Download Chapter 1 - Atoms: The Quantum World

Chapter 1 Atoms: The Quantum World
Observing Atoms (Sections 1.1–1.6)
Key Concepts
electronic structure, classical mechanics, quantum mechanics, spectroscopy,
electromagnetic radiation, oscillation, cycle, frequency, amplitude, intensity,
wavelength, speed of light, visible light, ultraviolet radiation, infrared radiation,
black body, black-body radiation, Stefan-Boltzmann law, Wien’s law, ultraviolet
catastrophe, quanta, Planck constant, photoelectric effect, photons, work function,
diffraction, constructive and destructive interference, duality of electromagnetic radiation
and matter, linear momentum, de Broglie relation, Heisenberg uncertainty principle
(complementarity of location and momentum), trajectories, wavefunction, Born
interpretation, node of the wavefunction, probability density, Schrödinger equation,
differential equation, Hamiltonian, particle in a box, standing wave, quantum number n,
quantized (discrete) energy values, zero-point energy, boundary conditions,
spectral lines, energy levels, Bohr frequency condition, Rydberg constant
1.1 The Characteristics of Electromagnetic Radiation
Example 1.1a
Calculate the range in frequency and period that corresponds to the wavelength
limits of vision, 400 to 700 nm.
Solution
First calculate each frequency by solving for ν in the equation λν = c.
Then calculate the period, which is 1 over the frequency.
ν=
c 3.00 × 108 m⋅ s −1
1
=
= 7.50 × 1014 s −1 & period = = 1.33 × 10−15 s = 1.33 fs
−9
λ
ν
400 × 10 m
ν=
c 3.00 × 108 m ⋅ s−1
1
=
= 4.29 × 1014 s −1 & period = = 2.33 × 10−15 s = 2.33 fs
9
−
λ
ν
700 × 10 m
Frequency range = 7.50 × 1014 s−1 to 4.29 × 1014 s−1
Period range = 1.33 fs to 2.33 fs
(decreases as λ increases)
Example 1.1b
Calculate the range in frequency and period that corresponds to the
wavelength limits of the infrared region, 700 nm to 1.00 mm.
Answers
Frequency range = 4.29 × 1014 s−1 to 3.00 × 1011 s−1
Period range = 2.33 fs to 3.33 ps
Note: The boundary between the visible and infrared is often chosen as either 700 nm or 800 nm.
1.2 Radiation, Quanta, and Photons
Example 1.2a
Calculate the wavelength of maximum intensity for a heated object that behaves as a black
body at 298 K (reference temperature used in thermochemistry) and 3680 K (the melting
point of elemental tungsten). Identify the wavelength region in which the maximum occurs.
Solution
Wien’s law is easily rearranged to evaluate the wavelength of maximum intensity.
Then, the wavelength region can be identified.
1
Chapter 1
Example 1.2b
Answers
λ max =
1.44 × 10−2 K ⋅m
= 9.66 × 10−6 m = 9.66 µm
5(298 K)
Infrared: 700 nm to 1 mm
λ max =
1.44 × 10−2 K ⋅m
= 7.82 × 10−7 m = 782 nm
5 (3680 K)
Infrared: 700 nm to 1 mm
Calculate the wavelength of maximum intensity for a heated object that
behaves as a black body at 194.5 K (dry ice temperature) and 6000 K
(Sun’s “color” temperature). Identify the wavelength region in which the
maximum occurs.
λ max = 1.48 × 10−5 m = 14.8 µm
Infrared: 700 nm to 1 mm
λ max = 4.80 × 10−7 m = 480 nm
Visible: 400 nm to 700 nm (appears yellow)
Example 1.2c
Calculate the photon energy range in Example 1.1a for one photon and one mole
of photons.
Solution
Use the frequencies calculated in Example 1.1a in the Planck equation for single
photon energies. Then multiply by the Avogadro constant to calculate energies
for one mole of photons.
Note: Energy changes in chemical reactions are usually given in kJ⋅mol–1.
E = hν = (6.63 × 10−34 J ⋅s)(7.50 × 1014 s −1 ) = 4.97 × 10−19 J
for single photon (400 nm)
E × N A = (4.97 × 10−19 J)(6.02 × 1023 mol −1 ) = 2.99 × 105 J ⋅mol −1 = 299 kJ ⋅mol −1
E = hν = (6.63 × 10−34 J ⋅s)(4.29 × 1014 s −1 ) = 2.84 × 10−19 J
for single photon (700 nm)
E × N A = (2.84 × 10−19 J)(6.02 × 1023 mol −1 ) = 1.71 × 105 J ⋅mol −1 = 171 kJ ⋅ mol −1
Thus, the photon energy range of vision (400 to 700 nm) is 299 kJ⋅mol−1 to 171 kJ⋅mol−1.
Example 1.2d
Calculate the photon energy range in Example 1.1b for one photon and one mole
of photons.
Answers
E = 2.84 × 10−19 J for single photon (700 nm) & 171 kJ⋅mol−1
E = 1.99 × 10−22 J for single photon (1.00 mm) & 120 J⋅mol−1
Thus, the photon energy range of the infrared (700 nm to 3.00 mm) is about 171 kJ⋅mol−1 to
120 J⋅mol−1.
1.3 Wave-Particle Duality of Matter
Example 1.3a
Solution
Calculate the de Broglie wavelength for an electron formed by photoelectric
emission from a metal surface (Φ = 7.20 × 10−19 J) exposed to light of wavelength
260 nm.
First find the energy of the photon. Then find the kinetic energy of the electron
from the photoelectric effect. Use the definition of kinetic energy to find the
electron speed. Calculate the de Broglie wavelength.
Note: 1 J = 1 kg⋅m2⋅s−2 and EK = (1/2)mv2 , where m = me = mass of electron
E = hν =
hc (6.63 × 10−34 J ⋅s)(3.00 × 108 m⋅s −1)
=
= 7.65 × 10−19 J (photon energy)
−9
λ
(260 × 10 m)
EK = hν − Φ = (7.65 × 10−19 J) − (7.20 × 10−19 J) = 4.5 × 10−20 J (electron kinetic energy)
2
Atoms: The Quantum World
v=
λ=
2(4.5 × 10−20 kg⋅m2 ⋅s−2 )
2 EK
=
me
(9.1094 × 10−31 kg)
= 9.88 × 1010 m 2 ⋅s −2 = 3.1 × 105 m ⋅ s −1 (speed)
h
(6.63 × 10−34 J ⋅s)
=
= 2.3 × 10−9 m = 2.3 nm
me v (9.1094 × 10−31 kg)(3.1 × 105 m⋅s −1 )
Example 1.3b
Repeat Example 1.3a for a different light source, but in reverse. Suppose that
the wavelength of the ejected electron is determined to be 232 pm. What is
the wavelength of a photon that ejects the electron from the same surface?
Answer
38.2 nm
1.5 Wavefunctions and Energy Levels
Example 1.5a
Calculate the probability of finding a particle in the box at x = L / 2 for n = 2. How
many nodes occur in the wavefunction?
Find ψ 2 for n = 2. Find values of x where ψ = 0 and ψ changes sign within the
box.
Solution
ψ 2  x = L  =  2  sin 2  2πx  =  2  sin 2  2 π( L / 2)  =  2  sin 2 ( π ) =  2  (0) = 0 (Node)
2 
 L  L

