Cover Page Chapter 8 Bonding: General Concepts

Cover Page Chapter 8 Bonding: General Concepts
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Table of Contents (8.1) Types of chemical bonds
(8.2) Electronegativity (8.3) Bond polarity and dipole moments (8.4) Ions: Electron configurations and sizes (8.5) Energy effects in binary ionic compounds (8.6) Partial ionic character of covalent bonds (8.7) The covalent chemical bond: A model (8.8) Covalent bond energies and chemical reactions

Table of Contents - Continued
(8.9) The localized electron bonding model (8.10) Lewis structures (8.11) Exceptions to the octet rule (8.12) Resonance (8.13) Molecular structure: The VSEPR model

An Introduction to Chemical Bonds
Forces that hold groups of atoms together and make them function as a unit Bond energy: Energy required to break a bond Ionic bonding: Results from electrostatic attractions of closely packed, oppositely charged ions

An Introduction to Chemical Bonds (continued)
Ionic compound: Results when a metal reacts with a nonmetal Example - When Na and Cl react to form NaCl, electrons are transferred from the Na atoms to the Cl atoms to form Na+ and Cl– ions Ions then aggregate to form solid sodium chloride

Used to calculate the energy of interaction between a pair of ions
Coulomb’s Law Used to calculate the energy of interaction between a pair of ions E has units of joules r - Distance between the ion centers in nanometers Q1 and Q2 - Numerical ion charges

Coulomb's Law (continued)
Used to calculate the repulsive energy when two like-charged ions are brought together Calculated value of the energy will have a positive sign

Development of Bonding Forces between Two Identical Atoms
Bond will form if the system can lower its total energy in the process System will act to minimize the sum of the positive (repulsive) energy terms and the negative (attractive) energy term Bond length: Distance between two atoms when potential energy is minimal

Figure 8.1 - (b) Energy Profile as a Function of the Distance between the Nuclei of Hydrogen Atoms
As atoms approach each other (right side of graph), the energy decreases until the distance reaches nm (74 pm) and then begins to increase again due to repulsions

Interaction between Two Hydrogen Atoms: Important Features
Energy terms involved are the: Net potential energy that results from the attractions and repulsions among the charged particles Kinetic energy due to the motions of the electrons Zero point of energy is defined with the atoms at infinite separation

Interaction between Two Hydrogen Atoms: Important Features (continued)
At very short distances the energy rises steeply because of the importance of the repulsive forces when the atoms are very close together Bond length is the distance at which the system has minimum energy

Two identical atoms share electrons equally
Covalent Bonding Two identical atoms share electrons equally Bonding results from the mutual attraction of the nuclei for the shared electrons Polar covalent bond Characterized by unequal sharing of electrons Example Bonding in the hydrogen fluoride (HF) molecule

Figure 8.2 - The Effect of an Electric Field on Hydrogen Fluoride Molecules

Join In (1) What type of bond is formed when electrons are unequally shared by nuclei? Ionic Covalent Polar ionic Polar covalent None of the above Correct answer Polar covalent Polar covalent bonds occur when two atoms are not so different that electron transfer occurs but are different enough that they do not share their electrons equally

Ability of an atom in a molecule to attract shared electrons to itself
Electronegativity Ability of an atom in a molecule to attract shared electrons to itself Can be determined by using Pauling’s model Consider a hypothetical molecule HX Relative electronegativities of the H and X atoms are determined by comparing the measured H—X bond energy with the expected H—X bond energy

Electronegativity (continued 1)
Difference (Δ) between actual and expected bond energies is given by the following equation: If H and X have identical electronegativities: Δ is 0 (H—X)act and (H—X)exp are the same

Electronegativity (continued 2)
If X has a greater electronegativity than H: Shared electron(s) will tend to be closer to the X atom Molecule will be polar, with the following charge distribution:

Electronegativity Trends
Attraction between the partially (and oppositely) charged H and X atoms will lead to a greater bond strength (H—X)act will be larger than (H—X)exp Greater the difference in the electronegativities of the atoms, the greater is the ionic component of the bond and the greater is the value of Δ

Figure 8.3 - The Pauling Electronegativity Values

Table 8.1 - The Relationship between Electronegativity and Bond Type

Interactive Example 8.1 - Relative Bond Polarities
Order the following bonds according to polarity: H—H O—H Cl—H S—H F—H

Interactive Example 8.1 - Solution
Polarity of bonds increases as difference in electronegativity increases From Pauling’s electronegativity values, the following variation in bond polarity is expected: H—H < S—H < Cl—H < O—H < F—H (2.1) (2.1) (2.5) (2.1) (3.0) (2.1) (3.5) (2.1) (4.0) (2.1)

Interactive Example 8.1 - Solution (continued)
Electronegativity difference

Exercise Without using Fig. 8.3, predict the order of increasing electronegativity in each of the following groups of elements C, N, O S, Se, Cl Si, Ge, Sn Tl, S, Ge C < N < O Se < S < Cl Sn < Ge < Si Tl < Ge < S

Critical Thinking We use differences in electronegativity to account for certain properties of bonds What if all atoms had the same electronegativity values? How would bonding between atoms be affected? What are some differences we would notice?

