Chapter 8 and Chapter 9
(Concepts of Chemical Bonding) (Molecular Geometries and Bonding Theories)
Name:__________________________________________
Date: ___________ Period:____
Three basic types of bonds:
• Ionic-Electrostatic attraction between ions
• Covalent-Sharing of electrons
• Metallic-Metal atoms bonded to several other atoms
Ionic Bonding-ionization energy, electron affinity, and lattice energy are the three energies needed when
ionic bonding takes place.
Ionization energy-Last chapter we discussed the energy needed to remove electrons when forming bonds. It
takes 495 kj/mol of energy to remove an electron from sodium and we get 349 kj/mol of energy back by giving
electrons to chlorine. This does not explain why this reaction is so exothermic.
Electron affinityWhat is as yet unaccounted for is
the electrostatic attraction between
the newly formed sodium cation
and chloride anion.
The third piece is “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 is the lattice energy between the two
objects, k = 1/4πεo is a constant equal
to 9 x 109, each q is the electric charge
of one of the two objects, and r is the
distance between the two objects.
• Lattice energy, then, increases with the
charge on the ions.
• It also increases with decreasing size
of ions.
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.
Covalent Bonding-In these bonds atoms share electrons.
There are several electrostatic interactions in these bonds:
Ø Attractions between electrons and nuclei
Ø Repulsions between electrons
Ø Repulsions between nuclei
Two types of covalent bonds are polar and nonpolar covalent bonds.
Polar Covalent Bonds-unequal sharing of electrons.
Nonpolar covalent Bonds-equal sharing of electrons.
1 Polar
•
•
•
Nonpolar
Although atoms often form compounds by sharing
electrons, the electrons are not always shared equally.
Fluorine pulls harder on the electrons it shares with
hydrogen than hydrogen does.
Therefore, the fluorine end of the molecule has more
electron density than the hydrogen end.
•
•
•
Sometimes the electrons are shared equally.
In the F2 molecule both of the fluorine’s are sharing the
electrons between them equally.
Therefore, both of the fluorine ends of the molecule
have equal electron density around them.
Electronegativity is the ability of atoms in a molecule to attract electrons to itself.
Reminder from the last chapter, The electronegativity increases as you go from the left to the right
across a row(period) and from the bottom to the top of a column (group).
Polar Covalent Bonds
•
•
•
When two atoms share electrons unequally, a bond dipole results.
The dipole moment, µ, produced by two equal but opposite charges separated by a distance, r, is calculated: µ = Qr
It is measured in debyes (D)
The greater the difference in
electronegativity, the more
polar is the bond.
Polarity Worksheet For each of the following pairs of molecules, determine which is most polar and
explain your reason for making this choice:
1)
carbon disulfide
OR
sulfur difluoride
2)
nitrogen trichloride
OR
oxygen dichloride
3)
boron trihydride
OR
ammonia
4)
chlorine
OR
phosphorus trichloride
5)
silicon dioxide
OR
carbon dioxide
6)
methane
OR
CH2Cl2
7)
silicon tetrabromide
OR
HCN
8)
nitrogen trifluoride
OR
phosphorus trifluoride
Ranking Molecules by Increasing Polarity
In each of the following problems, rank the molecules from lowest to highest polarity:
1)
PF3, LiOH, SF2, NF3
2 2)
Ni(OH)3, N2H2, CH3OH, C2H5OH
3)
B2F4, H2C2O4, CuCl2, CF2O
4)
PH3, PF3, NH3, NF3
5)
H2O, H2S, HF, H2
Lewis Structures
Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.
Writing Lewis Structures
1. Find the sum of valence electrons of all atoms in the
polyatomic ion or molecule.
Ø If it is an anion, add one electron for each negative
charge.
Ø If it is a cation, subtract one electron for each
positive charge.
2. The central atom is the least electronegative element that
isn’t hydrogen. Connect the outer atoms to it by single
bonds.
3. Fill the octets of the outer atoms.
4. Fill the octet of the central atom.
5. If you run out of electrons before the central atom has an
octet…
…form multiple bonds until it does.
• Then assign formal charges.
Ø For each atom, count the electrons in lone pairs and
half the electrons it shares with other atoms.
Ø Subtract that from the number of valence electrons
for that atom: The difference is its formal charge.
