Chapter 7: Atomic Structure

Section 7-1: Protons, Neutrons, and Electrons

Section 7-3: Atomic Orbitals and Electron Configurations

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.

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:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s¹

(b) Sb has 51 electrons, so the electron configuration is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p³

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 55-54 = 1 remaining electron, so the shorthand noble gas electron configuration will be [Xe] 6s¹. Sb has 51-36 = 15 remaining electrons, so the shorthand noble gas electron configuration will be [Kr] 5s² 4d¹⁰ 5p³.

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] 5s² 4d¹⁰ 5p³ 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) Ge²⁺ and Ge⁴⁺

Solution:

(a) P = 15 e^- : [Ne] 3s² 3p³

P^3- = 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

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

Solution:

(a) Sn = 50 e^- : [Kr] 5s² 4d¹⁰ 5p²

↑↓

↑

2 unpaired electrons, paramagnetic

(b) K = 19 e^- : [Ar] 4s¹

↑

1 unpaired electron, paramagnetic

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

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 ¹⁵N^3-

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^-

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 (a) barium and (b) nickel.

Click for Solution

7-4)

(a) Ba = 56 e^- : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² , 2 valence electrons

(b) Ni = 28 e^-: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸ , 2 valence electrons

7-5) Write the shorthand noble gas electron configuration and specify the number of valence electrons for (a) bromine and (b) cadmium.

Click for Solution

7-5)

(a) Br = 35 e^- : [Ar] 4s² 3d¹⁰ 4p⁵ , 7 valence electrons

(b) Cd = 48 e^- : [Kr] 5s² 4d¹⁰ , 2 valence electrons

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

(a) Te^2-

(b) In⁺ and In³⁺

Click for Solution

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

Te^2- = 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

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) O^2-

Click for Solution

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

↑↓

↑

2 unpaired electrons, paramagnetic

(b) O = 8 e^- : [He] 2s² 2p⁴

O^2- = 10e^- : [He] 2s²2p⁶

↑↓

0 unpaired electrons, diamagnetic

Ni²⁺ = 26 e^- : [Ar] 3d⁸

↑↓

↑

2 unpaired electrons, paramagnetic

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