Join In (2) Which of the following elements forms the most ionic bond with chlorine? K Al P Kr Br Correct answer K The most ionic bond results from the largest dipole moment and the greatest difference in electronegativity With chlorine, potassium has the greatest difference in electronegativity

As a general pattern, electronegativity is inversely related to:
Join In (3) As a general pattern, electronegativity is inversely related to: ionization energy atomic size the polarity of the atom the number of neutrons in the nucleus Correct answer atomic size Electronegativity increases from left to right and from bottom to top in the periodic table Atomic size increases from top to bottom in the periodic table Thus electronegativity and atomic size are inversely related.

Representation of dipolar character
Dipole Moment Molecules that have a center of positive charge and a center of negative charge are called dipolar Representation of dipolar character An arrow points to the negative charge center with the tail of the arrow indicating the positive center of charge

Electrostatic Potential Diagram
Alternative method for representing charge distribution in a molecule Colors of visible light are used to show the variation in charge distribution Red indicates the most electron-rich region Blue indicates the most electron-poor region

Figure 8.4 - An Electrostatic Potential Map of HF
Red indicates the most electron-rich area (the fluorine atom), and blue indicates the most electron-poor region (the hydrogen atom)

Dipole Moments of Hydrogen Halides (HX)
Depends on the electronegativity of X

Figure 8.5 - Molecular Charge Distribution in a Water (H2O) Molecule
Water molecule in an electric field Electrostatic potential diagram

Dipole Moment in Other Molecules
Some molecules have polar bonds but do not have a dipole moment Occurs when individual bond polarities are arranged in such a way that they cancel each other out Example - CO2

Figure 8.7 - Dipole Moment in a CO2 Molecule
The carbon dioxide molecule The molecule has no dipole moment because the opposed polarities cancel out Electrostatic potential diagram for carbon dioxide

Table 8.3 - Types of Molecules with Polar Bonds but No Resulting Dipole Moment

Example 8.2 - Bond Polarity and Dipole Moment
For each of the following molecules, show the direction of the bond polarities and indicate which ones have a dipole moment: HCl, Cl2, and SO3 (a planar molecule with the oxygen atoms spaced evenly around the central sulfur atom) CH4 (tetrahedral with the carbon atom at the center) H2S (V-shaped with the sulfur atom at the point)

The HCl molecule Example 8.2 - Solution
Electronegativity of chlorine (3.0) is greater than that of hydrogen (2.1) Thus the chlorine will be partially negative, and the hydrogen will be partially positive The HCl molecule has a dipole moment

Example 8.2 - Solution (continued 1)
The Cl2 molecule The two chlorine atoms share the electrons equally No bond polarity occurs, and the Cl2 molecule has no dipole moment The SO3 molecule The electronegativity of oxygen (3.5) is greater than that of sulfur (2.5) This means that each oxygen will have a partial negative charge, and the sulfur will have a partial positive charge

The bond polarities arranged symmetrically as shown cancel, and the molecule has no dipole moment This molecule is the second type shown in Table 8.3

The CH4 molecule Carbon has a slightly higher electronegativity (2.5) than does hydrogen (2.1) This leads to small partial positive charges on the hydrogen atoms and a small partial negative charge on the carbon

This case is similar to the third type in Table 8.3, and the bond polarities cancel The molecule has no dipole moment The H2S molecule Since the electronegativity of sulfur (2.5) is slightly greater than that of hydrogen (2.1), the sulfur will have a partial negative charge, and the hydrogen atoms will have a partial positive charge

This case is analogous to the water molecule, and the polar bonds result in a dipole moment oriented as shown:

Join In (4) Which of the following is the most polar bond without being considered ionic? C—O Mg—O N—O O—O O—F Correct answer C—O Given that O is common to all of the bonds, the relative difference in electronegativity for C, Mg, F, and N will determine the polarity Decreasing polarity results from lowering the difference in electronegativity While the greatest difference in electronegativity is between Mg and O, this bond is considered to be ionic, not polar

Join In (5) Which of the following is the least polar bond, yet still considered polar covalent? C—O Mg—O N—O O—O O—F Correct answer N—O and e. O—F Given that O is common to all of the bonds, the relative difference in electronegativity for C, Mg, F, and N will determine the polarity Decreasing polarity results from lowering the difference in electronegativity While the smallest difference in electronegativity is between O and O (both have the same electronegativity value, so the difference is zero), this bond is not considered to be polar, yet is perfectly covalent

Join In (6) What is the correct order of the following bonds in terms of decreasing polarity? N—Cl, P—Cl, As—Cl P—Cl, N—Cl, As—Cl As—Cl, N—Cl, P—Cl P—Cl, As—Cl, N—Cl As—Cl, P—Cl, N—Cl Correct answer As—Cl, P—Cl, N—Cl Given that Cl is common to all of the bonds, the relative difference in electronegativity for As, P, and N will determine the polarity Decreasing polarity results from lowering the difference in electronegativity

In which case is the bond polarity incorrect?
Join In (7) In which case is the bond polarity incorrect? +H—F– +K—O– +Mg—H– +Cl—I– +Si—S– Correct answer +Cl—I– Bond polarity results from the more electronegative element pulling the electron density toward it, resulting in a partial negative charge Because F, O, H, Cl, and S are more electronegative than H, K, Mg, I, and Si, respectively, they will carry the – charge