• The best Lewis structure…
Ø …is the one with the fewest charges.
Ø …puts a negative charge on the most
electronegative atom.
Lewis Structures Practice
Draw the Lewis structures for the following compounds: 1) PBr3 2) N2H2 3 3) 5) CH3OH 4) NO2-­‐1 C2H4 6)
BSF 7)
HBr
8)
C2H5OH (ethanol)
9)
N2F4
10)
SF6
(Do after you read exceptions to octet rule)
Exceptions to the Octet Rule
Odd Number of Electrons
Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd
number of electrons.
Fewer Than Eight Electrons
Consider BF3:
Ø
Ø
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 BF3.
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.
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.
More Than Eight Electrons
• The only way PCl5 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.
Ø Presumably d orbitals in these atoms participate in bonding.
4 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.
•
•
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.
Resonance
There are three types of ions or molecules that do not follow the octet rule:
Ø Ions or molecules with an odd number of electrons.
Ø Ions or molecules with less than an octet.
Ø Ions or molecules with more than eight valence electrons (an expanded octet).
This is the Lewis structure we would draw
for ozone, O3.
But this is at odds with the true, observed
structure of ozone, in which…
Ø …both O—O bonds are the same
length.
Ø …both outer oxygens have a charge
of −1/2.
• One Lewis structure cannot accurately
depict a molecule such as ozone.
• We use multiple structures, resonance
structures, to describe the molecule.
Just as green is a synthesis of blue and yellow…
•
…ozone is a synthesis of these two
resonance structures.
•
•
•
•
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.
The organic compound benzene, C6H6, has
two resonance structures.
It is commonly depicted as a hexagon with a
circle inside to signify the delocalized
electrons in the ring.
or
5 Resonance Structures Practice
Draw all of the possible resonance structures for the following ions or molecules:
1) nitrate ion:
2) formate ion (CHO2-1):
3) cyclobutadiene (C H ):
4
4
4) ozone (O ):
3
Covalent Bond Strength
Most simply, the strength of a bond is measured by determining how much energy is required to break
the bond.
• This is the bond enthalpy.
• The bond enthalpy for a Cl—Cl bond, Δ(Cl—Cl), is measured to be 242 kJ/mol.
Average Bond Enthalpies
•
This table lists the average bond enthalpies for
many different types of bonds.
• Average bond enthalpies are positive, because bond
breaking is an endothermic process.
NOTE: These are average bond enthalpies, not absolute
bond enthalpies; the C—H bonds in methane, CH4, will be
a bit different than the
C—H bond in chloroform, CHCl3.
NOTE: These are average bond enthalpies, not absolute
bond enthalpies; the C—H bonds in methane, CH4, will be
a bit different than the
C—H bond in chloroform, CHCl3.
•
Yet another way to estimate ΔH for a reaction is to compare the bond enthalpies of bonds
broken to the bond enthalpies of the new bonds formed.
• In other words,
ΔHrxn = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed)
•
CH4(g) + Cl2(g) ⎯⎯→ CH3Cl(g) + HCl(g)
In this example, one
C—H bond and one Cl—Cl bond are broken; one C—Cl and one H—Cl bond are formed.
So,
ΔHrxn = [Δ(C—H) + Δ(Cl—Cl) − [Δ(C—Cl) + Δ(H—Cl)
= [(413 kJ) + (242 kJ)] − [(328 kJ) + (431 kJ)]
= (655 kJ) − (759 kJ)
= −104 kJ
6 Practice: Estimate the change in enthalpy, ΔH, for the following reaction:
H2 (g) + Cl2 (g) → 2 HCl (g)
How much energy is absorbed or released (identify which) when each of the following reactions takes place?
Use stoichiometric amounts in each case (e.g. 1.0 mol of CH4 and 2.0 mol of O2 for problem #1)
The relevant bond energies are: C—H 414 kJ/mol O==O
502 kJ/mol C==O 730 kJ/mol O—H
464
kJ/mol H—H
435 kJ/mol Cl—Cl
243 kJ/mol H—Cl
431 kJ/mol C—Cl
331 kJ/mol
1.
CH4 (g) + 2O2 (g)àCO2 (g) + 2H2O(g)
2.
2H2(g)+O2(g)à2H2O(g)
3.
H2(g)+Cl2(g)à2HCl(g)
4.
