Solution 2.56
[He] 2s2
[Ne]3s1
[Ar]4s1
[He]2s1
[He]2s1, [Ne]3s1 and [Ar]4s1 all belong to group I. The atom that is located in the lowest period will
have the lowest IE1. Thus,
IE1 [Ar]4s1 = 419 KJ/mol
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
37
IE1 [Ne]3s1 = 496 KJ/mol
IE1 [He]2s1 = 520 KJ/mol
[He]2s2 belongs to the second group in period II. Thus, its IE1 must be higher than [HE]2s1. And so,
IE1 [He]2s1 = 899 KJ/mol
Problem 2.57 Similar to Figure 2.15 use (a) orbital diagrams and (b) Lewis symbols to explain the
formation of Na+ and O2- ions and the corresponding bonding. What is the formula of the
compound?
Solution 2.57
a) Orbital diagram
Na
1s2 2s2 2p6
+
Na
 Na+ O2-
+
Na+
2p4 3s1
1s2 2s
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
38
b) Lewis symbols
Chemical formula = Na2O
Problem 2.58 Calculate the attractive force (·¬·) between a pair of Ba2+ and S2- ions that
just touch each other. Assume the ionic radius of the Ba2+ ion to be 0.143 nm and that of the
S2- ion to be 0.174 nm.
Solution 2.58 The attractive force between the ion pair is found by applying Coulomb’s law,
2
1 2
attractive 4 2 o o
Z Z e
F
pe a
–
=
Where Z1 = +2 for Ba2+, Z2 = -2 for S 2–, and
2 2
-10
Ba S 0.143 nm 0.174 nm 0.317 nm 3.17 10 m o a r + r – = + = + = = ´
Substituting,
19 2
attractive -12 2 2 10 2
( 2)( 2)(1.60 10 C)
4 (8.85 10 C / N m )(3.17 10 m)
F
p
–
–
= – + – ´ =
´ ⋅ ´
9.16 × 10-9 N
Na
.
Na
.
Na+
Na+
+ +
2−
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
39
Problem 2.59 Calculate the net potential energy for a Ba2+ S2- ion pair by using the b constant
calculated from Prob. 2.58. Assume n = 10.5.
Solution 2.59 The repulsive energy constant b is:
1 1 9 10 11.5
Repulsive Attractive
-119 10
( ) ( 9.16 10 N)(3.17 10 m)
10.5
5.045 10 N m
n n
o o F a F a
b
n n
– + — + — ´ – ´ – = = =
= ´ ⋅
Thus the net potential energy between the ions is,
2 2
2
1 2
Ba S
19 2 119 10
12 2 2 10 10 10.5
18 19
4
( 2)( 2)(1.60 10 C) 5.045 10 N m
4 (8.85 10 C / (N m ))(3.17 10 m) (3.17 10 m)
( 2.905 10 J) (2.765 10 J)
n
o o o
Z Z e b
E
pe a a
p
+ –
– –
– – –
– –
= + +
= + + – ´ + ´ ⋅
´ ⋅ ´ ´
= – ´ + ´ = -2.63 × 10-18 J
Problem 2.60 If the attractive force between a pair of Cs+ and I- ions is 2.83 × 10-9 N and the ionic
radius of the Cs+ ion is 0.165 nm, calculate the ionic radius of the I- ion in nanometers.
Solution 2.60 From Coulomb’s law,
2 19 2
1 2
12 2 2 9
Attractive
10
( 1)( 1)(1.60 10 C)
4 4 (8.85 10 C /(N m ))(2.83 10 N)
2.852 10 m 0.2852nm
o
o
Z Z e
a
pe F p
–
– –
–
– – + – ´ = =
´ ⋅ ´
= ´ =
The ionic radius of iodine is thus,
0.285 nm 0.165 nm I o Cs r- a r + = – = – =0.120 nm
Problem 2.61 For the each pair of compounds presented below, determine which has the higher
lattice energy (more negative). Explain your answer. Also, which of the five ionic compound do
you think has the highest melting temperature and why? Verify your answers.
