Chapter 7: Atomic Structure
Section 71:Ê Protons, Neutrons, and Electrons
Section 72:Ê Average Atomic Mass
Section 73:Ê Atomic Orbitals and Electron Configurations
Section 74:Ê Orbital Diagrams
Chapter 7 Practice Exercises and Review Quizzes
Section 71: Protons, Neutrons, and Electrons
We can deduce the number of positivelycharged protons (p) and neutral neutrons (n) in
the central nucleus of an atom as well as the number of negativelycharged
electrons (e^{}) surrounding the nucleus if we are given a symbol in
the following format:
In the symbol above:
1. X is the symbol of the element.
2. Z is the atomic number for the element found on the periodic table
and is equal to the number of protons.
Note that if Z is omitted from the symbol, we can still determine the
number of protons from the periodic table.
3. A is the mass number and is equal to the number of protons plus
neutrons in the nucleus. Note that
if both A and Z are omitted from the symbol, we can still determine the number
of protons from the periodic table, but we cannot determine the number of
neutrons.
4. If the charge is omitted, we can
assume that the atom is neutral and, therefore, that the number of protons and
the number of electrons are equal.
If the number of protons and the number of electrons are not equal, the
charged species is known as an ion.
Isotopes are atoms of the same element that differ in the number of
neutrons, as demonstrated in the following problem:
Sample Exercise 7A:
How many protons, neutrons, and
electrons are in neutral atoms of the isotopes carbon12 and carbon13?
Solution:
First, we can translate the name of
each isotope into a symbol with the number after the hyphen in the isotope name
becoming the mass number, A, in the upper left of the symbol:
carbon12 = ^{12}C
carbon13 = ^{13}C
Since both are isotopes of carbon,
we know from the periodic table that each has 6 protons. Since both are neutral, we know that
each has 6 electrons as well. The
sum of protons plus neutrons equals the mass number 12 in ^{12}C, so
carbon12 has 12 – 6 protons = 6 neutrons. The sum of protons plus neutrons equals the mass number 13
in ^{13}C, so carbon13 has 13  6 protons = 7 neutrons.
A positively charged ion will have
more protons than electrons and is known as a cation,
while a negatively charged ion will have more electrons than protons and is
known as an anion, as demonstrated in the following problem:
Sample Exercise 7B:
How many protons, neutrons, and
electrons are in (a) the cation ^{204}Pb^{2+}
and (b) the anion ^{37}Cl^{} ?
Solution:
(a) We know from the periodic table
that lead has 82 protons. The mass
number 204 – 82 protons = 122 neutrons. From the charge, we know that the cation
has 2 more protons than electrons, so there are 82 protons – 2 = 80
electrons.
(b) We know from the periodic table
that chlorine has 17 protons. The
mass number 37 – 17 protons = 20 neutrons. From the charge, we know that the anion has 1 more electron
than the number of protons, so there are 17 protons + 1 = 18 electrons.
If we know the number of protons,
neutrons, and electrons in an atom or ion, we can deduce the symbol of the atom
or ion, as demonstrated in the following problem:
Sample Exercise 7C:
Write a symbol that includes atomic
number, mass number, and charge for the species with 16 protons, 18 neutrons,
and 18 electrons.
Solution:
We know from the periodic table
that the element with 16 protons and, therefore, an atomic number of 16 is
sulfur. The mass number = 16
protons + 18 neutrons = 34. Since
there are 2 more electrons than protons, the charge is 2. Therefore, the symbol of the anion is:
Section 72: Average Atomic Mass
The unit used for the masses of
different isotopes is the atomic mass unit (amu). The mass listed on the periodic table
for each element is an average of the masses of each naturally occurring
isotope that is weighted to take into account each isotopeÕs natural
abundance. To calculate the
average atomic mass of an element, we will multiply the first isotopeÕs mass by
the fraction of the elementÕs naturally occurring atoms belonging to the first
isotope and then do the same for all the other isotopes, after which we will
add together all the products:
average atomic mass =
(mass 1)(fraction 1) + (mass
2)(fraction 2) + . . .
Note that if percent natural
abundances are given for each isotope, we must first convert the percentages to
fractions in decimal form before using the average atomic mass equation above,
as demonstrated in the following problem:
Sample Exercise 7D:
Chlorine has two naturally
occurring isotopes, chlorine35 and chlorine37. Calculate the average atomic mass of chlorine using the
information in the table below:
isotope 
mass 
% natural abundance 
^{35}Cl 
34.969 amu 
75.76% 
^{37}Cl 
36.956 amu 
24.24% 
Solution:
average atomic mass =
(^{35}Cl
mass)(^{35}Cl fraction) + (^{37}Cl mass)(^{37}Cl
fraction) =
(34.969 amu)(0.7576) + (36.956 amu)(0.2424)
= 35.45 amu
Note that our calculated result
matches the average atomic mass listed for chlorine on the periodic table.
Section 73: Atomic Orbitals
and Electron Configurations
An atomic orbital is a
threedimensional space with the nucleus at the center wherein there is a high
probability of finding an electron.
A maximum of two electrons can occupy a single orbital, regardless of
the size of the orbital. The type
of orbital can be specified using the format nx, where:
1. n is the principle quantum
number essentially indicating the relative size of the orbital, with smaller
integers representing smaller orbitals.
2. x is a letter essentially indicating the shape of the orbital. For example, s indicates that the
orbital has a spherical shape and p indicates that the orbital has a dumbbell
shape:
There are also d and f orbitals.
For n = 1, there is only one orbital available labeled 1s that can accommodate up to 2 electrons, as will be the case for any other s orbitals.
