Chapter 4 – Ionic Bond

Introduction

Atoms can gain or lose valence electrons to become ions. Ions can be monatomic, such as Ca²⁺ and Cl^1–, or polyatomic, such as NH₄¹⁺ and CO₃^2–. An ionic bond is the electrostatic (Coulombic) force of attraction between two oppositely charged ions. Ions and how they bond are the topic of this chapter.

4.1 Ionic Bonding

Introduction

Ionic bonds are the electrostatic attraction of oppositely charged ions. The cation is usually a metal, and the anion is usually a nonmetal.

4.1-1. Introduction to Bonding Video

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4.1-2. Ionic Bonding

An ionic bond is the Coulombic attraction of two oppositely charged ions.

Compounds between metals and nonmetals are ionic.

The unfilled valence orbitals of nonmetals experience large effective nuclear charges, so they are low in energy, which makes nonmetals, such as chlorine, highly electronegative. The valence electrons of metals that lie on the left side of the periodic table experience very low effective nuclear charges, so they are characterized by low ionization energies. Therefore, metals, such as sodium, lose their valence electron easily. When sodium bonds to chlorine, the high-energy valence electron on sodium is transferred to the low-energy unfilled orbital on chlorine. The loss of an electron by sodium produces a sodium one-plus ion, while the gain of an electron by chlorine produces a chlorine one-minus ion. The resulting ions of opposite charge lower their energy by moving close to one another. The interaction between the two interacting ions is called an ionic bond, and NaCl is an ionic compound. We will assume that all compounds between metals and nonmetals are ionic. However, recall that late metals (those that lie to the right of the periodic table) are fairly electronegative, so they do not give up their valence electrons easily, and their compounds are not very ionic.

metals transfer electrons to nonmetals to make ionic bonds

Figure 4.1: Ionic Bond Formation

4.1-3. Ionic Structure

Ionic compounds are arrays of individual ions with no identifiable molecules.

Each ion in a crystal of table salt is surrounded by six identical ions of opposite charge, and all Na–Cl distances are identical. Consequently, there are no pairs of ions that are identifiable as NaCl molecules.

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4.1-4. Ionic Compounds vs. Covalent Compounds

Compounds between nonmetals are covalent.

Covalent compounds are arrays of individual molecules, i.e., they are molecular.

Although covalent bonding is the topic of the following two chapters, a brief introduction is given here to allow us to compare the two different compound types in this chapter. Covalent compounds contain only nonmetals, so the atoms have similar electronegativities. Neither atom in a covalent bond has a very high energy electron that is readily transferred, so the atoms share rather than transfer the bonding electrons. Covalent bonds involve an overlap of atomic orbitals, so they are very directional, while ionic bonds are simply electrostatic interactions between spherical ions with no directionality. While the distances between adjacent Na¹⁺ and Cl^1– ions in a crystal of NaCl are all identical, there are two distinct distances observed between O and H atoms in an ice crystal. The existance of two distinct distances in ice allow us to identify individual water molecules as an oxygen atom and the two hydrodgen atoms that are closest to it. The longer distance is then between the oxygen atom on one molecule and a hydrogen atom on an adjacent water molecule in the crystal. Thus, ice consists of individual water molecules that interact to produce the solid, while table salt consists of Na¹⁺ and Cl^1– ions with no NaCl molecules.

4.1-5. Ionic or Covalent Exercise

4.2 Naming Ions and Predicting their Charge

Introduction

Most ionic compounds form between a metal cation and a nonmetal anion. In this section, we examine the electrons that are lost by the metal and the orbitals that are filled by the nonmetal. Once we know which electrons are lost and which orbitals are filled, we can predict the probable charges on the ions and write their electron configurations.

Prerequisites

•
3.1 Valence Electrons (Write valence electron configurations for atoms.)
•
3.5 Ionization Energy (Predict relative ionization energies of atoms.)
•
3.6 Electronegativity (Predict relative electronegativities of atoms.)

Objectives

•
Predict the cation that a metal is likely to adopt and write its electron configuration.
•
Predict the charge on the anion that a nonmetal is likely to adopt and write its electron configuration.
•
Determine the orbital occupancy of an ion given the charge on the ion and the occupancy of its atom.

4.2-1. Metal Cations

Cations are produced by the loss of valence electrons with those with the highest n quantum number being lost first. Consequently, first row transition metals lose their 4s electrons before they lose any 3d electrons.

Metals are characterized by low ionization energies, so they lose electrons to become cations, and the charge on the cation is determined by the number of electrons that are lost. While all of the electrons in an atom can be removed, the ionization energy of each successive ionization increases (3.5 Ionization Energy). As a result, no more than three electrons can be removed in chemical processes. In addition, core electrons are very tightly bound and are never removed in chemical processes. Thus, only the valence electrons are lost. While some metals lose all of their valence electrons, others lose only some of them. The following rules help determine which electrons are lost in chemical processes:

•
Monatomic cations with charges greater than +3 do not form.
•
Electrons from the outermost shell (highest n quantum number) are lost first. This is important in determining the cations formed by transition metals.
•
Within a shell, electrons from the subshell with the highest l quantum number are lost first. This is important for the heavier metals in Groups 3, 4, and 5.