(a) LiCl and CsCl
(b) CsCl and RbCl
(c) LiF and MgO
(d) MgO and CaO
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
40
Solution 2.61
a) LiCl CsCl : Cl is common to both compounds. As the size of the cation increases
(Cs > Li), the lattice energy decreases. Thus, LiCl will have higher lattice energy than CsCl.
b) CsCl RbCl : Cl is common to both compounds. The size of Cs is larger than Rb
(Cs >Rb). Thus, RbCl has higher lattice energy than CsCl.
c) LiF MgO : Mg++ and O- – have a higher ionic charge than Li+ and F-. Thus, MgO will
have higher a lattice energy.
d) MgO CaO : O is common to both compounds. All ions have an ionic charge of 2:
Mg++, O- – and Ca++. However, Ca++ is larger in radius than Mg++. Thus, MgO will have higher
lattice energy.
Overall, MgO will have the highest lattice energy, and therefore has the highest melting point.
Problem 2.62 Calculate the lattice energy for the formation of solid NaF if the following
information is given. What does the calculated lattice energy tell you about the material?
109 kJ is required to convert solid Na to gaseous Na
243 kJ is required to convert gaseous F2 to two monatomic F atoms
496 kJ is required to remove the 3s1 electron of Na (form Na+ cation)
-349 kJ of energy (energy is released) to add an electron to the F (form Na- anion)
-411 kJ of energy to form gaseous NaF (heat of formation of NaF)
Solution 2.62
NaF
Atomization of Na, ΔH1 = +109kJ (energy input)
Atomization of F2, ΔH2 = +243kJ (energy input)
Removing 3s1 electrons, ΔH3 = +496kJ (energy input)
Adding electron to F, ΔH4 = -349 kJ (energy released)
To form ionic solid from
gaseous NaF, ΔH5 = lattice energy (unknown)
Heat of formation for NaF = -411 kJ (ΔH0)
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
41
ΔH0 = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5
-411 = +109 + 243 + 496 – 349 + ΔH5
ΔH5 = -910 kJ
As solid NaF is formed, 910 kJ is released.
Problem 2.63 Calculate the lattice energy for the formation of solid NaCl if the following
information is given. What does the calculated lattice energy tell you about the material?
109 kJ is required to convert solid Na to gaseous Na
121 kJ is required to convert gaseous Cl2 to two monatomic Cl atoms
496 kJ is required to remove the 3s1 electron of Na (form Na+ cation)
-570 kJ of energy (energy is released) to add an electron to the Cl
-610 kJ of energy to form gaseous NaCl (heat of formation of NaCl)
Solution 2.63
NaCl
Atomization of Na, ΔH1 = +109kJ (energy input)
Atomization of Cl2, ΔH2 = +121kJ (energy input)
Removing 3s1 electrons, ΔH3 = +496kJ (energy input)
Adding electron to Cl, ΔH4 = -570 kJ (energy released)
To form ionic solid from
gaseous NaCl, ΔH5 = lattice energy (unknown)
Heat of formation for NaCl = -610 kJ (ΔH0)
ΔH0 = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5
-610 = +109 + 121 + 496 – 570 + ΔH5
ΔH5 = -766 kJ
As solid NaF is formed, 766 kJ is released.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
42
Problem 2.64 For each bond in the following series of bonds, determine the bond order, rank
bond length, and rank bond strength. Use only the periodic table. Explain your answers.
(a) S-F; S-Br; S-Cl
(b) C-C; C=C; C=C–
Solution 2.64
a) S-F; S-Br; S-Cl
Note, S is the common element. All bonds have order “one” since all have one shared pair.
Since the diameter of Br is larger than Cl, and Cl is larger than F, the bond length of S-Br will be
greater than S-Cl, and these will be greater than S-F.
Bond strength will have an inverse relation with bond length.
Thus, S-F is stronger than S-Cl, and both are stronger than S-Br
b) C º C; C = C; C-C
The bond order is directly related to number of shared pairs.
C º C, bond order of +3
C = C, bond order of +2
C – C, bond order of +1
As the bond order increases, bond length decreases.
Thus, C º C has a smaller bond length than C = C, and both are smaller than than C – C.
Bond strength is inversely related to bond length. Therefore, C º C is the strongest, and C – C is
the weakest.
Problem 2.65 Rank the following covalently bonded atoms according to the degree of polarity:
C-N; C-C; C-H; C-Br.
Solution 2.65
Note: the electronegativities below the symbol
C – C non polar, no difference in electronegativities
2.5 2.5
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
43
C – N high polarity – electron shared more by N
2.5 3.0
C – H some polarity – electon shared by C
2.5 2.1
C – Br high polarity – electron shared more by C
2.5 2.0
Polarity
Increasing
C – C  C – H  C – N  C – Br
Problem 2.66 List the number of atoms bonded to a C atom that exhibits sp3, sp2, and sp
hybridization. For each, give the geometrical arrangement of the atoms in the molecule.
