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1/13/2015
Unit 4‐ Bonding II
Review
Unit 4‐ Bonding II
Determine the type of bond (Ionic, Covalent or Metallic) in the following compounds:
Compound
NaCl
CO
FeNi
SiS2
Bond Type
Compound
Bond Type
Ionic
NCl3
Covalent
Covalent
PF3
Covalent
Metallic
CaCl2
Ionic
Covalent
Fe2O3
Ionic
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Draw the Lewis Structures
•
1)
PBr3
4) NO2‐1
2)
N2H2
5) C2H4
3)
CH3OH
6) HBr
Draw the Lewis Structures
•
4) NO2‐1 [ ] ‐1
1)
PBr3
2)
N2H2
5) C2H4
3)
CH3OH
6) HBr
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Rules of writing Ionic formulas:
1. Positive ion is written first … this is usually a metal
2. Negative ion is written second … this is usually a nonmetal
3. Subscripts are used to show how many ions of each part are in
the compound. They are used to balance the charge of the ions.
The overall charge should be “0”
Criss-cross method:
Examples:
1. Sodium oxide
* sodium is the positive ion = +1
* oxide is the negative ion = -2
* therefore … it takes 2 sodium ions to balance
the charge of the oxide
Formula = Na2O
Rules of writing formulas:
2. Calcium nitrate
calcium is the positive ion = +2
nitrate is the negative ion = -1
therefore … it takes 2 nitrates to balance the charge of
calcium
Formula = Ca(NO3)2
3. Aluminum sulfide
aluminum is the positive ion = +3
sulfide is the negative ion = -2
therefore … it takes 2 aluminum ions and 3 sulfide
to balance the charge
Formula = Al2S3
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Unit 4‐ Bonding II
Covalent Naming
Binary covalent compounds are characterized by having two nonmetals. Naming these compounds involves the use of numerical prefixes:
Prefix
Number
Prefix
Number
monoditritetrapenta-
1
hexaheptaoctanonadeca-
6
2
3
4
5
7
8
9
10
Write the correct formulas for each covalent compound:
Compound Name
Oxidation States
Water
O (2)
H (1)
Carbon Dioxide
C (4)
O (2)
Chlorine (Diatomic Element)
Cl (1)
Methane (5 total atoms)
C (4)
H (1)
Ammonia (4 total atoms)
N (3)
H (1)
Carbon tetrabromide (5 total atoms)
C (4)
Br (1)
Phosphorous trichloride (4 total atoms)
P (3)
Cl (1)
Diphosphorous trioxide (5 total atoms)
P (3)
O (2)
Covalent Formula
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Write the correct formulas for each covalent compound:
Compound Name
Oxidation States
Covalent Formula
Water
O (2)
H (1)
H20
Carbon Dioxide
C (4)
O (2)
CO2
Chlorine (Diatomic Element)
Cl (1)
Methane (5 total atoms)
C (4)
H (1)
Ammonia (4 total atoms)
N (3)
H (1)
Carbon tetrabromide (5 total atoms)
C (4)
Br (1)
Phosphorous trichloride (4 total atoms)
P (3)
Cl (1)
Diphosphorous trioxide (5 total atoms)
P (3)
O (2)
Write the correct formulas for each covalent compound:
Compound Name
Oxidation States
Covalent Formula
Water
O (2)
H (1)
H20
Carbon Dioxide
C (4)
O (2)
CO2
Chlorine (Diatomic Element)
Cl (1)
Cl2
Methane (5 total atoms)
C (4)
H (1)
CH4
Ammonia (4 total atoms)
N (3)
H (1)
Carbon tetrabromide (5 total atoms)
C (4)
Br (1)
Phosphorous trichloride (4 total atoms)
P (3)
Cl (1)
Diphosphorous trioxide (5 total atoms)
P (3)
O (2)
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Write the correct formulas for each covalent compound:
Compound Name
Oxidation States
Covalent Formula
Water
O (2)
H (1)
H20
Carbon Dioxide
C (4)
O (2)
CO2
Chlorine (Diatomic Element)
Cl (1)
Cl2
Methane (5 total atoms)
C (4)
H (1)
CH4
Ammonia (4 total atoms)
N (3)
H (1)
NH3
Carbon tetrabromide (5 total atoms)
C (4)
Br (1)
CBr4
Phosphorous trichloride (4 total atoms)
P (3)
Cl (1)
PCl3
Diphosphorous trioxide (5 total atoms)
P (3)
O (2)
P2O3
Unit 4‐ Bonding II
Determine the type of bond (Ionic, Covalent or Metallic) in the following compounds:
Compound
NaCl
CO
FeNi
SiS2
Bond Type
Compound
Bond Type
Ionic
NCl3
Covalent
Covalent
PF3
Covalent
Metallic
CaCl2
Ionic
Covalent
Fe2O3
Ionic
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Balancing Charges:
Criss‐Cross rule
* Write out symbols and charge of elements
* Criss‐Cross charges as subscripts (Swap and Drop)
* Combine as a formula unit Equation Form of Balancing Charges
(Number of Cations)x(Cation Charge) + (Number of Anions)x(Anion Charge) = 0
EX: Aluminum and Oxygen
EX: Barium and Oxygen
Al3+ O2‐
AL2O3
Ba2+ O2‐
BaO
Lithium Iodide (LiI) Strontium Chloride (SrCl2) Sodium Sulfide (Na2S)
Balancing Charges Practice
• Lithium Iodide
so LiI
Li +1 I‐1
• Strontium Chloride
Sr+2 Cl‐1
so SrCl2
• Sodium Sulfide
Na +1 S‐2 so Na2S
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Balancing Charges Practice
Cl‐
S‐2
N‐3
F
P‐3
O
Mg+2
Cs+
Cr+3
Na
Zn+2
Al+3
K
Balancing Charges Practice
Cl‐
Mg+2
Cs+
Cr+3
Na
Zn+2
Al+3
K
S‐2
N‐3
F
P‐3
O
MgCl2
MgS
MgF2
Mg3N2
MgO
Mg3P2
CsCl
Cs2S
CsF
Cs3N
Cs2O
Cs3P
CrCl3
Cr2S3
CrF3
CrN
Cr2O3
CrP
NaCl
Na2S
NaF
Na3N
Na2O
Na3P
ZnCl2
ZnS
ZnF2
Zn3N2
ZnO
Zn3P2
AlCl3
Al2S3
AlF3
AlN
Al2O3
AlP
KCl
K2S
KF
K3N
K2O
K3P
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Valence Shell Electron Pair Repulsion Theory
• This is the way that we predict the geometry shape of molecules, A model was developed a qualitative model known as Valence Shell Electron Pair Repulsion Theory (_VSEPR__ Theory). • The basic assumptions of this theory are summarized below.
– 1) The electron pairs in the valence shell around the central atom of a molecule repel each other and tend to orient in space so as to minimize the repulsions and maximize the distance between them.
– 2) There are two types of valence shell electron pairs: __Bonded____ pairs and __Unbonded____ pairs
• Bond pairs are __Electrons shared__ by two atoms and are attracted by two nuclei. Hence they occupy less space and cause less repulsion.
• Lone pairs are pairs of electrons not involved in bond formation and are in attraction with only one nucleus. Hence they occupy more space. As a result, the lone pairs cause more repulsion.
• Note: The bond pairs are usually represented by a _line drawn between the two atoms_, whereas the lone pairs are represented by a lobe with two electrons.
Valence Shell Electron Pair Repulsion Theory
• 3) In VSEPR theory, the _double and triple_ bonds are
treated as if they were single bonds. The electron pairs in
multiple bonds are treated collectively as a single super
pair.
• 4) The shape of a molecule can be predicted from the
number and type of valence shell electron pairs around
the central atom.
• When the valence shell of central atom contains only
bond pairs, the molecule assumes symmetrical geometry
due to even repulsions between them.
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Unit 4‐ Bonding
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Unit 4‐ Bonding
• Steric number is the total number of atoms bonded to the central atom and plus the number of lone pairs on the central atom.
VSEPR Practice
Complete the table with the requested information.
Molecule
CClF3
Structural Diagram
Oxidation State of each element
Molecular Geometry
C (4) CL (1) F (1)
SF2
BF3
SiBr4
NH3
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VSEPR Practice
Complete the table with the requested information.
Molecule
CClF3
SF2
BF3
Structural Diagram
Oxidation State of each element
Molecular Geometry
C (4) CL (1) F (1)
S (2) F (1)
B (3) F (1)
Si (4) Br(1)
SiBr4
NH3
N (3) H (1)
Polyatomic Ions
• Polyatomic ions are groups of atoms that are covalently bonded, but carry an overall net charge.
