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Chapter 5
Ionic and Covalent Compounds
Ionic and Covalent Compounds
5.1 Compounds
5.2 Lewis Dot Symbols
5.3 Ionic Compounds and Bonding
5.4 Naming Ions and Ionic Compounds
Formulas of Ionic Compounds
Naming Ionic Compounds
5.5 Covalent Bonding and Molecules
Molecules
Molecular Formulas
Empirical Formulas
5.6 Naming Molecular Compounds
Specifying Numbers of Atoms
Compounds Containing Hydrogen
Organic Compounds
5
5.7 Covalent Bonding in Ionic Species
Polyatomic Ions
Oxoacids
Hydrates
Familiar Inorganic Compounds
5.8 Molecular and Formula Masses
5.9 Percent Composition of Compounds
5.10 Molar Mass
Interconverting Mass, Moles, and Number of Particles
Determination of Empirical Formula and Molecular Formula
from Percent Composition
Ionic and Covalent Compounds
5
Types of Chemical Bonding
1. Metal with nonmetal:
electron transfer and ionic bonding
2. Nonmetal with nonmetal:
electron sharing and covalent bonding
3. Metal with metal:
electron pooling and metallic bonding
The three models of chemical bonding.
A compound is a substance composed of two or more elements combined in a specific ratio and held together by chemical bonds.
Familiar examples of compounds are water and salt (sodium chloride).
Compounds
5.1
When atoms form compounds, it is their valence electrons that actually interact.
A Lewis dot symbol consists of the element’s symbol surround by dots.
Each dot represents a valence electron.
Boron 1s22s22p1
3 valence electrons
B



Lewis dot symbol for boron
B



B



B



other reasonable Lewis dot symbols for boron
Lewis Dot Symbols
5.2
Atoms combine in order to achieve a more stable electron configuration.
Maximum stability results when a chemical species is isoelectronic with a noble gas.
Na: 1s22s22p63s1
Na+: 1s22s22p6
10 electrons total, isoelectronic with Ne
Cl: 1s22s22p63s23p5
Cl : 1s22s22p63s23p6
18 electrons total, isoelectronic with Ar
Lewis Dot Symbols
Lewis Dot Symbols
Lewis dot symbols of the main group elements.
Lewis Dot Symbols
Dots are not paired until absolutely necessary.
Na

O




For main group metals such as Na, the number of dots is the number of electrons that are lost.
For nonmetals in the second period, the number of unpaired dots is the number of bonds the atom can form.
B



1s22s22p1
C




1s22s22p2
N




1s22s22p3
5 valence electrons; first pair formed in the Lewis dot symbol
Lewis Dot Symbols
Ions may also be represented by Lewis dot symbols.
Na

Na 1s22s22p63s1
Na+ 1s22s22p6
Na+
O




O 1s22s22p4
O2 1s22s22p6
O






2
Core electrons not represented in the Lewis dot symbol
Valence electron lost in the formation of the Na+ ion.
Remember the charge
Worked Example 5.1
Strategy Starting with the Lewis dot symbols for each element, add dots (for anions) or remove dots (for cations) as needed to achieve the correct charge on each ion. Don’t forget to include the appropriate charge on the Lewis dot symbol.
Write Lewis dot symbols for (a) fluoride ion (F-), (b) potassium ion (K+), and (c) sulfide ion (S2-).
Solution (a)
(b) K+
(c)
Think About It For ions that are isoelectronic with noble gases, cations should have no dots remaining around the element symbol, whereas anions should have eight dots around the element symbol. Note, too, that for anions, we put square brackets around the Lewis dot symbol and place the negative charge outside the brackets. Because the symbol for a common cation such as the potassium ion
has no remaining dots, square brackets are not necessary.
Ionic bonding refers to the electrostatic attraction that holds oppositely charged ions together in an ionic compound.
Na

Na+
+
e
+







Cl





Cl
e
Na

Na+
+
+







Cl





Cl
The attraction between the cation and anion draws them together to form NaCl
Ionic Compounds and Bonding
5.3
Ionic Compounds and Bonding
The resulting electrically neutral compound, sodium chloride, is represented with the chemical formula NaCl.
The chemical formula, or simply formula, of an ionic compound denotes the constituent elements and the ratio in which they combine.
Ionic Compounds and Bonding
A three-dimensional array of oppositely-charged ions is called a lattice. Lattice energy is the amount of energy required to convert a mole of ionic solid to its constituent ions in the gas phase.

+

+

+

+

+

+

+
NaCl(s) Na+(g) + Cl (g) Hlattice = +788 kJ/mol
Ionic Compounds and Bonding
The magnitude of lattice energy is a measure of an ionic compound’s stability.
Lattice energy depends on the magnitudes of the charge and on the distance between them.

