Chapter 8 Bonding and Molecular Structure: Fundamental Concepts

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Chapter 8Bonding and Molecular
StructureFundamental Concepts
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CHEMICAL BONDING

• Cocaine

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Chemical Bonding

• Problems and questions
• How is a molecule or polyatomic ion held
  together?
• Why are atoms distributed at strange angles?
• Why are molecules not flat?
• Can we predict the structure?
• How is structure related to chemical and physical
  properties?

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Structure Bonding
NN triple bond. Molecule is unreactive
Phosphorus is a tetrahedron of P atoms. Very
reactive!
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Red phosphorus, a polymer. Used in matches.
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Forms of Chemical Bonds

• There are 2 extreme forms of connecting or
  bonding atoms
• Ioniccomplete transfer of 1 or more electrons
  from one atom to another
• Covalentsome valence electrons shared between
  atoms
• Most bonds are somewhere in between.

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Ionic Compounds
Metal of low IE
Nonmetal of high EA
2 Na(s) Cl2(g) f 2 Na 2 Cl-
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Covalent Bonding

• The bond arises from the mutual attraction of 2
  nuclei for the same electrons. Electron sharing
  results.

Bond is a balance of attractive and repulsive
forces.
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Bond Formation

• A bond can result from a head-to-head overlap
  of atomic orbitals on neighboring atoms.

Overlap of H (1s) and Cl (2p)
Note that each atom has a single, unpaired
electron.
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Chemical Bonding Objectives

• Objectives are to understand
• 1. valence e- distribution in molecules and
  ions.
• 2. molecular structures
• 3. bond properties and their effect on
  molecular properties.

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Electron Distribution in Molecules

• Electron distribution is depicted with Lewis
  electron dot structures
• Valence electrons are distributed as shared or
  BOND PAIRS and unshared or LONE PAIRS.

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Bond and Lone Pairs

• Valence electrons are distributed as shared or
  BOND PAIRS and unshared or LONE PAIRS.

This is called a LEWIS ELECTRON DOT structure.
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Valence Electrons

• Electrons are divided between core and valence
  electrons
• B 1s2 2s2 2p1
• Core He , valence 2s2 2p1

Br Ar 3d10 4s2 4p5 Core Ar 3d10 ,
valence 4s2 4p5
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Rules of the Game

• No. of valence electrons of a main group atom
  Group number

For Groups 1A-4A, no. of bond pairs group
number.
For Groups 5A -7A, BPs 8 - Grp. No.
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Rules of the Game

• No. of valence electrons of an atom Group
  number
• For Groups 1A-4A, no. of bond pairs group
  number
• For Groups 5A -7A, BPs 8 - Grp. No.

Except for H (and sometimes atoms of 3rd and
higher periods), BPs LPs 4
This observation is called the OCTET RULE
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Building a Dot Structure

• Ammonia, NH3
• 1. Decide on the central atom never H.
• Central atom is atom of lowest affinity for
  electrons.
• Therefore, N is central
• 2. Count valence electrons
• H 1 and N 5
• Total (3 x 1) 5
• 8 electrons / 4 pairs

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Building a Dot Structure

• 3. Form a single bond between the central atom
  and each surrounding atom

4. Remaining electrons form LONE PAIRS to
complete octet as needed.
3 BOND PAIRS and 1 LONE PAIR.
Note that N has a share in 4 pairs (8 electrons),
while H shares 1 pair.
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Sulfite ion, SO32-

• Step 1. Central atom S
• Step 2. Count valence electrons S 6
• 3 x O 3 x 6 18
• Negative charge 2
• TOTAL 26 e- or 13 pairs
• Step 3. Form bonds

10 pairs of electrons are now left.
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Sulfite ion, SO32-

• Remaining pairs become lone pairs, first on
  outside atoms and then on central atom.

Each atom is surrounded by an octet of electrons.
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Carbon Dioxide, CO2

• 1. Central atom _______
• 2. Valence electrons __ or __ pairs
• 3. Form bonds.

This leaves 6 pairs.
4. Place lone pairs on outer atoms.
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Carbon Dioxide, CO2

• 4. Place lone pairs on outer atoms.

5. So that C has an octet, we shall form DOUBLE
BONDS between C and O.
The second bonding pair forms a pi (p) bond.
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Double and even triple bonds are commonly
observed for C, N, P, O, and S
H2CO
SO3
C2F4
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Sulfur Dioxide, SO2

• 1. Central atom S
• 2. Valence electrons 18 or 9 pairs

3. Form double bond so that S has an octet
but note that there are two ways of doing this.
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Sulfur Dioxide, SO2

• This leads to the following structures.

