Chemical
Bonding

WHAT IS A BOND?
• Bonds are attractive forces that hold groups of atoms together and make them function as a unit.
• Bonding relates to physical properties such as melting point, hardness and electrical and thermal conductivity as well as solubility characteristics.
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Bonding

WHAT IS A BOND?
• The system is achieving the lowest possible energy state by bonding.
• Being bound requires less energy than existing in the elemental form.
• It takes energy to break a bond, not make a bond!
• Energy is RELEASED when a bond is formed, therefore, it REQUIRES energy to break a bond.
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Bonding

CHEMICAL BONDS
• Three basic types of bonds:
 Ionic
• Electrostatic attraction between ions; electrons are transferred
 Covalent
• Sharing of electrons
 Metallic
• Metal atoms bonded to several other atoms
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Bonding

COULOMB’S LAW
• used to calculate the energy of an ionic bond.
• the energy interaction between a pair of ions.
• There is a (-) sign; indicated an attractive force – energy is lower.
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Bonding

BOND LENGTH
 the distance between the two nuclei where the system energy is at a minimum between the two nuclei.
 energy is given off when two atoms achieve greater stability together than apart.
 Attractive forces — proton - electron
 Repulsive forces — electron - electron
 Small energy decrease van der Waals IMFs
 Large energy decrease chemical bonds
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Bonding

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ELECTRONEGATIVITY
• The ability of atoms in a molecule to attract electrons to itself.
• On the periodic chart, electronegativity increases as you go…
 …from left to right across a row.
 …from the bottom to the top of a column.
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ELECTRONEGATIVITY
• Ionic : Eneg difference >1.7
• Polar Covalent: Eneg difference is ≥ 0.4 and ≤ 1.7
• Nonpolar Covalent: Eneg difference <0.4
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Bonding

PRACTICE ONE
• Order the following bonds according to polarity: H —H, O—H, Cl—H, S—H, and
F —H.
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Bonding

PRACTICE ONE - answer
• H—H 0.0 difference
• S—H 0.4 difference
• Cl—H 0.9 difference
• O—H 1.4 difference
• F—H 1.9 difference
Polarity increases
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Bonding

BOND POLARITY AND DIPOLE
MOMENTS
DIPOLAR MOLECULES
 A molecule with a somewhat positive end and a somewhat negative end.
 a dipole moment .
 also molecules with preferential orientation in an electric field; all diatomic molecules with a polar covalent bond are dipolar.
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Bonding

POLARITY OF WATER
The diagram shows the charge distribution in the water molecule (a), the water molecule in an electric field (b), and the electrostatic potential diagram of a water molecule (c).
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Bonding

IONS: configurations and sizes
• Goal is to achieve a noble gas configuration.
• COVALENT: two nonmetals share electrons so each has a noble gas configuration.
• IONIC: Metal and representative group metal form a binary ionic compound where electrons are transferred so each gets a noble gas configuration
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Bonding

IONIC COMPOUNDS
• The final result of ionic bonding is a solid, regular array of cations and anions called a
crystal lattice;
• This configuration limits (-) ion/(-) ion and
(+) ion/(+) ion interactions and maximizes
(+)ion and (-)ion interactions.
• *Ion size plays a role in determining the structure and stability of ionic solids and the properties of ions in aqueous solutions.
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Bonding

PRACTICE TWO
• Arrange the ions Se 2, Br , Rb + , and Sr +2 in order of decreasing size.
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Bonding

PRACTICE TWO - answer
In order of decreasing size:
(all have [Kr] electron configuration)
Se 2Br Rb + Sr +2
Largest smallest
Sr +2 has greatest Zeff and thus the strongest attractive force; metal ions are always smaller than their atoms.
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Bonding

PRACTICE THREE
Choose the largest ion in each of the following groups.
a. Li + , Na + , K + , Rb + , Cs + b. Ba 2+ , Cs + , I , Te 2-
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PRACTICE THREE - answer a. Li + , Na + , K + , Rb + , Cs +
- all in the same group; principle quantum number increases.
b. Ba 2+ , Cs + , I , Te 2-
- Isoelectronic with [Xe] e- configuration; smallest Zeff is the largest.
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Bonding