d
Q1
Q2
Q = amount of charge
d = distance of separation
Ionic Compounds and Bonding
Ionic Compounds and Bonding
The magnitude of lattice energy is a measure of an ionic compound’s stability.
Lattice energy depends on the magnitudes of the charge and on the distance between them.
Ionic Compounds and Bonding
Worked Example 5.2
Strategy Consider the charges on the ions and the distances between them. Apply Coulomb’s law to determine the relative lattice energies. All three compounds contain O2- and all three cations are +2. Recalling that lattice energy increases as the distance between ions decreases, we need only consider the radii of the cations as all three contain the same anion. From Figure 4.13, the ionic radii are 0.72 (Mg2+), 1.00 (Ca2+), and 1.18 (Sr2+).
Arrange MgO, CaO, and SrO in order of increasing lattice energy.
Solution MgO has the smallest distance between ions, whereas SrO has the largest distance between ions. Therefore, in order of increase lattice energy: SrO < CaO < MgO.
Think About It Mg, Ca, and Sr are all Group 2A metals, so we could have predicted this result without knowing their radii. Recall that ionic radii increase as we move down a column in the periodic table, and charges that are farther apart are more easily separated (meaning the lattice energy will be smaller.) The lattice energies of SrO, CaO, and MgO are 3217, 3414, and 3890 kJ/mol, respectively.
Electrostatic forces
and the reason ionic compounds crack.
Electrical conductance and ion mobility.
Solid ionic compound
Molten ionic compound
Ionic compound dissolved in water
A monatomic ion is named by changing the ending of the element’s name to –ide.
Cl– is chloride O2– is oxide
Some metals can form cations of more than one possible charge.
Fe2+ : ferrous ion [Fe(II)]
Fe3+ : ferric ion [Fe(III)]
Mn2+ : manganese(II) ion
Mn3+ : manganese(III) ion
Mn4+ : manganese(IV) ion
Naming Ions and Ionic Compounds
5.4
Naming Ions and Ionic Compounds
Naming Ions and Ionic Compounds
To name ionic compounds:
1) Name the cation
omit the word ion
use a Roman numeral if the cation can have more than one charge
2) Name the anion
omit the word ion
Examples:
NaCN sodium cyanide
FeCl2 iron(II) chloride
FeCl3 iron(III) chloride
Worked Example 5.3
Strategy Begin by identifying the cation and anion in each compound, and then combine the names for each, eliminating the word ion.
Name the following ionic compounds: (a) CaO, (b) Mg3N2, and (c) Fe2S3.
Solution (a) CaO is calcium oxide.
(b) Mg3N2 is magnesium nitride.
(c) Fe2S3 is iron(III) sulfide.
Think About It Be careful not to confuse the subscript in the formula with the charge in the metal ion. In part (c), for example, the subscript on Fe is 2, but this is an iron(III) compound.
Worked Example 5.4
Strategy Identify the ions in each compound, and determine their ratios of combination using the charges on the cation and anion in each.
Deduce the formulas of the following ionic compounds: (a) mercury(II) chloride, (b) lead(II) bromide, and (c) potassium nitride.
Solution (a) Mercury(II) chloride is a combination of Hg2+ and Cl-. To produce a neutral compound, these two ions must combine in a 1:2 ratio – HgCl2.
(b) Lead(II) bromide is a combination of Pb2+ and Cl-. These ions combine in a 1:2 ratio to give PbBr2.
(c) Potassium nitride is a combination of K+ and N3-. These ions combine in a 3:1 ratio to give K3N.
Think About It Make sure that the charges sum to zero in each compound formula. In part (a), for example, Hg2+ + 2Cl- = (+2) + 2(-1) = 0; in part (b), (+2) + 2(-1) = 0; and in part (c), 3(+1) + (-3) = 0.
When compounds form between elements with similar properties, electrons are not transferred from one element to another but instead are shared in order to give each atom a noble gas configuration.
This approach is known as the Lewis theory of bonding, named for it’s proponent, Gilbert Lewis.
Lewis theory depicts bond formation in H2 as
H + H → H:H
This type of arrangement, where two atoms share a pair of electrons, is known as covalent bonding, and the shared pair of electrons constitutes a covalent bond.
Covalent Bonding and Molecules
5.5
Diatomic molecules contain two atoms and may be either heteronuclear or homonuclear.
Polyatomic molecules contain more than two atoms.
Covalent Bonding and Molecules
Covalent Bonding and Molecules
A chemical formula denotes the composition of the substance.
A molecular formula shows the exact number of atoms of each element in a molecule.
Some elements have two or more distinct forms known as allotropes.
For example, oxygen (O2) and ozone (O3) are allotropes of oxygen.
A structural formula shows not only the elemental composition, but also the general arrangements.
Covalent Bonding and Molecules
Worked Example 5.5
Strategy Refer to the labels on the atoms (or see Table 5.3). There are two carbon atoms, six hydrogen atoms, and one oxygen atom, so the subscript on C will be 2 and the subscript on H will be 6, and there will be no subscript on O.
Write the molecular formula of ethanol based on its ball-and-stick model, shown here.
Solution C2H6O
Think About It Often the molecular formula for a compound such as ethanol (consisting of carbon, hydrogen, and oxygen) is written so that the formula more closely resembles the actual arrangement of atoms in the molecule. Thus, the molecular formula for ethanol is commonly written as C2H5OH.
Covalent Bonding and Molecules
Remember that binary molecular compounds are substances that consist of just two different elements.
Nomenclature:
1) Name the first element that appears in the formula.
2) Name the second element that appears in the formula, changing its ending to –ide.
Examples:
HCl hydrogen chloride
HI hydrogen iodide
Naming Molecular Compounds
5.6
Greek prefixes are used to denote the number of atoms of each element present.
Naming Molecular Compounds
The prefix mono- is generally omitted for the first element.
For ease of pronunciation, we usually eliminate the last letter of a prefix that ends in “o” or “a” when naming an oxide.
Example: N2O5 is dinitrogen pentoxide not dinitrogen pentaoxide
Naming Molecular Compounds
Worked Example 5.7
Strategy Each compound will be named using the systematic nomenclature including, where necessary, appropriate Greek prefixes.
Name the following binary molecular compounds: (a) NF3 and (b) N2O4.
Solution (a) nitrogen trifluoride
(b) dinitrogen tetroxide