Group 1A & 2A Metals	Group 3A Metals
Lose their valence electrons to form +1 and +2 ions, respectively.	Lose all of their valence electrons to form +3 ions. Tl forms both +3 and +1 ions, but not a +2 ion. The reason is that the heavier main group elements can lose only a portion of their valence shell. Tl is 6s² 6p¹. Both valence sublevels are in the same level, so the one with the highest l quantum number is emptied first. Thus, Tl can lose the 6p and not the 6s to form the +1 ion, but it cannot lose the 6s and not the 6p to form a +2 ion.
Group 4A Metals	Transition Metals
+4 monatomic ions do not exist, so the Group 4A metals cannot lose all of their valence electrons. However, the heavier metals in the group (Sn and Pb) can lose the electrons in the sublevel with the highest l quantum number, the outermost p sublevel, to form +2 ions.	Lose electrons in the level with the highest n quantum number. Thus, most transition elements lose their outermost s electrons to form +2 ions. Scandium is an exception because it loses all three valence electrons to form Sc³⁺ (no +2 ion). Silver forms only a +1 ion, and copper forms both +2 and +1 ions. In addition, several transition metals form a +3 ion in addition to a +2 ion.

Table 4.1: Cations Formed by Metals

4.2-2. Nonmetal Anions

Nonmetals form anions by gaining the number of electrons required to fill their valence shell (outermost s and p sublevels).

Nonmetals are electronegative, so they tend to gain electrons to become anions. The number of electrons gained equals the number required to fill their valence shell. A filled valence shell for all nonmetals except hydrogen contains eight electrons (two s and six p electrons). Thus,

charge on an anion = group number − eight

The resulting electron configuration is the same as that of the next noble gas, i.e., anions are isoelectronic with the next noble gas. Each successive electron is more difficult to add due to the Coulombic repulsion between the charge of the electron and that of the anion. Indeed, adding a fourth electron to a –3 anion is so difficult that –4 ions do not form in chemical processes.

7A	6A	5A	4A
–1 ions	–2 ions	–3 ions	Monatomic anions with charges of –4 do not exist, so the Group 4A nonmetals do not form anions.

Table 4.2: Anions Formed by Nonmetals

4.2-3. Predicting Charge and Electron Configuration Exercise

Exercise 4.2:

Each atom forms only one monatomic ion. What is the charge on that ion? (Express your answers as charge followed by magnitude. For example: +2, not 2+.)

S -2_0__ S has six valence electrons, so it needs two more to fill its valence shell. Adding two more electrons would produce the S^2– ion.

Mg +2_0__ Mg is a 2A metal, so it loses its two valence electrons to produce the Mg²⁺ ion.

Sc +3_0__ Sc is a 3A metal, so it loses its three valence electrons to produce the Sc³⁺ ion.

P -3_0__ P has five valence electrons, so it needs three more to fill its valence shell. Adding three more electrons would produce the P^3– ion.

Ag +1_0__ Ag is a transition metal, but it is an important exception to the +2 ion rule that applies to most transition metal ions. Ag forms only the +1 ion.

Zn +2_0__ The d-block of Zn is full, so it does not lose any d electrons. Thus, Zn forms only the +2 ion by losing its 4s electrons.

What are the electron configurations of the following ions? Use noble gas cores. (Separate all terms by a single space and indicate all superscripts with a carat (^). For example, [He] 2s^2 2p^2 for [He] 2s² 2p².)

Sc³⁺ o_[Ar]_s Scandium loses all of its valence electrons to become isoelectronic with argon.

N^3– o_[Ne]_s Nitrogen gains three electrons to become isoelectronic with neon.

Ni²⁺ o_[Ar] 3d^8_s Ni²⁺ is [Ar] 3d⁸. Remember that the electrons with the highest n quantum number are lost first. Thus, nickel loses its two 4s, not the 3d, electrons, to form the +2 ion.

Bi³⁺ o_[Xe] 6s^2 4f^14 5d^10_s Bi³⁺ is [Xe] 6s² 4f¹⁴ 5d¹⁰. The valence electron configuration of bismuth is 6s² 6p³. Recall that when electrons in the outermost shell are in two different sublevels, those with the highest l quantum number are lost first, so the three 6p electrons are lost before the 6s electrons.

K¹⁺ o_[Ar]_s K¹⁺ is [Ar] as the 4s electron is lost to produce the cation.

Br^1– o_[Kr]_s Br^1– is [Kr]. Monatomic anions are almost always isoelectronic with a noble gas because they almost always fill their valence shell.

4.2-4. Orbital Occupancies of Ions Exercise

4.3 Ionic vs. Atomic Size

Introduction

We saw in the previous chapter that the size of an atom depends upon the size of its outermost orbitals, and ions form when electrons leave or enter those orbitals. Consequently, the sizes of ions are different than those of the atoms. In this section, we compare the sizes of atoms, anions, and cations.

