Chapter 7:  Atomic Structure

 

 

Section 7-1:  Protons, Neutrons, and Electrons

Section 7-2:  Average Atomic Mass

Section 7-3:  Atomic Orbitals and Electron Configurations

Section 7-4:  Orbital Diagrams

Section 7-5:  Quantum Numbers

Chapter 7 Practice Exercises and Review Quizzes

 

 

Section 7-1:  Protons, Neutrons, and Electrons

 

We can deduce the number of positively-charged protons (p) and neutral neutrons (n) in the central nucleus of an atom as well as the number of negatively-charged 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 carbon-12 and carbon-13?

 

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:

 

carbon-12 = 12C                             carbon-13 = 13C

 

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 12C, so carbon-12 has 12 – 6 protons = 6 neutrons.  The sum of protons plus neutrons equals the mass number 13 in 13C, so carbon-13 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 204Pb2+ and (b) the anion 37Cl- ?

 

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 7-2:  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 isotope’s natural abundance.  To calculate the average atomic mass of an element, we will multiply the first isotope’s mass by the fraction of the element’s 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, chlorine-35 and chlorine-37.  Calculate the average atomic mass of chlorine using the information in the table below:

   

isotope

mass

% natural abundance

35Cl

34.969 amu

75.76%

37Cl

36.956 amu

24.24%

 

Solution:

 

average atomic mass =

(35Cl mass)(35Cl fraction) + (37Cl mass)(37Cl 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 7-3:  Atomic Orbitals and Electron Configurations

 

An atomic orbital is a three-dimensional 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.

 

For n = 4, in addition to one 4s orbital, three 4p orbitals, and five 4d orbitals, there are seven separate 4f orbitals, each of which can accommodate up to 2 electrons.  Therefore, the set of seven 4f orbitals can accommodate a maximum of 14 electrons, as will be the case for any other sets of seven f 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 4f 5d 6p 7s 5f 6d 7p

 

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 lead.

 

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, a maximum of 10 electrons for each different set of d orbitals, and a maximum of 14 electrons for each different set of f orbitals until we no longer have any electrons to add.

 

Pb has 82 electrons, so the electron configuration is:

 

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p2

 

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 Pb we would write the noble gas core [Xe] to represent the first 54 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, Pb is in row 6 from the top so we will continue after the noble gas core with the 6s orbital.  Pb has 82-54 = 28 remaining electrons, so the shorthand noble gas electron configuration will be [Xe] 6s2 4f14 5d10 6p2. 

 

Note that the electron configurations for some elements are exceptions and will not exactly follow the process above. Notable exceptions include:

 

Cr:  [Ar] 4s1 3d5

Mo:  [Kr] 5s1 4d5

 

Cu:  [Ar] 4s1 3d10

Ag:  [Kr] 5s1 4d10

Au:  [Xe] 6s1 4f14 5d10

 

 

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, Pb: [Xe] 6s2 4f14 5d10 6p2 has 2 + 2 = 4 valence electrons in the orbitals with the highest value of n = 6.

 

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) P3-

(b) Ge2+ and Ge4+

(c) Co2+ and Co3+

 

Solution:

   

(a) P = 15 e- :  [Ne] 3s2 3p3

      P3- = 18 e- :  [Ne] 3s2 3p6

 

(b) Ge = 32 e- :  [Ar] 4s2 3d10 4p2

remove two valence 4p electrons, so Ge2+ = 30 e- :  [Ar] 4s2 3d10

then remove two valence 4s electrons, so Ge4+ = 28 e- :  [Ar] 3d10

 

(c) Co = 27 e- :  [Ar] 4s2 3d7  

remove two valence 4s electrons since there are no valence p electrons, so Co2+ = 25 e- :  [Ar] 3d7

then remove one 3d electron outside noble gas core, so Co3+ = 24 e- :  [Ar] 3d6

        

 

 

Section 7-4:  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, a set of five d orbitals is represented by five adjacent boxes, and a set of seven f orbitals is represented by seven 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, d, or f 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 np3, 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:

np3

 

Incorrect:

 

np3

 

If an electron configuration ends in nd6, 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:

nd6

 

Incorrect:

 

 

nd6

 

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) Mn3+

 

Solution:

 

(a) Sn = 50 e- :  [Kr] 5s2 4d10 5p2

 

 

 

 

    

2 unpaired electrons, paramagnetic

 

(b) K = 19 e- :  [Ar] 4s1

 

                                                                                      

1 unpaired electron, paramagnetic

 

(c) Mn = 25 e- :  [Ar] 4s2 3d5

Mn3+ = 22 e- :  (no valence p electrons, so first remove two valence 4s electrons and then one 3d electron outside noble gas core) [Ar] 3d4

 

 

                                                                                                     

4 unpaired electrons, paramagnetic

 

 

 

 

Section 7-5:  Quantum Numbers

 

Each electron in an atom can be assigned a unique set of four quantum numbers that are typically expressed in the format (n, l, ml , ms ) where:

 

1. An electron with a larger value of the principle quantum number n will be farther from the nucleus on average.

 

2. The orbital quantum number l distinguishes the type of orbital.  For an s orbital, l = 0.  For a p orbital, l = 1.  For a d orbital, l = 2.  For an f orbital, l = 3.  For a given value of n, l is limited to 0 and the positive integers up to and including n - 1.