Generalizations That Apply to Electron Configurations in Stable Compounds
When two nonmetals react to form a covalent bond, they share electrons in a way that completes the valence electron configurations of both atoms Both nonmetals attain noble gas electron configurations

Generalizations That Apply to Electron Configurations in Stable Compounds (continued)
When a nonmetal and a representative-group metal react to form a binary ionic compound, the ions form so that the: Valence electron configuration of the nonmetal achieves the electron configuration of the next noble gas atom Valence orbitals of the metal are emptied Both ions achieve noble gas electron configurations

Solid State of Ionic Compounds
Solid ionic compounds contain large collection of positive and negative ions packed together in a way that: Minimizes the – .. – and repulsions Maximizes + .. – attractions

Gas Phase of an Ionic Substance
Ions are relatively far apart and would not contain large groups of ions

Predicting Formulas of Ionic Compounds - Example
Consider Ca and O Valence electron configuration for Ca is [Ar]s2 and for O is [He]2s22p4 Electronegativity of oxygen (3.5) is much greater than that of calcium (1.0) Electrons will be transferred from calcium to oxygen to form oxygen anions and calcium cations in the compound

Predicting Formulas of Ionic Compounds - Example (continued)
Oxygen needs two electrons to fill its 2s and 2p valence orbitals and to achieve the configuration of neon (1s22s22p6) By losing two electrons, calcium can achieve the configuration of argon Since there are equal numbers of Ca2+ and O2–, the empirical formula of the compound is CaO

Table 8.4 - Common Ions with Noble Gas Configurations in Ionic Compounds

Noble Gas Electron Configurations in Ionic Compounds - Trends
In losing electrons to form cations: Metals in Group 1A lose one electron Metals in Group 2A lose two electrons Metals in Group 3A lose three electrons In gaining electrons to form anions: Nonmetals in Group 7A (the halogens) gain one electron Nonmetals in Group 6A gain two electrons

Noble Gas Electron Configurations in Ionic Compounds - Exceptions
Representative elements Tin forms both Sn2+ and Sn4+ ions Lead forms both Pb2+ and Pb4+ ions Bismuth forms both Bi3+ and Bi5+ ions Thallium forms both Tl+ and Tl3+ ions

Plays an important role in determining:
Ion Size Plays an important role in determining: Structure and stability of ionic solids Properties of ions in aqueous solution Biologic effects of ions Ionic radii are determined from the measured distance between ion centers in ionic compounds Involves an assumption about how the distance should be divided up between the two ions

Factors That Influence Ionic Size
Relative sizes of an ion and its parent atom Positive ion is formed by removing one or more electrons from a neutral atom Resulting cation is smaller than its parent atom Negative ions are formed by adding electrons to a neutral atom Resulting anion is larger than its parent atom

Factors That Influence Ionic Size (continued)
Position of the parent element in the periodic table Ion sizes increase down a group A given period contains both elements that: Give up electrons to form cations Accept electrons to form anions

Series of ions containing the same number of electrons Examples Trend
Isoelectronic Ions Series of ions containing the same number of electrons Examples O2–, F–, Na+, Mg2+, and Al3+ Trend Size decreases as Z (nuclear charge) increases

Figure 8.8 - Sizes of Ions Related to Positions of the Elements on the Periodic Table

Ions have different radii than their parent atoms
Critical Thinking Ions have different radii than their parent atoms What if ions stayed the same size as their parent atoms? How would this affect ionic bonding in compounds?

Interactive Example 8.3 - Relative Ion Size I
Arrange the following ions in order of decreasing size: Se2–, Br–, Rb+, and Sr2+

This is an isoelectronic series of ions with the krypton electron configuration Since these ions all have the same number of electrons, their sizes will depend on the nuclear charge The Z values are 34 for Se2–, 35 for Br–, 37 for Rb+, and 38 for Sr2+

Since the nuclear charge is greatest for Sr2+, it is the smallest of these ions The Se2– ion is largest:

Which of the following has the largest radius?
Join In (8) Which of the following has the largest radius? S2– Cl– Ar K+ Ca2+ Correct answer S2– All of these ions have similar sizes as atoms, so the charge determines the largest radius Adding electrons increases the size

Defined as the energy released when an ionic solid forms from its ions
Lattice Energy Change in energy that takes place when separated gaseous ions are packed together to form an ionic solid Defined as the energy released when an ionic solid forms from its ions Lattice energy has a negative sign

Energy Changes in the Formation of Lithium Fluoride
Sublimation of solid lithium Sublimation involves taking a substance from the solid state to the gaseous state Enthalpy of sublimation for Li(s) is 161 kJ/mol

Energy Changes in the Formation of Lithium Fluoride (continued 1)
Ionization of lithium atoms to form Li+ ions in the gas phase Process corresponds to the first ionization energy for lithium, which is 520 kJ/mol Dissociation of fluorine molecules Form a mole of fluorine atoms by breaking the F—F bonds in a half mole of F2 molecules

Energy required to break this bond is 154 kJ/mol Since the bonds are broken in a half mole of fluorine, energy required is (154 kJ)/2 or 77 kJ

