Chapter 8: Modelling Atoms and Their Electrons
A: MULTIPLE CHOICE QUESTIONS
1. The atomic radii of the elements can be used to estimate bond-length. Referring to figure 8.4 what is
the estimated bond length of a bond between Na and P?
a) 186 pm
b) 115 pm
c) 301 pm
d) 93 pm
ANS: c
REF: Page 240
2. Choose the correct periodic trend in reactivity as electron donors (reducing agents) within the
periodic table, groups 1, 2, and 13 to 17, among the following.
a) Reducing agent strength increases from left to right within a period
b) Oxidizing agent strength increases from top to bottom in a group
c) Reducing agent strength increases from top to bottom within a group
d) Oxidizing agent strength decreases from left to right within a period
ANS: c
REF: Table 8.3 and pages 239, 240
3. Which one of the following is untrue when discussing ionic radii?
a) Cations are smaller than the neutral atom
b) Anions are larger than the neutral atom
c) Within groups 1,2, and 13, the left to right trend in size is decreasing
d) Radii are determined by analysis of ions in the gas phase
ANS: d
4.
REF: Page 243, 244
Which one of the following statements is true?
a) Electrons orbit the nucleus
b) Electrons are locked into fixed energy levels
c) Electrons can have any amount of energy
d) The energy of electrons is quantized
ANS: d
REF: Section 8.3
5. One of the following is not an allowable set of values for n, l, and m, respectively.
a) 3, 3,-2
b) 3, 2, +2
c) 2,1,0
d) 1,0,0
ANS: a
6.
REF: Page 255 and table 8.8
For any principal quantum number, n, the number of orbitals within that shell equals;
a) 2l +1
b) n2
Copyright © 2011 Nelson Education Limited
8-1
c) –l to +l
d) n
ANS: b
7.
REF: Table 8.8 & Pages 256 and 257
The Pauli Exclusion Principle says one of the following.
a) Lowest energy orbitals fill 1st
b) Two electrons in the same orbital must have opposite spins
c) Two electrons in the same orbital must have the same spin
d) Excited state atoms have more electrons
ANS: b
REF: Page 262
8. Hund’s rule and the Aufbau principle tell us that when filling orbitals, of equivalent energy, which of
the following is the case?
a) Electrons are paired spin-up to spin-down in the first available orbital before moving to the next,
where the orbital is again filled with a pair, if possible
b) Electrons are all spin down and added in any order
c) Electrons are all drawn spin-up and added in any order
d) Electrons are added one at a time to each orbital until all orbitals contain one electron, then
additional electrons are paired with electrons of opposite spin
ANS: e
REF: Page 263 and 264
9. Match the following spdf notation to the appropriate ground state element from the periodic table.
↑↓
↑↓
↑↓
↑↓
1s
2s 2px
2py
2pz
a)
b)
c)
d)
↑↓
3s
↑↓
↑↓
↑↓
3px
3py
3pz
↑
Br
O
Ar
F
ANS: d
REF: Page 265
10. The abbreviated spdf notation is given for an unidentified element. How many electrons does this
element have and if it acquired the electron configuration of the nearest noble gas, which gas would it
be?
[Kr]5s2
a) 10, Kr
b) 2, Ar
c) 31, Ar
d) 74, Kr
ANS: d
REF: Page 265
B: SHORT ANSWER
1. 2s orbitals and 2p-orbitals have nodal surfaces. How many each and what are they?
Copyright © 2011 Nelson Education Limited
8-2
ANS:
A 2s orbital has one node. Each of the three 2p orbitals has a node. Nodes can be defined
mathematically and theoretically. Mathematically they are solutions for the square of the
waveform equations at a null value. Theoretically, they are positions the electron can never be
found when in that orbital. There is zero probability of them being in that space.
REF: Section 8.4 of text
2. Calculate the wavelength of an electron travelling at the speed of light. The mass of an electron is
9.109 x 10-28 g.
