AP Chemistry Chapters 7-9 - Quia · The Lewis structure is used to represent the covalent ... AP...

AP Chemistry Chapters 79

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What is a Lewis Diagram?Lewis diagrams, also called electrondot diagrams, are used to represent paired and unpaired valence (outer shell) electrons in an atom.

What is a Lewis Structure?The Lewis structure is used to represent the covalent bonding of a molecule or ion.


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How to Build a Lewis Structure?For simple molecules, the most effective way to get the correct Lewis structure is to write the Lewis diagrams for all the atoms involved in the bonding and adding up the total number of valence electrons that are available for bonding.

Incorrect Structure Correct Structure


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Water has the chemical formula of H2O, which means there

S = N A = 12 8

= 4/2= 2 Bonds

Lone Pairs A what I used


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Example: Write the Lewis structure for methane (CH4).

S=NA


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Example: Write the Lewis structure for carbon dioxide (CO2).


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Lewis Structures of Polyatomic Ions

Building the Lewis Structure for a polyatomic ion can be done in the same way as with other simple molecules, but we have to consider that we will need to adjust the total number of electrons for the charge on the polyatomic ion.

• If the ion has a negative charge, the number of electrons that is equal to the charge on the ion should be added to the total number of valence electrons.

• If the ion has a positive charge, the number of electrons that is equal to the charge should be subtracted from the total number of valence electrons.

• After writing the structure, the entire structure should then be placed in brackets with the charge on the outside of the brackets at the upper right corner.

After counting the valence electrons, we have a total of 9 [5 from nitrogen + 4(1 from each oxygen)] = 9. • The charge of +1 means an electron should be subtracted, bringing the total electron count to 8. • Each hydrogen atom will be bonded to the nitrogen atom, using two electrons. • The four bonds represent the eight valence electrons with all octets satisfied, so your structure is complete. • (Do not forget your brackets and to put your charge on the outside of the brackets)


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Example: Write the Lewis structure for the hydroxide ion (OH).


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Formal ChargeSometimes when writing a Lewis structure you come across two different ways to write the molecule, both which look fine. Resonance can be shown using Lewis structures to represent the multiple forms that a molecule can exist. The molecule is not switching between these forms, but is rather an average of the multiple forms.

In this case, you should use formal charge to decide which structure is correct for the molecule.The formal charge is the difference in the number of valence electrons in the atom and the number of valence electrons in the Lewis structure.

Where we have certain definitions:

• FC formal charge • # of valence e in the atom the number of valence electrons in the free atom just as you've been calculating right along. • # of bonds the number of bonds to the atom in question where a single bond counts as 1 bond, a double bond counts as 2 bonds, and a triple bond counts as 3 bonds. • # of unshared electrons the total number of unshared electrons surrounding the atom in question (note that this is not the number of lone pairs).


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To decide if a given structure is correct, check the formal charge on some atoms in all possible structures. In general the most likely Lewis structure has: • all formal charges as close to zero as possible • negative formal charges on electronegative atoms like halogens or oxygen. For example, consider the methanol molecule CH3OH. This can be written two different ways:

In both cases the OctetRule is satisfied for all of the atoms in the structure. Which is correct?

To decide, work out the formal charges on the carbon and oxygen in each molecule. Carbon has 4 valence electrons in the base atom, oxygen 6.

For the leftmost structure • The carbon has four bonds, each worth 2 electrons, for a total of eight. It has no lone pairs. Thus, Cf = 4 (0 + 1/2*8) =0 • The oxygen has two bonds, each worth 2 electrons, for a total of four. It has two lone pairs. Thus, Cf = 6 (4 + 1/2*4) =0

For the rightmost structure • The carbon has three bonds, each worth 2 electrons, for a total of six. It has one lone pair. Thus, Cf = 4 (2 + 1/2*6) = 1 • The oxygen has three bonds, each worth 2 electrons, for a total of six. It has one lone pair. Thus, Cf = 6 (2 + 1/2*6) = +1

The leftmost structure has the formal charges closer to zero, and thus is probably the correct structure.


