Announcements Chem 7 Final Exam Wednesday, Oct 10 1:30-3:30AM Chapter 1-12 70 or 75 multiple choice questions Exam III (Chapter 7-10) Wednesday, October 3, 2012 Time: 6:00PM - 7:30PM SEC A 214A and 215A SKIPPING THIS STUFF 1. Classical distinction between energy and matter (p. 217) 2. Numerical problems involving the Rydberg equation (equations 7.3/4). 3. Spectral analysis in the laboratory (pp. 226-227) 4. Numerical problems involving the Heisenberg uncertainty principle (p. 231) 5. Trends among the transition elements (p. 261) 6. Trends in electron affinity (pp. 265-266) 7. Pseudo-noble gas configuration (p. 269) 8. Lattice energy (pp. 283-285) 9. IR spectroscopy (p. 292) 10. Numerical problems involving electronegativity (p. 296) 11. Electronegativity with oxidation number (p. 297) 12. Section 11.3: MO theory and electron delocalization 13. All sections in chapter 12 except 12.3 (types of intermolecular forces). Cadmium Bromine Metals Non-Metals Covalent compounds result from chemical reactions between non-metals. An ionic compound (salt) results from the chemical reaction between a metal and non-metal. Ionic Compounds Covalent Compounds 1. Solid: high melting and boiling points (>600˚C). 1. Gas, liquid or solid with low melting points (<300˚C) 2. Soluble in polar solvents (like water) 2. Insoluble in polar solvents 3. Insoluble in non-polar solvents (like organic solvents) 3. Soluble in organic solvents (not water soluble) 4. Molten compounds conduct electricity very well 4. Non-electrolytes as liquids 5. Aqueous solutions are electrolytes 5. No charged particles and (electrolytes) in water 6. Formed between elements with very different electronegativity. 6. Formed between elements of similar electronegativity. Ionic and covalent compounds have different physical properties. Valence electrons are the electrons in the outer shell (highest value of n quantum number) of the atom that dictate chemistry & in chemical bonding. 1A 1 ns 1 2A 2 ns 2 3A 3 ns 2 np 1 4A 4 ns 2 np 2 5A 5 ns 2 np 3 6A 6 ns 2 np 4 7A 7 ns 2 np 5 Group # of valence e - e - configuration Given by the Group Number for Group A • For B group elements, the valence electrons are in the highest value ns orbital and the (n-1)d orbitals. Lewis dot structures are used to depict valence electrons and bonding between atoms. • A chemical symbol represents the nucleus and all core e - . • A single dot around the symbol represents one valence e - . +1 +2 +3 -1 -2 -3 Lewis structures are used to depict how ionic bonds are formed by chemists. •• •• Mg • • Cl • •• •• Cl • •• •• •• Cl •• •• •• Mg 2+ - 2 Formation of sodium chloride O • •• • •• •• O •• •• •• Ba 2+ 2- Ba • • + Formation of magnesium chloride Formation of barium oxide noble gas electronic configuration
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Chapter 9 2012 · 6. Formed between elements with very different electronegativity. 6. Formed between elements of similar electronegativity. Ionic and covalent compounds have different
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AnnouncementsChem 7 Final Exam Wednesday, Oct 10 1:30-3:30AMChapter 1-1270 or 75 multiple choice questions
Exam III (Chapter 7-10)Wednesday, October 3, 2012Time: 6:00PM - 7:30PMSEC A 214A and 215ASKIPPING THIS STUFF
1. Classical distinction between energy and matter (p. 217)2. Numerical problems involving the Rydberg equation (equations 7.3/4). 3. Spectral analysis in the laboratory (pp. 226-227)4. Numerical problems involving the Heisenberg uncertainty principle (p. 231)5. Trends among the transition elements (p. 261)6. Trends in electron affinity (pp. 265-266)7. Pseudo-noble gas configuration (p. 269)8. Lattice energy (pp. 283-285)9. IR spectroscopy (p. 292)10. Numerical problems involving electronegativity (p. 296)11. Electronegativity with oxidation number (p. 297)12. Section 11.3: MO theory and electron delocalization13. All sections in chapter 12 except 12.3 (types of intermolecular forces).
Cadmium
Bromine
Iodine
Metals
Non-Metals
Covalent compounds result from chemical reactions between non-metals.
An ionic compound (salt) results from the chemical reaction between a metal and non-metal.
Ionic Compounds Covalent Compounds
1. Solid: high melting and boiling points (>600˚C).
1. Gas, liquid or solid with low melting points (<300˚C)
2. Soluble in polar solvents (like water) 2. Insoluble in polar solvents
3. Insoluble in non-polar solvents (like organic solvents)
3. Soluble in organic solvents (not water soluble)
4. Molten compounds conduct electricity very well 4. Non-electrolytes as liquids
5. Aqueous solutions are electrolytes
5. No charged particles and (electrolytes) in water
6. Formed between elements with very different electronegativity.
6. Formed between elements of similar electronegativity.
Ionic and covalent compounds have different physical properties.