ANS:
The de Broglie equation allows us to solve this problem. The equation is;  = h/mv
Lambda is the wavelength, h is Plancks constant, m is the mass of the electron, v is its velocity.
Mass must be converted to kilograms because the unit Joule is a N∙m and a Newton is a kg∙m/s2.
Units must match.
Wavelength =  = h/mv = (6.626 x 10 -34 J∙s) /(9.109 x 10 -31 kg)∙(3.0 x 108 m/s) = 2.42 x 1012m = 2.42 pm
REF: Example 8.2 of text
3. Energy of motion is kinetic energy. A particle with mass has kinetic energy, if it is moving relative to
another object. The equation for kinetic energy comes from classical Newtonian physics. It is KE =
½ mv2. Where m is the mass and v is the velocity. Notice that the symbol for velocity is v and the
symbol for frequency is v. Try not to confuse the two. Using the equation for kinetic energy, what is
the relative kinetic energy of the electron from question #2?
ANS:
KE = ½ mv2 = ½ ∙(9.109 x 10 -31 kg)∙( 3.0 x 10 8 m/s)2 = 4.09 x 10 -14 kg∙m2/s2 = 4.09 x 10 14 N.m = 4.09 x 10 -14 J
REF: Narrative B-3
4. From Planck’s equation, the energy of a photon, Ephoton = h∙v, where h is Planck’s constant and v is
the frequency of the radiation. Another useful relationship is that between wavelength and frequency.
The frequency of a wave is the number of cycles per second. It has units of s-1 and this unit is called
a hertz. The mathematical relationship between wavelength and frequency involves the constant for
the speed of light in a vacuum, c. The equation is; v = c/. Using the information given, what is the
energy of a photon having a wavelength of 349 nm?
ANS:
Ephoton = hc/
REF: Example 8.1 of text
-34 Js)(3.0 x 10 8 m/s)/349 x 10 -9 m = 5.7 x 10 -19 J
5. An excited hydrogen atom emits a photon with wavelength 410.2 nm. If transition was to energy
level 2, what level did the electron leave? Use the Rhydberg equation to solve.
ANS:
When dealing with the hydrogen atom, we can use the Rhydberg equation for solution of energy
transitions. This equation uses the Rhydberg constant, 2.179 x 10 -18, and the values of n, where
n is the principal quantum number. In hydrogen, n also refers to the quantized possible energy
levels for the electron.
The change in energy when the electron transitions is the difference in energy between the two
levels. The equation for the electrons energy at any given level, n, is;
Copyright © 2011 Nelson Education Limited
8-3
En = (- 2.179 x 10 -18/n2)J
We are told that the electron emitted a photon with a wavelength of 410.2 nm. From this, and the
equation for the energy of a photon, we can calculate the energy of the transition. The energy of
the photon emitted equals the energy difference between the two electron energy levels.
Ephoton = hc/ = (6.626 x 10 -34 Js)(2.998 x 10 8 m/s)/410.2 x 10 -9 m = 4.8427 x 10 -19J
Therefore the energy transition had an energy of 4.8427 x 10 -19J
Eelectron = En – E2 = (-2.179 x 10 -18/n2)J - (-2.179 x 10 -18/22) J = 4.8427 x 10 -19J
-2.179 x 10 -18 (1/n2 - 1/4) = 4.8427 x 10 -19
1/n2 – 1/4 = - 0.22239
0.02761 = 1/n2
n = 6.0182 = 6
REF: Text example 8.1
6. Draw the 3 p-orbitals 90% isodensity surface (probability distribution surface) and label each orbital.
Do the same for the 2s orbital.