Think About It Make sure that the prefixes match the subscripts in the molecular formulas and that the word oxide is not preceded immediately by an “a” or an “o”.
Worked Example 5.8
Strategy The formula for each compound will be deduced using the systematic nomenclature guidelines.
Write the chemical formulas for the following binary molecular compounds: (a) sulfur tetrafluoride and (b) tetraphosphorus decasulfide.
Solution (a) SF4
(b) P4S10
Think About It Double-check that the subscripts in the formulas match the prefixes in the compound names: (a) 4 = tetra and (b) 4 = tetra and 10 = deca.
The names of molecular compounds containing hydrogen do not usually conform to the systematic nomenclature guidelines.
Many are called by the common, nonsystematic names or by names that do not indicate explicitly the number of H atoms present.
Examples:
B2H6 Diborane
SiH4 Silane
NH3 Ammonia
PH3 Phosphine
H2O Water
H2S Hydrogen sulfide
Compounds Containing Hydrogen
Compounds Containing Hydrogen
One definition of an acid is a substance that produces hydrogen ions (H+) when dissolved in water.
HCl is an example of a binary compound that is an acid when dissolved in water.
To name these types of acids:
1) remove the –gen ending from hydrogen
2) change the –ide ending on the second element to –ic.
hydrogen chloride → hydrochloric acid
Compounds Containing Hydrogen
A compound must contain at least one ionizable hydrogen atom to be an acid upon dissolving.
Polyatomic ions consist of a combination of two or more atoms.
Formulas are determined following the same rule as for ionic compounds containing only monatomic ions: ions must combine in a ratio that give a neutral formula overall.
Calcium phosphate:
Covalent Bonding in Ionic Species
5.7
Ca2+
PO43–
Ca3(PO4)2
Sum of charges: 3(+2) + 2(–3) = 0
Covalent Bonding in Ionic Species
Covalent Bonding in Ionic Species
Worked Example 5.9
Strategy Begin by identifying the cation and anion in each compound, and then combine the names for each, eliminating the word ion.
Name the following ionic compounds: (a) Fe2(SO4)3, (b) Al(OH)3, and (c) Hg2O.
Solution (a) Fe2(SO4)3 is iron(III) sulfate.
(b) Al(OH)3 is aluminum hydroxide.
(c) Hg2O is mercury(I) oxide.
Think About It Be careful not to confuse the subscript in the formula with the charge in the metal ion. In part (a), for example, the subscript on Fe is 2, but this is an iron(III) compound.
Covalent Bonding in Ionic Species
Oxoanions are polyatomic anions that contain one or more oxygen atoms and one atom (the “central atom”) of another element.
Starting with the oxoanions that end in –ate, we can name these ions as follows:
The ion with one more O atom than the –ate ion is called the per…ate ion. Thus, ClO3- is the chlorate ion, so ClO4- is the perchlorate ion.
The ion with one less O atom than the –ate ion is called the –ite ion. Thus, ClO2- is the chlorite ion.
The ion with two fewer O atom than the –ate ion is called the hypo…ite ion. Thus, ClO- is the hypochlorite ion.
At minimum, memorize the oxoanions that end in –ate so you can apply these guidelines when necessary.
Covalent Bonding in Ionic Species
perchlorate ClO4-
chlorate ClO3-
chlorite ClO2-
hypochlorite ClO-
nitrate NO3-
nitrite NO2-
phosphate PO43-
phosphite PO33-
sulfate SO42-
sulfite SO32-
Worked Example 5.10
Strategy Each species is either an oxoanion or an oxoacid. Identify the “reference oxidation” (the one with the –ate ending) for each, and apply the rules to determine appropriate names.
Name the following species: (a) BrO4-, (b) HCO3-, and (c) H2CO3.
Solution (a) BrO4- has one more O atom than the bromate ion (BrO3-), so BrO4- is the perbromate ion.
(b) CO32- is the carbonate ion. Because HCO3- has one ionizable hydrogen atom, it is called the hydrogen carbonate ion.
(c) With two ionizable hydrogen atoms and no charge on the compound, H2CO3 is carbonic acid.
Think About It Make sure that the charges sum to zero in each compound formula. In part (a), for example, Hg2+ + 2Cl- = (+2) + 2(-1) = 0; in part (b), (+2) + 2(-1) = 0; and in part (c), 3(+1) + (-3) = 0.
Think About It Remembering all these names and formulas is greatly facilitated by memorizing the common ions that end in –ate.
chlorate ClO3- nitrate NO3-
iodate IO3- carbonate CO32-
bromate BrO3- oxalate C2O42-
sulfate SO42- chromate CrO42-
phosphate PO43- permanganate MnO4-
Worked Example 5.11
Strategy The –ous ending in the name of an acid indicates that the acid is derived from an oxoanion ending in –ite. The oxoanion must be sulfite, SO32-, so add enough hydrogen ions to make a neutral formula.
Determine the formula of sulfurous acid.
Solution The formula of sulfurous acid is H2SO3.
Think About It Remembering all these names and formulas is greatly facilitated by memorizing the common ions that end in -ate.
Hydrates
A hydrate is a compound that has a specific number of water molecules within its solid structure.
For example, in its normal state, copper(II) sulfate has five water molecules associated with it.
Systematic name: copper(II) sulfate pentahydrate
Formula: Cu(SO)4 5H2O
Some other hydrates are
BaCl2 2H2O
LiCl H2O
MgSO4 7H2O
Sr(NO3)2 4H2O
Hydrates
When the water molecules are driven off by heating, the resulting compound, Cu(SO)4, is sometimes called anhydrous copper(II) sulfate.
Anhydrous means the compound no longer has water molecules associated with it.
Familiar Inorganic Compounds
Chapter Summary: Key Points
5
Compounds
Lewis Dot Symbols
Ionic Compounds and Ionic Bonding
Chemical Formulas
Lattice Energy
Ionic Compound Nomenclature
Lewis Theory of Bonding
Covalent Bonding
Laws of Definite Proportions and Multiple Proportions
Molecular, Structural, and Empirical Formulas
Molecular Compound Nomenclature
Inorganic and Organic Compounds
Polyatomic Ions
Oxyanions
Oxoacids
Hydrates
Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 6
Representing Molecules
Chemical Bonding I: Basic Concepts
6
6.1 The Octet Rule
Lewis Structures
Multiple Bonds
6.2 Electronegativity and Polarity
Electronegativity
Dipole Moment, Partial Charges and Percent Ionic Character
6.3 Drawing Lewis Structures
6.4 Lewis Structures and Formal Charge
6.5 Resonance
6.6 Exceptions to the Octet Rule
Incomplete Octets
Odd Numbers of Electrons
Expanded Octets
According to the octet rule, atoms will lose, gain, or share electrons in order to achieve a noble gas electron configuration.
Only valence electrons contribute to bonding.












