Chapters 8 & 9: Covalent Bonding
Notes and images from instructor’s resource CD for Chemistry: The Central Science, 9 th Edition, Brown, LeMay, Bursten & Burdge, © 2003, Pearson Education, One Lake Street, Upper Saddle River, NJ 07485
8.3 Covalent Bonding
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The majority of chemical substances do not have characteristics of ionic
compounds.
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We need a different model for bonding between atoms.
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A chemical bond formed by sharing a pair of electrons is called a covalent bond.
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Both atoms acquire noble-gas electronic configurations.
•
This is the “glue” to bind atoms together.
Lewis Structures
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Formation of covalent bonds can be represented using Lewis symbols.
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The structures are called Lewis structures.
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We usually show each electron pair shared between atoms as a line and
show unshared electron pairs as dots.
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Each pair of shared electrons constitutes one chemical bond.
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

Example: H + H  H: has electrons on a line connecting the two H nuclei: H-H.
(a) The attractions and repulsions
among electrons and nuclei in the
hydrogen molecule.
(b) Electron distribution in the H2
molecule. The concentration of
electron density between the
nuclei leads to a net attractive
force that constitutes the covalent
bond holding the molecule
together.
Practice Exercise:
Write the Lewis symbols for the elements nitrogen and fluorine. Predict the formula of the stable binary
compound formed by the reaction of nitrogen with fluorine. Draw its Lewis structure.
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Multiple Bonds
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It is possible for more than one pair of electrons to be shared between two atoms
(e.g., multiple bonding):
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One shared pair of electrons is a single bond (e.g., H2);
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Two shared pairs of electrons is a double bond (e.g., O2);
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Three shared pairs of electrons is a triple bond (e.g., N2).
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Generally, bond distances decrease as we move from single through double to
triple bonds.
8.4 Bond Polarity and Electronegativity
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The electron pairs shared between two different atoms are usually unequally
shared.
•
Bond polarity describes the sharing of the electrons in a covalent bond.
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Two extremes:
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In a nonpolar covalent bond the electrons are shared equally.
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Example: Bonding between identical atoms (example: Cl2).
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In a polar covalent bond, one of the atoms exerts a greater
attraction for bonding electrons than the other (example: HCl)
•
If the difference is large enough, an ionic bond forms (example: NaCl).
Figure 8.7
Nonpolar Covalent Bond
Polar Covalent Bond
Ionic Bond
Electronegativity
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The ability of an atom in a molecule to attract electrons to itself is its electronegativity.
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The electronegativity of an element is related to its ionization energy and electron
affinity.
•
Pauling electronegativity scale: from 0.7 (Cs) to 4.0 (F).
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Electronegativity increases across a period and decreases down a group.
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Figure 8.6: Electronegativities of the Elements
Electronegativity and Bond Polarity
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Electronegativity differences close to zero result in nonpolar covalent bonds.
•The electrons are equally or almost equally shared.
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The greater the difference in electronegativity between two atoms, the more polar
the bond (polar covalent bonds)
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There is no sharp distinction between bonding types.
Dipole Moments
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Molecules like HF have centers of positive and negative charge that do not
coincide.
•
These are polar molecules.
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We indicate the polarity of molecules in two ways:
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The positive end (or pole) in a polar bond may be represented with a “+”
and the negative pole with a “-”.
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We can also place an arrow over the line representing the bond.
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The arrow points toward the more electronegative element and shows the
shift in electron density toward that atom.
Practice Problem
1. Which bond is more polar: (a) B-Cl or C-Cl; (b) P-F or P-Cl? Indicate in each case which atom
has the partial negative charge.
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We can quantify the polarity of the molecule.
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When charges are separated by a distance, a dipole is produced.
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The dipole moment is the quantitative measure of the magnitude of the
dipole (µ):
µ=Qr
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The magnitude of the dipole moment is
given in Debyes (D).
Practice Exercise:
1. The dipole moment of chlorine monofluoride is 0.88 D and its bond length is 1.63 angstroms.
Which atom is expected to have a negative charge? What is the charge on that atom in units of e?
Bond Types and Nomenclature
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Previously we used two different approaches to naming binary compounds
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One for ionic compounds and another for molecular compounds
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In both systems the less electronegative element is given first.