Prerequisites

•
3.3 Relative Atomic Size (Predict relative sizes of atoms from their positions in the periodic table.)

Objectives

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Explain why cations are smaller than their atoms, but anions are larger.
•
Predict the relative sizes of a series of ions.

4.3-1. Relative Ion Size

Cations are smaller than their atoms, but anions are larger than their atoms.

Figure 4.2

Relative Sizes of Cations and Their Atoms: A loss of electrons increases Z_eff and, if the valence shell is emptied, decreases the n quantum number. Consequently, cations are smaller than their parent atoms.

Figure 4.3

Relative Sizes of Anions and Their Atoms: A gain of electrons decreases Z_eff, so anions are larger than their atoms.

4.3-2. Ionic Size Exercise

Exercise 4.4:

Use only a periodic table to determine the largest ion in each group.

Cl^1– Ionic radii of ions with the same charge increase going down a group.
F^1– Ionic radii of ions with the same charge increase going down a group.
I^1–

Na¹⁺ Ionic radii increase in a period as the charge becomes more negative. The outermost electrons in the two metal ions are in the 2p sublevel, while those in the anion are in the 3p.
S^2–
Al³⁺ Ionic radii increase in a period as the charge becomes more negative. The outermost electrons in the two metal ions are in the 2p sublevel, while those in the anion are in the 3p.

Br^1–
K¹⁺ K¹⁺ is isoelectronic with Ar, and Br^1– is isoelectronic with Kr. Br^1– is larger than Se because the Br^1– ion has more electrons than protons, so the outermost electrons are shielded very well.
Se K¹⁺ is isoelectronic with Ar, and Br^1– is isoelectronic with Kr. Br^1– is larger than Se because the Br^1– ion has more electrons than protons, so the outermost electrons are shielded very well.

4.4 Oxidation States

Introduction

Electron counting (keeping track of where the electrons are in a compound) is a valuable aid in predicting the formulas of compounds, balancing certain types of chemical equations, predicting properties, and even predicting reactive centers. In this lesson, we introduce the oxidation state method for counting electrons. We then show how to determine the oxidation states of an atom in a molecule or ion and how to use oxidation states to predict formulas.

Prerequisites

•
3.4 Relative Orbital Energies (Predict relative valence orbital energies of atoms.)
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3.5 Ionization Energy (Predict relative valence orbital energies of atoms based on their ionization energies.)
•
3.6 Electronegativity (Predict relative valence orbital energies of atoms based on their electronegativities.)
•
2.3 Bohr Model

Objectives

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Define the term "oxidation state."
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Explain how the charge on a species (compound or ion) is related to the oxidation states of its atoms.
•
Determine the oxidation states of the atoms in an ion or molecule.
•
Predict chemical formulas for binary compounds.

4.4-1. Oxidation States from Chemical Formulas Video

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4.4-2. Oxidation State Definition

The oxidation state of an atom in a compound is the charge it would have if its bonds were ionic.

The oxidation state of an atom in a compound is the charge the atom would have if its bonds were ionic. However, it is used to account for the electrons in all compounds irrespective of the bond type. Oxidation states are obtained by assigning all bonding electrons to the more electronegative atom in each bond. Thus, we can conclude that the oxidation state of an atom is

•
negative if it is the more electronegative atom, and
•
positive if it is the less electronegative atom.

4.4-3. Oxidation States of Hydrogen and Chlorine

The most common oxidation state of Cl is –1 because most of its bonds are to less electronegative elements.

Cl has a zero oxidation state when bound to itself, and it can have positive oxidation states when bound to O or F.

As shown in Figure 4.4, the valence orbitals of hydrogen lie between those of the metals and the nonmetals. Thus, the electrons in a Metal–Hydrogen bond are assigned to the hydrogen, so it adopts a –1 oxidation state. However, in a Nonmetal–Hydrogen bond, the electrons are assigned to the lower energy nonmetal, and the hydrogen adopts a +1 oxidation state. In the H–H bond of H₂, both orbitals have the same energy, so the electrons must be shared to give each H an oxidation state of 0. The valence orbitals of chlorine lie below those of all other elements except those of oxygen and fluorine. Consequently, chlorine adopts a –1 oxidation state when bound to most elements. Like hydrogen, it has a zero oxidation state when bound to itself in Cl₂. However, it can adopt positive oxidation states when bound to O or F. In its binary compounds OCl₂ and FCl, it is assumed to lose the unpaired electron to adopt a +1 oxidation state, but if it is bound to more than one F or O, the appropriate number of electron pairs are also assigned to the O or F, so Cl can then adopt +3, +5, and +7 oxidation states. See Table 4.3 for some examples.