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Example: There are two possible Lewis structures for the sulfurous acid molecule, H2SO3. Which is probably correct?


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Solution: The difference between the two is the double bond in the leftmost one vs. the single bond + three lone pairs on the oxygen on the right.

Work out the formal charge on each: the oxygen and sulfur both have six valence electrons.

For the left structure • The oxygen has two bonds (total 4 electrons) and two lone pairs (total 4 electrons). Cf = 6 (4 + 1/2*4) = 0 • The sulfur has four bonds (total 8 electrons) and one lone pair (total 2 electrons). Cf = 6 (2 + 1/2*8) = 0

For the right structure • The oxygen has one bond (total 2 electrons) and three lone pairs (total 6 electrons). Cf = 6 (6 + 1/2*2) = 1 • The sulfur has three bonds (total 6 electrons) and one lone pair (total 2 electrons). Cf = 6 (2 + 1/2*6) = +1

Thus, the left structure is probably correct


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Let us look at how to build a nitrate ion (NO3).

Nitrogen is the least electronegative atom and should be the central atom.

After counting the valence electrons, we have a total of 23[5 from nitrogen + 3(6 from each oxygen)] = 23. The charge of 1 indicates an extra electron, bringing the total electron count to 24.

Each oxygen atom will be bonded to the nitrogen atom, using a total of six electrons. We then place the remaining 18 electrons initially as 9 lone pairs on the oxygen atoms (3 pairs around each atom).

Although all 24 electrons are represented in the structure (two electrons for each of the three bonds and 18 for each of the nine lone pairs), the octet for the nitrogen atom is not satisfied.• To satisfy the octet rule for the nitrogen atom, a double bond needs to be made between the nitrogen and one of the oxygen atoms. • Because of the symmetry of the molecule, it does not matter which oxygen atoms is chosen.• Because there are three different oxygen atoms that could form the double bond, there will be three different resonance structures showing each oxygen atom with a double bond to the nitrogen atom. • Doubleheaded arrows will be placed between these three structures. (Do not forget your brackets and to put your charge on the outside of the brackets)


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Example: What is the Lewis structure for the nitrite ion (NO2−)?

Nitrogen is the least electronegative atom and should be the central atom.


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Lewis Structures for Electronrich Compounds

Elements with atomic number greater than 13 often form compounds or polyatomic ions in which there are “extra” electrons.

Once all of the octets are satisfied, the extra electrons are assigned to the central atom either as lone pairs or an increase in the number of bonds.

(Never use multiple bonds with these compounds—you already have too many electrons.)

Example: Draw the Lewis structure for phosphorus pentafluoride, PF5.

Answer: The electronegativity of fluorine is greater than that of phosphorus—so the phosphorus atom is placed in the center of the molecule.


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Example: Draw the Lewis structure for phosphorus pentafluoride, PF5.


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Lewis Structures for Electronpoor Compounds

There is another type of molecule or polyatomic ion in which there is an electron deficiency of one or more electrons needed to satisfy the octets of all the atoms.

In these cases, the more electronegative atoms are assigned as many electrons to complete those octets first and then the deficiency is assigned to the central atom.

Example: Draw the Lewis structure for boron trifluoride, BF3.

Answer: The electronegativity of fluorine is greater than that of boron—so the boron atom is placed in the center of the molecule.


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The boron atom is two electrons shy of its octet. You may ask about the formation of a double bond (and even resonance). But, fluorine and boron are not in the list that can form double bonds (C, N, O, P, S) and so the compound is electron poor.


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Do For Homework!