Formation of F– ions from fluorine atoms in the gas phase Energy change for this process corresponds to the electron affinity of fluorine, which is –328 kJ/mol

Formation of solid lithium fluoride from the gaseous Li+ and F– ions This corresponds to the lattice energy for LiF, which is –1047 kJ/mol

Sum of the five processes yields the desired overall reaction, and the sum of the individual energy changes gives the overall energy change

Figure The Energy Changes Involved in the Formation of Solid Lithium Fluoride from Its Elements Numbers in parentheses refer to the reaction steps discussed in the text

Lattice Energy: Calculations
Represented by a modified form of Coulomb’s law k - Proportionality constant Depends on the structure of the solid and the electronic configurations of the ions Q1 and Q2 - Charges on the ions r - Shortest distance between the centers of the anions and the cations

Lattice Energy: Calculations (continued)
Lattice energy has a negative sign when Q1 and Q2 have opposite signs Result is expected because bringing cations and anions together is an exothermic process Process becomes more exothermic as: Ionic charges increase Distances between ions in the solid decrease

Most important factor involves the balancing of:
Factor Influencing the Composition and Structure of Solid Ionic Compounds Most important factor involves the balancing of: Energies required to form highly charged ions Energy released when highly charged ions combine to form the solid

Join In (9) Which of the following ionic compounds has the largest lattice energy (i.e., the lattice energy most favorable to a stable lattice)? CsI NaCl LiF CsF MgO Correct answer MgO Lattice energy is directly proportional to charge, and the species with the highest charge is MgO

Figure 8.12 - The Three Possible Types of Bonds
Covalent bond formed between identical F atoms Polar covalent bond of HF, with both ionic and covalent components An ionic bond with no electron sharing

Formula for Percent Ionic Character of a Bond
Totally ionic bonds between discrete pairs of atoms do not exist Evident from calculations of percent ionic character for bonds of various binary compounds in the gas phase

Figure Relationship between the Ionic Character of a Covalent Bond and the Electronegativity Difference of the Bonded Atoms Note that the compounds with ionic character greater than 50% (red) are normally considered to be ionic compounds

Complications in Identifying Ionic Compounds
Ionic character increases with electronegativity difference, but none of the bonds reaches 100% ionic character Therefore, no individual bonds are completely ionic Many substances contain polyatomic ions Bonds within the ions are covalent

Complications in Identifying Ionic Compounds (continued)
Avoided by adopting the operational definition of ionic compound Any compound that conducts an electric current when melted will be classified as ionic

Forces that cause a group of atoms to behave as a unit
Bonds Forces that cause a group of atoms to behave as a unit Result from the tendency of a system to seek its lowest possible energy Bonds occur when collections of atoms are more stable (lower in energy) than the separate atoms

Models that depict molecular stability
Chemical Bonds Models that depict molecular stability Example - Methane has a tetrahedron structure (four equal triangular faces) Bond concept is a human invention Provides a method for dividing up the energy evolved when a stable molecule is formed from its component atoms

Originate from observations of the properties of nature
Models Attempt to explain how nature operates on the microscopic level based on experiences in the macroscopic world Originate from observations of the properties of nature

Provides a framework to systematize chemical behavior
Bonding Model Provides a framework to systematize chemical behavior Molecules are perceived as collections of common fundamental components Physically sensible Makes sense that atoms can form stable compounds by sharing electrons Shared electrons give a lower energy state because they are simultaneously attracted by two nuclei

Single bond Double bond Triple bond
Types of Bonds Single bond One pair of electrons is shared Double bond Two pairs of electrons are shared Triple bond Three pairs of electrons are shared Bond length shortens with the increase in the number of shared electrons

Table 8.5 - Average Bond Energies (kJ/mol)

Table 8.6 - Bond Lengths and Bond Energies for Selected Bonds

Energy is released when bonds are formed (exothermic process)
Bond Energy Energy must be added to the system to break bonds (endothermic process) Energy terms associated with bond breaking have positive signs Energy is released when bonds are formed (exothermic process) Energy terms associated with bond making carry a negative sign

Enthalpy Change for a Reaction
Can be calculated as follows: Σ - Sum of terms D - Bond energy per mole of bonds Always has a positive sign n - Moles of a particular type of bond

Interactive Example 8.5 - ΔH from Bond Energies
Use the bond energies listed in Table 8.5, and calculate ΔH for the reaction of methane with chlorine and fluorine to give Freon-12 (CF2Cl2)

Break the bonds in the gaseous reactants to give individual atoms and then assemble these atoms into the gaseous products by forming new bonds Then combine the energy changes to calculate ΔH: ΔH = Energy required to break bonds – energy released when bonds form Minus sign gives the correct sign to the energy terms for the exothermic processes

Interactive Example 8.5 - Solution (continued 1)
Reactant bonds broken:

Product bonds formed:

We now can calculate ΔH: Since the sign of the value for the enthalpy change is negative, this means that 1194 kJ of energy is released per mole of CF2Cl2 formed

Join In (10) Given the average bond energies above, calculate ∆H for the following reaction: Bond Average Bond Energy (kJ/mol) H—H 432 F—F 154 H—F 565

Join In (10) (continued) –21 kJ 21 kJ 544 kJ -544 kJ 1151 kJ
Correct answer –544 kJ ∆H = [432 kJ kJ] – 2(565 kJ) = –544 kJ