• The names of polyatomic ions must be memorized to appropriately name these compounds.
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Polyatomic Ions
ClO3‐
Chlorate
Polarity
• Bond Polarity
• Electronegativity
• Ionic bonds have an electronegativity difference that is greater than 1.7. Covalent bonds have an electronegativity difference less than (or equal to) 1.7. Electronegativity differences between 0 and 0.4 indicate non‐polar covalent bonds. Electronegativity differences between 0.4 and 1.7 indicate polar covalent bonds.
• Polar Covalent Bond‐ a covalent bond in which the electrons are not shared equally because one atom attracts them more strongly than the other.
• Non‐polar Covalent Bond‐ a covalent bond in which the electrons are shared equally.
• Use the periodic table of electronegativities to answer the questions on electronegativity differences.
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Electronegativity
H
2.1
Li
1.0
Be
1.5
B
2.0
C
2.5
N
3.0
O
3.5
F
4.0
Na
0.9
Mg
1.2
Al
1.5
Si
1.8
P
2.1
S
2.5
Cl
3.0
K
0.8
Ca
1.0
Sc
1.3
Ti
1.5
V
1.6
Cr
1.6
Mn
1.5
Fe
1.8
Co
1.9
Ni
1.9
Cu
1.9
Zn
1.6
Ga
1.6
Ge
1.8
As
2.0
Se
2.4
Br
2.8
Rb
0.8
Sr
1.0
Y
1.2
Zr
1.4
Nb
1.6
Mo
1.8
Tc
1.9
Ru
2.2
Rh
2.2
Pd
2.2
Ag
1.9
Cd
1.7
In
1.7
Sn
1.8
Sb
1.9
Te
2.1
I
2.5
Cs
0.7
Ba
0.9
La‐Lu
1.0‐1.2
Hf
1.3
Ta
1.5
W
1.7
Re
1.9
Os
2.2
Ir
2.2
Pt
2.2
Au
2.4
Hg
1.9
Tl
1.8
Pb
1.9
Bi
1.9
Po
2.0
At
2.2
Fr
0.7
Ra
0.9
Ac
1.1
Th
1.3
Pa
1.4
U
1.4
Np‐No
1.4‐1.3
ELECTRONEGATIVITY
(electron attraction!)
Electronegativity
• Determine the type of bond that would form between the following two elements using differences in electronegativity.
Example: Mg – O
– O is 3.5 and Mg is 1.2, therefore, the difference is 3.5 – 1.2 = 2.3
IONIC
Example: Cl – Cl
– Cl is 3.0. The difference is 3.0 – 3.0 = 0
NON‐POLAR COVALENT
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Electronegativity
1.
C – N
2.
Li – F
3.
N – Cl
4.
Na ‐ Cl
5.
O – F
6.
B – H
7.
Ba – F
8.
C – H
Bond
Electronegativity Difference
Bond Type
Bond
Electronegativity Difference
Bond Type
Electronegativity
Bond
1.
C – N
2.
Li – F
Bond
Electronegativity Difference
Bond Type
0.5
Polar Covalent
3.0
Ionic
Electronegativity Difference
Bond Type
3.
N – Cl
0.0
Covalent
4.
Na ‐ Cl
2.1
Ionic
5.
O – F
0.5
Polar Covalent
6.
B – H
0.1
Covalent
7.
Ba – F
3.1
Ionic
8.
C – H
0.4
Covalent
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Molecular Polarity
• Dipole moment‐ a property of a molecule whereby the charge distribution can be represented by a center of positive charge and a center of negative charge.
•
• Polar Molecule‐ a molecule that has a permanent dipole moment.
• Determining if a molecule is polar.
• If ALL of the bonds are non‐polar, then the molecule is non‐polar.
• If some or all of the bonds are polar, you can consider the vectors. Vectors are arrows that point in the direction of the negative charge (the direction the electrons are pulled). Molecular Polarity
• Examples:
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Molecular Polarity
Structural formula Polar or non‐polar
Formula
*Diagram Formula
Polar/Non‐polar
NH3
SCl2
CF4
PCl3
H2 S
C2 H 2
* Bent must be draw as bent
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Structural formula Polar or non‐polar
Formula
*Diagram Formula
Polar/Non‐polar
NH3
Polar
SCl2
Polar
CF4
PCl3
H2 S
C2 H 2
Non Polar
Polar
Polar
NonPolar
* Bent must be draw as bent
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