For n = 2, there is a 2s orbital available that can accommodate up to 2 electrons. There are also three separate 2p orbitals oriented at right angles to each other, each of which can accommodate up to 2 electrons. Therefore, the set of three 2p orbitals can accommodate a maximum of 6 electrons, as will be the case for any other sets of three p orbitals.
For n = 3, in addition to one 3s orbital and three 3p orbitals, there are five separate 3d orbitals, each of which can accommodate up to 2 electrons. Therefore, the set of five 3d orbitals can accommodate a maximum of 10 electrons, as will be the case for any other sets of five d orbitals.
According to the Aufbau (or building up) principle, electrons are added to
the atomic orbitals outside the nucleus in the
following order:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s . . .
Note that the above order does
continue beyond 6s and will eventually include f orbitals
if extended.
An electron configuration describes the distribution of electrons
among the atomic orbitals in the order above using
superscripts to indicate the total number of electrons in each set of orbitals, as demonstrated in the following problem:
Sample Exercise 7E:
Write the complete electron
configuration for (a) cesium and (b) antimony.
Solution:
We will add a maximum of 2
electrons for each different s orbital, a maximum of 6 electrons for each
different set of p orbitals, and a maximum of 10
electrons for each different set of d orbitals until
we no longer have any electrons to add.
(a) Cs has 55 electrons, so the
electron configuration is:
1s^{2}
2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10}
4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{1}
(b) Sb
has 51 electrons, so the electron configuration is:
1s^{2}
2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10}
4p^{6} 5s^{2} 4d^{10} 5p^{3}
Note that the electron
configurations for some elements are exceptions and will not exactly follow the
process above.
For some of our purposes, we will
only need to know the distribution of electrons that occupy the larger orbitals in the atom and, therefore, that tend to be more
distant from the nucleus. In these
cases, we can write what are known as shorthand noble gas electron
configurations that focus only on the outermost electrons. (The noble gases are the elements in
Group 18 furthest to the right on the periodic table.) To write the shorthand noble gas
electron configuration for an element, we will do the following:
1. Represent the inner electrons in
the atom by writing the symbol of the noble gas at the end of the previous row
in brackets. For example, for Cs
we would write the noble gas core [Xe] to represent
the first 54 electrons in the atom, and for Sb we
would write the noble gas core [Kr] to represent the first 36 electrons in the
atom.
2. For the remaining electrons, continue
the electron configuration starting with the ns orbital, where n is equal to
the row number of your element.
For example, Cs is in row 6 from the top so we will continue after the
noble gas core with the 6s orbital, whereas Sb is in
row 5 from the top so we will continue after the noble gas core with the 5s
orbital. Cs has 5554 = 1
remaining electron, so the shorthand noble gas electron configuration will be [Xe] 6s^{1}.
Sb has 5136 = 15 remaining electrons, so the
shorthand noble gas electron configuration will be [Kr] 5s^{2} 4d^{10}
5p^{3}.
Valence electrons are the electrons in atomic orbitals
with the largest value of n. Since
valence electrons are outside the noble gas core, we can use the shorthand
noble gas electron configuration to determine the number of valence electrons
in an atom. For example, Sb: [Kr] 5s^{2} 4d^{10} 5p^{3}
has 2 + 3 = 5 valence electrons in the orbitals with
the highest value of n = 5.
To determine the electron
configuration of an anion, start by writing the electron configuration of the
neutral atom and then add the appropriate number of extra electrons.
To determine the electron
configuration of a cation, start by writing the
electron configuration of the neutral atom and then remove the appropriate
number of electrons in the following order:
1. valence p electrons
2. valence s electrons
3. d electrons outside the noble gas core
Sample Exercise 7F:
Write the shorthand noble gas
electron configuration for:
(a) P^{3}
(b) Ge^{2+} and Ge^{4+}
(c) Co^{2+} and Co^{3+}
Solution:
(a) P = 15 e^{}
: [Ne] 3s^{2} 3p^{3}
P^{3} = 18 e : [Ne] 3s^{2}
3p^{6}
(b) Ge =
32 e^{} : [Ar] 4s^{2} 3d^{10}
4p^{2}
remove two
valence 4p electrons, so Ge^{2+} = 30 e^{} : [Ar] 4s^{2}
3d^{10}
then remove two
valence 4s electrons, so Ge^{4+} = 28 e^{} : [Ar] 3d^{10}
(c) Co = 27 e^{}
: [Ar]
4s^{2} 3d^{7}
remove two
valence 4s electrons since there are no valence p electrons, so Co^{2+}
= 25 e^{} : [Ar] 3d^{7}
then remove one
3d electron outside noble gas core, so Co^{3+ }= 24 e^{}
: [Ar]
3d^{6}
Section 74: Orbital Diagrams
An orbital diagram uses boxes to represent
each individual atomic orbital outside the noble gas core. An s orbital is represented by one box,
a set of three p orbitals is represented by three adjacent boxes, and a set of
five d orbitals is represented by five adjacent boxes. Each box representing an orbital can be
occupied by either 0, 1, or 2 electrons. Each individual electron is represented by an arrow according to the
following:
1. The Pauli Exclusion Principle essentially states that two electrons occupying the same orbital must have opposite spins. Therefore, a pair of electrons in the same orbital must be represented as one up arrow and one down arrow:
↑↓ 
2. Hund’s Rule essentially states that electrons will occupy separate
p or d orbitals if possible in order to achieve the greatest number of
unpaired electrons with parallel (same direction) spins. For example, if an electron configuration ends in np^{3}, the
three electrons will spread out and occupy three separate p orbitals to achieve
three unpaired electrons with parallel spins rather than having one pair, one unpaired
electron, and one empty orbital:
Correct:
↑ 
↑ 
↑ 
np^{3}
Incorrect:
↑↓ 
↑ 

