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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 MgCl2 2326 LiCl 834 SrCl2 2127 LiI 730 MgO 3795 NaF 910 CaO 3414 NaCl 788 SrO 3217 NaBr 732 ScN 7547 KF 808 KCl 701 KBr 671

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-, SnCl42-

and PbI42- ions.

The amount of ligands joined is dependent on the coordination number for the transition metal. Ion Coordination

numbers Ion Coordination

number Cu+ 2,4 Mn2+ 4, 6 Ag+ 2 Fe2+ 6 Au+ 2,4 Cu2+ 4,6 Ni2+ 4,6 Sc3+ 6 Al+3 4, 6 Cr3+ 6 Au3+ 4 Examples AgCl + 2 NH3 → [Ag(NH3)2]+ Ni(CN)2 + 2 CN- → [Ni(CN)4]2-

Cu(NH3)42

, 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 p-orbitals. 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 # of hybrid

obitals Geometry example

sp 2 Linear sp2 3 Trigonal planar sp3 4 Tetrahedral dsp3 5 Trigonal

bipyramidal

d2sp3 6 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. BF4- e. XeF2

http://www.khanacademy.org/science/organic-chemistry/v/pi-bonds-and-sp2-hybridized-orbitals http://www.khanacademy.org/science/organic-chemistry/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 194 288 N--N 145 170 C--I 214 216 N≡N 110 945

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 Energy C - C 2 347 kJ/mol + 694 kJ C - H 8 413 kJ/mol +3,320 kJ O = O 5 498 kJ/mol +2,490 kJ

Total +6,488 kJ Bonds formed (products)

O - H 8 464 - 3,712 kJ C = O 6 805 - 4,830 kJ

Total - 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