 L
L
L
2   L 


  

  
 
ψ = 0, and ψ changes sign in the center of the box. There is only one node (x = L/2).
Example 1.5b
Repeat Example 1.5a for n = 3.
Answer
ψ 32  x = L  =  2  ( −1)2 = 2

L
2   L 

(Not a node, because ψ 2 ≠ 0)
ψ = 0, and ψ changes sign at x = L and 2 L (2 nodes)
3
3
Note: A pattern is suggested in which (n − 1) nodes exist for ψn. The existence of locations of zero
particle probability is a purely quantum effect.
1.6 Atomic Spectra and Energy Levels
Example 1.6a
Calculate the wavelength of light emitted when a hydrogen atom in the third excited state
reverts to the ground state (member of Lyman series).
(See Study Guide Section 1.10 for a presentation of ground and excited states.)
Solution
The third excited state corresponds to n2 = 4, the ground state to n1 = 1.
 1
1 
Thus, if we combine ν = ℜ  2 − 2  and λν = c , we obtain
 n1 n2 
λ=
c
=
ν
c
 1
1 
ℜ  2 − 2
 n1 n2 
=
(
3.00 × 108 m⋅s −1
= 9.73 × 10−8m = 97.3 nm
1 
− 1
3.29 × 1015s 1  2 − 2 
1
4