These equivalent structures are called RESONANCE
STRUCTURES. The true electronic structure is a
HYBRID of the two.
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Urea, (NH2)2CO
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Urea, (NH2)2CO

• 1. Number of valence electrons 24 e-
• 2. Draw sigma bonds.

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Urea, (NH2)2CO

• 3. Place remaining electron pairs in the
  molecule.

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Urea, (NH2)2CO

• 4. Complete C atom octet with double bond.

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Atom Formal Charges

• Atoms in molecules often bear a charge ( or -).
• The predominant resonance structure of a molecule
  is the one with charges as close to 0 as
  possible.
• Formal charge Group number 1/2 (no. of
  bonding electrons) - (no. of LP electrons)

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Carbon Dioxide, CO2
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Calculated Partial Charges in CO2
Yellow negative red positive Relative size
relative charge
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Thiocyanate Ion, SCN-
6 - (1/2)(2) - 6 -1
5 - (1/2)(6) - 2 0
4 - (1/2)(8) - 0 0
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Thiocyanate Ion, SCN-
Which is the most important resonance form?
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Calculated Partial Charges in SCN-
All atoms negative, but most on the S
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Violations of the Octet Rule

• Usually occurs with B and elements of higher
  periods.

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Boron Trifluoride

• Central atom _____________
• Valence electrons __________ or electron pairs
  __________
• Assemble dot structure

The B atom has a share in only 6 pairs of
electrons (or 3 pairs). B atom in many molecules
is electron deficient.
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Boron Trifluoride, BF3
1
-1
What if we form a BF double bond to satisfy the
B atom octet?
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Is There a BF Double Bond in BF3
F is negative and B is positive
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Sulfur Tetrafluoride, SF4

• Central atom
• Valence electrons ___ or ___ pairs.
• Form sigma bonds and distribute electron pairs.

5 pairs around the S atom. A common occurrence
outside the 2nd period.
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MOLECULAR GEOMETRY
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MOLECULAR GEOMETRY
Molecule adopts the shape that minimizes the
electron pair repulsions.

• VSEPR
• Valence Shell Electron Pair Repulsion theory.
• Most important factor in determining geometry is
  relative repulsion between electron pairs.

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Electron Pair GeometriesSee Active Figure 8.5
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Electron Pair GeometriesSee Active Figure 8.5
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Structure Determination by VSEPR

• Ammonia, NH3
• 1. Draw electron dot structure
• 2. Count BPs and LPs 4

3. The 4 electron pairs are at the corners of a
tetrahedron.
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Structure Determination by VSEPR

• Ammonia, NH3
• There are 4 electron pairs at the corners of a
  tetrahedron.

The ELECTRON PAIR GEOMETRY is tetrahedral.
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Structure Determination by VSEPR

• Ammonia, NH3
• The electron pair geometry is tetrahedral.

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The MOLECULAR GEOMETRY the positions of the
atoms is PYRAMIDAL.
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Structure Determination by VSEPR

• Water, H2O
• 1. Draw electron dot structure

2. Count BPs and LPs 4
3. The 4 electron pairs are at the corners of a
tetrahedron.
The electron pair geometry is TETRAHEDRAL.
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Structure Determination by VSEPR

• Water, H2O

The electron pair geometry is TETRAHEDRAL
The molecular geometry is BENT.
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Geometries for Four Electron PairsSee Figure 8.6
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Structure Determination by VSEPR

• Formaldehyde, CH2O
• 1. Draw electron dot structure

2. Count BPs and LPs at C
3. There are 3 electron lumps around C at the
corners of a planar triangle.
The electron pair geometry is PLANAR TRIGONAL
with 120o bond angles.
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Structure Determination by VSEPR

• Formaldehyde, CH2O

The electron pair geometry is PLANAR TRIGONAL
The molecular geometry is also planar trigonal.
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Structure Determination by VSEPR
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Methanol, CH3OH
Define H-C-H and C-O-H bond angles
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• H-C-H 109o
• C-O-H 109o
• In both cases the atom is surrounded by 4
  electron pairs.