ANS:
See figure 8.22 of your text, Page 260. Also check out the orbitron at;
http://winter.group.shef.ac.uk/orbitron/
REF: Pages 259 to 261
7. Calculate the effective nuclear charge for sulfur.
ANS:
Using the rules given on Page 268 of your text, Zeff = # of protons – (1 for every electron within
the n-2 and below) – (0.85 for each electron in the n-1 shell) – (0.35 for each electron in the
valence shell)
The electron configuration of carbon is; [Ne]3s23p4. There are 6 electrons in the valence shell.
An electron in this valence shell is thus shielded by all others in this same valence shell plus those
below. The atomic # of sulfur is 16. It has 16 electrons in total and a nuclear charge of +16 since
it has 16 protons.
`In the n-1 shell there are 2 electrons in a 2s orbital and 6 electrons in the 2p orbitals for a total of
8 electrons.
The remaining 2 electrons are in a 1s orbital (at the n-2 level).
Our solution is; Zeff = 16 – (1 x 2) – (0.85 x 8) – (0.35 x 5) = + 5.45
REF: Text example 8.4
8. In your text, it is presented that successive ionization energies for a given atom are an argument in
support of the wave model of electrons. Using your own words, explain why this is the case.
ANS:
The wave model for electrons enables us to derive the concepts of orbitals, of orbital shells,
valence shells, and sub-shells. By applying these concepts it is possible to derive a formula that
considers the shielding effect of electrons as a function of their orbital energies. This was applied
in the prior problem.
When we apply these calculations to a single atom and successively remove electrons, we find
that Zeff increases in a non-linear manner with removal of each successive electron. Also,
Copyright © 2011 Nelson Education Limited
8-4
removal of electrons in the n-1 shell is especially difficult as evidenced by the ionization energy.
This can be rationalized, using our model, as being due to their enhanced shielding, relative to
those in the shared valence shell. In contrast, electrons in the same valence shell are
comparatively easy to remove due to their lower shielding effect.
REF: Pages 269 to 271
9. Zeff for the halogens, in group 17, is high. In fact, from left to right within a period, the net trend is
an increase in Zeff. With reference to valence shell electrons and shielding, why is Zeff higher as one
traverses left to right within the period.
ANS:
As you move left to right, one atom at a time, you add one nuclear charge because the next atom
is more massive by one proton and some number of neutrons. This means that you also need to
add one more electron. One would expect that one negative electron would cancel one positive
proton but with respect to shielding, this is not the case. Inner electrons shield better than
electrons within the valence shell orbitals. This means that electrons added to a partially filled
valence shell are only weakly shielded by other electrons in that same shell. Consequently, as the
valence shell fills, when moving left to right, the effective nuclear charge for each successive
electron is greater than for the one before.
REF: Narrative B-4
10. On Page 264 of your text you’ll find figure 8.24. This diagram can be generated by hand at any time,
during an exam, a quiz, or while working on practice exercises. It’s easy to do. I prefer to do it
inverted as follows
After creating the text pattern,
draw parallel diagonal lines connecting the rows, as seen.
Follow the lines, tail to head then
move back up to the next tail to
arrive at the order of filling.
For example; start at 1s then move
to tail of next line, 2s, then move
to tail of next line, 2p, and move
down the line to 3s, then move to
tail of next line, 3p, etc.
You only have to remember the pattern of the diagram to regenerate it at any time. Notice; that
orbitals are added, one at a time, horizontal line by horizontal line until you reach the 4th line then
the 5th is the same length, then the lines of text decrease in length by one orbital per line as you move
through to the 8th horizontal line entry. Each horizontal line of text starts with your s-orbital, then if
required; a p, then d, then f.
For each line number, the principal quantum number remains the same for all orbitals on that
horizontal line. For instance the 5th line is 5s, 5p, 5d, 5f. In each case the line# matches the principal
quantum number. Line #1 is 1s, line #2 is 2s, 2p, line #3 is 3s, 3p, 3d, etc. Using this diagram,
generate the spdf shorthand notation electron configuration for the bromide ion. Do not abbreviate
further by employing the noble gas symbol.