Energies of Ionic Bonding
It takes 495 kJ/mol to remove electrons from sodium.
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Energies of Ionic Bonding
We get 349 kJ/mol back by giving electrons to chlorine.
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Energies of Ionic Bonding
• But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!
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Energies of Ionic Bonding
• There must be a third piece to the puzzle.
• What is as yet unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.
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Bonding

LATTICE ENERGY
• This third piece of the puzzle is the LATTICE
ENERGY :
The energy required to completely separate a mole of a solid ionic compound into its gaseous ions.
• The energy associated with electrostatic interactions is governed by Coulomb’s law:
E el
=

Q
1
Q
2 r
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LATTICE ENERGY
• Lattice energy increases with the charge on the ions.
• It also increases with decreasing size of ions.
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Energies of Ionic Bonding
By accounting for all three energies
(ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.
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Bonding

Energies of Ionic Bonding
• These phenomena also helps explain the
“octet rule.”
• Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy would be expended that cannot be overcome by lattice energies.
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Bonding

CALCULATING IONIC
CHARACTER
Ionic vs. Covalent
• Ionic compounds generally have greater than 50% ionic character; Eneg differences greater than 1.7
• Percent ionic character is difficult to calculate for compounds containing polyatomic ions.
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COVALENT BONDING
• Most compounds are covalently bonded, especially carbon compounds.
• Strengths of the Bond Model
- associates quantities of energy with the formation of bonds between elements
- allows the drawing of structures showing the spatial relationship between atoms in a molecule; provides a visual tool to understanding chemical structure
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Bonding

COVALENT BONDING
• Weaknesses of the Bond Model
- bonds are not actual physical structures; bonds cannot adequately explain some phenomena like resonance.
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Bonding

COVALENT BONDING
• Atoms form covalent bonds because they seek the lowest possible energy.
• We can calculate the ΔH for most chemical reactions by comparing the energy required versus energy released.
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Bonding

CALCULATING AVERAGE BOND
ENERGIES
When C and H combine to form CH
4
, 1652 kJ/mol is released (or 413kJ for each bond).
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Bonding

COVALENT BONDING
• Single and multiple bonds
 single bond - one pair of electrons shared
→ sigma (σ) bond
 Multiple bonds are most often formed by
C,N,O,P and S atoms — C-NOPS
 double bond - two pairs of electrons shared
→ one σ bond and one π bond
 triple bond - three pairs of electrons shared
→ one σ bond and two π bonds
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COVALENT BONDING
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COVALENT BONDING
***As the number of shared e- increases, the bond length ↓ and energy ↑.
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BOND ENERGY AND ENTHALPY
• using bond energy to calculate approximate energies for reactions.
• ΔH = sum of the energies required to break old bonds(endothermic) + sum of the energies released in forming new bonds (exothermic).
• Δ H =∑ D ( Bonds broken ) ∑ D ( Bonds formed )
• (D represents bond energy per mole of bonds and always has a positive sign)
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PRACTICE FOUR
• Using the bond energies in in the table in your text, calculate
Δ H for the reaction of methane with chlorine and fluorine to give Freon, CF
2
Cl
2
.
CH
4(g)
+ 2Cl
2(g)
+ 2F
2(g)
→ CF
2
Cl
2(g)
+ 2HF
(g)
+ 2HCl
(g)
• Break the bonds and then assemble with new bonds
Reactants → atoms → products
E required E released
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PRACTICE FOUR
Reactant Bonds Broken:
• CH
4
: 4 mol C – H
• 2Cl
2
: 2 mol Cl – Cl
• 2F
2
: 2 mol F – F
4(413kJ)
2(478kJ)
2(308kJ)
• Total 2438kJ
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PRACTICE FOUR
Product Bonds Formed:
• CF
2
Cl
2
: 2 mol C – F 2(485kJ)
2 mol C - Cl 2(339kJ)
• HF: 2 mol H – F
• HCl: 2 mol H – Cl
2(565kJ)
2(427kJ)
• Total 3632kJ
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PRACTICE FOUR
Δ H =∑ D broken
∑ D formed
Δ H =2438kJ 3632kJ = -1194kJ
This energy is released when CF
2
Cl
2 formed. is
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Bonding