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The other element follows with the ending -ide.
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Both approaches are sometimes used with the same substance!
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Metals with higher oxidation numbers tend to be molecular rather than
ionic.
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For example: TiO2
•
The names titanium(IV) oxide and titanium dioxide are used but
titanium dioxide is more commonly used.
8.5 Drawing Lewis Structures
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Some simple guidelines for drawing Lewis structures:
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Add up all of the valence electrons on all atoms.
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For an anion, add electrons equal to the negative charge.
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For a cation, subtract electrons equal to the positive charge.
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Identify the central atom.
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When a central atom has other atoms bound to it, the central atom
is usually written first.
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Example: In CO32– the central atom is carbon.
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Place the central atom in the center of the molecule and add all other
atoms around it.
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Place one bond (two electrons) between each pair of atoms.
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Complete the octets for all atoms connected to the central atom
(exception: hydrogen can only have two electrons).
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Complete the octet for the central atom; use multiple bonds if necessary.
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Practice Exercise:
Draw Lewis structures for the following: HCN, PCl3, BrO3-, C2H4, and PO43-
Formal Charge
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Sometimes it is possible to draw more than one Lewis structure with the octet rule
obeyed for all the atoms.
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To determine which structure is most reasonable, we use formal charge.
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The formal charge of an atom is the charge that an atom (in a molecule) would
have if all of the atoms had the same electronegativity.
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To calculate formal charge, electrons are assigned as follows:
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All nonbonding (unshared) electrons are assigned to the atom on which
they are found.
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Half of the bonding electrons are assigned to each atom in a bond.
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Formal charge is the number of valence electrons in the isolated atom,
minus the number of electrons assigned to the atom in the Lewis structure.
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For example: consider CN- (cyanide ion):
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For carbon:
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There are four valence electrons (from periodic table).
•
In the Lewis structure there are two nonbonding electrons and
three electrons from the triple bond.
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There are five electrons from the Lewis structure.
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Formal charge: 4 - 5 = -1.
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For nitrogen:
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There are five valence electrons.
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In the Lewis structure there are two nonbonding electrons and
three from the triple bond.
•
There are five electrons from the Lewis structure.
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Formal charge = 5 - 5 = 0.
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Using formal charge calculations to distinguish between alternative Lewis
structures:
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The most stable structure has the smallest formal charge on each atom and
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The most negative formal charge on the most electronegative atoms.
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It is important to keep in mind that formal charges do NOT represent REAL
charges on atoms.
Oxidation Numbers
Formal Charges
Distribution of
Electron Density
Practice Exercise:
Draw three possible Lewis structures for NCO-. Assign formal charges to the atoms in each structure.
Which Lewis structure should be the preferred one?
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8.6 Resonance Structures
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Some molecules are not well described by a single Lewis structure.
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Typically, structures with multiple bonds can have similar structures with
the multiple bonds between different pairs of atoms.
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Example: Experimentally, ozone has two identical bonds whereas
the Lewis structure requires one single (longer) and one double
bond (shorter).
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Resonance structures are attempts to represent a real structure that is a mix
between several extreme possibilities.
•
Resonance structures are Lewis structures that differ only with respect to
placement of the electrons.
•
The “true” arrangement is a blend or hybrid of the resonance structures.
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Example: In ozone the extreme possibilities have one double and one
single bond.
•
The resonance structure has two identical bonds of intermediate
character.
•
We use a double headed arrows () to indicate resonance.
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Common examples: O3, NO3–, SO3, NO2, and benzene.
Resonance in Benzene
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Benzene belongs to an important category of organic molecules called aromatic
compounds.
•
Benzene (C6H6) is a cyclic structure.
•
It consists of six carbon atoms in a hexagon.
•
Each carbon atom is attached to two other carbon atoms and one hydrogen
atom.
•
There are alternating double and single bonds between the carbon atoms.
•
Experimentally, the C-C bonds in benzene are all the same length.
•
Experimentally, benzene is planar.
•
To emphasize the resonance between the two Lewis structures (hexagons with
alternating single and double bonds), we often represent benzene as a hexagon
with a circle in it.
Practice Exercises:
1. Which is predicted to have the shorter sulfur-oxygen bonds, SO3 or SO32-?
2.
Draw two equivalent resonance structures for HCO2-.