Ox State of Cl	Examples
–1	CCl₄O, NCl₃, KCl
0	Cl₂
+1	ClF, Cl₂O, ClO^1–
+3	ClF₃, ClO₂^1–
+5	ClO₃^1–
+7	ClO₄^1–

Table 4.3

Figure 4.4

Oxidation States of Hydrogen and Chlorine

4.4-4. Oxidation State Sum

The oxidation states of all of the atoms in a chemical species must sum to the charge on the species.

The oxidation states of the atoms in a molecule or ion represent the charge they would have if all of the bonds were ionic. Consequently,

•
the oxidation states of all the atoms in an ion must sum to the charge of the ion, and
•
the oxidation states of all the atoms in a molecule must sum to zero.

Example:
For example, let us determine the charge on the carbonate ion (CO₃^x) given that the oxidation states of C and O are +4 and –2, respectively. The charge on the ion equals the sum of the oxidation states, so we write

charge = (1C)(+4/C) + (3O)(−2/O) = 4 − 6 = −2

The carbonate ion is the CO₃^2– ion.

4.4-5. Oxidation State Guidelines

The oxidation state of an atom lies between its group number and its group number minus eight.

The maximum oxidation state of an atom equals the number of electrons that can be lost from its valence shell, which equals its group number. The lowest oxidation state that an atom can achieve is equal to minus the number of electrons required to fill its valence shell, which is –(8 – group number).

+(group number) ≥ Oxidation State ≥ −(8 − group number)

The more electronegative the element, the more likely it is to be found in its lowest (most negative) oxidation state, and the lower its ionization energy, the more likely it is to be found in its highest oxidation state. The valence orbitals of metals are high in energy, so they seldom accept electrons to obtain negative oxidation states. Thus, the metal is always in a positive oxidation state in compounds with nonmetals.

•
The Group 1A, 2A, and 3A metals are usually found in their highest oxidation states (+1, +2, and +3) when bound to nonmetals.
•
Transition elements tend to adopt oxidation states of +2 and +3, but Ag is +1 and Cu is +1 or +2 in most of their compounds. Some transition elements can achieve oxidation states as high as their group number when surrounded by highly electronegative atoms (usually oxygen). For example, vanadium is +5 in VO₄^3–, chromium is +6 in CrO₄^2–, and Mn is +7 in MnO₄^1–.

The valence orbitals of nonmetals are much lower than those of metals, so nonmetals are usually in their lowest oxidation states when bound to metals. The more electronegative a nonmetal is, the more likely it is to assume its lowest oxidation state. However, nonmetals can achieve positive oxidation states when bound to more electronegative elements. They can obtain oxidation states as high as their group number when bound to highly electronegative elements (especially O and F). For example, P is +5 in PO₄^3– and PF₅, S is +6 in SO₃ and SF₆, and Cl is +7 in ClO₄^1–.

4.4-6. Oxidation State Rules

The oxidation state guidelines give us ranges for the oxidation states of the elements, but many atoms have the same oxidation state in almost all of their compounds. To determine the oxidation state of an atom in a molecule or ion, use the oxidation state rules given in Table 4.4. The oxidation state rules are listed in order of priority, so they should be used in the order given; that is, a rule takes precedence over any rule below it, or any rule can be violated only to satisfy a rule above it. Note that the oxidation state rules merely reflect what we already know about atomic properties: atoms with low ionization energies (Groups 1A and 2A) are generally found in the highest oxidation states, while highly electronegative atoms (F and O) are generally in their lowest oxidation states.

Rule	Reason
The oxidation states of the atoms in an element are all zero.	When the valence orbitals of the two atoms are the same, the bonding electrons are assumed to be shared, not transferred. For example, the oxidation state of Cu in metallic Cu is zero, both F atoms in F₂ are zero (the only time F is not –1), and all eight sulfur atoms in S₈ are zero.
F is –1.	The valence orbitals of F are lower than the valence orbitals on any other element, i.e., F is the most electronegative element. Consequently, it is always assigned the bonding electrons. There is only one compound in which an F atom is not –1. What is it? Hint: See the only rule that takes priority over this rule.
1A metals are +1, 2A metals are +2, and Al is +3.	The valence orbitals in these metals are very high in energy, i.e., these metals have low ionization energies. They also become isoelectronic with a noble gas when they form the ions.
H is +1.	The 1s orbital on H is lower in energy than the valence orbitals of most metals and higher in energy than those of most nonmetals. Therefore, H is +1 except when Rule 1 forces it to be 0 or Rule 3 (metals with higher energy valence orbitals) forces it to be –1.
O is –2.	Oxygen is the second most electronegative atom, so it is usually assigned the bonding electrons. However, it is not –2 when it is elemental O₂ (Rule 1) or bonds to F (Rule 2). In addition, it can be –1 if Rules 3 and 4 force it. Compounds in which the oxidation state of oxygen is –1 are called peroxides. Peroxides contain O₂^2–, which has an O–O bond. Hydrogen peroxide (H₂O₂) is a common peroxide.
7A elements are –1.	Halogens are electronegative, so they tend to fill their valence shell to attain –1 oxidation states. However, they can attain positive oxidation states when bound to more electronegative atoms, such as oxygen, or more electronegative halogens, e.g., Cl is +7 in ClO₄^1–, Br is +5 in BrO₃^1–, and I is +3 in IF₃.