Localized Electron (LE) Model
Assumes that a molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms Lone pairs: Pairs of electrons localized on an atom Bonding pairs: Pairs of electrons found in the space between atoms

Parts of the LE Model Description of the valence electron arrangement in the molecule using Lewis structures Prediction of the geometry of the molecule using the valence shell electron-pair repulsion (VSEPR) model Description of the type of atomic orbitals used by the atoms to share electrons or hold lone pairs

Depicts the arrangement of valence electrons among atoms in a molecule
Lewis Structure Depicts the arrangement of valence electrons among atoms in a molecule Named after G.N. Lewis Only valence electrons are included Most important requirement for the formation of a stable compound Atoms must attain noble gas electron configurations

Principle of Achieving a Noble Gas Electron Configuration - Hydrogen
Follows a duet rule Forms stable molecules where it shares two electrons Example - Two hydrogen atoms, each with one electron, combine to form the H2 molecule By sharing electrons, each hydrogen has a filled valence shell

Principle of Achieving a Noble Gas Electron Configuration - Helium
Does not form bonds because it is a noble gas Valence orbital is already filled Electron configuration - 1s2 Represented by the following Lewis structure: He:

Principle of Achieving a Noble Gas Electron Configuration - Second-Row Nonmetals (Carbon through Fluorine) Form stable molecules when surrounded by enough electrons to fill valence orbitals Obey the octet rule Octet rule: Elements form stable molecules when surrounded by eight electrons

Each fluorine atom in F2 is surrounded by eight electrons
Principle of Achieving a Noble Gas Electron Configuration - Second-Row Nonmetals (Carbon Through Fluorine) (continued) Each fluorine atom in F2 is surrounded by eight electrons Two electrons that are shared with the other atom constitute a bonding pair Each fluorine atom also has three pairs of electrons not involved in bonding These are the lone pairs

Principle of Achieving a Noble Gas Electron Configuration - Neon
Does not form bonds because it already has an octet of valence electrons Represented by the following Lewis structure: Only the valence electrons (2s22p6) are represented in the Lewis structure

Problem Solving Strategy - Steps for Writing Lewis Structures
Sum the valence electrons from all the atoms Use a pair of electrons to form a bond between each pair of bound atoms Arrange the remaining electrons to satisfy the duet rule for hydrogen and the octet rule for the second-row elements

Drawing the Lewis Structure of Water
Sum the valence electrons for H2O = 8 valence electrons Using a pair of electrons per bond, draw the O—H single bonds H—O—H A line is used to indicate each pair of bonding electrons H H O

Drawing the Lewis Structure of Water (continued)
Distribute the remaining electrons to achieve a noble gas electron configuration for each atom Dots represent lone electron pairs

Interactive Example 8.6 - Writing Lewis Structures
Give the Lewis structure for each of the following: HF N2 NH3 CH4 CF4 NO+

In each case, apply the three steps for writing Lewis structures Recall that lines are used to indicate shared electron pairs and that dots are used to indicate nonbonding pairs (lone pairs)

Join In (11) Which of the following does not contain at least one double bond in the Lewis structure? H2CO C2H4 CO2 C3H8 Correct answer C3H8 Choice (a) contains a C=O double bond Choice (b) contains a C=C double bond Choice (c) contains two C=O double bonds

Boron trifluoride (BF3)
Tends to form compounds in which the boron atom has fewer than eight electrons around it Boron trifluoride (BF3) Reacts energetically with molecules that have available lone pairs Boron atom is electron-deficient Has 24 valence electrons

Octet rule is satisfied by drawing the structure with a double bond
Boron (continued) Octet rule is satisfied by drawing the structure with a double bond Characteristic of boron to form molecules in which the boron atom is electron-deficient

Sulfur Hexafluoride (SF6)
Very stable molecule Sum of valence electrons 6 + 6(7) = 48 electrons

Sulfur Hexafluoride (SF6) (continued)
Sulfur exceeds the octet rule Localized electron model assumes that the empty 3d orbitals can be used to accommodate extra electrons Sulfur atom in SF6 can have 12 electrons around it by using the 3s and 3p orbitals to hold 8 electrons, with the extra 4 electrons placed in the formerly empty 3d orbitals

Write the Lewis structure for PCl5
Interactive Example Lewis Structures for Molecules That Violate the Octet Rule I Write the Lewis structure for PCl5

Sum the valence electrons 5 + 5(7) = 40 electrons Indicate single bonds between bound atoms P Cl

Distribute the remaining electrons In this case, 30 electrons (40 – 10) remain These are used to satisfy the octet rule for each chlorine atom Final Lewis structure is Phosphorus, a third-row element, has exceeded the octet rule by two electrons

Assume that the extra electrons should be placed on the central atom
Lewis Structure of Molecules with More Than One Atom That Exceed the Octet Rule Assume that the extra electrons should be placed on the central atom Example - Lewis structure of I3– (22 valence electrons) Central iodine exceeds the octet rule

Write the Lewis structure for the following:
Interactive Example Lewis Structures for Molecules That Violate the Octet Rule II Write the Lewis structure for the following: CIF3 RnCl2 ICl4–