F
F
+
F F














Each F counts both shared electrons to “feel” as though it has a Ne configuration.
F 1s22s22p5
valence electrons
The Octet Rule
6.1
The Octet Rule
Only two valence electrons participate in the formation of the F2 bond.
Pairs of valence electrons not involved in bonding are called lone pairs.
F F














F F














F F












or
Lewis Structures
A Lewis structure is a representation of covalent bonding.
Shared electron pairs are shown either as dashes or as pairs of dots.
Lone pairs are shown as pairs of dots on individual atoms.
H
O
H


H
O
H




O with 8 e
H with 2 e
H with 2 e
Shared electrons shown as dashes (bonds)
Multiple Bonds
In a single bond, atoms are held together by one electron pair.
In a double bond, atoms share two pairs of electrons.
H
O
H


C with 8 e
O with 8 e
O with 8 e
one shared pair of electrons results in a single bond

O
C
O







2 shared pairs of electrons result in double bonds
O
C
O




=
=
H
O
H




Multiple Bonds
A triple bond occurs when atoms are held together by three electron pairs.
each N has 8 e
3 shared pairs of electrons result in a triple bond

N
N




N
N


≡
Multiple Bonds
Bond length is defined as the distance between the nuclei of two covalently bonded atoms.
Multiple Bonds
Multiple bonds are shorter than single bonds.
For a given pair of atoms:
triple bonds are shorter than double bonds
double bonds are shorter than single bonds
<
<
N≡N
110 pm
N=N
124 pm
N N
147 pm
Multiple Bonds
The shorter multiple bonds are also stronger than single bonds.
We quantify bond strength by measuring the quantity of energy required to break it.
H2(g) → H(g) + H(g) Bond energy = 436.4 kJ/mol
We will examine this in more detail in Chapter 10.
Electronegativity and Polarity
There are two extremes in the spectrum of bonding:
covalent bonds occur between atoms that share electrons
ionic bonds occur between a metal and a nonmetal and involve ions
Bonds that fall between these extremes are polar.
In polar covalent bonds, electrons are shared but not shared equally.
M:X
Pure covalent bond
Neutral atoms held together by equally shared electrons
Mδ+Xδ
Polar covalent bond
Partially charged atoms held together by unequally shared electrons
M+X
Ionic bond
Oppositely charged ions held together by electrostatic attraction
6.2
Electronegativity and Polarity
Electron density maps show the distributions of charge.
Electrons spend a lot of time in red and very little time in blue.
Electrons are shared equally
nonpolar covalent
Electrons are not shared equally and are more likely to be associated with F
polar covalent
Electrons are not shared but rather transferred
from Na to F
ionic
Electronegativity
Electronegativity is the ability of an atom in a compound to draw electrons to itself.
Electronegativity and Polarity
Electronegativity varies with atomic number.
Electronegativity
There is no sharp distinction between nonpolar covalent and polar covalent or between polar covalent and ionic.
The following rules help distinguish among them:
A bond between atoms whose electronegativites differ by less than 0.5 is general considered purely covalent or nonpolar.
A bond between atoms who’s electronegativies differ by the range of 0.5 to 2.0 is generally considered polar covalent.
A bond between atoms whose electronegativities differ by 2.0 or more is generally considered ionic.
DEN
3.0
2.0
0.0
Boundary ranges for classifying ionic character of chemical bonds.
Classify the following bonds as nonpolar, polar, or ionic: (a) the bond in ClF, (b) the bond in CsBr, and (c) the carbon-carbon double bond in C2H4.
Worked Example 6.1
Solution
The difference between the electronegativies of F and Cl is 4.0 – 3.0 = 1.0, making the bond in ClF polar.
In CsBr, the difference is 2.8 – 0.7 = 2.1, making the bond ionic.
In C2H4, the two atoms are identical. (Not only are they the same element, but each C atom is bonded to two H atoms.) The carbon-carbon double bond is C2H4 is nonpolar.
Strategy Electronegativity values are: Cl (3.0), F (4.0), Cs (0.7), Br (2.8), C (2.5). Use this information to determine which bonds have identical, similar, and widely different electronegativities.
Think About It By convention, the difference in electronegativity is always calculated by subtracting the smaller number from the larger one, so the result is always positive.
Dipole Moment, Partial Charges, and Percent Ionic Character
An arrow is used to indicate the direction of electron shift in polar covalent molecules.
The consequent charge separation can be represented as:
Deltas (δ) denote a partial positive or negative charge.
Regions where electrons spend little time
Regions where electrons spend a lot of time
H
F