CH4 (g) + Cl2 (g)àCH3Cl (g) + HCl (g)
Bond Enthalpy and Bond Length
•
•
We can also measure an average bond length for
different bond types.
As the number of bonds between two atoms
increases, the bond length decreases.
7 Start of Chapter 9
Molecular Shapes
• The shape of a molecule plays an important role in its reactivity.
• By noting the number of bonding and nonbonding electron pairs we can easily predict the shape of the
molecule.
Simply put, electron pairs, whether they be bonding or nonbonding, repel each other.
By assuming the electron pairs are placed as far as possible from each other, we can predict the shape of
the molecule.
Electron domains
• We can refer to the electron pairs as electron domains.
• In a double or triple bond, all electrons shared between those two atoms are on the same side of the
central atom; therefore, they count as one electron domain.
•
•
This molecule has four electron domains.
(VSEPR)
Valence Shell Electron Pair
Repulsion Theory
“The best arrangement of a
given number of electron
domains is the one that
minimizes the repulsions
among them.”
Electron-Domain Geometries
These are the electrondomain geometries for two through
six electron domains around a
central atom.
•
•
All one must do is count
the number of electron
domains in the Lewis
structure.
The geometry will be that
which corresponds to that
number of electron
domains.
8 •
•
The electron-domain
geometry is often not the
shape of the molecule,
however.
The molecular geometry is
that defined by the
positions of only the atoms
in the molecules, not the
nonbonding pairs.
Molecular Geometries
Within each electron domain, then,
there might be more than one molecular
geometry.
•
•
•
In this domain, there is only one
molecular geometry: linear.
NOTE: If there are only two
atoms in the molecule, the
molecule will be linear no
matter what the electron domain
is.
There are two molecular
geometries:
Ø
Ø
Trigonal planar, if all the
electron domains are bonding
Bent, if one of the domains is
a nonbonding pair.
Tetrahedral Electron Domain
• There are three molecular
geometries:
Ø
Ø
Ø
Tetrahedral, if all are bonding
pairs
Trigonal pyramidal if one is a
nonbonding pair
Bent if there are two
nonbonding pairs
Trigonal Bipyramidal Electron Domain
• There are two distinct positions
in this geometry:
Ø
Ø
Axial
Equatorial
Lower-energy conformations result
from having nonbonding electron pairs
in equatorial, rather than axial,
positions in this geometry.
9 •
There are four distinct
molecular geometries in this
domain:
Ø
Ø
Ø
Ø
Trigonal bipyramidal
Seesaw
T-shaped
Linear
Octahedral Electron Domain
• All positions are equivalent in
the octahedral domain.
• There are three molecular
geometries:
Ø
Ø
Octahedral
Square pyramidal
Square planar
Ø
Larger Molecules
In larger molecules, it makes
more sense to talk about the geometry
about a particular atom rather than the
geometry of the molecule as a whole.
This approach makes sense, especially
because larger molecules tend to react
at a particular site in the molecule.
Nonbonding Pairs and Bond Angle
Nonbonding pairs are physically
larger than bonding pairs.
• Therefore, their repulsions are
greater; this tends to decrease
bond angles in a molecule.
Multiple Bonds and Bond Angles
• Double and triple bonds place
greater electron density on one
side of the central atom
therefore, they also affect bond
angles.
Book Problems 389 #12, 13, 15, 16,
These problems will take time. Work on in class and finish for
21, 22, 24, 25, &27
homework!
•
10 Polarity
• In Chapter 8 we discussed bond
dipoles.
• But just because a molecule possesses
polar bonds does not mean the
molecule as a whole will be polar.
By adding the individual bond dipoles, one
can determine the overall dipole moment for
the molecule.
Overlap and Bonding
• We think of covalent bonds forming
through the sharing of electrons by
adjacent atoms.
• In such an approach this can only
occur when orbitals on the two atoms
overlap.
Book Problems page 390 #32, 34, 35,& 38
Hybrid Orbitals
But it’s hard to imagine tetrahedral, trigonal
bipyramidal, and other geometries arising
from the atomic orbitals we recognize.
• Consider beryllium:
In its ground electronic state, it would not be
able to form bonds because it has no singlyoccupied orbitals.
But if it absorbs the small amount of energy
needed to promote an electron from the 2s to
the 2p orbital, it can form two bonds.