)
3
Chapter 1
Example 1.6b
Calculate the wavelength of light emitted when a hydrogen atom in the third excited state
undergoes a transition to the second excited state (member of the Paschen series).
Answers
n2 = 4, n1 = 3, and λ = 1.88 µm
Models of Atoms (Sections 1.7–1.10)
Key Concepts
Coulomb potential energy, atomic number, energy level/state, principal quantum number,
ground state, ionization, atomic orbital, radial wavefunction, angular wavefunction, Bohr
radius, shell, subshell, orbital angular momentum quantum number, orbitals, degenerate
orbitals, s-orbitals, p-orbitals, d-orbitals, f-orbitals, magnetic quantum number, boundary
surface, node, radial node, planar node, electron spin, spin magnetic quantum number
1.7 Principal Quantum Number
Example 1.7a
Calculate the coulomb potential energy for two electrons separated by a distance of 10−6 m.
Solution
Insert values into the equation for V ( r ) with q1 = q2 = the charge of the electron
and r = 10−6 m. Thus,
V (r) =
( − e)( − e)
( − 1.602 × 10−19 C)2
=
= 2.307 × 10−22 J
−12 2
−1 −2
−6
4πε0 r
4π(8.854 × 10
C ⋅ N ⋅ m )(10 m)
The energy is positive indicating a repulsive interaction, as expected.
Hint: When evaluating expressions with many terms, particularly ones containing
exponents, it is often useful to estimate the result first. This estimation can be
accomplished by factoring the exponents and estimating the pre-exponent and
exponent terms separately. With practice, the estimation can be done quickly and
in one’s head. Thus, in the present case, the expression above factors to
 ( −1.602)2   10−38 
−20
= 3 × 10−22 , which is a good estimate of the


 ≈ 0.03 × 10
 4 ⋅ π ⋅ 8.854   10−18 



correct answer.
Example 1.7b
Calculate the coulomb potential energy for four
electrons at the corner of a square of edge 10−6 m.
Solution
Label the electrons 1 through 4 as shown in the
diagram. All pairwise interactions must be
considered. There are 4 interactions (r1, r2, r3, r4) with
r = 10−6 m and two diagonal interactions (r5, r6) with
r = 1.414 × 10−6 m. Now,
V (r) =
4
( − e)( − e) 6 1  ( −1.602 × 10−19 )2   4
2

−21
J.
=
  −6 +
∑
 = 1.249 × 10
12
6
−
−


4 πε0 i =1 ri  (4 π)(8.854 × 10 )   10
1.414 × 10 
Atoms: The Quantum World
Notice that this repulsive energy is more than four times that in Example 1.7a
because of the cross terms. Electron-electron repulsions are important in
understanding the energy levels of atoms and molecules.
1.8 Atomic Orbitals (AOs)
Example 1.8a
The H atom 2s-orbital has one nodal surface [surface for which ψ( r, θ, φ) = 0]. Using the
wavefunction given below, determine the location and geometrical form of this surface.

r  − r / 2 a0
2 − e
a0 
R(r ) = 
(2a0 )3/ 2
Units: R ( r ) = m −3/2
Y (θ, φ) = (4π)−1/ 2 a0 =
Y (θ, φ) = none
a0 =
4 πε0! 2  5.291 77 × 10−11 m 
=

me e 2
 (Bohr radius) 
(C2 ⋅ N −1⋅m −2 ) ( J ⋅s) 2
=m
kg ⋅ C2
ψ 2 s ( r, θ, φ) = 0, if R2 s ( r ) = 0, which occurs when r = 2a0 = 2(5.29 × 10−11 m) and
r = 1.06 × 10–10 m. The locus of points for which r is constant defines the surface of
a sphere. The nodal surface is located 1.06 × 10−10 m (106 pm) from the proton.
Solution
Example 1.8b
For the H atom 2s orbital, calculate the electron density at r = a0 .
Solution
All s-orbitals have spherical symmetry and no angular dependence. The electron density
(probability of finding the electron) at a given radial distance, r, is given by the square of the
wavefunction.
2
ψ2 s2