np^{3}
If an electron configuration ends
in nd^{6}, five of the electrons will occupy five separate d orbitals,
but then the sixth electron will be paired with one of the other five to
achieve a total of four unpaired electrons with parallel spins rather than having three pairs
and two empty orbitals.
Correct:
↑↓ 
↑ 
↑ 
↑ 
↑ 
nd^{6}
Incorrect:
↑↓ 
↑↓ 
↑↓ 


nd^{6}
An atom or ion having unpaired
electrons is said to be paramagnetic and
will be attracted to a magnet. On
the other hand, an atom or ion having no unpaired electrons is said to be diamagnetic and will not be attracted
to a magnet.
Sample Exercise 7G:
For each of the following, write
the orbital diagram, determine the number of unpaired electrons, and state
whether the atom or ion is paramagnetic or diamagnetic:
(a) tin
(b) potassium
(c) Mn^{3+}
Solution:
(a) Sn = 50 e^{} :
↑↓


↑↓

↑↓

↑↓

↑↓

↑↓


↑

↑


2 unpaired
electrons, paramagnetic
(b) K = 19 e^{} : [Ar] 4s^{1}
↑

1 unpaired
electron, paramagnetic
(c) Mn = 25 e^{} : [Ar] 4s^{2} 3d^{5}
Mn^{3+} = 22 e^{} : (no valence p electrons, so
first remove two valence 4s electrons and then one 3d electron outside noble
gas core) [Ar] 3d^{4}
↑