H
F



δ+
δ
Dipole Moment, Partial Charges, and Percent Ionic Character
A quantitative measure of the polarity of a bond is its dipole moment (μ).
Q is the charge.
r is the distance between the charges.
μ is always positive and expressed in debye units (D).
μ = Q x r
1 D = 3.336×10 30 C m
Dipole Moment, Partial Charges, and Percent Ionic Character
A quantitative measure of the polarity of a bond is its dipole moment (μ).
μ = Q x r
Burns caused by hydrofluoric acid [HF(aq)] are unlike any other acid burns and present unique medical complications. HF solutions typically penetrate the skin and damage internal tissues, including bone, often with minimal surface damage. Less concentrated solutions actually can cause greater injury than more concentrated ones by penetrating more deeply before causing injury, thus delaying the onset of symptoms and preventing timely treatment. Determine the magnitude of the partial positive and partial negative charges in the HF molecule.
Worked Example 6.2
Strategy Solve for Q. Convert the resulting charge in coulombs to units of electronic charge. According to Table 6.2, μ = 1.82 D and r = 0.92 for HF. The dipole moment must be converted from debye to C m and the distance between ions must be converted to meters.
μ = 1.82 D ×
3.336×10-30 C m
1 D
= 6.07×10-30 C m
r = 0.92 ×
1×10-10 m
1
= 9.2×10-11 m
Worked Example 6.2 (cont.)
Solution In coulombs:
In units of electronic charge:
Therefore, the partial charges in HF are +0.41 and -0.41 on H and F, respectively.
Q =
6.07×10-30 C m
9.2×10-11 m
= 6.598×10-20 C
μ
r
=
6.598×10-20 C ×
1 e-
1.6022×10-19 C
= 0.41 e-
H
F