r  − r / a0
2 −  e
a0 
= R2 s2 Y2 s2 = 
4π(2a0 )3
With r = ao, this becomes ψ2s2 =
e −1
4π(2a0 )3
= 2.47 × 1028 electron ⋅m −3 .
Example 1.8c
For the H atom 2s-orbital, calculate the electron density at r = 3a0 .
Solution
For r = 3a0, the value is ψ2s2 =
e −3
4π(2a0 )3
= 4.95 × 1027 electron ⋅m −3 .
Note:
These values of the point electron density are huge. Expressing them on a scale
more nearly comparable to the size of the atom (a0) gives the values
electron
electron
Example 1.8b ⇒ 3.66 × 10−3
and Example 1.8c ⇒ 4.95 × 10−4
.
3
a0
a03
Note:
Integrating over all space:
2π π ∞
∫0 ∫0 ∫0
ψ22s r 2 sin θ d r dθ dφ = 1 electron,
where the volume element dx dy dz = r 2 sin θ dr dθ dφ.
5
Chapter 1
The Structures of Many-Electron Atoms
(Sections 1.11–1.13)
Key Concepts
many-electron atoms, effective nuclear charge, shielding, penetration,
building-up principle, electron configuration, Pauli exclusion principle, spin pairing,
closed shell, valence electron(s), parallel spins, Hund’s rule, excited state, valence shell,
periodic table, period, long period, group, block
1.11 Orbital Energies
Example 1.11
Write an expression for the coulomb potential energy of the Li atom (3 protons in the nucleus
plus 3 electrons). The atomic number Z is equal to the number of protons in the nucleus.
Which terms are attractive? Which are repulsive?
Solution
This is a 4-body problem similar to the one solved in Example 1.7b.
Label the nuclear charge Ze (Z = 3) and the electrons 1, 2, and 3 (charge = −e).
The potential energy is given as:
V (r) = ∑
=−
( Ze)e
e2
+∑
4πε0 ri
4πε0 rij
( Ze)e
( Ze)e
( Ze)e
e2
e2
e2
−
−
+
+
+
4πε0 r1 4πε0 r2 4πε0 r3 4πε0 r12 4πε0 r13 4πε0 r23
Here, ri is the distance of electron i from the nucleus and rij is the distance between electrons
i and j. The electron-nucleus terms (the first three terms in the equation above) are
attractive whereas the electron-electron terms (the last three terms) are repulsive.
1.12 The Building-Up Principle
Note: There are several different ways of writing electron configurations.
Examples 1.12a–d show four of them.
Example 1.12a Determine the electron configuration of carbon (atomic number Z = 6).
Solution
Carbon has six electrons in the neutral atom. The order of filling subshells is 1s, 2s, 2p.
The s-subshells contain a maximum of two electrons and the p-subshells each contain a
maximum of six (two electrons in each of three orbitals). The electron configuration is
then 1s22s22p2. In the ground state, carbon has two electrons in the 1s orbital, two in the
2s orbital and two in the 2p subshell. This notation does not show the individual orbitals
in subshells with " ≠ 0.
Example 1.12b Write the electron configuration of C (Z = 6) using noble-gas abbreviations for the core
electrons.
Solution
6
From Example 1.12a, the configuration is 1s22s22p2. The configuration of the nearest noble
gas atom of lower Z is [He] = 1s2. Therefore, the abbreviated configuration is [He]2s22p2.
This shorthand notation separates the core electrons ( [He] = 1s2 ) from the valence-shell
electrons. Core electrons are not involved in bonding and are relatively uninteresting.
Atoms: The Quantum World
Example 1.12c Write the electron configuration of C (Z = 6) showing orbital occupancy.
Solution
The 2p-subshell has two electrons. According to Hund’s Rule, they should be placed in
different orbitals. Therefore, the electron configuration showing orbital occupancy is:
1s 2 2 s 2 2 px1 2 py1 or [He]2s 2 2 px1 2 py1 .
This notation shows orbital occupancy ( n, ", m" quantum numbers). You may choose any two
p-orbitals (x and y, x and z, or y and z).
Example 1.12d Write the electron configuration of C (Z = 6) showing orbital occupancy and the spin states
