Upload
others
View
1
Download
0
Embed Size (px)
Citation preview
1/13/2015
1
Unit 4‐ Bonding IIReview
Unit 4‐ Bonding II
Compound Bond Type Compound Bond Type
NaCl Ionic NCl3 CovalentCO Covalent PF3 CovalentFeNi Metallic CaCl2 IonicSiS2 Covalent Fe2O3 Ionic
Determine the type of bond (Ionic, Covalent or Metallic) in the following compounds:
1/13/2015
2
Draw the Lewis Structures•
1) PBr3 4) NO2‐1
2) N2H2 5) C2H4
3) CH3OH 6) HBr
Draw the Lewis Structures•
1) PBr3 4) NO2‐1 [ ] ‐1
2) N2H2 5) C2H4
3) CH3OH 6) HBr
1/13/2015
3
1. Positive ion is written first … this is usually a metal2. Negative ion is written second … this is usually a nonmetal3. 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 oxideFormula = Na2O
Rules of writing Ionic formulas:
2. Calcium nitrate calcium is the positive ion = +2nitrate is the negative ion = -1therefore … it takes 2 nitrates to balance the charge of
calciumFormula = Ca(NO3)2
3. Aluminum sulfide aluminum is the positive ion = +3sulfide is the negative ion = -2therefore … it takes 2 aluminum ions and 3 sulfide
to balance the chargeFormula = Al2S3
Rules of writing formulas:
1/13/2015
4
Unit 4‐ Bonding II
Prefix Number Prefix Number
mono- 1 hexa- 6
di- 2 hepta- 7
tri- 3 octa- 8
tetra- 4 nona- 9
penta- 5 deca- 10
Covalent NamingBinary covalent compounds are characterized by having two nonmetals. Naming these compounds involves the use of numerical prefixes:
Write the correct formulas for each covalent compound:Compound Name Oxidation States Covalent Formula
WaterO (2)
H (1)
Carbon DioxideC (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)
1/13/2015
5
Write the correct formulas for each covalent compound:Compound Name Oxidation States Covalent Formula
WaterO (2)
H (1) H20
Carbon DioxideC (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
WaterO (2)
H (1) H20
Carbon DioxideC (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)
1/13/2015
6
Write the correct formulas for each covalent compound:Compound Name Oxidation States Covalent Formula
WaterO (2)
H (1) H20
Carbon DioxideC (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
Compound Bond Type Compound Bond Type
NaCl Ionic NCl3 CovalentCO Covalent PF3 CovalentFeNi Metallic CaCl2 IonicSiS2 Covalent Fe2O3 Ionic
Determine the type of bond (Ionic, Covalent or Metallic) in the following compounds:
1/13/2015
7
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‐ Ba2+ O2‐
AL2O3 BaO
Lithium Iodide (LiI) Strontium Chloride (SrCl2) Sodium Sulfide (Na2S)
Balancing Charges:
Balancing Charges Practice
• Lithium Iodide
Li +1 I‐1 so LiI
• Strontium Chloride
Sr+2 Cl‐1 so SrCl2• Sodium Sulfide
Na +1 S‐2 so Na2S
1/13/2015
8
Balancing Charges Practice
Cl‐ S‐2 F N‐3 O P‐3
Mg+2
Cs+
Cr+3
Na
Zn+2
Al+3
K
Balancing Charges Practice
Cl‐ S‐2 F N‐3 O P‐3
Mg+2MgCl2 MgS MgF2 Mg3N2 MgO Mg3P2
Cs+CsCl Cs2S CsF Cs3N Cs2O Cs3P
Cr+3CrCl3 Cr2S3 CrF3 CrN Cr2O3 CrP
NaNaCl Na2S NaF Na3N Na2O Na3P
Zn+2ZnCl2 ZnS ZnF2 Zn3N2 ZnO Zn3P2
Al+3 AlCl3 Al2S3 AlF3 AlN Al2O3 AlP
K KCl K2S KF K3N K2O K3P
1/13/2015
9
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.
1/13/2015
10
Unit 4‐ Bonding
1/13/2015
11
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.
Molecule Structural Diagram Oxidation State of each
element
Molecular Geometry
CClF3C (4) CL (1) F (1)
SF2
BF3
SiBr4
NH3
VSEPR PracticeComplete the table with the requested information.
1/13/2015
12
Molecule Structural Diagram Oxidation State of each
element
Molecular Geometry
CClF3C (4) CL (1) F (1)
SF2 S (2) F (1)
BF3B (3) F (1)
SiBr4
Si (4) Br(1)
NH3N (3) H (1)
VSEPR PracticeComplete the table with the requested information.
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.
1/13/2015
13
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.
1/13/2015
14
Electronegativity
H
2.1
Li
1.0
Be
1.5 ELECTRONEGATIVITY
(electron attraction!)
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
• 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
1/13/2015
15
ElectronegativityBond Electronegativity Difference Bond Type
1. C – N
2. Li – F
Bond Electronegativity Difference Bond Type
3. N – Cl
4. Na ‐ Cl
5. O – F
6. B – H
7. Ba – F
8. C – H
ElectronegativityBond Electronegativity Difference Bond Type
1. C – N 0.5 Polar Covalent
2. Li – F 3.0 Ionic
Bond 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
1/13/2015
16
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:
1/13/2015
17
Molecular Polarity
Structural formula Polar or non‐polar
Formula *Diagram Formula Polar/Non‐polar
NH3
SCl2
CF4
PCl3
H2S
C2H2
* Bent must be draw as bent
1/13/2015
18
Structural formula Polar or non‐polar
Formula *Diagram Formula Polar/Non‐polar
NH3
Polar
SCl2Polar
CF4 Non Polar
PCl3Polar
H2SPolar
C2H2NonPolar
* Bent must be draw as bent