+0.41
-0.41
Think About It Calculated partial charges should always be less than 1. If a “partial” charge were 1 or greater, it would indicate that at least one electron had been transferred from one atom to the other. Remember that polar bonds involve unequal sharing of electrons, not a complete transfer of electrons.
Dipole Moment, Partial Charges, and Percent Ionic Character
Although the designations “covalent,” “polar covalent,” and “ionic” can be useful, sometimes chemists wish to describe and compare chemical bonds with more precision.
Comparing the calculated dipole moment with the measured values gives us a quantitative way to describe the nature of a bond using the term percent ionic character.
percent ionic character = ×100%
μ (observed)
μ (calculated assuming discrete charges)
Dipole Moment, Partial Charges, and Percent Ionic Character
The figure below demonstrates the relationship between percent ionic character and the electronegativity difference in a heteronuclear diatomic molecule.
Using data from Table 6.2, calculate the percent ionic character of the bond in HI.
Worked Example 6.3
Strategy Calculate the dipole moment in HI assuming that the charges on H and I are +1 and -1, respectively; then calculate the percent ionic character. The magnitude of the charges must be expressed in coulombs (1 e- = 1.6022×10-19 C); the bond length (r) must be expressed as meters (1 = 1×10-10 m); and the calculated dipole moment should be expressed as debyes (1 D = 3.336×10-30 C m).
Setup From Table 6.2, the bond length in HI is 1.61 (1.61×10-10 m) and the measured dipole moment of HI is 0.44 D.
Worked Example 6.3 (cont.)
Solution The dipole we would expect if the magnitude of the charges were 1.6022×10-19 C is
Converting to debyes gives
The percent ionic character of the H–I bond is ×100% = 5.7%.
μ = Q × r = (1.6022×10-19 C) × (1.61×10-10 m)
= 2.58×10-29 C m
2.58×10-29 C m ×
1 D
3.336×10-30 C m
= 7.73 D
0.44 D
7.73 D
Think About It A purely covalent bond (in a homonuclear diatomic molecule such as H2) would have 0 percent ionic character. In theory, a purely ionic bond would be expected to have 100 percent ionic character, although no such bond is known.
Drawing Lewis Structures
Follow these steps when drawing Lewis structure for molecules and polyatomic ions.
Draw the skeletal structure of the compound. The least electronegative atom is usually the central atom. Draw a single covalent bond between the central atom and each of the surrounding atoms.
Count the total number of valence electrons present; add electrons for negative charges and subtract electrons for positive charges.
For each bond in the skeletal structure, subtract two electrons from the total valence electrons.
Use the remaining electrons to complete octets of the terminal atoms by placing pairs of electrons on each plete the octets of the most electronegative atom first.
Place any remaining electrons in pairs on the central atom.
If the central atom has fewer than eight electrons, move one or more pairs from the terminal atoms to form multiple bonds between the central atom and terminal atoms.
6.3
Drawing Lewis Structures
Draw the Lewis Structure for ClO3 .
Solution:
Step 1 Draw the skeletal structure of the compound. The least electronegative atom is usually the central atom. Draw a single covalent bond between the central atom and each of the surrounding atoms.
chlorine’s electronegativity = 3.0
oxygen’s electronegativity = 3.5
Drawing Lewis Structures
Draw the Lewis Structure for ClO3 .
Solution:
Step 2 Count the total number of valence electrons present; add electrons for negative charges and subtract electrons for positive charges.
Cl: 1 x 7 = 7 e
O: 3 x 6 = 18 e
Charge: 1 e
Total: 26 valence e
Drawing Lewis Structures
Drawing Lewis Structures
Draw the Lewis Structure for ClO3 .
Solution:
Step 4 Use the remaining electrons to complete octets of the terminal atoms by placing pairs of electrons on each plete the octets of the most electronegative atom first.
6 electrons used in single bonds
20 electrons must be placed in the structure starting with terminal atoms
The structure now has a total of 24 valence electrons of the available 26.
Drawing Lewis Structures
Draw the Lewis Structure for ClO3 .
Solution:
Step 5 Place any remaining electrons on the central atom.
The structure now has a total of 24 valence electrons of the available 26.
2 e remain to be added to the structure.
Lewis structures for polyatomic ions are enclosed by square brackets.
Drawing Lewis Structures
Draw the Lewis Structure for ClO3 .
Solution:
Step 6 If the central atom has fewer than eight electrons, move one or more pairs from the terminal atoms to form multiple bonds between the central atom and terminal atoms.
All atoms already have an octet in this structure.
Drawing Lewis Structures
Draw the Lewis structure for carbon disulfide (CS2).
Worked Example 6.4
Setup
Step 1: C and S have identical electronegativities. We will draw the skeletal structure with the unique atom, C, at the center.
Step 2: The total number of valence electrons is 16: 6 from each S atom and 4 from the C atom [2(6) + 4 = 16].
Step 3: Subtract 4 electrons to account for the bonds in the skeletal structure, leaving us 12 electrons to distribute.
Step 4: Distribute the 12 remaining electrons as 3 lone pairs on each S atom.
Draw the Lewis structure for carbon disulfide (CS2).
Worked Example 6.4 (cont.)
Setup
Step 5: There are no electrons remaining after step 4, so step 5 does not apply.
Step 6: To complete carbon’s octet, use one lone pair from each S atom to make a double bond to the C atom.
Think About It Counting the total number of valence electrons should be relatively simple to do, but it is often done hastily and is therefore a potential source of error in this type of problem. Remember that the number of valence electrons for each element is equal to the group number of that element.
Solution
Lewis Structures and Formal Charge
Formal charge can be used to determine the most plausible Lewis Structure when more than one possibility exists for a compound.
To determine associated electrons:
1) All the atom’s nonbonding electrons are associated with the atom.
2) Half the atom’s bonding electrons are associated with the atom.
Formal charge = valence electrons – associated electrons
6.4
Lewis Structures and Formal Charge
Determine the formal charges on each oxygen atom in the ozone molecule (O3).
O
O
O




=



e associated with atom
Valence e
Difference (formal charge)
6
6
6
6
5
7
0
+1
1
4 unshared + 4 shared = 6 e
2
2 unshared + 6 shared = 5 e
2
6 unshared + 2 shared = 7 e
2
The widespread use of fertilizers has resulted in the contamination of some groundwater with nitrates, which are potentially harmful. Nitrate toxicity is due primarily to its conversion in the body to nitrite (NO2-), which interferes with the ability of hemoglobin to transport oxygen. Determine the formal charges on each atom in the nitrate ion (NO3-).
Worked Example 6.5
Strategy Follow the six steps to draw the Lewis structure of (NO3-). For each atom, subtract the associated electrons from the valence electrons.
Setup
The N atom has five valence electrons and four associated electrons (one from each single bond and two from the double bond). Each singly bonded O atom has six valence electrons and seven associated electrons (six in three lone pairs and one from the single bond). The doubly bonded O atom has six valence electrons and six associated electrons (four in two lone pairs and two from the double bond.)
Worked Example 6.5 (cont.)
Solution
The formal charges are as follows: +1 (N atom), -1 (singly bonded O atoms), and 0 (doubly bonded O atom).
Think About It The sum of formal charges (+1) + (-1) + (-1) + (0) = -1 is equal to the overall charge on the nitrate ion.
Lewis Structures and Formal Charge
When there is more than one possible structure, the best arrangement is determined by the following guidelines:
1) A Lewis structure in which all formal charges are zero is preferred.
2) Small formal charges are preferred to large formal charges.
3) Formal charges should be consistent with electronegativities.
O
C
O