of each electron. Use ↑ for ms =
Solution
+
1
2
and ↓ for
−
1
.
2
The 2p-subshell has two electrons. According to Hund’s rule, they should be placed in
different orbitals with parallel spins. Therefore, the electron
configuration is ↑↓ ↑↓
1s 2s
↑↓ ↑↓
1s 2s
↑ ↑ __ or
2px 2py 2pz
↓ ↓ __ or four other possibilities.
2px 2py 2pz
This notation shows orbital occupancy and electron spin ( n, ", m", ms quantum numbers).
Note: In accordance with Hund’s rule, the two electrons in the 2p-subshell occupy separate orbitals
with parallel spins. Any two of the three 2p-orbitals may be occupied and both spins could
also be pointing down; such configurations are all degenerate.
Example 1.12e Write the ground-state electron configuration (subshells only) of germanium, Ge. How many
unpaired electrons does Ge have?
Solution
Ge has atomic number 32 and therefore has 32 protons and 32 electrons. The electrons will fill
the subshells in the order given above: 1s , 2s, 2p, 3s, 3p, 4s, 3d , 4p, 5s, 4d, … Recognizing
that s-, p-, and d-subshells can hold 2, 6, and 10 electrons, respectively, we fill the subshells
from lowest to highest energy with 32 electrons.
Ge: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2 filling order
or 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p2 final energy order
or [Ar] 3d10 4s2 4p2
noble-gas abbreviations
Note: The valence electrons are in the shell with n = 4. From the electron configuration, only the
4p-subshell is partially filled and therefore only it can contain unpaired electrons. Using
Hund’s rule, the two electrons in the 4p-subshell should occupy separate orbitals with spins
parallel. So, Ge has two unpaired electrons:
↑ ↑
.
4px 4py 4pz
Example 1.12f
In the ground state, how many unpaired electrons does selenium, Se, have?
Solution
Se, atomic number 34, has two more electrons than Ge. The additional two electrons will enter
the 4p-subshell. The first will enter the vacant 4pz-orbital and the second must pair with one of
the three. The 4p-subshell will appear as
↑↓ ↑ ↑ . Like Ge, the Se atom also has two unpaired electrons in the ground state.
4px 4py 4pz
7
Chapter 1
Example 1.12g An exception to the rule. Determine the ground-state electron configuration of chromium, Cr.
Solution
Cr has 24 electrons, and we expect the electron configuration to be
Cr: 1s2 2s2 2p6 3s2 3p6 4s2 3d 4 filling order
However, the correct configuration is
Cr: 1s2 2s2 2p6 3s2 3p6 4s1 3d 5 or [Ar] 3d 5 4s1 final energy order
Note: To understand Example 1.12g, look at the orbitals of the outermost two subshells, 4s and 3d.
We expect configuration (A) with four unpaired electrons in the 3d subshell; we find
configuration (B) with six unpaired electrons, one in the 4s-orbital and 5 in the 3d-subshell.
(A)
↑
↑
↑
↑↓ ↑
4s 3dxy 3dxz 3dyz 3d x 2 − y 2 3d z 2
(B) ↑
4s
↑
↑
↑
↑
↑
3dxy 3dxz 3dyz 3d x 2 − y 2 3d z 2
We can rationalize configuration (B) by assuming that half-filled subshells have additional
stability and lower energy. Configuration (B) has two half-filled subshells (4s, 3d) while
configuration (A) has none. There are a number of such anomalies in the periodic table.
(See Section 1.13 in the Study Guide.)
1.13 Electronic Structure and the Periodic Table
Example 1.13a Write the ground-state electron configuration (core plus valence electrons) of arsenic,
As (Z = 33).
Solution
The element As in Period 4 has the Ar (Z = 18) core and additional subshells 4s, 3d, and 4p.
According to its position in Group 15, the 4s- and 3d-subshells are filled and the 4p-subshell
has 3 electrons. It is not an exception. Thus, the configuration is
As: [Ar] 4s2 3d10 4p3 (filling order) or [Ar] 3d10 4s2 4p3 (final energy order)