Solution 2.66 sp3 hybridization: Four atoms are bonded to a central carbon atom in a tetrahedral
arrangement. An example is methane, CH4.
sp2 hybridization: Three atoms are bonded to a carbon atom in a planar arrangement. An
example is ethylene, CH2: CH2.
sp hybridization: Two atoms are bonded to a carbon atom in a linear arrangement. An
example is acetylene, CH:CH.
H C º C H
Methane
Tetrahedron
Acetylene
Linear
Ethylene
H H
C
Plane
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
44
Problem 2.67 Is there a correlation between the electron configurations of the elements
potassium
(Z = 19) through copper (Z = 29) and their melting points? (See Tables 2.7).
Solution 2.67 A possible correlation between the melting points and the electron configurations
of the elements from scandium (Z = 21) through copper (Z = 29) is that unpaired 3d electrons
cause covalent hybridized bonds, and hence give higher melting points to these transition
metals.
Problem 2.68 Compare the percentage ionic character in the semiconducting compounds CdTe
and InP.
Solution 2.68 Applying Pauling’s equation to CdTe and InP compounds can help calculate ionic
character. This equation is:
( ) ( ) 2 % 1 0.25 Xa Xb 100% ionic character e=⎛⎜ − − − ⎞⎟⎜⎜⎝ ⎠⎟⎟
where Xa and Xb represent the electronegativities of the atoms. Looking at Figure 2.14, we see
the eletronegativities are as follows:
Cd= 1.7, Te= 2.1, In= 1.7, P= 2.1
Therefore, when solving the equation using these electronegativities, the %ionic character would
be the same for the two compounds.
( ) ( ) 2 0.25 1.7 2.1 %ionic character 1 e 100% 4% =⎜⎛ − − − ⎞⎟ = ⎜⎜⎝ ⎠⎟⎟
Problem 2.69 39K, 40K, and 41K are the three isotopes of potassium. If 40K has the lowest
%abundance, which other isotope has the highest?
Solution 2.69
Note, from the periodic table, that the relative atomic mass of K is 39.10 grams. This number is
much closer to 39K than to 40K. Thus, given that 41K is the last abundant, 39K must be the most
abundant.
Problem 2.70 Most modern scanning electron microscopes (SEMs) are equipped with energy
dispersive x-ray detectors for the purpose of chemical analysis of the specimens. This x-ray
analysis is a natural extension of the capability of the SEM because the electrons that are used to
form the image are also capable of creating characteristic x-rays in the sample. When the
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
45
electron beam hits the specimen, x-rays specific to the elements in the specimen are created.
These can be detected and used to deduce the composition of the specimen from the wellknown
wavelengths of the characteristic x-rays of the elements. For example:
Element
Wavelength
of Ka x-rays
Cr 0.2291 nm
Mn 0.2103 nm
Fe 0.1937 nm
Co 0.1790 nm
Ni 0.1659 nm
Cu 0.1542 nm
Zn 0.1436 nm
Suppose a metallic alloy is examined in an SEM and three different x-ray energies are detected. If
the three energies are 7492, 5426, and 6417 eV, what elements are present in the sample? What
would you call such an alloy? (Look ahead to Chap. 9 in the textbook.)
Solution 2.70 The elements may be identified by calculating their respective wavelengths.
34 8
-19
10
(6.63 10 J s)(3.00 10 m/s)
(a)
(7492 eV)(1.60 10 J/eV)
1.659 10 m 0.1659 nm
hc
E
l
–
–
= = ´ ⋅ ´
´
= ´ =  Ni
From the table provided, a wavelength of 0.1659 nm corresponds to nickel, Ni.
34 8
-19
10
(6.63 10 J s)(3.00 10 m/s)
(b)
(5426 eV)(1.60 10 J/eV)
2.291 10 m 0.2291 nm
hc
E
l
–
–
= = ´ ⋅ ´
´
= ´ =  Cr
From the table provided, a wavelength of 0.2291 nm corresponds to chromium, Cr.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
46
34 8
-19
10
(6.63 10 J s)(3.00 10 m/s)
(c)
(6417 eV)(1.60 10 J/eV)
1.937 10 m 0.1937 nm
hc
E
l
–
–
= = ´ ⋅ ´
´
= ´ =  Fe
From the table provided, a wavelength of 0.1937 nm corresponds to iron, Fe.