• Mixing the s and p orbitals yields two
degenerate orbitals that are hybrids of
the two orbitals.
Ø These sp hybrid orbitals have
two lobes like a p orbital.
Ø One of the lobes is larger and
more rounded as is the s
orbital.
• These two degenerate orbitals would
align themselves 180° from each other.
• This is consistent with the observed
geometry of beryllium compounds:
linear.
• With hybrid orbitals the orbital
diagram for beryllium would look like
this.
• The sp orbitals are higher in energy
than the 1s orbital but lower than the
2p.
Using a similar model for boron leads to…
11 …three degenerate sp2 orbitals.
With carbon we get four sp3 orbitals
For geometries involving expanded octets on
the central atom, we must use d orbitals in our
hybrids. This leads to five degenerate sp3d
orbitals or six degenerate sp3d2 orbitals.
Book problems page 391 #42, 44, 45, 47
•Hybridization is a major player in this approach to bonding.
•There are two ways orbitals can overlap to form bonds between atoms. They are either sigma(σ) or pi (π)
bonds
•Sigma bonds are characterized by
•
•
Head-to-head overlap.
Cylindrical symmetry of electron density about the internuclear (bond takes place between the nuclei) axis.
•Pi bonds are characterized by
•
•
Side-to-side overlap.
Electron density above and below the internuclear axis.
Single bonds are always σ bonds, because σ overlap is greater, resulting in a stronger bond and more energy
lowering.
In a multiple bond one of the bonds is a σ bond and the rest are π bonds.
Book problems pages 391 #49,51,55
We will not cover molecular orbital theory in chapter 9
12 Polarity Worksheet Answers
For each of the following pairs of molecules, determine which is most polar and explain your reason
for making this choice:
1)
carbon disulfide
OR
carbon disulfide is nonpolar
2)
nitrogen trichloride
OR
oxygen dichloride
both are polar, but oxygen dichloride is less symmetric than nitrogen trichloride,
making it more polar.
3)
boron trihydride
OR
boron trihydride is nonpolar.
ammonia
4)
chlorine
chlorine is nonpolar
phosphorus trichloride
5)
silicon dioxide
OR
carbon dioxide
It’s a tie, because both are nonpolar
6)
methane
methane is nonpolar
7)
silicon tetrabromide
OR
silicon tetrabromide is nonpolar
8)
nitrogen trifluoride
OR
phosphorus trifluoride
Both are polar and equally symmetric, but the difference in electronegativity between NF is less than that between P-F
OR
OR
sulfur difluoride
CH2Cl2
HCN
Ranking Molecules by Increasing Polarity Solutions
In each of the following problems, rank the molecules from lowest to highest polarity:
1)
PF3, LiOH, SF2, NF3
NF3 < PF3 < SF2 < LiOH
2)
Ni(OH)3, N2H2, CH3OH, C2H5OH
N2H2 < C2H5OH < CH3OH < Ni(OH)3
3)
B2F4, H2C2O4, CuCl2, CF2O
B2F4 < H2C2O4 < CF2O < CuCl2
4)
PH3, PF3, NH3, NF3
PH3 < NH3 < NF3 < PH3
5)
H2O, H2S, HF, H2
H2 < H2 S < H2 O <
Lewis Structures Practice Worksheet - Solutions
13 Please forgive my very poor drawing skills! 1) PBr3 2) N2H2 3) CH3OH 4) NO2-­‐1 6)
BSF
7)
HBr
8)
C2H5OH (ethanol)
9)
N2F4
10)
SF6
14 11) C2H4 Resonance Structures Solutions
Draw all of the possible resonance structures for the following ions or molecules:
Solution To work this problem, think of the reaction in terms of simple steps: Step 1 The reactant molecules, H2 and Cl2, break down into their atoms H2(g) → 2 H(g) Cl2(g) → 2 Cl(g) Step 2 These atoms combine to form HCl molecules 2 H (g) + 2 Cl (g) → 2 HCl (g) In the first step, the H-­‐H and Cl-­‐Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-­‐H and Cl-­‐Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction: ΔH1 = +(436 kJ + 243 kJ) = +679 kJ
The answers are:
1. 2. 3. 4.
656 kJ released 484 kJ released 184 kJ released 105 kJ released
15