• 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. Electron pairs are assumed to be localized on a particular atom lone pairs - or in the space between two atoms bonding pairs .
• Lone electron pairs - electrons localized on an atom (unshared)
• Bonding electron pairs - electrons found in the space between atoms (shared pairs)
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LE BONDING MODEL
Derivations of the Localized Model
1. Lewis Structures describe the valence electron arrangement
2. Geometry of the molecule is predicted with VSEPR
3. Description of the type of atomic orbitals
“blended” by the atoms to share electrons or hold lone pairs (hybrids —next chapter).
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LEWIS STRUCTURES
• "the most important requirement for the formation of a stable compound is that the atoms achieve noble gas configurations
• Duet rule - hydrogen, lithium, beryllium, and boron form stable molecules when they share two electrons (helium configuration)
• Octet Rule - elements carbon and beyond form stable molecules when they are surrounded by eight electrons
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LEWIS STRUCTURES
1. H is always a terminal atom and connected to only one other atom.
2. Lowest electronegativity element is central atom in molecule.
3. Add the TOTAL number of valence electrons from all atoms.
4.
Place one pair of electrons, a σ bond, between each pair of bonded atoms.
5. Arrange the remaining atoms to satisfy the duet rule for hydrogen and the octet rule for the
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Bonding second row elements.

Lewis Structures
Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.
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Writing Lewis Structures
PCl
3
5 + 3(7) = 26
1. Find the sum of valence electrons of all atoms in the polyatomic ion or molecule.
 If it is an anion, add one electron for each negative charge.
 If it is a cation, subtract one electron for each positive charge.
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Writing Lewis Structures
Keep track of the electrons:
2. The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds.
26

6 = 20
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Writing Lewis Structures
3. Fill the octets of the outer atoms.
Keep track of the electrons:
26

6 = 20

18 = 2
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Bonding

Writing Lewis Structures
4. Fill the octet of the central atom.
Keep track of the electrons:
26

6 = 20

18 = 2

2 = 0
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Writing Lewis Structures
5. If you run out of electrons before the central atom has an octet…
…form multiple bonds until it does.
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PRACTICE FIVE – do this now
• Give the Lewis structure for each of the following: a. HF c. NH e. CF
4
3 b. N
2 d. CH
4 f. NO +
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PRACTICE FIVE
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Exceptions to the Octet Rule
• There are three types of ions or molecules that do not follow the octet rule:
 Ions or molecules with less than an octet.
 Ions or molecules with more than eight valence electrons (an expanded octet).
 Ions or molecules with an odd number of electrons.
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LESS THAN OCTET
• H at most only two electrons (one bond)
• BeH
2
, only 4 valence electrons around
Be (only two bonds)
• Boron compounds, only 6 valence electrons (three bonds)
• ammonia boron trifluoride is a classic
Lewis A/B reaction.
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LESS THAN OCTET
Example: Boron Trifluoride
1. Note that boron only has six electrons around it
2. BF
3 is electron deficient and acts as a Lewis acid (electron pair acceptor)
3. Boron can form molecules that obey the octet rule
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MORE THAN EIGHT ELECTRONS
• can only happen if the central element has d -orbitals which means it is from the
3rd period or greater and can thus be surrounded by more than four valence pairs in certain compounds.
• The number of bonds depends on the balance between the ability of the nucleus to attract electrons and the repulsion between the pairs.
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MORE THAN EIGHT ELECTRONS
Example: Sulfur Hexafluoride
1. Note that sulfur has 12 electrons around it, exceeding the octet rule
2. Sulfur hexafluoride is very stable
3. SF
6 fills the 3s and 3p orbitals with 8 of the valence electrons, and places the other 4 in the higher energy 3d orbital
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Bonding

Odd Number of Electrons
Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons.
NO, NO
2
, ClO
2
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Bonding

More About the Octet Rule
1. Second row elements C, N, O and F should always obey the octet rule
2. B and Be (second row) often have fewer than eight electrons around them, and form electron deficient, highly reactive molecules
3. Second row elements never exceed the octet rule
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Bonding

More About the Octet Rule
4. Third row and heavier elements often satisfy (or exceed) the octet rule
5. Satisfy the octet rule first. If extra electrons remain, place them on elements having available d orbitals.
When necessary to exceed the octet rule for one of several third row elements, assume that the extra electrons be placed on the central atom.
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Bonding

Some atoms, such as N and P, tend to share a lone pair with another atom that is short of electrons, leading to the formation of a coordinate covalent bond .
These bonds are in all coordination compounds and Lewis Acids/Bases.
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N is sharing the lone PAIR of electrons by drawing an arrow from it to the H+, remember
H+ has NO electrons to contribute to the bond.
Note that all four bonds are actually identical in length and strength.
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PRACTICE SIX
• Write the Lewis structure for PCl
5
.
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Bonding