The elements present, nickel, chromium, and iron, are the primary constituents of the austenitic
stainless steels.
Problem 2.71 According to section 2.5.1, in order to form monatomic ions from metals and
nonmetals, energy must be added. However, we know that primary bonds form because the
involved atoms want to lower their energies. Why then do ionic compounds form?
Solution 2.71
Although energy is added during atomization of metals and non metals, when solid ionic
materials are formed, a significant amount of energy is released. The activity will result in a net
release of energy (lattice energy). Thus, the energy of the involved atoms is lowered.
Problem 2.72 Of the noble gases Ne, Ar, Kr, and Xi, which should be the most chemically
reactive?
Solution 2.72 Xenon should be most reactive since its outermost electrons (5s26p6 ) are further
away from the nucleus than the other noble gases, and thus easier to remove.
Problem 2.73 The melt temperature of Na is (89oC) is higher than the melt temperature of K
(63.5oC). Can you explain this in terms of the differences in electronic structure?
Solution 2.73
The electronic structure of Na is 1s2 2s2 2p6 3s1 while that of K is 1s2 2s2 2p6 3s2 3p6 4s1.
The outer electrons in K are farther from the nucleus and thus easier to remove. For this reason,
K has a lower melting temp than Na.
Problem 2.74 The melt temperature of Li (180C) is significantly lower than the melt temperature
of its neighbor Be (1287C). Can you explain this in terms of the differences in electronic
structure?
Solution 2.74 The electronic structure of Li is 1s2 2s1 and that of Be is 1s2 2s2. There are more +
particles in the nucleus of Be than in Li. Thus, the attraction force between the valence electrons
and the nucleus is larger in Be than in Li. Therefore, the melt temperature of Be is higher due to
its higher nucleus charge.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
47
Problem 2.75 The melting point of the metal potassium is 63.5C, while that of titanium is
1660C. What explanation can be given for this great difference in melting temperatures?
Solution 2.75 Both K & Ti are located in period IV of the periodic table. The electronic structure
of K is 1s2 2s2 2p6 3s2 3p6 4s1 and that of Ti is 1s2 2s2 2p6 3s2 3p6 3d24s2. First, as the number of
valence electrons increases, such as in Ti, the attraction force between the electron and the
nucleus also increases. As 3d electrons are introduced (in Ti), the melt temperature further
increases. This is due to more covalent bonding being introduced.
Problem 2.76 Cartridge brass is an alloy of two metals: 70 wt% copper and 30 wt% zinc. Discuss
the nature of the bonds between copper and zinc in this alloy.
Solution 2.76
Both Copper and Zinc are in period IV of the periodic table.
The electronic structure of Copper is 1s2 2s2 2p6 3s2 3d104s1 and that of Zn is 1s2 2s2 2p6 3s2 3p6
3d104s2. The 3d and 4s orbitals of Zinc are full. Thus, it has significantly less tendency for covalent
bonding. When Cu is replaced by Zn, the overall percentage of covalent bonds is dropped. Thus,
brass (70 wt % Cu – 30 wt % Zi) will have a higher percentage of metallic bonds when compared
to Cu. Therefore, Brass will have a lower melt temperature than Cu, and a higher melt
temperature than Zn.
Problem 2.77 After ionization, why is the sodium ion smaller than the sodium atom? After
ionization, why is the chloride ion larger than the chlorine atom?
Solution 2.77 After ionization to the Na+, the Na atom becomes smaller because the electron-toproton
ratio of the Na atom is decreased when the Na+ ion forms. Also, the outer third shell no
longer exists once the 3s 1 electron is lost by the Na atom.
After ionization, the Cl – ion is larger because the electron-to-proton ratio of the chlorine atom is
decreased by the ionization process.
Problem 2.78 Regardless of the type of primary bond, why does this tendency exist for atoms to
bond?
Solution 2.78 The tendency of an atom to bond with other atom is to lower its energy level, and
to exist in the most stable state.
Problem 2.79 Pure aluminum is a ductile metal with low tensile strength and hardness. Its oxide
Al2O3 (Alumina) is extremely strong, hard, and brittle. Can you explain this difference from an
atomic bonding point of view?
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
48
Solution 2.79
Pure aluminum is a metal with mostly metallic bonds. Alumina (Al2O3) is a ceramic with mostly
ionic bonds in a 3-d network. The strength of the ionic bonds is significantly higher than the
metallic bonds. Thus, more energy is required to break the bonds between Al and O than
between Al and Al.