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8.7 Exceptions to the Octet Rule
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There are three classes of exceptions to the octet rule:
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Molecules with an odd number of electrons.
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Molecules in which one atom has less than an octet.
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Molecules in which one atom has more than an octet.
Odd Number of Electrons
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Most molecules have an even number of electrons and complete pairing of
electrons occurs although some molecules have an odd number of electrons.
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Examples: ClO2, NO, and NO2.
Less than an Octet
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Molecules with less than an octet are also relatively rare.
•
Most often encountered in compounds of boron or beryllium.
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A typical example is BF3.
More than an Octet
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This is the largest class of exceptions.
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Atoms from the third period on can accommodate more than an octet.
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Examples: PCl5, SF4, AsF6–, and ICl4–.
•
Elements from the third period and beyond have unfilled d orbitals that can be
used to accommodate the additional electrons.
•
Size also plays a role.
•
The larger the central atom, the larger the number of atoms that can
surround it.
•
The size of the surrounding atoms is also important.
•
Expanded octets occur often when the atoms bound to the central atom are
the smallest and most electronegative (e.g., F, Cl, O).
Practice Exercises:
1. Draw Lewis structures for IBr4- and XeF2.
2.
Which of the following atoms is never found with more than an octet of electrons: S, C, P, or Br?
8.8 Strengths of Covalent Bonds
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The energy required to break a covalent bond is called the bond enthalpy, D.
•
That is, for the Cl2 molecule, D(Cl-Cl) is given by H for the reaction:
Cl2(g)  2Cl(g).
•
When more than one bond is broken:
CH4(g)  C(g) + 4H(g)
H = 1660 kJ
•
The bond enthalpy is a fraction of H for the atomization reaction:
D(C–H) = ¼ H = ¼(1660 kJ) = 415 kJ.
•
Bond enthalpy is always a positive quantity.
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Bond Enthalpies and the Enthalpies of Reactions
•
We can use bond enthalpies to calculate the enthalpy for a chemical reaction.
•
We recognize that in any chemical reaction bonds need to be broken and then new
bonds form.
•
The enthalpy of the reaction is given by:
•
The sum of bond enthalpies for bonds broken less the sum of bond
enthalpies for bonds formed.
•
Where Hrxn is the enthalpy for a reaction,
Hrxn = D(bonds broken) - D(bonds formed)
•
We illustrate the concept with the reaction between methane, CH4, and chlorine:
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) (See illustration below.)
•
In this reaction one C-H bond and one Cl-Cl bond are broken while one
C-Cl bond and one H-Cl bond are formed.
•
So Hrxn = [D(C-H) + D(Cl-Cl)] - [D(C-Cl) + D(H-Cl)] = -104 kJ.
•
The overall reaction is exothermic which means than the bonds formed are
stronger than the bonds broken.
•
The above result is consistent with Hess’s law.
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Practice Excercises:
1. Using Table 8.4, estimate ΔH for the complete combustion of one mole of C2H6.
2. Using Table 8.4, estimate ΔH for the decomposition of N2H4 to form nitrogen gas and hydrogen gas.
Bond Enthalpy and Bond Length
•
The distance between the nuclei of the atoms involved in a bond is called the
bond length.
•
Multiple bonds are shorter than single bonds.
•
We can show that multiple bonds are stronger than single bonds.
•
As the number of bonds between atoms increases, the atoms are held
closer and more tightly together.
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9.1
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9.2
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Molecular Shapes
Lewis structures give atomic connectivity: they tell us which atoms are physically connected to which.
The shape of a molecule is determined by its bond angles.
•
The angles made by the lines joining the nuclei of the atoms in a molecule are the bond angles.
Consider CCl4:
•
Experimentally we find all Cl–C–Cl bond angles are 109.5o.
•
Therefore, the molecule cannot be planar.
•
All Cl atoms are located at the vertices of a tetrahedron with the C at its center.
In order to predict molecular shape, we assume that the valence electrons repel each other.
•
Therefore, the molecule adopts the three-dimensional geometry that minimizes this repulsion.
•
We call this model the Valence Shell Electron Pair Repulsion (VSEPR) model
The VSEPR Model
A covalent bond forms between two atoms when a pair of electrons occupies the space between the
atom.
