Lecture 28 (Slides)REVISED October 25

Report
Lewis Structures –Bond Pairs and
Lone Pairs
• In Lewis structures with five or six pairs of
electrons around the central atom we need to
distinguish between lone pairs and bonding
pairs. Lone pairs on average are located closer
to the central atom than bonding pairs. Why?
Coulombs law would suggest that the
magnitude of electron pair repulsions will vary
and be strongest when two lone pairs are
involved.
Order of Lone Pair Repulsions
• The magnitude of electron repulsions follows
the order
• Lone pair : lone pair ˃ lone pair : bond pair ˃
bond pair : bond pair
• We attempt (as does nature!) to place the lone
pairs so as to minimize repulsive interactions.
We maximize the angles between lone pairs
and other pairs where possible.
Class Example
• 3. Draw Lewis structures for SF4, IF5 and ClF3.
Determine the shape of each of these
molecules and which, if any, of these
molecules are polar.
Single and Multiple Bonds – Strengths
and Lengths
• We have drawn Lewis structures with single,
double and triple bonds. The Lewis structures
do reflect physical reality (at least when
properly drawn!). Multiple bonds between a
given pair of atoms (e.g. C and O) are shorter
than single bonds. Multiple bonds between a
given pair of atoms require more energy to
break than do single bonds.
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 10
Slide 5 of 48
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 10
Slide 6 of 48
Bond Energies and Enthalpies of
Reaction – ΔH’s
• Bond energies can be combined to calculate
very crude values for heats of reaction. Care
must be taken to use bond energies for single
bonds, double bonds or triple bonds as
appropriate. Often one must draw Lewis
structures to decide what types of bonds to
consider. ΔH values calculated in this way are
very approximate since bond energies for a
particular atom pair vary between molecules.
Bond Energies
FIGURE 10-16
•Some bond energies compared
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 10
Slide 8 of 48
Calculating an Enthalpy of Reaction from Bond Energies.
ΔHrxn =  ΔH(bond breakage) + ΔH(bond formation)
≈  BE(reactants) -  BE(products)
Does this equation look
familiar?
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 10
Slide 9 of 48
Copyright  2011 Pearson
Canada Inc.
10 - 10
Class Example
• 4. Use bond energy values from Petrucci in
order to calculate the enthalpy change for the
reaction N2(g) + 3H2(g) → 2 NH3(g). Compare
this crude ΔH value to that calculated from
tabulated heat of formation data.
• Key ideas: Bond breaking is an endothermic
process. Bond formation is an exothermic
process.
Class Example
• 5. Draw a Lewis structure or OPCl3 using (a)
four single bonds and (b) one double bond
between P and O. Assign formal charges to all
atoms in both structures. Which structure is
more reasonable? Determine the shape of the
molecule. Is this molecule polar?
Chemical Bonds – Molecular Orbitals
• The electronic structures of atoms and
molecules have many features in common.
Individual atoms usually possess unpaired
electrons. These atoms are often chemically
unstable. Two such atoms (or more!) can
combine to form molecules with (usually) no
unpaired electrons. The process involves the
formation of chemical bonds and is highly
exothermic.
Chemical Bonds – Energetics
(Homonuclear Diatomics)
• H(g) + H(g) → H:H(g)
ΔH = -436 kJ
• N(g) + N(g) → :N:::N:(g)
ΔH = -946 kJ
• Note on signs: The above reactions are highly
exothermic. Energy is released when the atoms
combine to form molecules. The bond energy tells
us how much energy is required to break a mole
of bonds (usually!). The bond energy of the bond
in N2 is +946 kJ mol-1.
Atomic and Molecular Orbitals
• In isolated atoms most electrons are found in
pairs in a number of different atomic orbitals.
When atoms combine valence shell electrons
are rearranged. Two electrons from different
atoms can, for example, “pair up” to form a
single covalent bond where the bonding
molecular orbital is associated with more than
one atom (and often more than two atoms!).
Molecular Orbitals – Wave Properties
of Electrons
• We will need to consider the wave properties
of electrons when discussing molecular
orbitals. By analogy to constructive and
destructive interference in conventional waves
we will see atomic orbitals combining
constructively to form bonding molecular
orbitals and combining destructively to form
anti-bonding molecular orbitals.