Problem 2.80 Graphite and diamond are both made from carbon atoms. a) List some of the
physical characteristics of each. b) Give one application for graphite and one for diamond.
c) If both materials are made of carbon, why does such a difference in properties exist?
Solution 2.80
a) Graphite: Dark color that can have powder (flake-like) form or fiber form. It is an electrical
conductor. Strength is high in transverse plane but not high in other directions
(anisotropic). It has a layered hexagonal structure. The bonds inside the layer are
covalent, and the bonds between the layers are of secondary (electrostatic) nature.
Diamond: Transparent, hard, strong, very high melt temperature. It is an electrical
insulator. It has a 3-D networked structure. All bonds are covalent.
b) Graphite is used as a lubricant because various layers are weakly bonded and can slide
on top of each other.
Diamond (synthetic type) is used as abrasive and cutting tool materials.
c) The bonding structure for diamond is 3-D covalent network (very strong) but graphite is
layered with very weak forces between the layers.
Problem 2.81 Silicon is extensively used in the manufacturing of integrated circuit devices such
as transistors and light emitting diodes. It is often necessary to develop a thin oxide layer (SiO2)
on silicon wafers. a) What are the differences in properties between the silicon substrate and the
oxide layer? b) Design a process that produces the oxide layer on a silicon wafer. c) Design a
process that forms the oxide layer only in certain desired areas.
Solution 2.81
a) Si is a metalloid (refer to the periodic table, page 32). It has a structure similar to diamond
(fig 2.2). It has excellent semiconductor characteristics. SiO2 is a ceramic. It has excellent
insulative characterstics. The bonding is mostly ionic.
b) Exposing Si to an oxygen atmosphere at high temperatures will create a thin oxide layer
on the surface of Si.
c) After oxidation, we can expose certain regions to chemical etchants and remove the
oxide material. This is often done by applying a mask with a specific pattern to the surface
of Si (see figure 14.44).
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
49
Problem 2.82 How can the high electrical and thermal conductivities of metals be explained by
the “electron cloud” model of metallic bonding? Ductility?
Solution 2.82 The high electrical and thermal conductivities of metals are explained by the
mobility of their outer valence electrons in the presence of an electrical potential or thermal
gradient. The ductility of metals is explained by the bonding “electron gas” which enables atoms
to pass over each other during deformation, without severing their bonds.
Problem 2.83 Describe fluctuating dipole bonding among the atoms of the noble gas neon. Of a
choice between the noble gases krypton and xenon, which noble gas would be expected to
have the strongest dipole bonding and why?
Solution 2.83 A fluctuating electric dipole exists in the atoms of noble gases, such as neon,
because there is, at any instant, an asymmetrical distribution of electrical charge among their
electrons. The noble gas xenon would be expected to have a stronger fluctuating dipole
moment than krypton since it has an additional electron shell; the krypton atom has four electron
shells while the xenon atom has five. The electrons of this fifth shell, being further away from the
xenon nucleus, are able to fluctuate more and, thus, create greater asymmetry of charge.
Problem 2.84 Carbon tetrachloride (CCl4) has a zero dipole moment. What does this tell us about
the C—Cl bonding arrangement in this molecule?
Solution 2.84 Since the molecule CCl4 has a zero dipole moment, the C—Cl bonding
arrangement must be symmetrical about the carbon nucleus.
Problem 2.85 Methane (CH4) has a much lower boiling temperature than does water (H2O).
Explain why this is true in terms of the bonding between molecules in each of these two
substances.
Solution 2.85 The methane molecules are bonded together by weak –C–H dipoles. The water
molecules are bonded together by the much stronger –O–H hydrogen bonded dipoles.
Problem 2.86 For each of the following compounds, state whether the bonding is essentially
metallic, covalent, ionic, van der Waals, or hydrogen: (a) Ni, (b) ZrO2, (c) graphite, (d) solid Kr, (e) Si,
(f) BN, (g) SiC, (h) Fe2O3, (i) MgO, (j) W, (k) H2O within the molecules, (l) H2O between the
molecules. If ionic and covalent bonding are involved in the bonding of any of the compounds
listed, calculate the percentage ionic character in the compound.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
50
Solution 2.86
(a) Ni: Nickel bonding is primarily metallic.