=
=
O
C
O




≡

0
0
0
+1
0
1
Formal charge
better structure (based on formal charge)
Lewis Structures and Formal Charge
Based on formal charge, identify the best and the worst structures for the isocyanate ion below:
Solution:
Step 1 Assign formal charges on each atom using the formula
4
5
6
2
+1
0
Ve
Formal charge = valence electrons – associated electrons
Ae
FC
4
5
6
4
5
6
6
4
6
5
4
7
7
4
5
1
+1
1
3
+1
+1
Lewis Structures and Formal Charge
Based on formal charge, identify the best and the worst structures for the isocyanate ion below:
Solution:
Step 2 Determine the best and worst structure
2
+1
0
FC
1
+1
1
3
+1
+1
Worst structure:
Large formal charges
Formal charges inconsistent with electronegativities
Best structure:
Small formal charges
Formal charges more consistent with electronegativities
Classify the following bonds as nonpolar, polar, or ionic: (a) the bond in ClF, (b) the bond in CsBr, and (c) the carbon-carbon double bond in C2H4.
Worked Example 6.5
Solution
The difference between the electronegativies of F and Cl is 4.0 – 3.0 = 1.0, making the bond in ClF polar.
In CsBr, the difference is 2.8 – 0.7 = 2.1, making the bond ionic.
In C2H4, the two atoms are identical. (Not only are they the same element, but each C atom is bonded to two H atoms.) The carbon-carbon double bond is C2H4 is nonpolar.
Strategy Electronegativity values are: Cl (3.0), F (4.0), Cs (0.7), Br (2.8), C (2.5). Use this information to determine which bonds have identical, similar, and widely different electronegativities.
Think About It By convention, the difference in electronegativity is always calculated by subtracting the smaller number from the larger one, so the result is always positive.
Formaldehyde (CH2O), which can be used to preserve biological specimens, is commonly sold as a 37 percent aqueous solution. Use formal charges to determine which skeletal arrangements of atoms shown here is the best choice for the Lewis structure of CH2O.
Worked Example 6.6
Strategy The complete Lewis structures for the skeletons shown are:
Worked Example 6.6 (cont.)
Strategy The complete Lewis structures for the skeletons shown are:
In the structure on the left, the formal charges are as follows:
Both H atoms: 1 valence e- – 1 associated e- (from single bond) = 0
C atom: 4 valence e- – 5 associated e- (two in the lone pair, one from the single bond, and two from the double bond) = -1
O atom: 6 valence e- – 5 associated e- (two from the lone pair, one from the single bond, and two from the double bond) = +1
0
0
+1
-1
Formal charges
Worked Example 6.6 (cont.)
Strategy The complete Lewis structures for the skeletons shown are:
In the structure on the right, the formal charges are as follows:
Both H atoms: 1 valence e- – 1 associated e- (from single bond) = 0
C atom: 4 valence e- – 4 associated e- (one from each single bond, and two from the double bond) = 0
O atom: 6 valence e- – 6 associated e- (four from the two lone pairs and two from the double bond) = 0
Formal charges all zero
Worked Example 6.6 (cont.)
Solution Of the two possible arrangements, the structure on the left has an O atom with a positive formal charge, which is inconsistent with oxygen’s high electronegativity. Therefore, the structure on the right, in which both H atoms are attached directly to the C atoms and all atoms have a formal charge of zero, is the better choice for the Lewis structure of CH2O.
Think About It For a molecule, formal charges of zero are preferred. When there are nonzero formal charges, they should be consistent with the electronegativities of the atoms in the molecules. A positive formal charge on oxygen, for example, is inconsistent with oxygen’s high electronegativity.
Resonance
A resonance structure is one of two or more Lewis structures for a single molecule that cannot by represented accurately by only one Lewis structure.
Resonance structures are a human invention.
Resonance structures differ only in the positions of their electrons.
O
O
O




=



O
O
O





=


6.5
High oil and gasoline prices have renewed interest in alternative methods of producing energy, including the “clean” burning of coal. Part of what makes “dirty” coal dirty is its high sulfur content. Burning dirty coal produces sulfur dioxide (SO2), among other pollutants. Sulfur dioxide is oxidized in the atmosphere to form sulfur trioxide (SO3), which subsequently combines with water to produce sulfuric acid – a major component of acid rain. Draw all possible resonance structures of sulfur trioxide.
Worked Example 6.7
Strategy Following the steps for drawing Lewis structures, we determine that a correct Lewis structure for SO3 contains two sulfur-oxygen single bonds and one sulfur-oxygen double bond.
But the double bond can be put in any one of three positions in the molecule.
Worked Example 6.7 (cont.)
Solution
Think About It Always make sure that resonance structures differ only in the position of the electrons, not in the positions of the atoms.
Exceptions to the Octet Rule
Exceptions to the octet rule fall into three categories:
1) The central atom has fewer than eight electrons due to a shortage of electrons.
Elements in group 3A also tend to form compounds surrounded by fewer than eight electrons.
Boron, for example, reacts with halogens to form compounds of the general formula BX3 having six electrons around the boron atom.
H
Be
H
only 4 total valence electrons in the system
6.6
2) The central atom has fewer than eight electrons due to an odd number of electrons.
Molecules with an odd number of electrons are sometimes referred to as free radicals.
O
N
O