PRACTICE SIX - answer
Structure of PCl
5
VE = 5 + 5(7) = 40
Connect the 5 Cl atoms around P
40-10 = 30
Add 6 e- to each chlorine
30 – 30 = 0
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Bonding

PRACTICE SEVEN
Write the Lewis structure for each molecule or ion.
a. ClF
3 b. XeO
3 c. RnCl
2 d. BeCl
2 e. ICl
4
-
Note that this is a negative one ion.
Correct it please
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Bonding

PRACTICE SEVEN - answers a. ClF
3
VE = 7 + 3(7) = 28
Attach all fluorine atoms to chlorine
28 – 6 = 22
Give each fluorine 3 pairs of electrons
22 – 18 = 4
Give chlorine 2 more pairs
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PRACTICE SEVEN - answers b. XeO
3
VE = 8 + 3(6) = 26
Attach three oxygen to xenon
26 – 6 = 20
Give each oxygen 3 pairs of electrons
20 – 18 = 2
Xenon gets one more pair of electrons
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PRACTICE SEVEN - answers c. RnCl
2
VE = 8 + 2(7) = 22
Attach chlorine atoms to radon
22 – 4 = 18
Give each chlorine 3 pairs of electrons
18 – 12 = 6
Radon gets three more pairs of electrons
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PRACTICE SEVEN - answers d. BeCl
2
VE = 2 + 2(7) = 16
Attach two chlorine to Be
16 – 4 = 12
Give each chlorine 3 pairs of electrons
12 – 12 = 0
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Bonding

PRACTICE SEVEN - answers e. ICl
4
-
VE = 7 + 4(7) +1 = 36
Attach the four chlorine atoms to iodine
36 – 8 = 28
Give each chlorine 3 pairs of electrons
28 – 24 = 4
Add two pairs of electrons to iodine
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Bonding

RESONANCE
This is the Lewis structure we would draw for ozone, O
3
.
-
+
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RESONANCE
• But this is at odds with the true, observed structure of ozone, in which…
 …both O—O bonds are the same length.
 …both outer oxygens have a charge of

1/2.
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RESONANCE
• One Lewis structure cannot accurately depict a molecule such as ozone.
• We use multiple structures, resonance structures, to describe the molecule.
The same but different!
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Bonding

RESONANCE
Just as green is a synthesis of blue and yellow…
…ozone is a synthesis of these two resonance structures.
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RESONANCE
• In truth, the electrons that form the second C—O bond in the double bonds below do not always sit between that C and that O, but rather can move among the two oxygens and the carbon.
• They are not localized , but rather are delocalized .
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RESONANCE
• The organic compound benzene, C
6
H
6
, has two resonance structures.
• It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring.
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RESONANCE
KEY POINTS
1. Resonance structures differ only in the assignment of electron pair positions,
NEVER atom positions.
2. Resonance structures differ in the number of bond pairs between a given pair of atoms.
3. The actual structure is an average of the depicted resonance structures.
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PRACTICE EIGHT
• Describe the electron arrangement in the nitrite anion, NO
2
, using the localized electron model.
• Follow what you know to obtain the
Lewis Structure for the ion.
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Bonding

PRACTICE EIGHT
• Describe the electron arrangement in the nitrite anion, NO
2
, using the localized electron model.
• 5 + 2(6) + 1 = 18 valence e-
• Single bonded structure: O – N – O
• 18 valence - 4 = 14 valence e-
• Make one double bond – now 12 valence
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Bonding

PRACTICE EIGHT
• Rest of the electrons go around the terminal atoms to complete an octet.
• The two N – O bonds are equivalent – each intermediate between a single and double bond.
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Bonding