Valence electrons are the electrons in the outershell (highest value of n quantum number) of the atom that dictate chemistry & in chemical bonding.
1A 1ns1
2A 2ns2
3A 3ns2np1
4A 4ns2np2
5A 5ns2np3
6A 6ns2np4
7A 7ns2np5
Group # of valence e-e- configuration
Given by the Group Number for
Group A
• For B group elements, the valence electrons are in the highest value ns orbital and the (n-1)d orbitals.
Lewis dot structures are used to depict valence electrons and bonding between atoms.
• A chemical symbol represents the nucleus and all core e-.
• A single dot around the symbol represents one valence e-.
+1 +2 +3 -1-2-3
Lewis structures are used to depict how ionic bonds are formed by chemists.
••
••
Mg••
Cl• ••
••
Cl•••
••
••Cl••
••••
Mg2+ -2
Formation of sodium chloride
O•••
•••
••O••
••
••Ba
2+ 2-Ba• • +
Formation of magnesium chloride
Formation of barium oxide
noble gas electronic configuration
1. Using spdf electron configurations
2. Orbital box diagrams
3. Lewis electron-dot symbols
Li (1s22s1) + F (1s22s22p5) Li+ 1s2 + F- (1s22s22p6)
Li1s 2s 2p
F1s 2s 2p
+
Li+
1s 2s 2p
F-
1s 2s 2p+
.+ F: ::Li. Li+ +
.F: ::
.
Chemists use different ways to conceptualize and draw ionic bonding. Know both!
NON-METALSMetals: Charge of Cation = Group Number (1A, 2A, 3A)
Non-metals: Charge of Anion = Group Number - 8
Ni2+
The for main group or representative elements the group number in the periodic table can tell us the charge of the metal cation and the charge of the non-metal anion that will be formed in an ionic reaction.
Transition Metalsstarts with filling of d-orbitals
Use partial orbital diagrams and Lewis symbols to depict the formation of Na+ and O2- ions from the atoms, and determine the formula of the compound.
SOLUTION:
3s 3p
Na
3s 3p
2s 2p
O2s 2p
O2-
2 Na+
:Na
Na+ O
.
:
..
.
2Na+ + O 2-
::
::
Na
.
1. Covalent bonding results from sharing of one or more electrons between non-metals atom.
share electrons
Occurs generally between non-metal elements forming molecules or molecular compounds.
• Every covalent bond has a characteristic bond length that leads to maximum stability.
Energy Released When Bond Formed
The formation of a covalent bond releases energy and reduces the potential energy (i.e minimizes energy).
Atoms Far Apart
Atoms to Close Together
F F lone pairslone pairs
single covalent bond
F F
single covalent bond
F F+
7e- 7e-
F F
8e- 8e-
8 e- or an “octet” around each atom
Lewis Dot Structure
Lewis Skeletal Structure
We also use Lewis structures and the “octet rule” to show covalent bonds and bonding between atoms.
Note the outer electrons are not shown!
• •
OH H H O H H O H
Dash FormulaLewis Dot
C••
••
O ••• • O•
•
• •
• •CO O
•
••
•••
••
••
••
•CO O ••
•••
••
••
••CO O••
••
••
••
What is the Lewis dot structure and the dash formula for carbon dioxide?
What is the Lewis dot structure and dash formula for water, H2O and for carbon dioxide?
Bonding Pairs (double bond)
Non-bonding Pairs
• Bond Order is the number of bonds between two atoms.– Single bond, order = 1– Double bond, order = 2– Triple bond, order = 3
• Higher bond orders give rise to:– Shorter bond lengths between bonded atoms– Stronger bond energies (more energy needed to break)
H O HCO O••
••
••
••
Bond order is the number of covalent bonds between two atoms.
Higher bond orders give shorter bond lengths and require more energy to break a bond.
Bond LengthsTriple bond < Double Bond < Single Bond
Comparing Bond Length and Bond StrengthThe trends in atomic radius can be extended to the bond length in molecules in a family of compounds.
(a) S - F, S - Br, S - Cl
(b) C = O, C - O, C O
(c) C-Br, C-I, C-Cl
Using the periodic table, but not Tables 9.2 and 9.3, rank the bonds in each set in order of increasing bond length and bond strength:
Comparing Bond Length and Bond StrengthUsing the periodic table, but not Tables 9.2 and 9.3, rank the bonds in each set in order of decreasing bond length and bond strength:
(a) S - F, S - Br, S - Cl
(b) C = O, C - O, C O PLAN: (a) The bond order is one for all and sulfur is bonded to halogens;
bond length should increase and bond strength should decrease with increasing atomic radius. (b) The same two atoms are bonded but the bond order changes; bond length decreases as bond order increases while bond strength increases as bond order increases.
