iB chemistry Bonding

IB Chemistry 4 — Bonding

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4.1 Ionic Bonds Ionic compounds are composed of positive ions (cations) and negative ions (anions) that are held together because opposite charges attract. There is no specific cations connected to a specific anion. It forms a lattice. The force that attracts the cations and anions increases with charge and decreases with distance.

We measure the strength of a bond by the energy required to pull the ions apart. More highly charged ions will have a stronger ionic bonds (most important). Smaller ions will have stronger ionic bonds (considered after charge). Ex 1: Rank the following ionic compounds in order of decreasing bond strength. LiCl NaCl MgCl2 MgF2 MgO CaO The energy to break an ionic bond is the energy required to take the ions in a compound and pull them apart to individual ions. This will be an endothermic process (breaking bonds required energy). This will be the same as the energy released if individual ions form an ionic solid (forming bonds in an exothermic process). The energy of the ionic bonds is called the lattice energy. A reaction to make a solid from gaseous ions would be very difficult.

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According to Hess’s Law, we can find a different set of steps that have the same overall result, we can find the energy. This process is called the Born-Haber Cycle. This cycle uses the enthalpy of formation and a series of steps to find the lattice energy Born-Haber Cycle 1. Sublimation energy 2. Bond strength (covalent) 3. Ionization energy 4. Electron affinity 5. Lattice energy Net reaction: enthalpy of formation. Ex 2: Using data from your text (p.362), Data Booklet and The diagram above, to find the lattice energy for KF. What are the differences between this Born-Haber cycle and the one for NaCl? Problems

9.17, 22, 24, 26, 28, 30, 33


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4.2 Lewis Structures Covalent bonds are formed by molecules sharing electrons. Usually one electron is donated to the bond from each atom in the bond. (If both electrons are from one species it forms a ligand or a coordinate covalent bond.) To determine the way electrons are shared we draw Lewis structures. Each dot is an electron. Lines between atoms represent a pair of bonding electrons. This is a polyatomic ions with a charge of -2: SO4

2-

Generally atoms will have an octet about them to become more stable. To draw a Lewis Structure with octets:

1. Sum the valence electrons of all atoms. Account for ions. 2. Connect the atoms with single bonds. (Atoms are often

written in order or if there is a central atom with a group of atoms, the central atom is often written first.)

3. Complete octets for most atoms (H is a duet). 4. If short of electrons form multiple bonds. (sharing 4 or 6

electrons) Ex 1: Draw the Lewis structures for: CH4 H2O CO2 SO3

2- O2 N2 HCN


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There can be times when a double bond can equally be placed at two positions; the choice of which is arbitrary. In the Lewis Structure for NO2

- the double bond can be placed with either oxygen. The result is half of each. This is a resonance structure. Draw the Lewis structure for: SO3 O3

CH3COO- There are times when atoms do not form octets. Boron is small and can form stable compounds with six electrons. Beryllium can form compounds with four electrons. These exceptions are included because they have been found to exist. There are also compounds that can have more than octets. An atom that has empty “d” orbitals can have 10 or 12 electrons around it. This is only done if it is the only way to place all the electrons in the Lewis Structure. Ex 3: Draw the Lewis Structure for: a) I3

- b) XeF4 c) ICl5 Follow Up Problems 10.1, 3, 4, 5 Problems 10.1, 3, 5, 7, 9, 11, 19, 21


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4.3 Covalent Bond Energies Covalent molecules share electrons. The atomic orbitals of the bonded atoms overlap. The electrons orbit throughout the molecule, not just in the place that they are drawn in a Lewis Structure. Every molecule will have unique bonds between its atoms. To simplify the bonding model, we assume that all bonds between similar atoms are similar. We can then use average bonds. Average bond energies are given in your Data Booklet. This is defined as the energy required to break the bond per mole. Stronger bonds are a result of better overlap of atomic orbitals. Stronger bonds are shorter bonds. Compare the bond strength and bond length on the chart in your Data Booklet. The enthalpy of a reaction can be estimated from the energy changes in the bonds that are broken and the bonds that are formed. This is assuming that the bonds are average covalent bonds. Example 1: Calculate the energy for the following reactions: C2H4 + H2 C2H6 Combustion of acetylene (C2H2). Follow Up Problems 9.2, 3 Problems 9.44, 46, 48, 50, 51