FORMAL CHARGE
• Use formal charge to determine the most favored resonance structure.
• Formal Charge The number of valence electrons on the free element minus the number of electrons assigned to the atom in the molecule.
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FORMAL CHARGE
• lone pair (unshared electrons belong completely to the atom in question.
• shared electrons are divided equally between the sharing atoms
• Atom’s formal charge = group number −
[# of lone electrons − 2 (# of bonding electrons)]
• The sum of the formal charges must equal an ion’s charge.
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FORMAL CHARGE
• Use formal charges along with the following to determine resonance structure:
• Atoms in molecules (or ions) should have formal charges as small as possible —as close to zero as possible - called principle of electroneutrality.
• A molecule (or ion) is most stable when any negative formal charge resides on the most electronegative atom.
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FORMAL CHARGES
Assigning formal charges:
 For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms.
 Subtract that from the number of valence electrons for that atom: The difference is its formal charge.
Carbon dioxide resonance structures
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FORMAL CHARGES
• So since we want formal charges to be as close to zero as possible, we choose the left structure as the likely structure.
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Bonding

FORMAL CHARGES
The best Lewis structure…
 …is the one with the fewest charges.
 …puts a negative charge on the most electronegative atom.
VE - 5 4 6
Assign 7 4 5
5 4 6 5 4 6
6 4 6 5 4 7
-2 0 +1 -1 0 0 0 0 -1
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Bonding

Fewer Than Eight Electrons
• Consider BF
3
:
 Giving boron a filled octet places a negative charge on the boron and a positive charge on fluorine.
 This would not be an accurate picture of the distribution of electrons in BF
3
.
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Bonding

Fewer Than Eight Electrons
Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.
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Fewer Than Eight Electrons
The lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom.
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More Than Eight Electrons
• The only way PCl
5 can exist is if phosphorus has 10 electrons around it.
• It is allowed to expand the octet of atoms on the 3rd row or below.
 Presumably d orbitals in these atoms participate in bonding.
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More Than Eight Electrons
• This eliminates the charge on the phosphorus and the charge on one of the oxygens.
• The lesson is: When the central atom is on the
3rd row or below and expanding its octet eliminates some formal charges, do so.
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More Than Eight Electrons
Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, the better structure puts a double bond between the phosphorus and one of the oxygens.
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PRACTICE NINE
• Draw all possible structures for the sulfate ion. Decide which is the most plausible using formal charges.
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PRACTICE NINE - answer
VE 6 6 6 6 6 6 6 6 6
Assign 6 7 6 6 7 6 6 7 6
0 -1 0 0 -1 0 0 -1 0
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FORMAL CHARGES
REMINDER
• Formal charges are only estimates and should not be taken as the actual atomic charges.
• Using formal charges can often lead to erroneous structures, so tests based on experiments must be used to make the final decisions on the correct description of bonding.
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PRACTICE TEN
• Give possible Lewis structures for
XeO
3
, an explosive compound of xenon.
Which Lewis structure or structures are most appropriate according to the formal charges?
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Bonding

PRACTICE TEN - answer
• XeO
3
• VE = 26
• Structure with lowest formal charge is most appropriate.
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VSEPR
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Introduction
Lewis structures show us the basic , 2-D structure of compounds.
This is the structure that we draw on a piece of paper.
..
H-N-H
H
However, we are also interested in the 3-D structure of the compounds.
The dashed
..
line, - -, is a of electrons on the central bond behind atom.
plane of the ..
In this drawing, the solid bond in front of the plane of the
H
H
N
H line, , is a bond in the plane of the surface.
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Bonding surface.

Introduction
We use Lewis structures to look at how the atoms of the compound are bound together.
Lewis structures will tell us:
 the number of atoms bound to a central atom.
 the number of lone pairs on the central atom.
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Introduction
We will use the VSEPR model to help us determine the 3-D structure of compounds.
VSEPR stands for V alence S hell E lectron P air R epulsion.
In this model, we will place the bonding and non-bonding electron pairs as far apart as possible in 3-D space.
Then, we can look at the position of the atoms of the compound and determine the 3-D structure.
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Introduction
 Electronic (structural pair)geometry – all the electron pairs surrounding a central atom are considered.
 Molecular geometry - the arrangement in space of the atoms bonded to a central atom and nonbonding electrons become “invisible”.
This is not necessarily the same as the electronic geometry.
 ANY time lone pairs are present , the electronic and molecular geometries will be different.
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Bonding

Introduction
 Each lone pair or bond pair repels all other lone pairs and bond pairs.
 Lone pairs have a different repulsion since they are experiencing an attraction or “pull” from only one nucleus as opposed to two nuclei.
• works well for elements of the s and p -blocks
• VSEPR does not apply to transition element compounds (exceptions).
• They do not cause distortion when bond angles are
120º or greater.
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LONE PAIRS
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Application - TWO
Two orbitals around a central atom:
B A B
We arrange the orbitals around the atom as far apart as possible - 180 ° apart.
The orbitals are in a linear structure.
With AB
2
, there are two bonding orbitals and zero lone pairs.
The atoms are in a linear structure.
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Bonding