Table 4.4

4.4-7. Oxidation State Exercise

Exercise 4.5:

Oxidation State Rules

1
Atoms in elements are zero.
2
F is –1.
3
Group 1A metals are +1, 2A metals are +2, and Al is +3.
4
H is +1.
5
O is –2.
6
Group 7A elements are –1.

Determine the oxidation state of the first element in each of the following.

FeCl₃ +3_0__ Cl is –1 by Rule 6, and FeCl₃ has no charge, so x + 3(–1) = 0, or x = +3. The oxidation state of Fe is +3 in FeCl₃. Many transition metals are found in the +3 oxidation state.

MnO₄^1– +7_0__ O is –2 by Rule 5, and the charge on the MnO₄ ^1– ion is –1, so x + 4(–2) = –1, or x = +7. The oxidation state of Mn is +7 in MnO₄^1–. This very high oxidation state is stabilized by the highly electronegative oxygen atoms.

Cr₂O₇^2– +6_0__ O is –2 by Rule 5, and the charge on the Cr₂O₇^2– ion is –2, so 2x + 7(–2) = –2, or 2x = +12, or
x = +6. The oxidation state of Cr is +6 in Cr₂O₇^2–. Cr attains its highest oxidation state when surrounded by the highly electronegative oxygen atoms.

CaO₂ +2_0__ Ca is a Group 2A metal, so Ca has priority over oxygen and is assigned a +2 oxidation state by Rule 2. The oxidation state of O is determined as +2 + 2x = 0, or 2x = –2, or x = –1. This compound contains the O₂^2– ion, which is the peroxide ion. Thus, CaO₂ is called calcium peroxide. Calcium oxide is CaO.

NH₃ -3_0__ H is +1 by Rule 4, and there is no charge on NH₃, so x + 3(+1) = 0, or x = –3. The oxidation state of N is –3 in NH₃. This is one of the few cases where the element with the negative oxidation state is written first. The H is not written first because hydrogen atoms written at the beginning of a formula are assumed to be acidic (Chapter 12).

4.4-8. Using Oxidation States to Determine Charge Exercise

Exercise 4.6:

Determine the charge on each ion. Use the following oxidation states. (Express your answers as charge followed by magnitude. For example: +2, not 2+.)

O = –2 Cr = +6 P = +5 Mn = +7 N = +5

CrO₄^x -2_0__ The charge must equal the sum of the oxidation states: (1 Cr)(+6/Cr) + (4 O)(–2/O) = 6 – 8 = –2. The ion is the chromate ion: CrO₄^2–.

PO₄^x -3_0__ The charge must equal the sum of the oxidation states: (1 P)(+5/Cr) + (4 O)(–2/O) = 5 – 8 = –3. The ion is the phosphate ion: PO₄^3–.

MnO₄^x -1_0__ The charge must equal the sum of the oxidation states:
(1 Mn)(+7/Mn) + (4 O)(–2/O) = 7 – 8 = –1. The ion is the permanganate ion: MnO₄^1–.

NO₃^x -1_0__ The charge must equal the sum of the oxidation states: (1 N)(+5/Mn) + (3 O)(–2/O) = 5 – 6 = –1. The ion is the nitrate ion: NO₃^1–.

4.4-9. Chemical Formulas from Oxidation States Video

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4.4-10. Predicting Formulas

The likely formula of a compound can be predicted by assigning likely oxidation states to each of the atoms.

Compounds carry no net charge, so the total positive charge must equal the negative of the total negative charge. The total positive charge equals the positive oxidation state (OX_pos) times the number of atoms with that oxidation state (N_pos), and the total negative oxidation state equals the negative oxidation state (OX_neg) times the number of atoms that have that oxidation state (N_neg). We can therefore write the following:

N_posOX_pos = −N_negOX_neg

Rearranging, we obtain

N_pos

N_neg

OX_neg

OX_pos

Thus, the subscripts of the atoms in a chemical formula are inversely proportional to their oxidation state. To determine the formula of a binary compound (only two elements) follow these steps:

1
Assign oxidation states to the two elements. If the elements are not listed in the oxidation state rules, assign +Group number to the less electronegative element and –(8 – Group number) to the more electronegative element. However, recall that nonmetals can empty only their p sublevel, leaving two s electrons. In this case, OX_pos = Group number – 2.
2
Determine the ratio of oxidation states as a ratio of smallest whole numbers.
3
The ratio of the subscripts in the formula varies inversely with the ratio of oxidation states, so the whole numbers in step 1 are the subscripts.
4
Write the formula, but be sure to always write the less electronegative element first.

4.4-11. Predicting Formulas Exercise

Exercise 4.7:

Predict the formula of the compound formed between each pair of elements below. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH₄¹⁺.)

oxidation state of 1^st element

Na and O +1_0__ Na is a Group 1A metal and is always found in the +1 oxidation state in its compounds.