=



17 valence electrons in the system
Odd Numbers of Electrons
Draw the Lewis structure of chlorine dioxide (ClO2).
Worked Example 6.8
Strategy The skeletal structure is
This puts the unique atom, Cl, in the center and puts the more electronegative O atoms in terminal positions.
There are a total of 19 valence electrons (7 from the Cl and 6 from each of the two O atoms). We subtract 4 electrons to account for the two bonds in the skeleton, leaving us with 15 electrons to distribute as follows: three lone pairs on each O atom, one lone pair on the Cl atom, and the last remaining electron also on the Cl atom.
Solution
Think About It ClO2 is used primarily to bleach wood pulp in the manufacture of paper, but it is also used to bleach flour, disinfect drinking water, and deodorize certain industrial facilities. Recently, it has been used to eradicate the toxic mold in homes in New Orleans that were damaged by the devastating floodwaters of Hurricane Katrina in 2005.
Strategy The skeletal structure is
This puts the unique atom, Cl, in the center and puts the more electronegative O atoms in terminal positions.
There are a total of 19 valence electrons (7 from the Cl and 6 from each of the two O atoms). We subtract 4 electrons to account for the two bonds in the skeleton, leaving us with 15 electrons to distribute as follows: three lone pairs on each O atom, one lone pair on the Cl atom, and the last remaining electron also on the Cl atom.
Think About It ClO2 is used primarily to bleach wood pulp in the manufacture of paper, but it is also used to bleach flour, disinfect drinking water, and deodorize certain industrial facilities. Recently, it has been used to eradicate the toxic mold in homes in New Orleans that were damaged by the devastating floodwaters of Hurricane Katrina in 2005.
Expanded Octets
3) The central atom has more than eight electrons.
Atoms in and beyond the third period can have more than eight valence electrons.
In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding.
Sulfur has 6 bonds corresponding to 12 electrons
Worked Example 6.9
Strategy The skeletal structure is
There are a total of 24 valence electrons (3 from the B and 7 from each of the three I atoms). We subtract 6 electrons to account for the three bonds in the skeleton, leaving 18 electrons to distribute in lone pairs on each I atom.
Solution
Draw the Lewis structure of boron triiodide (BI3).
Worked Example 6.9 (cont.)
Think About It Boron is one of the elements that does not always follow the octet rule. Like BF3, however, BI3 can be drawn with a double bond in order to satisfy the octet of boron. This gives rise to a total of four resonance structures.
Worked Example 6.10
Strategy The skeletal structure already has more than an octet around the As atom.
There are 40 total valence electrons [5 from As (Group 5A) and 7 from each of the five F atoms (Group 7A)]. We subtract 10 electrons to account for the five bonds in the skeleton, leaving 30 to be distributed. Next, we place three lone pairs on each F atom, thereby completing all their octets and using up all the electrons.
Solution
Draw the Lewis structure of arsenic pentafluoride (AsF5).
Think About It Always make sure that the number of electrons represented in your final Lewis structure matches the total number of valence electrons you are supposed to have.
Worked Example 6.11
Strategy Begin by drawing the skeletal structure and counting the total number of valence electrons. Use the steps outlined in Section 6.4 to draw the first structure and reposition one or more lone pairs to adjust the formal charges for the second structure.
Note that each hydrogen in an oxoacid is attached to an oxygen atom, and not directly to the central atom. The total number of valence electrons is 26 (6 from S, 6 from each O, and 1 from each H).
Draw two resonance structures for sulfurous acid (H2SO3): one that obeys the octet rule for the central atom, and one that minimizes the formal charges. Determine the formal charges on each atom in both structures.
Worked Example 6.11 (cont.)
Solution Following the steps we get the first structure:
From the top O atom, which has three lone pairs, we reposition one lone pair to create a double bond between O and S to get the second structure.
Incorporating the double bond results in every atom having a formal charge of zero.
+1
-1
Think About It In some species, such as the sulfate ion, it is possible to incorporate too many double bonds. Structures with three and four double bonds to sulfur would give formal charges on S and O that are inconsistent with the electronegativities of these elements. In general, if you are trying to minimize formal charges by expanding the central atom’s octet, only add enough double bonds to make the formal charge on the central atom zero.
Worked Example 6.12
Strategy Follow the steps for drawing Lewis structures. The skeletal structure is
There are 36 total valence electrons (8 from Xe and 7 from each of the four F atoms). We subtract 8 electrons to account for the bonds in the skeleton, leaving 28 to distribute. We first complete the octets of all four F atoms. When this is done, 4 electrons remain, so we place two lone pairs on the Xe atom.
Solution
Draw the Lewis structure of xenon tetrafluoride (XeF4).
Think About It Atoms beyond the second period can accommodate more than an octet of electrons, whether those electrons are used in bonds or reside on the atoms as lone pairs.
Chapter Summary: Key Points
6
Lewis Structures
Octet Rule
Bond Order
Electronegativity and Polarity
Dipole Moment
Percent Ionic Character
Drawing Lewis Structures
Formal Charge
Resonance Structures
Exceptions to the Octet Rule
Incomplete Octets
Odd Numbers of Electrons
Expanded Octets

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