(a) Atomic size increases going down a group.
Bond length: S - Br > S - Cl > S - F
Bond strength: S - F > S - Cl > S - Br
(b) Using bond orders we get
Bond length: C - O > C = O > C O
Bond strength: C O > C = O > C - O
Bond length depends on the size of the bonded atoms.
Internuclear distance(bond length)
Internuclear distance(bond length)
Internuclear distance(bond length)
Internuclear distance(bond length)
Covalent radius
72 pm
Covalent radius
114 pm
Covalent radius
133 pm
Covalent radius
100 pm
F2
Cl2
Br2
I2
72pm
100pm
114pm
133pm
VSEPRT explains the geometry of molecules but NOT how covalent bonds are formed with that geometry.
Molecular formula
Lewis structure
VSEPRT
GeometryHybrid orbitals
VSEPRT Valence BondTheoryVSEPRTLewis Structure
Write the Lewis dot and skeletals structure of nitrogen trifluoride (NF3).
Write the Lewis dot and skeletal structures of the carbonate ion (CO3
2-).
Write the Lewis dot and skeletal structures structure of the carbonate ion (BrO3
-).
Write the Lewis dot and skeletal structures structure of the carbonate ion HCN?
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F --> N is central atom!
F N F
F
A = 5 + 21 = 26 valence electrons
Step 3 - Write structure with N central and three bonds and rest non-bonding octet electrons around the central atom.
Step 4 - Write structure with N central and three bonds and rest non-bonding octet electrons.
]BrO3[–
Valence e- = 7 + 3(6) + 1 = 26
O Br O
O
Look at the formula sometimes it gives clues to the central atom
H C N
HCNValence e- = 1 + 4 + 5 = 10
Carbon is central atom, watch for hydrogen--1 bond
–
Write the Lewis structure of the carbonate ion (BrO3-).
Bond Enthalpies and ReactionsHess’s Law allows us to analyze the energetics of a reaction. This provides insight into the relative stability of products and reactants.
Exothermic-energy releasedEndothermic
The bond energy, BE is the positive amount of energy (enthalpy) required to break or make one mole of bonds in a gaseous covalent compound to form products in the gaseous state at constant T and P.
!H0 = +436.4 kJ/mol
!H0 = +242.7 kJ/mol
H2 (g) H (g) + H (g)
Cl2 (g) Cl (g)+ Cl (g)
HCl (g) H (g) + Cl (g) !H0 = +431.9 kJ/mol
Bond Dissociation Energy
Example
All chemical reactions give off a net heat of reaction that can be either (!Hrxn) absorbed or gained.
The source of heat is the breaking and making of chemical bonds!
∆Hrxn = ΣBEreactant bonds broken − ΣBEproduct bonds formed
Calculating Bond Enthalpies In A ReactionCalculate the !Hrxn for the chlorination of methane to form chloromethane gas using bond energies.
CH4(g) + Cl2(g) " CH3Cl(g) + HCl(g) !Hrxn = ?1. Write the balanced chemical equation and set up table below.2. Write Lewis structures break all bonds in products and reform all bonds in products. 3. Use the simple equation below to calculate !Hrxn
∆Hrxn = ΣBEreactant bonds broken − ΣBEproduct bonds formed
Type of bonds formed
Number of bonds formed
Bond energy (kJ/mol)
Energy change (kJ)
Type of bonds broken
Number of bonds broken
Bond energy (kJ/mol)
Energy change (kJ)
+ +
Type of bonds broken
Number of bonds broken
Bond energy (kJ/mol)
Energy change (kJ)
Type of bonds formed
Number of bonds formed
Bond energy (kJ/mol)
Energy change (kJ)
41
11
339431 431
339H-ClC-ClC-H 3 414 1232
414243
C-HCl-Cl 243
1656
Calculate the !Hrxn for the chlorination of methane to form chloromethane gas using bond energies.
Use Table 9.2 to calculate !H0rxn for the following reaction:
CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)
1. Write the balanced chemical equation2. Write Lewis structures to see bonds broken and bonds formed. 3. Use the given Bond energies in the problem and a table account for bonds broken/formed to calculate !Hrxn
∆Hrxn = ΣBEreactant bonds broken − ΣBEproduct bonds formed
Type of bonds formed
Number of bonds formed
Bond energy (kJ/mol)
Energy change (kJ)
Type of bonds broken
Number of bonds broken
Bond energy (kJ/mol)
Energy change (kJ)
Calculating Enthalpy Changes from Bond Energies
Use Table 9.2 to calculate !H0rxn for the following reaction:
CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)
Calculate the number of bonds broken and bond formed using the bond energies found in Table 9.2.