Download Chemical Bonding Ionic bonds result from an electrostatic attraction

Chemical Bonding
Ionic bonds result from an electrostatic attraction
between ions.
Ionic bonds occur between two atoms that have a
large electronegativity difference between them.
(metals and nonmetals)
The larger the Electronegativity difference the more
ionic is the bond.
There is no such thing as a molecule of sodium
chloride(or any other ionic compound).
Ionic compounds exist in crystal lattice structures as
solids. A formula unit of an ionic compound is the
smallest ratio of ions. For sodium chloride that is
NaCl
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Properties of Ionic Compounds
(Explain using the crystal lattice)
i. Solid ionic compounds do not conduct an
electric current.
ii. Molten samples of ionic compounds can
conduct an electric current due to the
mobility of the ions which are free to move
to the electrodes and react.
iii.
Ionic compounds have high melting and
boiling points.
iv.
Many ionic compounds can dissolve in
water.
v. Dissolved ionic compounds separate into
cations and anions in solution.
vi.
The mobile ions move to the electrodes
and react to accept and release electrons
creating a flow of electricity in the outer
circuit.
vii. Ionic compounds that are water soluble
are strong electrolytes.
viii. Ionic compounds are brittle
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Lattice Energy (ΔHlattice)
The energy required to completely separate a
mole of a solid ionic compound into its gaseous ions.
Example: The lattice energy of NaCl is the energy given off when Na+ and Cl- ions in the
gas phase come together to form the lattice of alternating Na+ and Cl- ions in the NaCl
crystal shown in the figure below.
Na+(g) + Cl-(g) NaCl(s)
Ho = -787.3 kJ/mol
Lattice energy = k Q1Q2
r
Q1 and Q2 are the charges of the two ions, r is the
distance between the center of the two ions.
The charge has a larger effect on the lattice energy
compared to the size of the ions.
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Lattice energies (kj/mol)
LiF
1030
LiCl
834
LiI
730
NaF
910
NaCl 788
NaBr 732
KF
808
KCl
701
KBr
671
MgCl2
SrCl2
MgO
CaO
SrO
ScN
2326
2127
3795
3414
3217
7547
Lattice energies are higher for smaller atoms (r) and
higher for larger charges(q)
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Electron configuration of Variable ions in ionic
compounds
The electrons are removed from the s or p orbitals
before the d-orbitals
Give the electron configuration for the following ions
Cu+1
Fe+2
Pb+2
Cu+2
Fe+3
Pb+4
Predicting ionic compounds formulas using
transfer of electrons
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Covalent Bonds
Formed when two atoms share one or more pairs of
electrons.
Lewis Diagrams
Diagrams showing the valence electrons and the
arrangement of atoms in covalent compounds
Constructing Lewis Diagrams
1. Sum up the valence electrons in the molecule.
If it is an ion add or subtract electrons
depending on the charge.
2. Put the structure so that on atom is in the center
and the rest bonded to it. Usually the first
atoms if it is not hydrogen is in the center. Do
not make chains with the atoms.
3. Place two electrons between each two atoms
4. Place the remaining electrons so that all atoms
have complete octets.
5. If there is not enough electrons more the pairs
of electrons around to give double or triple
bonds.
6. If it is an ion put the structure in brackets with
the charge in the right top corner.
Draw the Lewis diagram for CBr4
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Draw the Lewis diagram for H2O
Draw the Lewis diagram for O2
Draw the Lewis diagram for N2
Draw the Lewis diagram for the nitrate ion.
Exceptions to the Octet Rule
1. When the species contains an odd number of
electrons one atom will have only seven
electrons around it.
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2. When the compound contains a central atom
from groups 2 or 13, the number of electrons
around the central atom is twice the group
number. Examples BeI2, BCl3
3. When a central atom has more than four other
atoms bonded to it, it will have more than eight
electrons around it. This is known as an
expanded octet. Examples PCl5, SF6
An expanded octet can occur if the atomic
number of the central atom is greater than 10.
Examples SF4, BrI3, I3 Occurs with elements
with d-orbitals.
Exercise 1 Give the lewis diagram for PCl5.