Application - TWO
Two orbitals around a central atom:
B A
We arrange the orbitals around the atom as far apart as possible - 180 ° apart.
The orbitals are in a linear structure.
With AB , there is one bonding orbital and one lone pair.
The atoms are in a linear structure.
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Bonding

Application - THREE
Three orbitals around a central atom:
A
We arrange the orbitals around the atom as far apart as possible - 120 ° apart.
The orbitals are in a trigonal planar structure.
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Bonding

Application - THREE
Three orbitals around a central atom:
B
A
B B
We arrange the orbitals around the atom as far apart as possible - 120 ° apart.
With AB
3
, there are 3 bonding orbitals and no lone pairs.
The atoms are in a trigonal planar structure.
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Bonding

Application - THREE
Three orbitals around a central atom:
A
B B
We arrange the orbitals around the atom as far apart as possible - 120 ° apart.
With AB
2
, there are 2 bonding orbitals and 1 lone pair.
The atoms are in a bent structure.
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Bonding

Application - THREE
Three orbitals around a central atom:
A
B
We arrange the orbitals around the atom as far apart as possible - 120 ° apart.
With AB , there is 1 bonding orbital and 2 lone pairs.
The atoms are in a linear structure.
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Bonding

Application - FOUR
Four orbitals around a central atom:
A
We arrange the orbitals around the atom as far apart as possible - into the shape of a tetrahedron – 109.5°.
The orbitals are in a tetrahedral structure.
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Bonding

Application - FOUR
Four orbitals around a central atom:
B
B
B
A
B
We arrange the orbitals around the atom as far apart as possible - into the shape of a tetrahedron.
With AB
4
, there are 4 bonding orbitals and no lone pairs.
The atoms are in a tetrahedral structure.
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Bonding

Application - FOUR
Four orbitals around a central atom:
B
B
A
B
We arrange the orbitals around the atom as far apart as possible - into the shape of a tetrahedron.
With AB
3
, there are 3 bonding orbitals and 1 lone pair.
The atoms are in a trigonal pyramidal structure.
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Bonding

Application - FOUR
Four orbitals around a central atom:
A
B
B
We arrange the orbitals around the atom as far apart as possible - into the shape of a tetrahedron.
With AB
2
, there are 2 bonding orbitals and 2 lone pairs.
The atoms are in a bent structure.
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Bonding

Application - FOUR
Four orbitals around a central atom:
A
B
We arrange the orbitals around the atom as far apart as possible - into the shape of a tetrahedron.
With AB , there is 1 bonding orbital and 3 lone pairs.
The atoms are in a linear structure.
Chemical
Bonding

Application - FIVE
Five orbitals around a central atom:
A
We arrange the orbitals around the atom as far apart as possible - into the shape of a trigonal bipyramid.
The orbitals are in a trigonal bipyramidal structure.
Chemical
Bonding

Application - FIVE
Five orbitals around a central atom:
B
B
A B
B
B
We arrange the orbitals around the atom as far apart as possible - into the shape of a trigonal bipyramid.
With AB
5
, there are 5 bonding orbitals and no lone pairs.
Chemical
Bonding
The atoms are in a trigonal bipyramidal structure.

Application - FIVE
Five orbitals around a central atom:
B
A B
B
B
We arrange the orbitals around the atom as far apart as possible - into the shape of a trigonal bipyramid.
With AB
4
, there are 4 bonding orbitals and 1 lone pair.
Chemical
Bonding
The atoms are in a see-saw structure.

Application - FIVE
Five orbitals around a central atom:
B
A B
B
We arrange the orbitals around the atom as far apart as possible - into the shape of a trigonal bipyramid.
With AB
3
, there are 3 bonding orbitals and 2 lone pairs.
Chemical
Bonding
The atoms are in a T-shaped structure.

Application - FIVE
Five orbitals around a central atom:
B
A
B
We arrange the orbitals around the atom as far apart as possible - into the shape of a trigonal bipyramid.
With AB
2
, there are 2 bonding orbitals and 3 lone pairs.
Chemical
Bonding
The atoms are in a linear structure.