Zn and Cl +2_0__ Zn is a Group 2B metal and is always found in the +2 oxidation state in its compounds.

Al and S +3_0__ Al is a Group 3A metal and is always found in the +3 oxidation state in its compounds.

C and O +4_0__ C is the less electronegative element, so we assume that it is in its highest oxidation state. It is a Group 4A nonmetal, so its highest oxidation state is +4.

oxidation state of 2^nd element

Na and O -2_0__ Oxygen is usually found in the –2 oxidation state in its compounds.

Zn and Cl -1_0__ Chlorine is usually found in the –1 oxidation state in its compounds.

Al and S -2_0__ Sulfur is usually found in the –2 oxidation state in its compounds.

C and O -2_0__ Oxygen is usually found in the –2 oxidation state in its compounds.

LCM of the two oxidation states

Na and O 2_0__ The lowest common multiple of 2 and 1 is 2.

Zn and Cl 2_0__ The lowest common multiple of 2 and 1 is 2.

Al and S 6_0__ The lowest common multiple of 3 and 2 is 6.

C and O 4_0__ The lowest common multiple of 4 and 2 is 4.

formula of compound

Na and O o_Na_2O_s The LCM = 2, so two +1 atoms and one –2 atom are required. Sodium oxide is Na₂O.

Zn and Cl o_ZnCl_2_s The LCM = 2, so one +2 atom and two –1 atoms are required. Zinc chloride is ZnCl₂.

Al and S o_Al_2S_3_s The LCM = 6, so two +3 atoms and three –2 atoms are required. Aluminum sulfide is Al₂S₃.

C and O o_CO_2_s The LCM = 4, so one +4 atom and two –2 atoms are required. The compound formed between carbon and oxygen is CO₂.

4.4-12. Writing Formulas of Binary Compounds Exercise

Exercise 4.8:

Use the highest and lowest oxidation states of nonmetals and common oxidation states of metals to predict the formulas of the compounds that would form between each pair of elements. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH₄¹⁺.)

phosphorus and chlorine

o_PCl_5_s Phosphorus is a 5A element and chlorine is a 7A. Chlorine is more electronegative, so it is assigned the lowest oxidation state (7 – 8 = –1), while phosphorus is assigned its highest oxidation state (+5). The lowest common multiple is 5, so the compound is PCl₅. Phosphorus also forms a +3 oxidation state, in which case, the compound would be PCl₃

calcium and oxygen

o_CaO_s Calcium is a 2A metal, so it is +2. Oxygen is a 6A nonmetal, so it is assigned an oxidation state of –2. The compound is CaO.

carbon and fluorine

o_CF_4_s Fluorine is the most electronegative element, so it is always assigned the negative oxidation state. Thus, F is –1. Carbon is a 4A element, so it is assigned a +4 oxidation state when combined with fluorine. Four fluorine atoms at –1 each are required to balance the +4 carbon atom, so the formula is CF₄.

copper and sulfur

o_CuS_s Copper is a transition element, and it commonly adopts the +2 oxidation state. Sulfur is a 6A nonmetal, so its oxidation state is –2. The formula is CuS. However, copper also adopts the +1 oxidation state, in which case, the formula would be Cu₂S.

4.4-13. Orbital Diagrams and Oxidatation State Exercise

Exercise 4.9:

Use the energy level diagram for the valence electrons of X, Y, and Z to determine each answer. Always write the more electronegative element last. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH₄¹⁺. When more than one oxidation state is possible, choose the highest oxidation state.)

oxidation state of X when it bonds to Y or Z 1_0__ The electron on X is at higher energy than the unfilled orbitals on Y and Z, so the electron would be lost to form a +1 ion.

oxidation state of Y when it bonds to X -3_0__ The unfilled orbitals on Y are lower in energy than the electrons on X, so it would gain three electrons from X to fill its valence shell.

oxidation state of Y when it bonds to Z 5_0__ The electrons on Y are at higher energy than the unfilled orbitals on Z, so it would either lose the three p electrons or all five electrons.

oxidation state of Z when it bonds to Y -2_0__ The unfilled orbitals on Z are at the lowest energy, so three electrons would be donated to Z from either X or Y to fill its valence shell.

formula of the compound formed between X and Y o_X_3Y_s X is +1 and Y is –3, so the compound is X₃Y.

formula of the compound formed between X and Z o_X_2Z_s X is +1 and Z is –2, so the compound is X₂Z.

formula of the compound formed between Y and Z o_Y_2Z_5_s Z is –2 and Y is either +3 or +5. Y is less electronegative, so it is written first. If Y is +3, the compound is Y₂Z₃, and if it is +5, the compound is Y₂Z₅.

What are the identities (symbols) of X, Y, and Z if they are all in the second period?

X = o_Li_s X has only one valence electron, so it is in Group 1A. X is Li.