Exercise 2. Write the Lewis structure for each molecule or ion.
a. ClF3
b. XeO3
c. RnCl2
d. BeCl2
e. ICl4-
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Resonance
Seen in molecules when more than one Lewis
structure can be drawn for the same arrangement of
atoms. The molecule is a hybrid of the different
resonance structures.
Draw the resonance structures for the nitrate ion.
Draw the resonance structures for the azide ion ( N3-).
Formal Charge
Assigned to atoms in Lewis diagrams. Formal charge
is equal the number of valence electrons on an atom –
the number of valence electrons assigned to an atom.
The resonance structure that contributes the most
contains all atoms with formal charges closest to
zero.
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Coordinate Covalent Bonds
Atoms Such as nitrogen and oxygen share a lone pair
of electrons with another atom.
These atoms are coordinate compounds and Lewis
acids/bases. Both electrons in the electron pair
comes from the Lewis base
Coordination Compounds
Lewis-Acid Base Complexes
Formed from a transition metal and ligands
The ligands can be a substance with an anion or a
molecules with a lone pair of electrons.
Water and ammonia act as ligands or any substance
with a lone electron pair
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Although coordination complexes are particularly important in the chemistry of
the transition metals, some main group elements also form complexes.
Aluminum, tin, and lead, for example, form complexes such as the AlF63-, SnCl42and PbI42- ions.
The amount of ligands joined is dependent on the
coordination number for the transition metal.
Ion
Cu+
Ag+
Au+
Coordination
numbers
2,4
2
2,4
Sc3+
Cr3+
6
6
Ion
Mn2+
Fe2+
Cu2+
Ni2+
Al+3
Coordination
number
4, 6
6
4,6
4,6
4, 6
Au3+
4
Examples
AgCl
+
2 NH3
→
[Ag(NH3)2]+
2Ni(CN)2
+ 2 CN- → [Ni(CN)4]
2
Cu(NH3)4 , Cu(H2O)62+.
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Molecular Structure
A molecules three-dimensional structure has a lot to
do with its properties.
Valence Electron Pair Repulsion (VSEPR)
A model to determine the three dimensional structure
of a molecule
In this model interactions of valence electrons on
different atoms is minimized. They are kept as far
apart as possible.
By looking at the central carbon the structure can be
labeled using AB notation and the structure can be
determined
To determine the molecular geometry you must
draw the Lewis diagram. Make sure you do this in
the exam.
A molecule with two atoms off a central atom is
known as AB2
The shape is linear and the angle is 180 oC
Examples are CO2 and BeH2
Electronic geometry: Trigonal planar,
AB3 contains three atoms off a central atom. The
molecular shape is tigonal planar and the angles are
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120 oC
Examples are BF3 and CH2O
AB2E example is SO2, two atoms and a lone pair of
electrons, shape is bent, angle is 120o
Electronic geometry: tetrahedral,
AB4 Contains 4 atoms off the central atom. The
angles are 107.5 o and the shape is tetrahedral.
Examples are CH4 or C2H6
AB2E2 example is H2O, two atoms and two lone pairs
of electrons, shape is bent, angle is 109o
AB3E example is NH3, Three atoms and a lone pair
of electrons, shape is trigonal pyramidal. Angle is
107o
Electronic geometry: Trigonal bipyramidal
AB5 example is PF5, 5 atoms off the central atom.
Angles are 120 o and 90 o and the shape is trigonal
bipyramidal.
AB2E3 example is I3- , Shape is t shaped, angles are
90o
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Electronic geometry octahedral
AB6 example is SF6, 6 atoms off the central atom,
shape is ocatahedral, angles are 90o
AB4E2 example is XeF4, shape is square planar,
angles are 90o
Exercise 3
Describe the molecular structure of the water molecule.
Exercise 4
When phosphorus reacts with excess chlorine gas, the compound
phosphorus pentachloride (PCl5) is formed. In the gaseous and liquid state,
this substance consists of PCl5 molecules, but in the solid state it consists of
a 1:1 misture of PCl4+ an PCl6- ions. Predict the geometric structures of
PCl5, PCl4+, and PCl6-.
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Covalent bonds are described based on the
electronegativity difference between the two bonds.