Application - FIVE
Five orbitals around a central atom:
B
A
We arrange the orbitals around the atom as far apart as possible - into the shape of a trigonal bipyramid.
With AB , there is 1 bonding orbital and 4 lone pairs.
Chemical
Bonding
The atoms are in a linear structure.

Application - SIX six orbitals around a central atom:
A
We arrange the orbitals around the atom as far apart as possible - into the shape of an octahedron.
The orbitals are in an octahedral structure.
Chemical
Bonding

Application - SIX six orbitals around a central atom:
B
B
B
A
B
B
B
We arrange the orbitals around the atom as far apart as possible - into the shape of an octahedron.
With AB
6
, there are 6 bonding orbitals and no lone pairs.
Chemical
Bonding
The atoms are in an octahedral structure.

Application - SIX six orbitals around a central atom:
B
B
A
B
B
B
We arrange the orbitals around the atom as far apart as possible - into the shape of an octahedron.
With AB
5
, there are 5 bonding orbitals and 1 lone pair.
Chemical
Bonding
The atoms are in a square pyramidal structure.

Application - SIX six orbitals around a central atom:
B
A
B
B
B
We arrange the orbitals around the atom as far apart as possible - into the shape of an octahedron.
With AB
4
, there are 4 bonding orbitals and 2 lone pairs.
Chemical
Bonding
The atoms are in a square planar structure.

Application - SIX six orbitals around a central atom:
B
A
B
B
We arrange the orbitals around the atom as far apart as possible - into the shape of an octahedron.
With AB
3
, there are 3 bonding orbitals and 3 lone pairs.
Chemical
Bonding
The atoms are in a T-shaped structure.

Application - SIX six orbitals around a central atom:
B
A
B
We arrange the orbitals around the atom as far apart as possible - into the shape of an octahedron.
With AB
2
, there are 2 bonding orbitals and 4 lone pairs.
Chemical
Bonding
The atoms are in a linear structure.

Application - SIX six orbitals around a central atom:
B
A
We arrange the orbitals around the atom as far apart as possible - into the shape of an octahedron.
With AB , there is 1 bonding orbital and 5 lone pairs.
Chemical
Bonding
The atoms are in a linear structure.

total orbitals
2
3
4
# bonding
2
1
4
3
3
2
2
1
1
Summary
# nonbonding
2
3
0
1
0
1
0
1
2 structure linear linear trigonal planar bent linear tetrahedral trigonal pyramidal bent linear
Chemical
Bonding

total orbitals
5
6
# bonding
5
2
1
4
3
6
5
2
1
4
3
Summary
# nonbonding
0
3
4
1
2
0
1
4
5
2
3 structure trigonal bipyramidal see-saw
T-shaped linear linear octahedral square pyramidal square planar
T-shaped linear linear
Chemical
Bonding

• Arrangements of Electron
Pairs Around an Atom
Yielding
Minimum
Repulsion
Chemical
Bonding

REMEMBER
1. structural pairs σ bond pairs about an atom
2. The presence of lone pairs alters the six basic MOLECULAR geometries , but the electronic or structural pair geometry remains one of the six basic types.
Chemical
Bonding

Structures of
Molecules that
Have Four
Electron Pairs
Around the
Central Atom
8 –134
Chemical
Bonding

Structures of
Molecules with
Five Electron
Pairs Around the Central
Atom
Chemical
Bonding

DETERMINING GEOMETRY
1. Sketch the Lewis dot structure – do not skip this step!
2. Determine the number of electron pair groups surrounding the central atom(s). Remember that double and triple bonds are treated as a single group.
3. Determine the geometric shape that maximizes the distance between the electron groups. This is the geometry of the electron groups.
4. Mentally allow nonbonding pairs to become invisible.
Determine the molecular geometry by looking at the remaining arrangement of atoms (as determined by the bonding electron groups) around the central atom.
Chemical
Bonding

Molecular shapes for central atoms with normal valence
• This will be no more than four structural pairs if the atom obeys the octet rule.
• Since no lone pairs are present, the molecular and structural pair [or electron ic] geometry is the same.
109.5
° bond angle.
• Ignore lone pairs AFTER you’ve determined the angles and only the relative positions of the atoms are important in molecular geometry
• Examples are ammonia (107.5
° bond angle) and water
(104.5
° bond angle).
Chemical
Bonding