ANS:
Copyright © 2011 Nelson Education Limited
8-5
Bromide ions arise from bromine atoms through acquisition of one electron. Elemental bromine
has 35 electrons.
Following the sequence in the diagram, the order of orbital filling is;
1s22s22p63s23p64s23d103p6
REF: Page 264 of text
C: LONG-ANSWER QUESTIONS
1
a) What is the energy of an electron in the 6th energy level of hydrogen, and in the 5th.
b) Which is higher in energy?
c) Why?
d) Judging from the trend as distance from the nucleus increases, what, do you expect, is the
energy of a free electron?
ANS:
5th = -8.72 x 10 -20 (from figure 8.13 of your text) 6th = -6.06 x 10 -20
The 6th energy level is higher in energy. It is a negative number of smaller magnitude (closer to
zero).
The 6th level is higher energy because the electron’s average distance from the nucleus is greater
and such charge separation requires energy.
As n tends to infinity, the energy of the electron will approach zero.
REF: Figure 8.13 of text
2.
Fill in the blanks for the following table without referring to your notes or your text
n
1
2
l
0
Orbital
1s
ml
0
degeneracy
1
1
3
4
2
0
4
3
4f
-3-,2,-1,0,1,2,3
7
n
1
2
l
0
0
1
0
1
2
0
Orbital
1s
2s
2p
3s
3p
3d
4s
ml
0
0
-1, 0, +1
0
-1,0, +1
-2,-1,0,+1+2
0
degeneracy
1
1
3
1
3
5
1
ANS:
3
4
Copyright © 2011 Nelson Education Limited
8-6
4
1
2
3
4p
4d
4f
-1,0,+1
-2,-1,0,+1,+2
-3-,2,-1,0,1,2,3
3
5
7
REF: Section 8.4 of text
3. In the photoelectric effect, electrons are ejected from a metal by a light source having an adequately
high frequency. Frequencies decrease with increasing wavelength and increase with decreasing wavelength. Higher frequency light is more effective than lower frequency at ejecting electrons. Higher
frequency equates with shorter wavelength. The shorter the wavelength, the higher the energy of the
photons. The electron that is ejected has kinetic energy = ½ mv2. Since mass of the electron is a fixed
value, only velocity can vary for the ejected electrons. The velocity and thus the kinetic energy are determined by the energy of the photon that hit the metal. Part of the photon energy is used to remove
the electron. This is the threshold energy, also called the work function, and it varies with the metal.
Any energy beyond the threshold limit goes to kinetic energy of the electron since energy is always
conserved (1st law).
This suggests that total energy of the photon equals the threshold energy + the energy transferred to
the electron. Since the energy given the electron is KE then the total energy of the photon, which we
know equals hv, equals KE + work function, .
I.e.; hv photon = KE electron + metal
An electron is ejected with a speed of 6.68 x 10 5 m/s from potassium metal when it is irradiated
with light. The work function of potassium is 2.29 eV. (one eV, is an electron volt. It is 1.60218 x
10 -19 J)
a) what is the energy of the photon?
b) what is the wavelength of the photon?
c) what is the longest wavelength of light that could have resulted in ejection of an electron with just
greater than zero kinetic energy?
ANS:
a) Since KE = ½ mv2 and the mass of an electron is 9.10938 x 10 -31kg then;
KE = ½ (9.10938 x 10 -31 kg)(6.68 x 10 5 m/s)2 = 4.0648 x 10 -19 J
The work function is given but must be converted to joules.
2.29 eV x 1.60218 x 10 -19 J/eV = 3.66899 x 10 -19 J
Then; hv = 4.0648 x 10 -19 J + 3.66899 x 10 -19 J = 7.7338 x 10 -19J
b) The wavelength of the light can be calculated using E photon= hc/
 = hc/hv = hc/(7.7338 x 10 -19) = (6.626 x 10 -34Js) (3.0 x 10 8 m/s) / (7.7338 x 10 -19) =
2.57 x 10 -9 m = 2.57 nm
c) The maximum wavelength of the light can be calculated from the minimum energy required.