(b) ZrO2: From Pauling’s equation, the Zr–O bond is 73.4% ionic and
26.6% covalent, where xA and xB are the electronegativities
of zirconium and oxygen, respectively.
(c) Graphite: The bonding is covalent within the layers and secondary
between the layers.
(d) Solid Kr: The bonding represents van der Waals due to fluctuating
dipoles.
(e) Si: Silicon bonding is covalent.
(f) BN: The B–N bond, from Pauling’s equation, is 26.1% ionic and
73.9% covalent.
(g) SiC: The Si–C bond is 11% ionic and 89% covalent.
(h) Fe2O3: The Fe-O bond is 55.5% ionic and 44.5% covalent.
(i) MgO: The Mg-O bond is 70.2% ionic and 29.8% covalent.
(j) W: Tungsten bonding primarily consists of metallic bonding
with some covalent character.
(k) H2O within the molecules: The H–O bond is 38.7% ionic and 61.3% covalent.
(l) H2O between the molecules: Hydrogen bonding exists between H2O molecules.
Problem 2.87 In the manufacturing of a light bulb, the bulb atmosphere is evacuated and then
filled with argon gas. What is the purpose of this?
Solution 2.87 Because, the tungsten filament in a light bulb becomes hot, it can react (form
bonds) with impurities in the atmosphere of a light bulb. This will weaken and damage the
filament. Argon is a noble gas and is non-reactive (inert). It does not react with tungsten and
keeps the integrity of the wire.
Problem 2.88 Stainless steel is a corrosion resistant metal because it contains large amounts of
chromium in its composition. How does chromium protect the metal from corrosion? Hint:
Chromium reacts with oxygen.
Solution 2.88
Chromium easily reacts with oxygen in the atmosphere, and it forms chromium oxide Cr2O3.
Chromium oxide is a ceramic with high hardness. The chromium on the surface forms a thin
protective ceramic layer that is chemically stable and protects material inside. This is called a
passivity layer.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
51
Problem 2.89 Robots are used in auto industries to weld two components at specific locations.
Clearly, the end-position of the arm must be determined accurately in order to weld the
components at the precise position. a) In selecting the material for the arm of such robots, what
factors must be considered? b) Select a proper material for this application.
Solution 2.89
a) One would have to choose materials that are very stiff (little elastic deformation under
load).
The stronger the bonds are between the atoms in the material, the stiffer the material will
behave.
b) Thus, ceramics, advanced composites and some metals will be good choices (economic
considerations are important).
Problem 2.90 A certain application requires a material that is lightweight, an electrical insulator,
and has some flexibility. a) Which class of materials would you search for this selection?
b) Explain your answer from a bonding point of view.
Solution 2.90
a) The characteristics required are found in polymers.
b) Polymers are insulators because all valence electrons are involved in covalent bonds
within the polymeric molecule.
– The bonding between molecules however is of secondary type (electrostatic
attraction). This allows the material to be flexible.
– Polymers are also low density because there exist large empty spaces (gaps) in the
structure between molecules.
Problem 2.91 A certain application requires a material that is electrically non-conductive
(insulator), extremely stiff, and lightweight. Which classes of materials would you search for this
selection? b) Explain your answer from a bonding point of view.
Solution 2.91
a) The only class of materials that possess the properties is structural ceramics.
b) Structural ceramics are mostly made of ionically-bonded atoms in a 3-D network (there is
often a mix of ionic and covalent). They are insulators because valence electrons are
tightly held in ionic bonds. Ionic bonds are very strong and require high temperatures for
melting.
Problem 2.92 Solid potassium and solid calcium have densities 0.862g/cm3 and 1.554g/cm2.
Compare the mass of individual potassium and calcium atoms; what is your observation? How do
you explain the difference in the solid density between the two elements?
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written consent of McGraw-Hill Education.
52
Solution 2.92
Potassium and calcium are in the same period, but calcium has a higher atomic number. Calcium
has 2 electrons in its 4s orbital , whereas potassium only has 1.
potassium: 39g/mol
23
23
39 6.48 10
6.022 10
g mol x g
of potassium
mol x atoms atom
–
´ =
Calcium: 40g/mol
23
23
40 6.64 10
6.022 10
g mol x g
of calcium
mol x atoms atom
–
´ =
Atoms of calcium weight more than atoms of potassium due to having a higher atomic number
(an additional electron and proton).
Calcium has a higher density because the calcium has smaller atomic radii due to having a larger
nuclear charge (more protons in the nucleus).