Molecular shapes for central atoms with expanded valence
• only elements with a principal energy level of 3 or higher can expand their valence and violate the octet
• rule on the high side.
• d orbitals are needed for the expansion to a 5th or 6th bonding location —the combination of 1 s and 3 p ’s provides the four bonding sites that make up the octet rule.
• seems to be a limit of three lone pairs about the central atom
• Example is XeF
4
(equatorial and axial)
Chemical
Bonding

PRACTICE ELEVEN
Prediction of Molecular Structure I
• Describe the molecular structure of the water molecule.
Chemical
Bonding

PRACTICE ELEVEN
• Lewis Structure:
• Four pairs of e-
 Two bonding pairs and two lone pairs.
Chemical
Bonding

PRACTICE TWELVE
Prediction of Molecular Structure II
• When phosphorus reacts with excess chlorine gas, the compound phosphorus pentachloride (PCl
5
) is formed. In the gaseous and liquid states, this substance consists of PCl
5 molecules, but in the solid state it consists of a 1:1 mixture of PCl
4
+ and
PCl
6
ions. Predict the geometric structures of PCl
5
, PCl
4
+ , and PCl
6
.
Chemical
Bonding

PRACTICE TWELVE
• PCl
5
• VE = 5 + 5(7) = 40
• Five bonded e-
• trigonal bipyramidal
Chemical
Bonding

PRACTICE TWELVE
• PCl
4
+
• VE = 5 + 4(7) - 1 = 32
• Four bonded e-
• tetrahedral
Chemical
Bonding

PRACTICE TWELVE
• PCl
6
-
• VE = 5 + 6(7) +1= 48
• Six bonded e-
• octahredral
Chemical
Bonding

MOLECULAR POLARITY
• Polar bonds can be polar while the entire molecule isn’t and vice versa.
• dipole moment - separation of the charge in a molecule; product of the size of the charge and the distance of separation – learned this earlier.
Chemical
Bonding

• align themselves with an electric field.
• align with each other as well in the absence of an electric field
• water — two lone pairs establish a strong negative pole
• ammonia — one lone pair which establishes a negative pole
• note that the direction of the arrow indicating the dipole moment always points to the negative pole
Chemical
Bonding with the cross hatch on the arrow at the positive pole.

MOLECULAR POLARITY
Nonpolar molecule octet rule is obeyed AND all the surrounding bonds are the same; the molecule is nonpolar since all the dipole moments cancel each other out.
Example: CO
2
Chemical
Bonding

MOLECULAR POLARITY
Polar Molecule octet rule is obeyed and all the surrounding bonds are NOT the same; the molecule is polar since all the dipole moments do not cancel each other out.
Example: NH
3
. Methane is a great example. Replace one H with a halogen and it becomes polar. Replace all and it
Chemical is nonpolar again!
Bonding

DO NOW
• Pick up handout
• Get out your notes and the last two homework sheets
Chemical
Bonding

PRACTICE THIRTEEN
Drawing molecules
• Draw CH
4
, CH
3
Cl, CH
2
Cl
2
, CHCl
Indicate dipole moment(s) where
3
, CCl
4. necessary.
Chemical
Bonding

CH
4
PRACTICE THIRTEEN
CH
3
Cl
Chemical
Bonding

PRACTICE THIRTEEN
CH
2
Cl
2
CHCl
3
CCl
4.
Chemical
Bonding

PRACTICE FOURTEEN
A team at the Argonne National
Laboratory produced the stable colorless compound xenon tetrafluoride (XeF
4
).
Predict its structure and whether it has a dipole moment.
Chemical
Bonding

PRACTICE FOURTEEN
• XeF
4
• VE = 8 + 4(7) = 36
• Four bonded pairs and two lone pairs
• Square planar
Chemical
Bonding

PRACTICE FIFTEEN
Structures of Molecules with Multiple
Bonds
• Predict the molecular structure of the sulfur dioxide molecule. Is this molecule expected to have a dipole moment?
Chemical
Bonding

PRACTICE FIFTEEN
• SO
2
• VE = 6 + 2(6) = 18
• Double bond counts as one so there are three pairs around sulfur
• Bent
• yes – dipole moment
Chemical
Bonding

BONDING SUMMARY
• Read on your own.
Chemical
Bonding