The minimum energy required is the threshold energy. The photon would have to have the same
energy as the value of the work function in joules.
From part “a”, the work function = 3.66899 x 10 -19 J
Therefore Ephoton = hc/ = 3.66899 x 10 -19 J
And; 
10 -19 J = (6.626 x 10 -34J.s)(3.0 x 10 8 m/s)/( 3.66899 x 10 -19 J) =
5.41 x 10 -7 m = 541nm
Copyright © 2011 Nelson Education Limited
8-7
REF: Narrative
4. Light can be used to activate chemical reactions. One of the effects of light of sufficient wavelength
is to break bonds homolytically. Homolytic bond dissociation is the formation of radicals. This
occurs regularly in the upper atmosphere and leads to ozone formation. For each of the following,
calculate the wavelength of the photon required to break the bonds. The bond energy is given.
Assume that the energy required equals the bond energy. Also indicate in which region of the
spectrum electromagnetic radiation of this wavelength would appear.
Bond
Bond Energy
C-C
345
O-O
495
Br-Br
190
Wavelength of
Photons required
Region of
Electromagnetic
spectrum
ANS:
Bond
Bond Energy in
kJ/mole
C-C
345
Wavelength of
Photons required
(nm)
343
O-O
495
242
Region of
Electromagnetic
spectrum
Ultraviolet, almost
visible
ulraviolet
Br-Br
190
630
visible
Sample Calculation:
C-C bond; (345 kJ/mole of bonds) x (1 mole of bonds/(6.022 x 10 23 bonds)) = 5.79 x 10 -22
kJ/bond
The energy of the photon must be at least the energy of this bond = Ephoton = hc/
 = hc/Ephoton = 6.626 x 10 -34J.s  3.0 x 10 8 m/s/ 5.79 x 10 -19 J = 3.43 x 10 -7 m = 343 nm
REF: Fig. 3.20 & example 8.1of text
NARR: C-4
5.
On page 250 of your text is a photo of sodium chloride, strontium chloride, and boric acid, that have
been doused in ethanol and ignited. Each substance has a flame of characteristic color. The color is
not due to the ethanol, it is due to the nature of the salts. In each case, it is the cation that is creating
the color. Answer the following questions, based upon the narrative, and what you have learned about
the energetic of electrons.
a) What quantum effect is responsible for the flame colors?
b) How could you accurately determine the wavelength of the light being produced in each case?
c) Do you think this is a narrow frequency of light or broad, and why?
d) Are other wavelengths and their corresponding frequencies being produced by these same salts and
how would you know?
e) How could you increase the intensity of the light being produced and what is meant by this term?
ANS:
Copyright © 2011 Nelson Education Limited
8-8
This is atomic emission. The high temperature produced through combustion of the ethanol,
transfers energy to the electrons of the sodium, strontium, and boron. The electrons are promoted
to orbitals of higher energy level than are found in the ground state atoms or ions. The electrons
spontaneously drop back down to the lower energy levels and emit photons of characteristic
wavelength in the process. The wavelengths emitted are characteristic of the element and can be
used to identify them.
If the emitted light was shone through a diffraction grating or a prism, the light could be divided
into bands of specific wavelength on a detector or a photographic plate.
It is narrow frequency and of specific wavelengths. This is because the electrons exhibit discrete
quantized energy transitions within the atom or ion.
Yes, most elements produce emission bands at multiple wavelengths. Some are visible, some are
uv and some are longer than visible light.
Intensity is the number of photons being produced per unit area or over a given amount of time.
Intensity could be increased by increasing the efficiency of heating the salts or by increasing the
quantity of salts heated.
REF: http://en.wikipedia.org/wiki/Emission_spectrum
Copyright © 2011 Nelson Education Limited
8-9