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Structure Determination by VSEPR

• Acetonitrile, CH3CN

Define unique bond angles
H-C-H 109o C-C-N 180o
180
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One C is surrounded by 4 electron lumps and the
other by 2 lumps
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Phenylalanine, an amino acid
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Phenylalanine
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Structures with Central Atoms with More Than or
Less Than 4 Electron Pairs
Often occurs with Group 3A elements and with
those of 3rd period and higher.
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Boron Compounds

• Consider boron trifluoride, BF3

The B atom is surrounded by only 3 electron pairs.
Bond angles are 120o
Geometry described as planar trigonal
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Compounds with 5 or 6 Pairs Around the Central
Atom
5 electron pairs
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Molecular Geometries for Five Electron PairsSee
Figure 8.8
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Sulfur Tetrafluoride, SF4

• Number of valence electrons 34
• Central atom S
• Dot structure

Electron pair geometry is trigonal bipyramid
(because there are 5 pairs around the S)
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Sulfur Tetrafluoride, SF4

• Lone pair is in the equator because it requires
  more room.

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Molecular Geometries for Six Electron PairsSee
Figure 8.8
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Compounds with 5 or 6 Pairs Around the Central
Atom
6 electron pairs
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Bond Properties

• What is the effect of bonding and structure on
  molecular properties?

Free rotation around CC single bond
No rotation around CC double bond
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Bond Order of bonds between a pair of atoms
Acrylonitrile
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Bond Order

• Fractional bond orders occur in molecules with
  resonance structures.
• Consider NO2-

The NO bond order 1.5
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Bond Order

• Bond order is proportional to two important bond
  properties
• (a) bond strength
• (b) bond length

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Bond Length

• Bond length is the distance between the nuclei of
  two bonded atoms.

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Bond Length

• Bond length depends on size of bonded atoms.

Bond distances measured in Angstrom units where 1
A 10-2 pm.
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Bond Length

• Bond length depends on bond order.

Bond distances measured in Angstrom units where 1
A 10-2 pm.
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Bond Strength

• measured by the energy reqd to break a bond.
  See Table 8.9

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Bond Strength

• measured by the energy reqd to break a bond.
  See Table 8.9.
• BOND Bond dissociation enthalpy (kJ/mol)
• HH 436
• CC 346
• CC 602
• C?C 835
• N?N 945

The GREATER the number of bonds (bond order) the
HIGHER the bond strength and the SHORTER the bond.
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Bond Strength

• Bond Order Length Strength
• HOOH
• OO

142 pm
210 kJ/mol
1
2
121
498
1.5
128
?
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Using Bond Dissociation Enthalpies

• Estimate the energy of the reaction
• HH(g) ClCl(g) f 2 HCl(g)
• Net energy ?rH
• energy required to break bonds
• - energy evolved when bonds are made

HH 436 kJ/mol ClCl 242 kJ/mol HCl
432 kJ/mol
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Using Bond Dissociation Enthalpies

• Estimate the energy of the reaction
• HH ClCl ----gt 2 HCl

HH 436 kJ/mol ClCl 242 kJ/mol HCl
432 kJ/mol
Sum of H-H Cl-Cl bond energies 436 kJ 242
kJ 678 kJ
2 mol H-Cl bond energies 864 kJ
Net ?rH 678 kJ - 864 kJ -186 kJ
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Using Bond Dissociation Enthalpies

• Estimate the energy of the reaction
• 2 HOOH f OO 2 HOH
• Is the reaction exo- or endothermic?
• Which is larger
• A) energy reqd to break bonds
• B) or energy evolved on making bonds?

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Using Bond Dissociation Enthalpies

• 2 HOOH f OO 2 HOH
• Energy required to break bonds
• break 4 mol of OH bonds 4 (463 kJ)
• break 2 mol OO bonds 2 (146 kJ)

TOTAL ENERGY to break bonds 2144 kJ
TOTAL ENERGY evolved on making OO bonds and 4
O-H bonds bonds 2350 kJ
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Using Bond Dissociation Enthalpies

• 2 HOOH f OO 2 HOH

Net energy 2144 kJ - 2350 kJ - 206 kJ
The reaction is exothermic!
More energy is evolved on making bonds than is
expended in breaking bonds.
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Molecular Polarity
Why do water and methane differ so much in their
boiling points?

• Why do ionic compounds dissolve in water?

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Bond Polarity

• HCl is POLAR because it has a positive end and a
  negative end.

Cl has a greater share in bonding electrons than
does H.
Cl has slight negative charge (-d) and H has
slight positive charge ( d)
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Bond Polarity

• Three molecules with polar, covalent bonds.
• Each bond has one atom with a slight negative
  charge (-d) and and another with a slight
  positive charge ( d)

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Bond Polarity

• This model, calcd using CAChe software for
  molecular calculations, shows that H is (red)
  and Cl is - (yellow). Calcd charge is or -
  0.20.