Draw the Lewis structure for the following:

Hydronium ion (H3O+)

Hypochlorite ion (ClO)

Carbonate ion (CO32)

Ammonia (NH3)

Hydrogen fluoride (HF)

Ozone (O3)

Xenon difluoride (XeF2)


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Resonance in Benzene

• Each carbon is joined to each of its neighbors by a oneandhalf bond. • Each bond in the benzene ring has the same number of electrons and is the same length. • This picture is in complete accord with experiments which show that all carboncarbon bonds in benzene are the same length, with no hint of shorter (double) or longer (single) bonds. • It also helps explain why benzene does not undergo addition reactions: there are no simple pi bonds.

Resonance has another important feature: when resonance is involved, the real structure is more stable than we would expect from any of the structures we write using the one line = two electrons symbolism.

This extra lowering of energy, which for benzene is about onethird as much as making a typical covalent bond, is quite important in the reactions of benzene and other aromatic compounds. As we will see, reactions of the benzene ring almost always result in products which in which the benzene ring persists an outcome of its stability.

aromatic moleculeshave a smell; more importantly means they contain a benzene ring. These are found in many compounds, aspirin, morphine, nicotine, proteins, saccharin and many plastics. Benzene and many other aromatic compounds are considered carinogens; however, a large number of beneficial compounds with aromatice character are not carinogenic.


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THE RULESLewis ("electron dot") Structures

1. Consult the molecular formula and sum up all the valence electrons from the separate atoms.

The Group number in the Periodic Table = the number of valence electrons for an atom.

• Add one for each () charge (extra electron);

• Subtract one for each (+) charge (missing electron).

• For example: CH3NH2 = 5(1) + 4 + 5 = 14 e POCl3 = 5 + 6 + 3(7) = 32 e NO3

= 5 + 3(6) + 1 = 24 e


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2. Choose the central atom(s).

• Almost always the least electronegative atom is the central atom. • For example, in ClO2, the Cl is the central atom; in SF5 the S is the central atom. • Occasionally, you will need to choose the unique atom, even when it is the most electronegative: e.g., the O in Cl2O. • A wrong choice usually will be signaled by your being unable to write a valid structure. Arrange the other atoms around the central atom, in accord with the normal valences of the atoms. That is, do not place more atoms around a central one than it normally can bond to. • For first and second row elements, the maximum valence = the Group number through Group IV; after that, it is 8 (the Group number). • The difference is that up through IV, the atoms tend to donate electrons to get an octet, whereas beyond IV, they accept electrons. • Recognition of exceptions will come with experience. • You may find that you will have to place fewer atoms than normal around a central one; this is taken care of later. Hydrogen never is the central atom. It forms only one bond, so it must generally be in the outer layer of atoms. Therefore, place hydrogen atoms last.


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3. Insert pairs of electrons between all pairs of atoms that are to be bonded together. If this uses up all available electrons, go to Rule 6.


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4. Place any remaining electrons on peripheral atoms as unshared pairs, starting with the most electronegative such atom. • Fill this atom up to an octet. • Then proceed to the next most electronegative, and so on. • Remember that hydrogens can only have two electrons, and so cannot have any unshared pairs.

5. If electrons still remain unused, place them on the central atoms as unshared pairs, again beginning with the most electronegative atom.

Fill that atom to an octet. Then proceed to the next most electronegative, and so on.

By application of Rules 4 and 5, the structures of our examples become:


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6. Examine the resulting structure. You will observe one of five situations:

a. All of the atoms in the structure will have octets, except the hydrogens, which will have two electrons each. Go to Rule 7.

b. The central atom is a Be with 4 electrons, or a B or Al with 6. These elements do not obey the octet rule. Go to Rule 7.

c. The molecule has an odd number of electrons, which results in one of the central atoms having only 7 electrons. Go to Rule 7.

d. The central atom has Z > 11, and has other than an octet. Go to Rule 7.

e. The central atom is C, N, or O and i. the number of electrons in the molecule is even and the central atom lacks an octet; or

ii. the number of electrons in the molecule is odd and the central atom has fewer than 7 electrons.