!H = Dbroken " Dformed##


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4.4 VSEPR The shapes of covalent molecules can be explained using the VSEPR theory: Valence Shell Electron Pair Repulsion. Electron pairs (bonding and non-bonding) will arrange themselves to maximize the distance between them so to minimize their potential energy. The shapes that the atoms adopt depend on the number of electron pairs around the central atom. These arrangements allow the electron pairs to be as far apart as possible. You need to know the names and angles for these VSEPR geometries. In describing a molecule’s geometry, we only describe the bonds that are present. Lone pairs must be accounted for in determining the VSEPR geometry, but are ignored in the molecular geometry. Lone pairs repel more than a bonding pair: They will repel other bonding pairs farther away from themselves. If there are two or more lone pairs they will be arranges far from each other. Note that bent geometry can result from Triganal Planer and Tetrahedral geometries. What are the angles for these molecular bonds?


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4.4 VSEPR Note the position of the lone pairs in the triganol bipyramidal geometry. All positions are equal in the octahedral arrangements. Example 1: Give the VSEPR geometry and molecular geometry (including angles) for: NH3

XeF4 SeCl4 SO2 O3 I3

- Follow Up Problems 10.6, 7, 8 Problems 10.28, 30, 32, 34, 36, 40, 46


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4.5 Bond Hybridization VSEPR Theory is an empirical theory that lets you determine what the angles in a molecule or ion should be. Bond hybridization gives us the explanation of how the molecules for the bonds in the way that they do. The “s” and the “p” orbitals are combined in ways that give us the molecular geometries that we find in nature. Covalent bonds come in two flavours. A sigma bond (σ)is a bond that forms from atomic orbitals that overlap directly between the atoms. Single bonds are sigma bonds. Methane has four sigma bonds. A molecule that is tetrahedral is combining its “s” and three “p” orbitals to make the four tetrahedral orbitals. These new four degenerate orbitals are each called an “sp3” orbital. Their geometry is tetrahedral. Each of these “sp3” orbitals is of equal energy. They are higher energy than an “s” orbital, but lower than the “p” orbitals.

Methane has four sigma bonds. Each bond is between an “sp3” orbital for the carbon and an “s” orbital from the hydrogen.


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The second type of covalent bond is a pi “π” bond. This bond forms if a double or triple bond is necessary. In a multiple bond the first bond is a sigma bond and the second and third bond are pi bonds. These bonds are outside of the axis between the two atoms

To make a pi bond, an atom must have a “p” orbital that is not hybridized. The atom is combining one “s” orbital and two “p” orbitals.

The “p” orbital is perpendicular to the plane of the “sp2” bonds.


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Notice that the geometry of the “sp2” hybrids is trigonal planer. This is where the trigonal planer geometry comes from. Pi bonds are also weaker than sigma bonds. This is why a double bond is not twice as strong as a single bond. A bond is stronger if the orbital overlap is greater and if the electrons can spend more time in the shared orbital. To make two pi bonds, an atom must have two “p” orbitals that are not hybridized. This can occur if an atom is making a triple bond or two separate double bonds. To have two “p” orbitals for pi bonding, there is only one “p” for hybridization.

The “p” orbitals for pi bonding are perpendicular to each other.


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To determine the “p” orbitals that are hybridized, subtract any needed for pi bonds, hybridize the rest. Example 1: Draw the following molecules indicating the bonding orbitals:

a) Ammonia (NH3)

b) Oxygen

c) C2H4

d) HCN

e) H2C=C=CH2


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The last aspect we will examine is bond delocalization. This allows us to explain resonance. In this diagram of NO2

–, what is the hybridization of each atom? Each atom must be able to form a pi bond. The electrons in the pi bond are actually shared among all of the “p” orbitals. Bond delocalization adds stability to molecules. Breaking a delocalized bond is harder, because it involved more atoms. Example 2: Draw the bonding for the following molecules: a) Ozone (O3) b) NO3