Bonding Theory Objectives
• A detailed bonding theory should account
quantitatively for experimentally observed
molecular shapes, bond distances, electrical
polarity, bond strength and so on. It should
also have predictive capability. A quantitative
treatment is reserved for higher level courses.
In a qualitative manner we begin by reminding
ourselves that coulombic interactions (between
particles with wave character) are key.
Simplest example – the H2 molecule.
What a Bonding Theory Should Do
FIGURE 11-1
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 11
Slide 18 of 57
The H2 Molecule – Coulombic
Interactions
• In physics you may already have calculated the
size of coulombic forces and associated
potential energies. For static point charges this
is a simple exercise. For a group of charged
particles in rapid motion (and wavelike!) life is
more complex. Nevertheless the charged
particles are “capable” of finding the lowest
energy configuration. Why?
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 11
Slide 19 of 57
FIGURE 11-2
Energy
of interaction of two hydrogen atoms plotted for internuclear
separations from zero to infinity
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 11
Slide 20 of 57
Valence Bond Theory
• Provides a simple picture of covalent bond
formation from overlapping atomic orbitals
(each containing one electron). One assumes
that both core electrons and lone pair valence
shell electrons keep the orbital locations that
they had in the separated atoms. The valence
bond method does account for some common
valences. We’ll consider O, S and N atoms
reacting with H atoms.
Reaction of O, S and N with H Atoms
• The complete electron configurations for O, N
and H are:
• Oxygen 1s22s22px22py12pz1
• Nitrogen 1s22s22px12py12pz1
• Hydrogen 1s1
• We can “pair up” all electrons if O combines
with two H atoms and N combines with three
H atoms to form H2O and NH3 respectively.
Reaction of O, S and N with H Atoms
• The complete electron configurations for S could
be written as
• Sulfur 1s22s22p63px23py13pz1
• Again, we can “pair up” all electrons if S and two
H atoms combine to form H2S. The valence bond
picture suggests that all bond angles in H2O, NH3
and H2S should be 90o. This is close to the value
seen in H2S (92o) but significantly underestimates
bond angles in H2O (105o) and NH3 (107o).
FIGURE 11-3
Bonding in H2S represented by atomic orbital overlap
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 11
Slide 24 of 57
The Methane “Problem”
• The ground state configurations for C can be
written as 1s22s22px12py1 Using the valence
bond picture and the concept of paired
electrons in molecular orbitals we might
expect C to react with H atoms to form CH2.
The CH2 molecule does form but is unstable (a
transient species). However, carbon “happily”
reacts with H to form the methane, CH4.
Methane and Hybridization
• By experiment, as previously discussed,
methane has a regular tetrahedral geometry –
four equal bond distances and all bond angles
of 109.5o. The regular geometry of methane
and its ability to form four bonds can be
explained using the concept of hybridization.
How have we explained carbon’s tendency to
form four bonds previously?
Methane and Hybridization – cont’d:
• In the hybridization picture we imagine
methane being formed from C and H atoms in
three steps. In the first step we take a ground
state C atom and excite one electron (from the
2s orbital) to form the lowest lying ( or first)
excited state.
• Carbon Ground State: 1s22s22px12py1
• Carbon Excited State: 1s22s12px12py12pz1
Methane and Hybridization – cont’d:
• In the second step we imagine “combining”
the single occupied 2s orbital and the three
occupied 3p orbitals in the excited to form four
equivalent sp3 hybrid orbitals (each containing
a single unpaired electron). In step three the
“hybridized C atom” reacts with four H atoms
to form a CH4 molecule. The process is
represented on the next few slides.
Hybridization of Atomic Orbitals
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 11
Slide 29 of 57
FIGURE 11-6
The sp3 hybridization scheme
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 11
Slide 30 of 57
FIGURE 11-7
Bonding and structure of CH4
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 11
Slide 31 of 57
Organic Compounds and Structures:
An Overview
FIGURE 26-1
•Representations of the methane molecule
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 26
Slide 32 of 75
Organic Chemistry
• The next two slides illustrate what starts to
happen when two or more sp3 hybridized
carbons are linked. The chemistry of carbon is
infinitely varied and organic compounds are
part of all of living things, important energy
sources, key pharmaceuticals and so on.
FIGURE 26-2
The ethane molecule C2H6
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 26
Slide 34 of 75
FIGURE 26-3
The propane molecule, C3H8
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 26
Slide 35 of 75

similar documents