These are often called free radicals. The unpaired electron makes the molecule reactive. Free radicals have been implicated in aging and cancer. In an effort to pair up the single electron, free radicals may also form dimers (molecule pairs). For example, NO2 dimerizes to produce N2O2

In either case, move an unshared pair from a peripheral atom to make a double bond to the central atom. • If the central atom still has too few electrons, move another pair from the same atom to make a triple bond, or a pair from another atom to make a second double bond. When this action can be taken in more than one way, write all possible ways as separate structures. • You have discovered resonance; the actual structure is a hybrid of all of the individual structures.

Our two remaining example structures thus become:


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7. Examine every atom in the structure and assign it a formal charge as follows:

formal charge = (number of valence electrons on the neutral, uncombined atom) (number of covalent bonds to the atom in the current structure) (the number of unshared electrons [not pairs] on the atom in the current structure)

• When counting covalent bonds, count double bonds (two pairs of electrons) as 2 and triple bonds (three pairs of electrons) as 3.

• The sum of all formal charges of all atoms must equal the given charge on the molecule or ion. • No unreasonable charges should result from this process. That is, electronegative atoms should not get two (+) charges, metals should not get () charges, and so on. If this happens, you surely have done something wrong. Go back to Rule 1 and start over. • If you get a lot of nonzero formal charges in a molecule that is neutral overall, you probably have misplaced the hydrogens. Go back to Rule 2.

For example, for the O in POCl3, we have: 6 (1 covalent bond) (6 unshared electrons) = 1 whereas for the P: 5 (4 covalent bonds) (0 unshared electrons) = +1 The full set of formal charges for the example molecules is shown on the structures above.


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8. Rules 17 may allow some collections of atoms to form several valid structures with differing arrangements of atoms. This is OK; the alternative arrangements of atoms are called isomers. Isomers are particularly common among compounds of C, N, and O. For example, two isomeric Lewis structures can be written for C2H6O and four can be written for HCNO.

9. Although four of the five cases in Rule 6 require no alteration of the structure, each case represents a normal bonding condition for the atoms involved. The Octet Rule actually applies only to C, N, O, and F. Bonding conditions unusual with respect to the Octet Rule will be found to be perfectly satisfactory if the orbital interactions are considered instead of forcing the molecule to use electron pair bonds.

10. Application of Rule 17 also may lead to multiple structures having the same arrangement of atoms but different placement of electrons. This is OK too. These are resonance structures: the molecule actually resembles an average of all of the structures rather than any single one. Like the cases in Rule 9, these also will go away when we consider orbital interactions.


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Dipole MomentsYou can estimate the polarity of a molecule by using the difference in EN.

The element with the higher EN will have the partial charge and the other will have a partial + charge.

A better method to be more accurate is by calculating the Dipole Moment.

The Dipole Moment is a measure of the diffenece in charge, q, on two covalently bonded atoms and the distance, r, between the two nuclei.

Dipole Moment = q x r; dipole moment is an absolute value.

The units for the dipole moment are coulombmeter (Cm) and a common unit for the dipole moment is the debye.

1 debye = 3.34 x 1030 Cm


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How does MO theory explain bond length trends for O2 and its ions?

Are these bond lengths consistent with the bond order determined from a molecular orbital diagram?

O2+ = 112 pm

O2 = 121 pmO2

= 128 pmO2

2 = 149 pm

Bond strength depends on how stretchy or stiff the bonds are (with stiffer bonds being stronger). • Stretchier bonds tend to be longer bonds. Since an increase in bond order leads to stronger bonds, there is an expectation that higher bond order means shorter bonds. It's a reasonable expectation when you're comparing bonds between a particular pair of atoms, as you are.