Y = o_N_s Y has five valence electrons, so it is in Group 5A and the second period. It is N.

Z = o_O_s Z has six valence electrons, so it is in Group 6A and the second period. It is O.

4.5 Polyatomic Ions

Introduction

So far, we have considered only monatomic ions (ions composed of a single atom), but many ions consist of several atoms and are called polyatomic ions. The bonds between the atoms in a polyatomic ion are not ionic, but a polyatomic ion does have a net charge and bonds to other ions via an ionic bond.

Objectives

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Identify the common polyatomic ions.

4.5-1. Polyatomic Ions

A number of ionic compounds are composed of polyatomic ions, which are charged groups of covalently bound atoms. While the polyatomic ion forms ionic bonds with oppositely charged ions, the atoms within a polyatomic ion are nonmetals held together by covalent bonds. Many of the polyatomic ions are oxoanions, i.e., they are negative ions that contain oxygen atoms covalently bound to another element. In the common polyatomic ions listed in Table 4.5, the only cations are ammonium and hydronium, and the only anions that are not oxoanions are hydroxide and cyanide. Recall that ionic compounds can be identified as those that contain metals because metals represent almost all of the common monatomic cations. However, ions can also be polyatomic. Ammonium is by far the most common polyatomic cation in compounds. Thus, NH₄Cl, NH₄NO₃, and (NH₄)₂SO₄ are also ionic compounds. We conclude that ionic compounds are those that contain either a metal or a polyatomic cation such as ammonium.

Cations
NH₄¹⁺	ammonium ion	H₃O¹⁺	hydronium ion
Anions
C₂H₃O₂^1–	acetate ion	OH^1–	hydroxide ion
CO₃^2–	carbonate ion	NO₃^1–	nitrate ion
ClO₄^1–	perchlorate ion	NO₂^1–	nitrite ion
ClO₃^1–	chlorate ion	MnO₄^1–	permanganate ion
ClO₂^1–	chlorite ion	O₂^2–	peroxide ion
ClO^1–	hypochlorite ion	PO₄^3–	phosphate ion
CrO₄^2–	chromate ion	SO₄^2–	sulfate ion
Cr₂O₇^2–	dichromate ion	SO₃^2–	sulfite ion
CN^1–	cyanide ion

Table 4.5: Some Common Polyatomic Ions

4.6 Naming Ionic Compounds

Introduction

Understanding how ionic compounds are named helps you better communicate about chemistry. In this section, we explain how binary ionic compounds and ionic compounds containing polyatomic ions are named.

Prerequisites

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1.11 Dimitri Mendeleev and The Periodic Law

Objectives

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Name ions that contain one or more protons.
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Name ionic compounds containing polyatomic ions.

4.6-1. Binary Compounds

The oxidation state of the metal is given as a Roman numeral in parentheses when naming inorganic compounds that contain a metal that can attain more than one oxidation state.

The name of a binary ionic compound is simply the name of the cation (name of the metal atom) followed by the name of the anion (name of nonmetal with ending changed to -ide). If the metal has the same oxidation state in all of its compounds, the oxidation state is not indicated. The metals for which the oxidation state is usually omitted are the metals of Groups 1A and 2A, Al, Sc, Zn, Ag, and Cd. The remaining metals have more than one possible oxidation state, and so the oxidation state of the metal is indicated with a Roman numeral in parentheses after the name of the metal. Note that there is no space between the name of the metal and the Roman numeral in parentheses. Some examples:

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NaCl = sodium chloride (Group 1A are always +1.)
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Mg₃N₂ = magnesium nitride (Group 2A are always +2.)
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ZnBr₂ = zinc bromide (Zn is always +2.)
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AgF = silver fluoride (Ag is always +1.)
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Sc₂O₃ = scandium oxide (Sc is always +3.)
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CoO is cobalt(II) oxide to distinguish it from Co₂O₃, which is cobalt(III) oxide.
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TlCl is thallium(I) chloride to distinguish it from TlCl₃, which is thallium(III) chloride.

4.6-2. Naming Exercise

Exercise 4.10:

Name the following compounds.

AlBr₃ o_aluminum bromide_s Al is always +3.

ZnCl₂ o_zinc chloride_s Zn is always +2.

Ag₂O o_silver oxide_s Ag is always +1.

FeCl₃ o_iron(III) chloride_s Fe can be +2 or +3.

CuCl o_copper(I) chloride_s Cu can be +1 or +2.

PbO₂ o_lead(IV) oxide_s Pb can be +2 or +4.

Hg₂Cl₂ o_mercury(I) chloride_s Hg can be +1 or +2.

MnO₂ o_manganese(IV) oxide_s Manganese can adopt many oxidation states.

ZnBr₂ o_zinc bromide_s Zinc is always in the +2.

SnO₂ o_tin(IV) oxide_s Tin can be +2 and +4.

CoCl₃ o_cobalt(III) chloride_s Cobalt is a transition metal that can adopt several oxidation states.