The chlorine atom (3rd row) accepts the extra electrons Radon, a noble gas in Period 6, accepts the extra electrons

Iodine exceeds the octet rule

Which of the following has a Lewis structure most like that of CO32–?
Join In (12) Which of the following has a Lewis structure most like that of CO32–? CO2 SO32– NO3– O3 NO2 Correct answer NO3– The number of atoms and the electron count are the same for NO3– and CO32–; thus these two ions have similar Lewis structures

Which molecule or ion violates the octet rule?
Join In (13) Which molecule or ion violates the octet rule? H2O NO3– PF3 I3– None of these Correct answer I3– Violations of the octet rule involve having more than 8 electrons around an atom, which is the case with I3–

Resonance: An Introduction
Invoked when more than one valid Lewis structure can be written for a particular molecule Resulting electron structure of the molecule is given by the average of these resonance structures Represented by double-headed arrows

Resonance: An Introduction (continued)
Arrangement of nuclei is the same across all structures Placement of electrons differs Arrows show that the actual structure is an average of the three resonance structures Concept is necessary to compensate for the defective assumption of the LE model LE model postulates that electrons are localized between a given pair of atoms

Example 8.9 - Resonance Structures
Describe the electron arrangement in the nitrite anion (NO2–) using the localized electron model

NO2– possesses 18 valence electrons (5 + 2(6) + 1 = 18)
Example Solution NO2– possesses 18 valence electrons (5 + 2(6) + 1 = 18) Indicating the single bonds gives the structure O—N—O Remaining 14 electrons (18 – 4) can be used to produce these structures:

Example 8.9 - Solution (continued)
This is a resonance situation, and two equivalent Lewis structures can be drawn The electronic structure of the molecule is correctly represented not by either resonance structure but by the average of the two There are two equivalent N—O bonds, each one intermediate between a single and a double bond

Odd-Electron Molecules
Few molecules formed from nonmetals contain odd numbers of electrons Example - Nitric oxide (NO) Emitted into the air where it reacts with oxygen to form NO2 (g), which is another odd-electron molecule

Used to evaluate nonequivalent Lewis structures
Formal Charge Difference between the number of valence electrons on the free atom and the number of valence electrons assigned to the atom in the molecule Used to evaluate nonequivalent Lewis structures Nonequivalent Lewis structures contain different numbers of single and multiple bonds

Determining Formal Charge of an Atom in a Molecule
Information required Number of valence electrons on the free neutral atom Free neutral atom has zero net charge because the number of electrons equals the number of protons Number of valence electrons belonging to the atom in a molecule

Determining Formal Charge of an Atom in a Molecule (continued 1)
Assign valence electrons in the molecule to the various atoms, making the following assumptions: Lone pair electrons belong entirely to the atom in question Shared electrons are divided equally between the two sharing atoms

Determining Formal Charge of an Atom in a Molecule (continued 2)
Number of valence electrons assigned to a given atom is calculated as follows:

Fundamental Assumptions about Formal Charges to Evaluate Lewis Structures
Atoms in molecules try to achieve formal charges as close to zero as possible Any negative formal charges are expected to reside on the most electronegative atoms

Rules Governing Formal Charge
To calculate the formal charge on an atom: Take the sum of the lone pair electrons and one-half the shared electrons This is the number of valence electrons assigned to the atom in the molecule Subtract the number of assigned electrons from the number of valence electrons on the free, neutral atom to obtain the formal charge

Rules Governing Formal Charge (continued)
Sum of the formal charges of all atoms in a given molecule or ion must be equal to the overall charge on that species Nonequivalent Lewis structures Species with formal charges closest to zero and with any negative formal charges on the most electronegative atoms are considered to best describe the bonding in the molecule or ion

Example 8.10 - Formal Charges
Give possible Lewis structures for XeO3, an explosive compound of xenon Which Lewis structure or structures are most appropriate according to the formal charges?

Example Solution For XeO3 (26 valence electrons) we can draw the following possible Lewis structures:

Based on the ideas of formal charge, we would predict that the Lewis structures with the lower values of formal charge would be most appropriate for describing the bonding in XeO3

Cautions about Formal Charge
Formal charges provide estimates of charge Should not be considered as actual atomic charges Evaluation of Lewis structures using formal charge ideas can lead to erroneous predictions Experimental evidence must be used to make the final decisions on the correct description of the bonding in a molecule

How many resonance structures does the molecule SO2 have?
Join In (14) How many resonance structures does the molecule SO2 have? 1 2 3 4 Correct answer 2 The two resonance structures (without the lone pairs) are O—S=O and O=S—O

Join In (15) Which of the following statements concerning resonance structures is correct? The concept of resonance is used because the Lewis structure model is incomplete when describing bonding in a molecule For a species having three resonance structures, it is best to think of the species as existing as each of these structures one-third of the time

Join In (15) (continued) All charged molecules have resonance structures The octet rule must not be violated in writing resonance structures Correct answer The concept of resonance is used because the Lewis structure model is incomplete when describing bonding in a molecule. Lewis structures show static, localized electrons These structures are incomplete because electrons are better thought of as delocalized in the molecule