•
This is a bonding pair of electrons.
•
Such a region is an electron domain.
A nonbonding pair or lone pair of electrons defines an electron domain located principally on one atom.
Example: NH3 has three bonding pairs and one lone pair.
VSEPR predicts that the best arrangement of electron domains is the one that minimizes the repulsions
among them.
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The arrangement of electron domains about the central atom of an ABn molecule is its electrondomain geometry.
•
There are five different electron-domain geometries:
•
Linear (two electron domains), trigonal planar (three domains), tetrahedral (four
domains), trigonal bipyramidal (five domains) and octahedral (six domains).
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The molecular geometry is the arrangement of the atoms in space.
•
To determine the shape of a molecule we distinguish between lone pairs and bonding pairs.
•
We use the electron domain geometry to help us predict the molecular geometry.
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Draw the Lewis structure.
•
Count the total number of electron pairs around the central atom.
•
Arrange the electron pairs in one of the above geometries to minimize electron-electron
repulsion.
•
Next, determine the three-dimensional structure of the molecule:
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We ignore lone pairs in the molecular geometry.
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Describe the molecular geometry in terms of the angular arrangement of the bonded
atoms.
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Multiple bonds are counted as one electron domain.
The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles
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We refine VSEPR to predict and explain slight distortions from “ideal” geometries.
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Consider three molecules with tetrahedral electron domain geometries: CH4, NH3, and H2O.
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By experiment, the H–X–H bond angle decreases from C (109.5o in CH4) to N (107o in NH3) to
O (104.5o in H2O).
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A bonding pair of electrons is attracted by two nuclei. They do not repel as much as lone pairs.
which are primarily attracted by only one nucleus.
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Electron domains for nonbonding electron pairs thus exert greater repulsive forces on adjacent
electron domains.
•
They tend to compress the bond angles.
•
The bond angle decreases as the number of nonbonding pairs increases.
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Similarly, electrons in multiple bonds repel more than electrons in single bonds. (e.g. in Cl2CO
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the O–C–Cl angle is 124.3o, and the Cl–C–Cl bond angle is 111.4o).
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We will encounter the following basic molecular shapes:
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Three atoms (AB2): Linear, Bent
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Four atoms (AB3): Trigonal planar, trigonal pyramidal
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Five atoms (AB4): Tetrahedral
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Six atoms (AB5): Trigonal bipyramida
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Seven atoms (AB6): Octahedral
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Practice Exercises:
1. Predict the electron domain geometry, molecular geometry, and bond angles for the following:
O3, SnCl3-, SeCl2, and CO32-.
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Molecules with Expanded Valence Shells
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Atoms that have expanded octets have AB5 (trigonal bipyramidal) or AB6 (octahedral) electron domain
geometries.
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Trigonal bipyramidal structures have a plane containing three electron pairs.
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The fourth and fifth electron pairs are located above and below this plane.
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In this structure two trigonal pyramids share a base.
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For octahedral structures, there is a plane containing four electron pairs.
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Similarly, the fifth and sixth electron pairs are located above and below this plane.
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Two square pyramids share a base.
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Consider a trigonal bipyramid:
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The three electron pairs in the plane are called equatorial;
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The two electron pairs above and below this plane are called axial;
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The axial electron pairs are 180o apart and 90o to the equatorial electrons;
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The equatorial electron pairs are 120o apart;
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To minimize electron–electron repulsion, nonbonding pairs are always placed in equatorial
positions and bonding pairs in either axial or equatorial positions.
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Consider an octahedron:
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The four electron pairs in the plane are at 90o to each other;
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The two axial electron pairs are 180o apart and at 90o to the electrons in the plane.
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Because of the symmetry of the system, each position is equivalent.
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If we have five bonding pairs and one lone pair, it doesn’t matter where the lone pair is placed.
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The molecular geometry is square pyramidal.
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If two lone pairs are present, the repulsions are minimized by pointing them toward opposite
sides of the octahedron.
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The molecular geometry is square planar.
Practice Exercise:
Predict the electron domain geometry and the molecular geometry of SF4, IF5, ClF3, and ICl4-. Include
bond angles for each species.
Shapes of Larger Molecules
•
In acetic acid, CH3COOH, there are three interior atoms: two C and one O.