This is known as polarity
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A nonpolar covalent bond has a difference below
0.4
a polar covalent bond has a difference between 0.4
and 1.0
a very polar covalent bond has a difference between
1 and 2
bonds with difference greater than 2.0 are ionic
Don’t need to memorize the numbers
A polar molecule contains a dipole. A dipole
contains one end having a slight positive charge and
the other end having a slight negative charge due to
differences in electronegativity.
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Dipole moment is a quantitative measure of a dipole.
The greater the dipole moment, the more polar is the
molecule
Molecular polarity
A molecule is polar if its bonds are polar and the
dipoles do not cancel out in its three-dimensional
structure.
HCl, CO, HF are polar because the bond is polar
CO2 is not polar because although the bonds are polar
they cancel out because it is a linear molecule.
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Which molecules are polar?
SF6
PH3
PF3
CCl4
SH2
Draw CH4 , CH3Cl, CH2Cl2, CHCl3, CCl4 Indicate dipole moment(s) where necessary.
Exercise 2 Bond Polarity and Dipole Moment For each of the following molecules, show the direction of the bond polarities and indicate which ones have
a dipole moment:HCL, Cl2, SO3 (a planar molecule with the oxygen atoms spaced evenly around the central
sulfur atom), CH4 [tetrahedral (see
Table 8.2) with the carbon atom at the center], and H2S (V-shaped with the sulfur atom at the point).
Exercise 13 Prediction of Molecular Structure III Because the noble gases have filled s and p valence orbitals, they were not expected to be chemically
reactive. In fact, for manyyears these elements were called inert gases because of this supposed inability to
form any compounds. However, in the early1960s several compounds of krypton, xenon, and radon were
synthesized. For example, a team at the Argonne National
Laboratory produced the stable colorless compound xenon tetrafluoride (XeF4). Predict its structure and
whether it has a dipole
moment.
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Exercise 14 Structures of Molecules with Multiple Bonds
Predict the molecular structure of the sulfur dioxide molecule. Is this molecule expected to have a dipole
moment?
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Hybridization model for molecular bonding
Bonding orbitals are a composite of the orbitals that
the electrons came from
A sigma bond (σ) is formed by the end-to end
overlap of orbitals All single bonds are sigma bonds
A pi (π) bond results
• from side-to-side overlap of orbitals
• A double bond consists of one sigma and one pi
bond.
• A triple bond results from one sigma and two pi
bonds.
• Occurs with sp or sp2 hybridization never sp3
Delocalized electrons are spread out over a number
of atoms in a molecule. These are seen when
multiple double or triple bonds are seen in molecules.
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Sp3 hybridization
All bonding orbitals are have equal energy
This gives four equal energy bonding orbitals which
results in a tetrahedral shape.
Examples
CH4 and NH3
How many sigma bonds are in each?
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Sp2 Hybridization
Gives a trigonal planar shape
Examples are CH2O, CH3CHCH3
How many sigma and pi bonds are in each?
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SP hybridization
Gives a linear shape
Example is C2H2
How many sigma and pi bonds are present?
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dsp3 Hybridization
Uses an empty d-orbitals, a s- orbital and 3 p-orbitals
Giving five equal energy orbitals giving a trigonal
bipyramidal shape
Examples PCl5, I3
How many sigma and pi bonds are present?
d2sp3
Uses two empty d-orbitals, a s orbital and 3 porbitals. This gives six equal energy orbitals giving
an octahedral arrangement.
Examples SF6, XeF4
How many sigma and pi bonds are present?
.
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Hybridization
sp
sp2
sp3
dsp3
# of hybrid
obitals
2
3
4
5
d2sp3
6
Geometry
example
Linear
Trigonal planar
Tetrahedral
Trigonal
bipyramidal
octhedral
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Exercise 1
Describe the bonding in the ammonia molecule using the localized electron
model.
Exercise 2
Describe the bonding in the N2 molecule.
Exercise 3
Describe the bonding in the triiodide ion (I3-)
Exercise 4
How is the xenon atom in XeF4 hybridized?