Valence Shell Electron-Pair Repulsion (VSEPR) Model
Used to predict the molecular structure of molecules formed from nonmetals Molecular structure: Three-dimensional arrangement of molecules in an atom Main postulate Structure around a given atom is determined principally by minimizing electron-pair repulsions Bonding and nonbonding pairs around a given atom will be placed as far apart as possible

Molecular Structure: Types
Linear structure Molecule has a 180-degree bond angle Example - BeCl2

Molecular Structure: Types (continued 1)
Trigonal planar structure Molecule has a planar (flat) and triangular structure with 120° bond angles Example - BF3

Tetrahedral structure Molecule has bond angles of degrees Example - CH4 Whenever four pairs of electrons are present around an atom, they should always be arranged tetrahedrally

Problem-Solving Strategy - Steps to Apply the VSEPR Model
Draw the Lewis structure for the molecule Count the electron pairs and arrange them in the way that minimizes repulsion Determine the positions of the atoms from the way the electron pairs are shared Determine the name of the molecular structure from the positions of the atoms

Example 8.11 - Prediction of Molecular Structure I
Describe the molecular structure of the water molecule

The Lewis structure for water is
Example Solution The Lewis structure for water is There are four pairs of electrons: two bonding pairs and two nonbonding pairs To minimize repulsions, these pairs are best arranged in a tetrahedral array

Although H2O has a tetrahedral arrangement of electron pairs, it is not a tetrahedral molecule The atoms in the H2O molecule form a V shape

Addition to the Original Postulate of the VSEPR Model
Lone pairs require more room than bonding pairs and tend to compress the angles between the bonding pairs

Table 8.7 - Arrangements of Electron Pairs around an Atom Yielding Minimum Repulsion

Table 8.7 - Arrangements of Electron Pairs around an Atom Yielding Minimum Repulsion (continued)

Table 8.8 - Structures of Molecules with Four Electron Pairs around the Central Atom

Trigonal bipyramidal structure Produces minimum repulsion for molecules that contain five pairs of electrons around the central atom Bond angles - 90° and 120° Octahedral structure Produces minimum repulsion for molecules that contain six pairs of electrons around the central atom Bond angle - 90°

Table 8.9 - Structures of Molecules with Five Electron Pairs around the Central Atom

You and a friend are studying for a chemistry exam
Critical Thinking You and a friend are studying for a chemistry exam What if your friend tells you that all molecules with polar bonds are polar molecules? How would you explain to your friend that this is not correct? Provide two examples to support your answer

Interactive Example 8.12 - Prediction of Molecular Structure II
When phosphorus reacts with excess chlorine gas, the compound phosphorus pentachloride (PCl5) is formed In the gaseous and liquid states, this substance consists of PCl5 molecules, but in the solid state it consists of a 1:1 mixture of PCl4+ and PCl6– ions Predict the geometric structures of PCl5, PCl4+, and PCl6–

The Lewis structure for PCl5 is shown Five pairs of electrons around the phosphorus atom require a trigonal bipyramidal arrangement When chlorine atoms are included, a trigonal bipyramidal molecule results:

The Lewis structure for the PCl4+ ion [5 + 4(7) – 1 = 32 valence electrons] is shown below

There are four pairs of electrons surrounding the phosphorus atom in the PCl4+ ion, which requires a tetrahedral arrangement of the pairs Since each pair is shared with a chlorine atom, a tetrahedral PCl4+ cation results

The Lewis structure for PCl6– [5 + 6(7) + 1 = 48 valence electrons] is shown below

Since phosphorus is surrounded by six pairs of electrons, an octahedral arrangement is required to minimize repulsions, as shown below Since each electron pair is shared with a chlorine atom, an octahedral PCl6– anion is predicted

Interactive Example 8.13 - Prediction of Molecular Structure III
Because the noble gases have filled s and p valence orbitals, they were not expected to be chemically reactive In fact, for many years these elements were called inert gases because of this supposed inability to form any compounds However, in the early 1960s several compounds of krypton, xenon, and radon were synthesized

Interactive Example 8.13 - Prediction of Molecular Structure III (continued)
For example, a team at the Argonne National Laboratory produced the stable colorless compound xenon tetrafluoride (XeF4) Predict its structure and whether it has a dipole moment

The Lewis structure for XeF4 is The xenon atom in this molecule is surrounded by six pairs of electrons, which means an octahedral arrangement

The structure predicted will depend on the arrangement of lone pairs and bonding pairs Consider the following possibilities:

The bonding pairs are indicated by the presence of the fluorine atoms Since the structure predicted differs in the two cases, we must decide which of these arrangements is preferable, and the key is to look at the lone pairs In the structure in part (a), the lone pair–lone pair angle is 90 degrees In the structure in part (b), the lone pairs are separated by 180 degrees

Since lone pairs require more room than bonding pairs, a structure with two lone pairs at 90 degrees is unfavorable Thus the arrangement (b) is preferred, and the molecular structure is predicted to be square planar Note that this molecule is not described as being octahedral There is an octahedral arrangement of electron pairs, but the atoms form a square planar structure

Although each Xe—F bond is polar (fluorine has a greater electronegativity than xenon), the square planar arrangement of these bonds causes the polarities to cancel Thus XeF4 has no dipole moment

The VSEPR Model and Multiple Bonds
Consider the NO3– ion Requires three resonance structures to describe its electronic structure:

The VSEPR Model and Multiple Bonds (continued 1)
Known to be planar with 120-degree bond angles Structure is expected for three pairs of electrons around a central atom, which means that a double bond should be counted as one effective pair in using the VSEPR model

The VSEPR Model and Multiple Bonds (continued 2)
Rules Multiple bonds count as one effective electron pair When a molecule exhibits resonance, any one of the resonance structures can be used to predict the molecular structure using the VSEPR model

Interactive Example 8.14 - Structures of Molecules with Multiple Bonds
Predict the molecular structure of the sulfur dioxide molecule Is this molecule expected to have a dipole moment?