•
We assign the molecular (and electron-domain) geometry about each interior atom separately:
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The geometry around the first C is tetrahedral;
•
The geometry around the second C is trigonal planar; and
•
The geometry around the O is bent (tetrahedral).
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9.3
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Molecular Shape and Molecular Polarity
Polar molecules interact with electric fields.
We previously saw that binary compounds are polar if their centers of negative and positive charge do
not coincide.
•
If two charges, equal in magnitude and opposite in sign, are separated by a distance d, then a
dipole is established.
•
The dipole moment, μ, is given by
μ = Qr
•
Where Q is the magnitude of the charge.
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We can extend this to polyatomic molecules.
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For each bond in a polyatomic molecule, we can consider the bond dipole.
•
The dipole moment due only to the two atoms in the bond is the bond dipole.
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Because bond dipoles and dipole moments are vector quantities, the orientation of these
individual dipole moments determines whether the molecule has an overall dipole moment.
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Examples:
•
In CO2 each C−O dipole is canceled because the molecule is linear.
•
In H2O, the H−O dipoles do not cancel because the molecule is bent.
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It is possible for a molecule with polar bonds to be either polar or nonpolar.
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Example:
•
For diatomic molecules:
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Polar bonds always result in an overall dipole moment.
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For triatomic molecules:
If the molecular geometry is bent, there is an overall dipole moment;
If the molecular geometry is linear, and the B atoms are the same, there is no overall
dipole moment;
If the molecular geometry is linear and the B atoms are different, there is an overall
dipole moment.
For molecules with four atoms:
If the molecular geometry is trigonal pyramidal, there is an overall dipole moment;
If the molecular geometry is trigonal planar, and the B atoms are identical, there is no
overall dipole moment;
If the molecular geometry is trigonal planar and the B atoms are different, there is an
overall dipole moment.
Pracatice Exercise:
Determine whether the following molecules are polar or nonpolar: NF3, BCl3, BrCl, and SF6. For the
polar molecules, determine which atom(s) carry the partial negative charge.
9.4
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Covalent Bonding and Orbital Overlap
Lewis structures and VSEPR theory give us the shape and location of electrons in a molecule.
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•
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They do not explain why a chemical bond forms.
How can quantum mechanics be used to account for molecular shape? What are the orbitals that are
involved in bonding?
We use valence-bond theory:
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A covalent bond forms when the orbitals on two atoms overlap.
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The shared region of space between the orbitals is called the orbital overlap.
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There are two electrons (usually one from each atom) of opposite spin in the orbital
overlap.
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As two nuclei approach each other their atomic orbitals overlap.
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9.5
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As the amount of overlap increases, the energy of the interaction decreases.
At some distance the minimum energy is reached.
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The minimum energy corresponds to the bonding distance (or bond length).
As the two atoms get closer, their nuclei begin to repel and the energy increases.
At the bonding distance, the attractive forces between nuclei and electrons just balance the
repulsive forces (nucleus-nucleus, electron-electron).
Hybrid Orbitals
We can apply the idea of orbital overlap and valence-bond theory to polyatomic molecules.
sp Hybrid Orbitals
•
Consider the BeF2 molecule:
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Be has a 1s22s2 electron configuration.
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There is no unpaired electron available for bonding.
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We conclude that the atomic orbitals are not adequate to describe orbitals in molecules.
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We know that the F–Be–F bond angle is 180o (VSEPR theory).
We also know that one electron from Be is shared with each one of the unpaired electrons from F.
We assume that the Be orbitals in the Be–F bond are 180o apart.
We could promote an electron from the 2s orbital on Be to the 2p orbital to get two unpaired electrons
for bonding.
•
BUT the geometry is still not explained.
We can solve the problem by allowing the 2s and one 2p orbital on Be to mix or form two new hybrid
orbitals (a process called hybridization).
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The two equivalent hybrid orbitals that result from mixing an s and a p orbital and are called sp
hybrid orbitals.
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The two lobes of an sp hybrid orbital are 180o apart.
According to the valence-bond model, a linear arrangement of electron domains implies sp
hybridization.
Since only one of 2p orbitals of Be has been used in hybridization, there are two unhybridized p
orbitals remaining on Be.