Exercise 5
For each of the following molecules or ions, predict the hybridization of
each atom, and describe the molecular structure.
a. CO
b. BF4e. XeF2
http://www.khanacademy.org/science/organicchemistry/v/pi-bonds-and-sp2-hybridized-orbitals
http://www.khanacademy.org/science/organicchemistry/v/newman-projections
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Molecular Orbitals
Formed from atomic orbitals
We use the following procedure when drawing
molecular orbital diagrams.
Determine the number of electrons in the molecule.
We get the number of electrons per atom from their
atomic number on the periodic table. (Remember
to determine the total number of electrons, not just
the valence electrons.)
Fill the molecular orbitals from bottom to top until
all the electrons are added. Describe the electrons
with arrows. Put two arrows in each molecular
orbital, with the first arrow pointing up and the
second pointing down.
Orbitals of equal energy are half filled with
parallel spin before they begin to pair up.
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Molecular orbital diagram
We describe the stability of the molecule with bond
order.
Bonding Orbital
An orbital of lower energy than the atomic
orbitals
Antibonding Orbital
An orbital of higher energy than the atomic
orbitals
Bond Order
The difference between the number of
bonding electrons and the number of
antibonding electrons divided by two. Indicates
bond strength.
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The higher the bond order the stronger is the bond.
bond order = 1/2 (#e- in bonding MO's - #e- in
antibonding MO's)
We use bond orders to predict the stability of
molecules.
If the bond order for a molecule is equal to zero,
the molecule is unstable.
A bond order of greater than zero suggests a
stable molecule.
The higher the bond order is, the more stable the
bond.
We can use the molecular orbital diagram to predict
whether the molecule is paramagnetic or
diamagnetic. If all the electrons are paired, the
molecule is diamagnetic. If one or more electrons are
unpaired, the molecule is paramagnetic.
Paramagnetism
Certain substances exhibit magnetism when placed in
a magnetic field. Paramamagnetism causes the
substance to be attracted into the inducing magnetic
field. Diamagnetism causes the substance to be
repelled from the inducing magnetic field.
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A substance with an unpaired electron exhibits
paramagnetism
The molecular orbital diagram for a diatomic hydrogen molecule, H2, is
The bond order is _____.
The bond order suggests that H2 is _______.
H2 is diamagnetic/paramagnetic.
2. The molecular orbital diagram for a diatomic helium molecule, He2, shows the
following.
The bond order is_____.
The bond order for He2 suggests that He2 is _______.
If He2 did form, it would be diamagnetic/magnetic.
3. The molecular orbital diagram for a diatomic oxygen molecule, O2, is
O2 has a bond order of ____.
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The bond order suggests that the oxygen molecule is ______.
O2 is paramagnetic/dimagnetic.
4. The molecular orbital diagram for a diatomic fluorine molecule, F2, is
F2 has a bond order of ____.
F2 has a bond order of _______.
F2 is diamagnetic/ paramagnetic.
5. The molecular orbital diagram for a diatomic neon molecule, Ne2, is
Ne2 has a bond order of ____.
The bond order for Ne2 suggests that Ne2 is_______.
If Ne2 did form, it would be diamagnetic/paramagnetic.
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Exercise 6 For the species O2, O2+, O2-, give the electron configuration and the bond order for each.
Which has the strongest bond?
Exercise 7 Use the molecular orbital model to predict the bond order and magnetism of each of the
following molecules.
a) Ne2
b) P2
Exercise 8 Use the MO Model to predict the magnetism and bond order of the NO+ and CN- ions.
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Bond Length
Shorter bonds have greater energy. Triple bonds are
shorter than double bonds which are shorter than
single bonds.