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Bond Polarity

• Due to the bond polarity, the HCl bond energy is
  GREATER than expected for a pure covalent bond.

BOND ENERGY pure bond 339 kJ/mol calcd real
bond 432 kJ/mol measured
Difference 92 kJ. This difference is
proportional to the difference in
ELECTRONEGATIVITY, ?.
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Electronegativity, ?

• ? is a measure of the ability of an atom in a
  molecule to attract electrons to itself.

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Linus Pauling, 1901-1994
PLAY MOVIE

• The only person to receive two unshared Nobel
  prizes (for Peace and Chemistry).
• Chemistry areas bonding, electronegativity,
  protein structure

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ElectronegativitySee Figure 8.11
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Electronegativity, ?
See Figure 8.11

• F has maximum ?.
• Atom with lowest ? is the center atom in most
  molecules.
• Relative values of ? determine BOND POLARITY
  (and point of attack on a molecule).

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Bond Polarity

• Which bond is more polar (or DIPOLAR)?
• OH OF
• ? 3.5 - 2.1 3.5 - 4.0
• ? 1.4 0.5
• OH is more polar than OF

and polarity is reversed.
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Molecular Polarity

• Moleculessuch as HI and H2O can be POLAR (or
  dipolar).

They have a DIPOLE MOMENT. The polar HCl molecule
will turn to align with an electric field.
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Molecular Polarity

• The magnitude of the dipole is given in Debye
  units. Named for Peter Debye (1884 - 1966). Recd
  1936 Nobel prize for work on x-ray diffraction
  and dipole moments.

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Dipole Moments
Why are some molecules polar but others are not?
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Molecular Polarity

• Molecules will be polar if
• a) bonds are polar
• b) AND the molecule is NOT symmetric

All above are NOT polar
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Polar or Nonpolar?

• Compare CO2 and H2O. Which one is polar?

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Carbon Dioxide

• CO2 is NOT polar even though the CO bonds are
  polar.
• CO2 is symmetrical.

Positive C atom is reason CO2 and H2O react to
give H2CO3
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Polar or Nonpolar?

• Consider AB3 molecules BF3, Cl2CO, and NH3.

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Molecular Polarity, BF3
B atom is positive and F atoms are negative.
BF bonds in BF3 are polar.
But molecule is symmetrical and NOT polar
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Molecular Polarity, HBF2
B atom is positive but H F atoms are negative.
BF and BH bonds in HBF2 are polar. But molecule
is NOT symmetrical and is polar.
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Is Methane, CH4, Polar?

• Methane is symmetrical and is NOT polar.

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Is CH3F Polar?
CF bond is very polar. Molecule is not
symmetrical and so is polar.
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CH4 CCl4Polar or Not?

• Only CH4 and CCl4 are NOT polar. These are the
  only two molecules that are symmetrical.

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Substituted Ethylene

• CF bonds are MUCH more polar than CH bonds.
• Because both CF bonds are on same side of
  molecule, molecule is POLAR.

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Substituted Ethylene

• CF bonds are MUCH more polar than CH bonds.
• Because both CF bonds are on opposing ends of
  molecule, molecule is NOT POLAR.

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Visualizing Charges and PolarityElectrostatic
Potential Surfaces

• Electrostatic potential surfaces (EPS) can be
  used to visualize
• Where the charges lie in molecules
• The polarity of molecules

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Visualizing Charges and Polarity
F
H
The boundary surface around the molecule is
made up of all the points in space where the
electron density is a given value (here 0.002
e-/A3). The colors indicate the potential
experienced by a H ion on the surface. More
attraction (a negative site) is red, and
repulsion (a positive site) is blue.
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Visualizing Charges and Polarity
As expected, the surface near O in H2O and the
N is CH3-NH2 is red because that is the more
electronegative atom. H2O is polar - with O
more negative and H more positive. CH3-NH2 is
also polar.
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Visualizing Charges and Polarity
Cl2CO
BF3
The EP surfaces show BF3 is not polar (but the
B is slightly positively charged), whereas Cl2CO
is polar with the O more negative than Cl.
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Visualizing Charges and Polarity
The EP surfaces show cis-C2H2Cl2 (left) is
polar whereas trans-C2H2Cl2 (right) is not polar.