K₂O o_potassium oxide_s Potassium is a 1A metal, so it can attain only the +1 oxidation state.

4.6-3. Naming Oxoanions of the Elements of Groups 4, 5, and 6

Suffixes are used to indicate the oxidation state of the central atom in most oxoanions:

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-ate implies that the central atom is in its highest oxidation state (group number).
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-ite tells us that the oxidation state of the central atom is lower than the highest oxidation state by two because the ion has one less oxygen atom than the corresponding ion that ends in -ate.

Removing an oxygen atom from an oxoanion reduces the oxidation state of the central atom by two, but it does not change the charge on the ion. Some examples:

Group 4	CO₃^2– is the carbonate ion because Group 4A carbon is in the +4 oxidation state.
Group 5	PO₄^3– is the phosphate ion because Group 5A phosphorus is in the +5 oxidation state. However, NO₃^1– is the nitrate ion. Removing a single oxygen atom produces NO₂^1–, the nitrite ion.
Group 6	SO₄^2– is the sulfate ion because Group 6A sulfur is in the +6 oxidation state. Removal of one oxygen atom produces the sulfite ion, SO₃^2–.

Table 4.6

4.6-4. Naming Oxoanions of Group 7

The oxoanions of the Group 7A elements are an exception because, unlike the others, they each form four oxoanions. Consequently, both prefixes and suffixes must be used (see Table 4.7). In the perchlorate ion (ClO₄^1–), the chlorine is in its highest oxidation state (+7). The chlorate ion (ClO₃^1–) has one less oxygen, so the oxidation state of Cl is two less, or +5. The chlorite ion (ClO₂^1–) has one less oxygen than chlorate, which lowers the oxidation state of Cl to +3. Finally, the hypochlorite ion (ClO^1–) has one less oxygen than chlorite, and the oxidation state of the Cl is reduced to +1. Similarly, perbromate is BrO₄^1–, bromate is BrO₃^1–, etc. Note, however, that fluorine is the most electronegative element, so it never has a positive oxidation state. Consequently, fluorine forms no oxoanions.

Prefix	Suffix	Oxidation State of Halogen	Formula
Per-	-ate	+7	XO₄^1–
	-ate	+5	XO₃^1–
	-ite	+3	XO₂^1–
Hypo-	-ite	+1	XO^1–

Table 4.7: Prefixes and Suffixes of the Oxoanions Formed by the Halogens

4.6-5. Protonated Ions

Ions that have charges of –2 and –3 pick up protons to produce protonated anions, which are named by placing hydrogen (or dihydrogen) and a space in front of the name of the ion. An older, but still common, method for naming some of these ions is to replace the "hydrogen and a space" with simply "bi" with no space. Thus, HS^1– is either the hydrogen sulfide ion or the bisulfide ion.

HCO₃^1–	hydrogen carbonate or bicarbonate ion
HPO₄^2–	hydrogen phosphate ion
H₂PO₄^1–	dihydrogen phosphate ion
HSO₄^1–	hydrogen sulfate or bisulfate ion
HSO₃^1–	hydrogen sulfite or bisulfite ion

Table 4.8: Common Protonated Oxoanions

4.6-6. Writing Formulas for Compounds with Polyatomic Ions Exercise

Exercise 4.11:

Use the fact that the names of the polyatomic ions are used without change when naming compounds that contain one or more polyatomic ions to write formulas for the following compounds. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH₄¹⁺.)

ammonium bromide o_NH_4Br_s The ammonium ion is NH₄¹⁺. The bromide ion is Br^1–. Remember that the ending is -ate or -ite for oxoanions. The charges cancel in the 1:1 compound, so ammonium bromide is NH₄Br.

potassium chlorate o_KClO_3_s The potassium (Group 1A) ion is K¹⁺. The chlorate ion is ClO₃^1–. The charges cancel in the 1:1 compound, so potassium chlorate is KClO₃.

cobalt(III) nitrate o_Co(NO_3)_3_s The cobalt(III) ion is Co³⁺. The nitrate ion is NO₃^1–. Three nitrate ions are required for each cobalt(III) ion, so cobalt(III) nitrate is Co(NO₃)₃.

scandium phosphate o_ScPO_4_s The scandium ion is Sc³⁺. The phosphate ion is PO₄^3–. The +3 and –3 charges cancel in a 1:1 ratio, so scandium phosphate is ScPO₄.

ammonium sulfate o_(NH_4)_2SO_4_s The ammonium ion is NH₄¹⁺. The sulfate ion is SO₄^2–. Two ammonium ions are required to balance the charge of one sulfate ion, so the formula of ammonium sulfate is (NH₄)₂SO₄.

calcium cyanide o_Ca(CN)_2_s The calcium ion is Ca²⁺. The cyanide ion is CN^1–. Note that this is one of only two anions that end in -ide but are not monatomic ions. The other is the hydroxide ion, OH^1–. The formula of calcium cyanide is Ca(CN)₂.

4.7 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.

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End-of-Chapter Exercises
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Solutions to Odd Exercises