First, we must determine the Lewis structure for the SO2 molecule, which has 18 valence electrons Expected resonance structures are To determine the molecular structure, we must count the electron pairs around the sulfur atom

In each resonance structure the sulfur has one lone pair, one pair in a single bond, and one double bond Counting the double bond as one pair yields three effective pairs around the sulfur A trigonal planar arrangement is required, which yields a V-shaped molecule

Thus the structure of the SO2 molecule is expected to be V-shaped, with a 120-degree bond angle The molecule has a dipole moment directed as shown Since the molecule is V-shaped, the polar bonds do not cancel

Molecules Containing No Single Central Atom
Consider methanol (CH3OH) Molecular structure can be predicted from the arrangement of pairs around the carbon and oxygen atoms

Figure 8.22 - The Molecular Structure of Methanol
The arrangement of electron pairs and atoms around the carbon atom The arrangement of bonding and lone pairs around the oxygen atom The molecular structure

VSEPR Model: Advantages
Correctly predicts the molecular structures of most molecules formed from nonmetallic elements Used to predict the structures of molecules with hundreds of atoms

Limitation of the VSEPR Model
Does not always provide an accurate prediction Example - Phosphine (PH3) Lewis structure of PH3 is analogous to that of ammonia (NH3) Predicted molecular structure of PH3 is similar to that of NH3, with bond angles of approximately 107°

Limitation of the VSEPR Model (continued)
Bond angles of phosphine are 94°

Which of the following molecules is polar?
Join In (16) Which of the following molecules is polar? XeF4 XeF2 BF3 NF3 Correct answer NF3 A polar molecule results from a species with a nonzero dipole moment The square planar (XeF4), linear (XeF2), and trigonal planar (BF3) geometries have zero dipole moments because the atoms and lone pairs are symmetrically arranged The trigonal pyramidal geometry (NF3) has a nonzero dipole moment because there is one lone pair in the structure

Which of the following molecules does not have a dipole moment?
Join In (17) Which of the following molecules does not have a dipole moment? H2S H2O H2Xe All of these have a dipole moment None of these has a dipole moment Correct answer H2Xe Dipole moments result from differences in electronegativity and geometry Both H2S and H2O are bent molecules, whereas H2Xe is linear

Which of the following molecules has a dipole moment?
Join In (18) Which of the following molecules has a dipole moment? CF4 SF4 XeF4 All of these have a dipole moment None of these has a dipole moment Correct answer SF4 Dipole moments result from differences in electronegativity and geometry Both the tetrahedral (CF4) and square planar (XeF4) geometries result in dipole moments of zero The see-saw geometry of SF4 produces a dipole moment

Which of the following molecules has a dipole moment?
Join In (19) Which of the following molecules has a dipole moment? BCl3 SiCl4 PCl3 Cl2 Correct answer PCl3 The trigonal planar (BCl3), tetrahedral (SiCl4), and linear (Cl2) geometries result in dipole moments of zero The trigonal pyramidal geometry (PCl3) results in a nonzero dipole moment

Join In (20) Which of the following statements about the species N2, CO, CN–, and NO+ is false? All are isoelectronic Each contains a triple bond All are linear The bond in each species is polar Correct answer The bond in each species is polar The species N2, CO, CN–, and NO+ all have 10 valence electrons, which requires using a triple bond Because they all involve only two atoms, these species are linear Because the atoms are the same in N2, it does not have polar bond and is not a polar molecule

Join In (21) What is the approximate measure of the bond angles about the carbon atom in the formaldehyde molecule, H2CO? 120° 60° 109° 180° 90° Correct answer 120° According to the VSEPR model, C in formaldehyde has three electron pairs, which are oriented 120° from one another

What type of structure does the XeOF2 molecule have?
Join In (22) What type of structure does the XeOF2 molecule have? Pyramidal Tetrahedral T-shaped Trigonal planar Octahedral Correct answer T-shaped According to the VSEPR model, Xe has five electron pairs with three bonding pairs, which means its geometry is T-shaped

Join In (23) Which of the following statements best describes BF3 and NF3? (Note: Geometry refers to the electron pair arrangement, and shape refers to the atom arrangement) They have variable geometries and shapes due to their potential resonance structures They have the same geometry and different shapes

Join In (23) (continued) They have the same geometry and the same shape They have different geometries and the same shape They have different geometries and different shapes Correct answer They have different geometries and different shapes BF3 has a trigonal planar geometry and a trigonal planar shape NF3, owing to the lone pair on the nitrogen, has a tetrahedral geometry and a trigonal pyramidal shape

Cover Page Chapter 8 Bonding: General Concepts

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