↑

↑

↑


4 unpaired
electrons, paramagnetic
Chapter 7 Practice Exercises and Review Quizzes:
71) Determine
the number of protons, neutrons, and electrons in:
(a) a
neutral silver109 atom
(b) the
anion ^{15}N^{3}
(c) the cation ^{41}K^{+}
Click for Solution
71) (a) silver109 = ^{109}Ag: silver = 47 p, 109 – 47 p = 62 n, neutral = 47 e^{}
(b) nitrogen = 7 p, 15 – 7 p = 8 n, 7 p + 3 = 10 e^{}
(c) potassium = 19 p, 41 – 19 p = 22 n, 19 p – 1 = 18 e^{}
72) Write
a symbol that includes atomic number, mass number, and charge for the species
with 49 protons, 64 neutrons, and 46 electrons.
Click for Solution
72) 49 p = indium, mass number = 49 p + 64 n = 113, charge = 49 p – 46 e^{} = 3+
cation symbol =
73) Magnesium has three naturally
occurring isotopes. Calculate the
average atomic mass of magnesium using the information in the table below:
isotope 
mass 
% natural abundance 
^{24}Mg 
23.985 amu 
78.99% 
^{25}Mg 
24.986 amu 
10.00% 
^{26}Mg 
25.983 amu 
11.01% 
Click for Solution
73)
average atomic mass =
(^{24}Mg
mass)(^{24}Mg fraction) + (^{25}Mg mass)(^{25}Mg
fraction) +
(^{26}Mg
mass)(^{26}Mg fraction) =
(23.985 amu)(0.7899) + (24.986 amu)(0.1000)
+ (25.983 amu)(0.1101) = 24.31 amu
74) Write the complete electron
configuration and specify the number of valence electrons for (a) barium and
(b) nickel.
Click for Solution
74) (a) Ba = 56 e^{} : 1s^{2} 2s^{2} 2p^{6}
3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2}
4d^{10} 5p^{6} 6s^{2} , 2 valence electrons
(b) Ni = 28 e^{ }: 1s^{2} 2s^{2} 2p^{6}
3s^{2} 3p^{6} 4s^{2} 3d^{8} , 2 valence
electrons
75) Write the shorthand noble gas
electron configuration and specify the number of valence electrons for (a) bromine
and (b) cadmium.
Click for Solution
75) (a)
Br = 35 e^{} : [Ar] 4s^{2} 3d^{10} 4p^{5}
, 7 valence electrons
(b) Cd = 48 e^{} : [Kr] 5s^{2} 4d^{10} , 2 valence electrons
76) Write
the shorthand noble gas electron configuration for:
(a) Te^{2}
(b) In^{+} and In^{3+}
(c) Fe^{2+} and Fe^{3+}
Click for Solution
76) (a)
Te = 52 e^{} : [Kr] 5s^{2}
4d^{10} 5p^{4}
Te^{2}
= 54 e^{} : [Kr] 5s^{2} 4d^{10} 5p^{6}
(b) In = 49 e^{}
: [Kr] 5s^{2}
4d^{10} 5p^{1}
remove one
valence 5p electron, so In^{+} = 48 e^{} : [Kr] 5s^{2} 4d^{10}
then remove two
valence 5s electrons, so In^{3+} = 46 e^{} : [Kr] 4d^{10}
(c) Fe = 26 e^{}
: [Ar]
4s^{2} 3d^{6}
remove two
valence 4s electrons since there are no valence p electrons, so Fe^{2+}
= 24 e^{} : [Ar] 3d^{6}
then remove one
3d electron outside noble gas core, so Fe^{3+ }= 23 e^{}
: [Ar]
3d^{5}
77) For
each of the following, write the orbital diagram, determine the number of
unpaired electrons, and state whether the atom or ion is paramagnetic or
diamagnetic:
(a) selenium
(b) O^{2}
(c) Ni^{2+}
Click for Solution
77) (a) Se = 34 e^{} :
[Ar] 4s^{2} 3d^{10} 4p^{4}
↑↓


↑↓

↑↓

↑↓

↑↓

↑↓


↑↓

↑

↑

2 unpaired
electrons, paramagnetic
(b) O = 8 e^{} : [He] 2s^{2} 2p^{4}
O^{2} = 10e^{} : [He] 2s^{2 }2p^{6}
↑↓


↑↓

↑↓

↑↓

0 unpaired
electrons, diamagnetic
(c) Ni = 28 e^{} : [Ar] 4s^{2} 3d^{8}
Ni^{2+} = 26 e^{} :
↑↓

↑↓

↑↓

↑

↑

2 unpaired
electrons, paramagnetic