 

3. The magnetic quantum number ml distinguishes the relative orientation of an orbital.  For a given value of l, ml is limited to the set of integers between and including -l and l.  For an s orbital with l = 0, ml can only have a value of 0.  For a p orbital with l = 1, ml can only have a value -1, 0, or 1.  For a d orbital with l = 2, ml can only have a value of -2, -1, 0, 1, or 2.  For an f orbital with l = 3, ml can only have a value of -3, -2, -1, 0, 1, 2, or 3.

 

4. The spin quantum number ms distinguishes the spin of the electron and can have a value of -1/2 or +1/2.  Since two electrons paired in the exact same orbital must have opposite spins, the first three quantum numbers will be the same but one electron will have ms = +1/2 while the other will have ms = -1/2.  Since all unpaired electrons in an atom occupy different orbitals and must have parallel spins, each must have a different value of ml , and the group of unpaired electrons will either all have ms = +1/2 or all have ms = -1/2.   

 

Sample Exercise 7H:

 

State whether each set of quantum numbers is acceptable or not acceptable.  For those that are not acceptable, explain why:

 

a. (1, 0, 0, -1)

b. (2, 1, 2, -1/2)

c. (3, 3, -1, +1/2)

d. (4, 2, -2, -1/2)

 

Solution:

 

a. not acceptable, ms = -1 but should be +1/2 or -1/2

b. not acceptable, ml = 2 but should be -1, 0, or 1

c. not acceptable, l = 3 but should be either 0, 1, or 2 (no 3f orbital)

d. acceptable

 

 

 

 

Chapter 7 Practice Exercises and Review Quizzes:

 

7-1) Determine the number of protons, neutrons, and electrons in:

 

(a) a neutral silver-109 atom

(b) the anion 15N3-

(c) the cation 41K+

Click for Solution

 

7-1) (a) silver-109 = 109Ag:  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-

 

 

 

 

7-2) 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

 

7-2) 49 p = indium, mass number = 49 p + 64 n = 113, charge = 49 p – 46 e- = 3+

 

cation symbol =

 

 

 

 

7-3) Magnesium has three naturally occurring isotopes.  Calculate the average atomic mass of magnesium using the information in the table below:

 

isotope

mass

% natural abundance

24Mg

23.985 amu

78.99%

25Mg

24.986 amu

10.00%

26Mg

25.983 amu

11.01%

Click for Solution

 

7-3)

average atomic mass =

(24Mg mass)(24Mg fraction) + (25Mg mass)(25Mg fraction) +

(26Mg mass)(26Mg fraction) =

(23.985 amu)(0.7899) + (24.986 amu)(0.1000) + (25.983 amu)(0.1101) = 24.31 amu

 

 

 

7-4) Write the complete electron configuration and specify the number of valence electrons for thallium.

 

Click for Solution

 

7-4) Tl = 81 e- : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p1, 3 valence electrons

 

 

 

7-5) Write the shorthand noble gas electron configuration and specify the number of valence electrons for astatine.

 

Click for Solution

 

7-5) (a) At = 85 e- :  [Xe] 6s2 4f14 5d10 6p5, 7 valence electrons

 

 

 

 

 

7-6) Write the shorthand noble gas electron configuration for:

 

(a) Te2-

(b) In+ and In3+

(c) Fe2+ and Fe3+

Click for Solution

 

7-6) (a) Te = 52 e- :  [Kr] 5s2 4d10 5p4

           Te2- = 54 e- :  [Kr] 5s2 4d10 5p6 

 

(b) In = 49 e- :  [Kr] 5s2 4d10 5p1

remove one valence 5p electron, so In+ = 48 e- :  [Kr] 5s2 4d10

then remove two valence 5s electrons, so In3+ = 46 e- :  [Kr] 4d10

 

(c) Fe = 26 e- :  [Ar] 4s2 3d6  

remove two valence 4s electrons since there are no valence p electrons, so Fe2+ = 24 e- :  [Ar] 3d6

then remove one 3d electron outside noble gas core, so Fe3+ = 23 e- :  [Ar] 3d5

 

 

 

7-7) 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) O2-

(c) Ni2+

Click for Solution

 

7-7) (a) Se = 34 e- :  [Ar] 4s2 3d10 4p4

 

 

 

                                    

2 unpaired electrons, paramagnetic

 

(b) O = 8 e- :  [He] 2s2 2p4

  O2- = 10e- :  [He] 2s2 2p6

 

                                                                   

0 unpaired electrons, diamagnetic

 

(c) Ni = 28 e- :  [Ar] 4s2 3d8

Ni2+ = 26 e- :  [Ar] 3d8

 

2 unpaired electrons, paramagnetic

 

 

 

7-8) State whether each set of quantum numbers is acceptable or not acceptable.  For those that are not acceptable, explain why:

 

a. (1, 1, 1, -1/2)

b. (2, 1, 1, 1)

c. (3, 0, 0, +1/2)

d. (3, 2, -3, -1/2)

Click for Solution

7-8)

a. not acceptable, l = 1 but should be 0 (no 1p orbital)

b. not acceptable, ms = 1 but should be +1/2 or -1/2

c. acceptable

d. not acceptable, ml = -3 but should be -2, -1, 0, 1, or 2

 

 

 

Click for Review Quiz 1

Click for Review Quiz 1 Answers