The electrons in the sp hybrid orbitals form shared electron bonds with the two fluorine atoms.
sp2 and sp3 Hybrid Orbitals
•
Important: When we mix n atomic orbitals we must get n hybrid orbitals.
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Three sp2 hybrid orbitals are formed from hybridization of one s and two p orbitals.
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Thus, there is one unhybridized p orbital remaining.
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The large lobes of the sp2 hybrids lie in a trigonal plane.
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Molecules with trigonal planar electron-pair geometries have sp2 orbitals on the central atom.
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Four sp3 hybrid orbitals are formed from hybridization of one s and three p orbitals.
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Therefore, there are four large lobes.
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Each lobe points towards the vertex of a tetrahedron.
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The angle between the large lobes is 109.5o.
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Molecules with tetrahedral electron pair geometries are sp3 hybridized.
Hybridization Involving d Orbitals
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Since there are only three p orbitals, trigonal bipyramidal and octahedral electron-pair geometries must
involve d orbitals.
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Trigonal bipyramidal electron pair geometries require sp3d hybridization.
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Octahedral electron pair geometries require sp3d2 hybridization.
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Note that the electron pair VSEPR geometry corresponds well with the hybridization.
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Use of d orbitals in making hybrid orbitals corresponds well with the idea of an expanded octet.
Practice Exercises:
1. Predict the electron domain geometry and the hybridization of the central atom in SF6.
2. Indicate the hybridization of the central atom in NH2- and SF4.
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Summary
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We need to know the electron-domain geometry before we can assign hybridization.
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To assign hybridization:
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Draw a Lewis structure. Use VSEPR theory to assign electron domain geometry.
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Specify the hybridization required for the electron pairs based on their geometric
arrangement.
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Name the geometry by the positions of the atoms.
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9.6
Multiple Bonds
• In the covalent bonds we have seen so far the electron density has been concentrated symmetrically about the
internuclear axis.
•
Sigma () bonds: electron density lies on the axis between the nuclei.
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All single bonds are  bonds.
•
What about overlap in multiple bonds?
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Pi () bonds: electron density lies above and below the plane of the nuclei.
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A double bond consists of one  bond and one  bond.
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A triple bond has one  bond and two  bonds.
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Often, the p orbitals involved in  bonding come from unhybridized orbitals.
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For example: ethylene, C2H4, has:
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One  and one  bond.
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Both C atoms sp2 hybridized.
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Both C atoms with trigonal planar electron-pair and molecular geometries.
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For example: acetylene, C2H2:
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The electron-domain geometry of each C is linear.
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Therefore, the C atoms are sp hybridized.
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The sp hybrid orbitals form the C–C and C–H  bonds.
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There are two unhybridized p orbitals on each C atom.
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Both unhybridized p orbitals form the two  bonds;
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One  bond is above and below the plane of the nuclei;
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One  bond is in front and behind the plane of the nuclei.
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When triple bonds form (e.g., N2), one  bond is always above and below and the other is in front and
behind the plane of the nuclei.
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Practice Exercise:
Draw the Lewis structure for CH2O. (a) Predict the bond angles around the central carbon. (b) Give the
hybridization around the central atom. (c) Determine the number of pi and sigma bonds in the molecule.
Delocalized  Bonding
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So far all the bonds we have encountered are localized between two nuclei.
•
In the case of benzene:
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There are six C–C  bonds and six C–H  bonds
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Each C atom is sp2 hybridized.
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There is one unhybridized p orbital on each carbon atom, resulting in six unhybridized carbon p
orbitals in a ring.
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In benzene there are two options for the three  bonds:
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Localized between carbon atoms or
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Delocalized over the entire ring (i.e., the  electrons are shared by all six carbon atoms).
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Experimentally, all C–C bonds are the same length in benzene.
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Therefore, all C–C bonds are of the same type (recall single bonds are longer than double bonds).
Practice Exercise:
Which of the following will exhibit delocalized bonding: SO3, SO32-, H2CO, O3, or NH4+?
General Conclusions
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Every pair of bonded atoms shares one or more pairs of electrons.
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Two electrons shared between atoms on the same axis as the nuclei are  bonds.
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 Bonds are always localized in the region between two bonded atoms.
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If two atoms share more than one pair of electrons, the additional pairs form  bonds.
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When resonance structures are possible, delocalization is also possible.
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