Bondlength (pm) and bond energy (kJ/mol)
Bond Length Energy Bond Length Energy
H--H
74
436 H--C
109
413
C--C
154
348 H--N
101
391
N--N
145
170 H--O
96
366
O--O
148
145 H--F
92
568
F--F
142
158 H--Cl
127
432
Cl-Cl
199
243 H--Br
141
366
Br-Br
228
193 H--I
161
298
I--I
267
151
C--C
154
348
C--C
154
348 C=C
134
614
C--N
147
308 C≡C
120
839
C--O
143
360
C--S
182
272 O--O
148
145
C--F
135
488 O=O
121
498
C--Cl
177
330
C--Br
194
288 N--N
145
170
C--I
214
216 N≡N
110
945
Bondlength (pm) and bond energy (kJ/mol)
Bond Length Energy Bond Length Energy
H--H
74
436 H--C
109
413
C--C
154
348 H--N
101
391
N--N
145
170 H--O
96
366
O--O
148
145 H--F
92
568
F--F
142
158 H--Cl
127
432
Cl-Cl
199
243 H--Br
141
366
Br-Br
228
193 H--I
161
298
I--I
267
151
C--C
154
348
C--C
154
348 C=C
134
614
C--N
147
308 C≡C
120
839
C--O
143
360
C--S
182
272 O--O
148
145
C--F
135
488 O=O
121
498
C--Cl
177
330
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C--Br
C--I
194
214
288 N--N
216 N≡N
145
110
170
945
2H2 (g) + O2 (g) à 2H2O(g)
Calculate the energy change in this reaction. Is it
exothermic or endothermic?
2. H2 (g) + Br2 (g) à 2HBr (g)
Calculate the energy change in this reaction. Is it
exothermic or endothermic
Bond Energies (for chemical reactions)
When chemical change occurs bonds are rearranged.
The bonds in the reactants are broken and new bonds
are formed to produce the products.
Calculating ΔH using bond energies.
To calculate ΔH using bond energies we can use
the following procedure.
Step 1) Determine the number (in moles) and
type of bonds broken and formed from the
balanced equation. This may require you to
draw the molecules to identify how many of
each bond type are present.
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Step 2) Multiply the number of bonds by the
average bond energy given in the table to
determine the energy change.
Step 3) Plug values into the above formula to
calculate the ΔH of the overall reaction.
Example : Calculate the energy of the reaction
for the burning of methane in oxygen to form
carbon dioxide gas and water gas, using heats
of formation. The balanced equation is given
below. Use the following list of bond energies
C3H8(g) + 5 O2(g) à 3 CO2 (g) + 4 H2O(g)
Steps 1 and 2
1) Determine the number and types of bonds
broken and formed.
2) Determine the energy change.
Bonds Broken (reactants)
Type
#
Bond Energy
C-C
2
347 kJ/mol
C-H
8
413 kJ/mol
O=O
5
498 kJ/mol
Total
Bonds formed (products)
O-H
8
464
C=O
6
805
Total
Energy
+ 694 kJ
+3,320 kJ
+2,490 kJ
+6,488 kJ
- 3,712 kJ
- 4,830 kJ
- 8,542 kJ
Step 3)
H = Energy of Bonds Broken - Energy of Bonds formed
H = 6,488 kJ - 8,542 kJ
H = - 2,054 kJ
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Intermolecular Forces
§ Forces of attraction between molecules.
§ Sometimes referred to bonds although they are
much weaker than ionic or covalent bonds
§ Responsible for the different properties of
different molecular compounds.
§ Absent in a gas present in a liquid and solid.
Stronger in solid than liquid.
§ Molecules with stronger intermolecular forces
have relatively higher boiling and melting points.
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Dipole-Dipole forces
The attraction of the opposite end of polar
molecules.
The stronger the dipole the stronger is the forces.
Rate the strength of the dipole-dipole force
SH2, HCl, HBr, PH3. PCl3
Hydrogen bonding
A strong dipole-dipole force. Exists only in
molecules containing a hydrogen bonded to a
nitrogen, fluorine or nitrogen
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London Dispersion Forces
§ Results from attractions between induced
dipoles.
§ It is the only intermolecular force present in
nonpolar molecules.
§ Increases with increasing size.
§ Occurs in the Noble gases
§ The weakest intermolecular force
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Flash Card List
Ionic bond
Properties of ionic compounds
Lattice energy
Variable ions electron configuration
Covalent bonds
Plarity and dipole moment
Lewis diagrams
Exceptions to the octet rule
Resonance and formal charge
VSEPR (know geometry and shapes from page 20 of notes)
Molecular polarity
Hybridizaion model
Sigma and pi bonds
Bond length
Intermolecular forces A( H-bonding, dipole-dipole, london dispersion forces)
Properties of covalent compounds
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