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Valence Bond Model (VSEPR) vs. Molecular Orbital Theory

Because arguments based on atomic orbitals focus on the bonds formed between valence electrons on an atom, they are often said to involve a valencebond theory.The valencebond model can't adequately explain the fact that some molecules contains two equivalent bonds with a bond order between that of a single bond and a double bond. The best it can do is suggest that these molecules are mixtures, or hybrids, of the two Lewis structures that can be written for these molecules.

This problem, and many others, can be overcome by using a more sophisticated model of bonding based on molecular orbitals. Molecular orbital theory is more powerful than valencebond theory because the orbitals reflect the geometry of the molecule to which they are applied. But this power carries a significant cost in terms of the ease with which the model can be visualized.


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Bond Order

Bond orderis based on the MO theory.

It is a term that refers to the average number of bonds that an atom makes in all of its bonds to other atoms.

Instead of using formula, easy way...look at lewis structure.

Bond order is given by

Bond order = [number of bonding electrons – number of antibonding electrons]/2


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The bond length is the distance between those two atoms. The greater the number of electrons between two atoms, the closer the atoms can be brought towards one another, and the shorter the bond.

The BO is an indication of the bond length, the greater the bond order, the shorter the bond.


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In the table below, the bond orders are predicted from electron configurations by subtracting the number of electrons in bonding orbitals from the number of electrons in antibonding (*) orbitals, and dividing by two.


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In what way does a bond change as the bond order increases?

Both the length and strength of a bond change with bond order.


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One analogy you can use is to think about atoms as nerf balls and bonds as rubber bands. The rubber bands act as the force which holds the balls together, as we increase the number of rubber bands the balls are squished closer together and it takes more force to pull them apart.

In a molecule as you increase the number of electrons shared between two atoms, you increase the bond order, increase the strength of the bond, and decrease the distance between nuclei.

Bond strength by measuring the amount of energy needed to break a bond


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Calculating Bond EnergyThis basically means that you add up all the energies of the broken bonds add up all the energies of the bonds that are reformed and subtract one from the other.

Its another version of Hess' Law . Similar to the one that can be used when you know all the enthalpies of formation for the substances in a reaction.


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Step 1 Write and Balance the equation.

Calculating Bond EnergyQuestion: Calculate the bond energy for the complete combustion of propane.

Step 2 Draw the structural formula under equation.

This allows you to see all the different types of bonds.


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Step 3 Determine the number of bonds broken and formed. (reactants and products)

We now need to work out how many of each bond type we have broken. • 8xCH • 2xCC • 5xO=O

And then how many bonds have been formed!• 6xC=O • 8xHO


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Step 4 Use a chart to determine the energy for each type of bond.

Notice they are all endothermic.

So we can now do the sum, remember, sum of bonds broken sum of bonds formed.

= [(8x413)+(2x347)+(5x498)] [(6x805)+(8x464)]

= 2054 kJ mol18x CH 2x CC 5x O=O 6x C=O 8x HO

The value for the enthalpy of combustion of propane from the data table is 2219kJ mol1 . means exothermic

*Note: BE =

Step 5 Plug and Chug


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Key Concepts• Bond energy (enthalpy) is the energy (enthalpy) required per mole of gaseous compound to break a particular bond to produce gaseous fragments. • Bond energies (enthalpies) are positive. • delta H is positive. • Breaking bonds is an endothermic reaction. • Bond formation releases energy. • delta H is negative. • Bond formation is an exothermic reaction. • Bond energies (enthalpies) can be used to indicate how stable a compound is or how easy it is break a particular bond. • The more energy that is required to break a bond, the more stable the compound will be. • A larger bond energy implies the bond is harder to break, so the compound will be more stable. • Bond energies (enthalpies) can be used to estimate the heat (enthalpy) of a reaction.

AP Chemistry Chapters 7-9 - Quia · The Lewis structure is used to represent the covalent ... AP...

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Transcript of AP Chemistry Chapters 7-9 - Quia · The Lewis structure is used to represent the covalent ... AP...