– c) Benzene (C6H6 ring)


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Different types of bonding can be seen in the allotropes of carbon: diamond, graphite and fullerene (bucky ball). A diamond has carbon atoms that are sp3 hybridized. Each carbon is bonded to four other carbons in a three dimensional matrix. What properties could this give to carbon? Graphite is made of carbon atoms that are sp2 hybridized. They are like benzene rings that continue in large layers. The electrons are delocalized between the layers of carbon atoms. Properties? If a graphite layer is curled into a ball, it forms a fullerene (bukminsterfullerene). Properties? Follow Up Problems 11.2 Problems 11.7, 9, 11, 20, 21, 23, 40, 42, 50, 58


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4.6 Intermolecular Bonding The strength of bonding between separate molecules can be seen through boiling points, melting points, and other physical properties. Stronger intermolecular bonds generate higher boiling points and higher melting points. Ionic compounds do not have discreet molecules. They form a lattice. To melt this lattice, the ions must be separated by breaking the ionic bonds. Are those ionic bonds intermolecular bonds, or intramolecular bonds? They are both. Regardless, these bonds are hard to break, and ionic compounds usually have very high melting points. For covalent compounds, intermolecular bonds fall into three categories: Dipole-dipole forces are an attraction between molecules that have a permanent dipole (positive and negative end). Hydrogen bonds are a special type of dipole. Van der Waals forces that involve temporary dipoles. The first step is to determine if a molecule has a dipole (a positive and a negative end). This is done by:

1) Identifying polar bonds by comparing the electronegativity of the bonded atoms.

2) Finding lone pairs of electrons. 3) Determining if symmetry in the molecule balances out the

positive and negative parts of the molecule.

Example 1: Determine if the following have a dipole: CH4 NH3 H2O CO2 SO2 Cl2


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Dipole bonding attracts molecules together. The stronger the bonding, the higher the temperature must be before the substance will boil or melt. Solid Liquid

Hydrogen bonding is a dipole that is formed with H—N, H—O, H—F bonds. The small size of these atoms produces a very strong dipole bond. Van der Waals forces are due to a temporary dipole that forms when the electrons are temporarily unbalanced. This produces a weak, short lived dipole attraction. (also called London dispersion Forces) The temporary dipoles are stronger in larger atoms. Larger atoms have a greater chance for electron imbalances. The Van der Waals attraction will also be stronger in larger (longer) molecules. There are more places for attraction to occur. The temporary dipoles occur in molecules with permanent dipoles, but they are most important in non-polar molecules. To determine the strength of the inter-molecular bonding:

1) Determine if there is Hydrogen bonding. 2) Determine if there is a dipole.

3) Determine the extent of the Van der Waals forces.

These are in priority order.


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Example 2: Explain the trends in the graph below.

Example 3: Explain the following trends: B.P of noble gasses B.P. of halogens Predict the strength of the inter molecular bonding for: CO2, SO3, O2, CH4, H2CO Follow Up Problems 10.9, 12.2, 3 Problems 10.52, 55, 57, 12.5, 32, 34, 37, 39, 41, 43, 45(polarizability is ease of

forming dispersion forces), 47, 49, 51, 53


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4.6 Metallic Bonding Bonding in metals is different than ionic or covalent solids. Generally, metals are:

• Malleable • Ductile • Electrically conductive • Thermally conductive

Can these be explained by ionic or covalent bonding? Metallic bonding is modeled as positive ions with a “sea” of electrons among the cations. How does this explain metallic properties. To melt a metal, the cations must be pulled from the electrons. This should be harder for metals that form cations with a greater charge. Larger cations should also have lower forces and lower melting points. Check this model with the melting points of some first and second row metals. Check with other metal properties.


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Some compounds can form large, covalently bonded molecules, macro-molecules.

Diamond is an example of a network covalent solid. Melting a diamond involves breaking the covalent bonds between carbon atoms.

Are these inter- or intra-molecular bonds? Silicon dioxide also forms a network covalent solid. Be able to explain trends in m.p. and b.p. for metals. Explain how the metallic bonding model explains the properties of metals.

iB chemistry Bonding

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