The 4s- and 3d-subshells are filled whereas the 4p-subshell is half filled, with one electron in
each of the 4px-, 4py- and 4pz-orbitals. The filled 3d-subshell becomes part of the core, as
shown in the second configuration. There are five valence electrons.
Example 1.13b Write the ground-state electron configuration (core plus valence electrons) of zirconium,
Zr (Z = 40).
Solution
The element Zr is in Period 5 and Group 4. It has the Kr (Z = 36) core and is in the period with
additional subshells 5s, 4d, and 5p. It is not an exception. Thus, we have
Zr: [Kr] 5s2 4d 2 (filling order) or [Kr] 4d 2 5s2 (final energy order)
The 5s-subshell is filled; the two 4d-electrons occupy separate orbitals, with spins parallel
according to Hund’s rule. The second configuration shows the correct energy pattern after the
subshells are filled.
Example 1.13c Write the ground-state electron configuration (core plus valence electrons) of gold,
Au (Z = 79).
Solution
8
The element Au is in Period 6 and Group 11. It has the Xe (Z = 54) core and is in the period
with additional subshells 6s, 5d and 6p. We also observe that the 4f subshell (corresponding to
the lanthanides) fills between Ba and Hf. Finally, we note that Au is a one-electron exception
to the subshell-filling rule. Using the rule first, we find
Au: [Xe] 6s2 4f 14 5d 9, which is incorrect.
The correct configuration is
Au: [Xe] 6s1 4f 14 5d10 (filling order) or [Xe] 4f 14 5d10 6s1 (final energy order)
Atoms: The Quantum World
is obtained by moving one electron from the 6s- to the 5d-subshell. Thus, gold has a complete
5d-subshell and a half-filled 6s-subshell. The 4f-electrons are lower in energy (penetration)
and are not involved in the chemistry of gold. The other two coinage metals, silver (Ag) and
copper (Cu), show similar one-electron anomalies in their electron configurations.
The Periodicity of Atomic Properties
(Sections 1.14–1.19)
Key Concepts
atomic radius, covalent radius, van der Waals radius, ionic radius, isoelectronic
atoms and ions, ionization energy, first ionization energy, second ionization energy,
electron affinity, metallic character, inert-pair effect, diagonal relationship
1.15 Ionic Radius
Example 1.15
Consider the isoelectronic species of the elements sulfur, chlorine, argon, potassium and
calcium with 18 electrons. Determine the charge on each species and arrange them in order of
decreasing size using the wavefunction definition of size.
Solution
The atomic numbers of S, Cl, Ar, K, and Ca are 16, 17, 18, 19, and 20. In this problem, all
species have 18 electrons. Sulfur with 16 protons has a charge of −2 (sulfide anion). Chlorine
with 17 protons has a charge of −1 (chloride anion). Argon with 18 protons is neutral.
Potassium with 19 protons has a charge of +1. Calcium with 20 protons has a charge of +2.
Ionic size decreases as nuclear charge increases, therefore the expected order with decreasing
size is
r (S2–) > r (Cl–) > r (Ar) > r (K+) > r (Ca2+)
1.16 Ionization Energy ( I )
Example 1.16
Consider the first three ionization energies (I1, I2, and I3) of the argon atom (Group 18 noble
gases). Arrange the ionization energies in order of increasing magnitude. Look up the
values on the Periodic Table web site:
http://www.webelements.com/webelements/elements/text/Ar/ionz.html .
Solution
All ionization energies are positive numbers, and each successive value is larger than the
preceding one. Therefore, the order is I3 > I2 > I1. The reactions and values taken from the
web site are
first ionization energy:
Ar(g) → Ar+(g) + e−(g)
I1 = 1520.6 kJ⋅mol−1
+
−
2+
second ionization energy:
Ar (g) → Ar (g) + e (g)
I2 = 2665.8 kJ⋅mol−1
−
2+
3+
I3 = 3931 kJ⋅mol−1
third ionization energy:
Ar (g) → Ar (g) + e (g)
9
Chapter 1
1.17 Electron Affinity ( Eea )
Example 1.17a Write equations showing the chemical reactions occurring when two electrons are added
sequentially to an atom of sulfur, S. Show the energy change, ∆E, for each reaction and the
value of Eea.
Solution
Unlike ionization energy values, electron affinity values may be positive or negative numbers.
Energy release requires a change in sign: Eea = −∆E. If the electron-attachment reaction
releases energy, then Eea > 0 (anion stable). If the electron-attachment reaction consumes
energy, then Eea < 0 (anion unstable). The values of Eea for sulfur are taken from the Appendix
section on The Elements in the text.
first electron affinity:
second electron affinity:
S(g) + e−(g) → S−(g)
−
−
Eea = −∆E = +200 kJ⋅mol−1
Eea = −∆E = –532 kJ⋅mol−1
S (g) + e (g) → S (g)
2−
The anion, S−, is stable in the gas phase, whereas the anion, S2−, is unstable.
Example 1.17b Write an expression that combines the two reactions in Example 1.17a. Calculate the value
of Eea for the attachment of two electrons to S in one step.
Solution
Add the two reactions together to obtain the overall reaction for the addition of two electrons
to S. Combine the energy changes for the two reactions as well.
combined reaction: S(g) + 2e−(g) → S2−(g)
Eea = 200 + (−532) = −332 kJ⋅mol−1
Notes: Only mononegative anions are stable in the gas phase.
Some mononegative anions are unstable: nitrogen in Group 15, beryllium and magnesium
in Group 2, and all of the elements in Group 18 are the major examples.
In ionic solids, many polynegative anions are stable: O2−(oxide), S2−(sulfide), N3−(nitride),
C4−(carbide: methanide). There are other carbides: C22− (carbide: acetylide).
1.18 The Inert-Pair Effect
Example 1.18
Answer
List the elements in the main-group that exhibit the inert-pair effect.
Group 13
Group 14
Group 15
In+ and In3+
Sn2+ and Sn4+
Sb3+ and Sb5+
Tl+ and Tl3+
Pb2+ and Pb4+
Bi3+ and Bi5+
1.19 Diagonal Relationships
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Example 1.19
Give several examples of elements in the main-group that are diagonally related in that they
display similar chemical properties.
Answer
Diagonal band of metalloids dividing metals from nonmetals
Similarity of Li and Mg (react directly with N2 to form nitrides)
Similarity of Be and Al (both react with acids and bases)
Atoms: The Quantum World
The Impact on Materials (Sections 1.20–1.21)
Key Concepts
metals, metalloids, nonmetals, s-block elements, p-block elements,
d-block elements, transition metals, lanthanides
1.20 The Main-Group Elements (Groups 1–2, 13–18)
Example 1.20
List the chemical reactions of sodium metal with water and oxygen.
Solution
Sodium metal reacts violently with water to form a basic solution, a metal cation, and
hydrogen gas. Sodium metal reacts with gaseous oxygen to form the peroxide salt. Write
the overall reactions and balance by inspection.
Answer
2 Na(s) + 2 H2O(l) → 2 Na+(aq) + 2 OH−(aq) + H2(g) (See Section H.1)
2 Na(s) + O2(g) → Na2O2(s)
[O22− (basic peroxide ion)]
1.21 The Transition Metals
Example 1.21
Chromium forms oxides with the oxidation states of +2, +3, and +6. Write the formula of
each oxide compound. Show the electron configuration of each cation.
Solution
Use the oxidation number rules and procedures in Section K.2. The electron configuration of
Cr is [Ar] 3d 5 4s1. When forming cations, remove the outer 4s electron first.
Answer
Oxidation state
II
III
VI
Oxide compound
CrO
Cr2O3
CrO3
Cation configuration
Cr2+ [Ar]3d 4
Cr3+ [Ar]3d 3
Cr6+ [Ar]
Note: Common anion salts incorporating chromium (VI) are CrO42− (chromate) and
Cr2O72− (dichromate).
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