bonding, molecular geometry, intermolecular forces, classification of · pdf...

UNIT 6 1

HONORS CHEMISTRY

HARVARD-WESTLAKE

UNIT 6

Bonding,

Molecular Geometry,

Intermolecular Forces,

Classification of Substances

A group of organic molecules were having a party, when a group of robbers broke into the room and stole all of the guest's joules. A tall, strong man, armed with a machine gun came into the room and killed the robbers one by one. The guests were very grateful to this man, and they

wanted to know who he was. He replied:

My name is BOND, Covalent Bond.

UNIT 6 2

Reference Sheet

IA

VIIIA

H

2.1 IIA Electronegativities of the Elements IIIA IVA VA VIA VIIA

Li

1.0

Be

1.5

B

2.0

C

2.5

N

3.0

O

3.5

F

4.0

Na

0.9

Mg

1.2 IIIB IVB VB VIB VIIB VIII IB IIB

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

UNIT 6 3

Drawing Lewis Dot Structures

Gilbert N. Lewis first proposed a conceptual framework for covalent bonding in 1915 and suggested a system of dots to represent in a simple manner how these bonds might be treated.

In the examples we have seen so far, we have used a pictorial process of "fitting" atoms in to make octets. This works well for very simple molecules but it is sometimes difficult with more complex structures. For that reason we can introduce a series of steps which will generally result in a "correct" Lewis structure for most molecules. 1) Draw a skeleton structure for the molecule or ion, joining atoms by single pairs of electrons

(bonds). You need to know the central atom or centers of symmetry to do this – the center is usually the atom of which there is only one and/or the first atom in the formula. Keep shapes simple: linear, trigonal planar, T-shaped, square planar, etc.

2) Sum up the number of electrons (dots) available from all of the atoms in the molecule.

Remember to ADD electrons for negative charges on ions and SUBTRACT for positive ions. 3) Deduct the electrons already used in setting up the diagram and fill in the remainder so that

each atom has an octet, beginning with the "outer" atoms and ending with the center. 4) If there are not enough electrons for the central atom to have a full octet, you may use double

or triple bonds by sharing more than one pair of electrons between atoms. CHCl3 (chloroform) 26 total valence electrons - 8 bonding electrons 18 electrons to distribute

CO2 16 total valence electrons - 4 bonding electrons 12 electrons to distribute

ClO- 14 total valence electrons - 2 bonding electrons 12 electrons to distribute

NH4

+ CH2O (formaldehyde) SO32-

UNIT 6 4

PCl3 CN- Cl2O Of course, there are always the exceptions. Sometimes there is no reason to prefer one possible Lewis structure over another equivalent but different one. CO3

2-

There are also some molecules that don't obey the simple rules. One group of those molecules manages to remain intact with fewer than 8 electrons in the valence level of the central atom. BF3 24 total valence electrons - 6 bonding electrons 18 electrons to distribute

BeH2 4 total valence electrons - 4 bonding electrons 0 electrons to distribute

BeH2 4 total valence electrons - 4 bonding electrons 0 electrons to distribute

Another group of exceptions is only possible with more than 8 electrons in the valence level of the central atom. This is generally not seen until phosphorus and beyond. PCl5 I3

- XeF4

X-ray measurements of bond lengths show that all of the C-O bonds are identical: shorter than single bonds but longer than double bonds. They are about what one might expect for 1 ½ bonds. So, we use three diagrams to represent this structure, recognizing that none of the pictures actually represents the ion as it really exists. We say the electrons from the double bond are delocalized and we call the phenomenon resonance.

UNIT 6 5

VSEPR (Valence Shell Electron Pair Repulsion)

Type Bond Angle Example Lewis Diagram Structure

AB2: Linear 180o BeH2


AB3: Trigonal Planar 120o AlH3


AB4: Tetrahedral 109.5o CH4


AB5: Trigonal Bipyramidal

90o, 120o, 180o

PCl5


AB6: Octahedral 90o, 180o SF6

1. Draw the Lewis structure 2. Determine the number of electron pairs on the central atom 3. Match the number of electron pairs with the shape category 4. Draw the 3-D structure, placing un-bonded or "lone pairs" appropriately

UNIT 6 6

UNIT 6 7

Practice with VSEPR

Lewis Diagram e- pairs Type Structure

BF3


O3


H2O


SF4


NH3

UNIT 6 8

Practice with VSEPR (continued)


ClF3


XeF2


XeF4


IF5

UNIT 6 9

Summary of Weak Intermolecular Forces

UNIT 6 10

Summary of crystalline solids

Metallic Network covalent (macromolecular)

Ionic Atomic /

Molecular

Examples Particles

Na, Fe, Cu “almost” cations

diamond, SiO2 atoms

NaCl, CaO ions

Ne, I2, CH4, H2O atoms/molecules

Forces of attraction

“sea” of mobile valence electrons binds positive nuclei by strong bonds

atoms linked throughout 3D structure by very strong covalent bonds

positive ions attract negative ions—strong ionic bonds

atoms/molecules held next to each other by weak intermolecular forces

Properties Volatility State at RT Hardness Conductivity Solubility

non-volatile high mp and bp usually solid some hard, most malleable liquid and solid only in liquid metals

non-volatile VERY high mp and bp solid VERY hard and brittle non-conductors insoluble in most everything

non-volatile high mp and bp solid hard and brittle conductors when melted and in water solution most soluble in polar solvents

often volatile med. to low mp and bp often gases or volatile liquids soft non-conductors some soluble in polar solvents, some in non-polar solvents

UNIT 6 11

LAB: Gas Chromatography and Intermolecular Forces You know from your studies that structure plays an important part in the strength of intermolecular forces such as hydrogen bonding, dipole forces and dispersion forces. In turn, the strength of such forces influences physical properties such as phase change temperature, the energy required to complete a phase change, and solubility in various solvents. One way to compare the intermolecular forces among various molecules is to measure the rate of vaporization. At a constant temperature and pressure higher evaporation rates indicate weaker forces if all other factors are roughly equal. There is a modern analytical method closely related to this behavior and based on intermolecular forces which illustrates the relationship between structure and weak forces: gas chromatography. You may have done experiments in paper chromatography before, separating mixtures of colored compounds. All the various chromatographic separation methods have some principles in common. In each there is a stationary phase. In gas chromatography this phase is generally a compound which has been used to coat a solid, or perhaps the solid itself. This stationary phase is packed into a column or tube. In all but the simplest gas chromatographs (GCs) this column can be heated. The sample to be analyzed is generally injected on to the column where the sample molecules adhere to the stationary phase. How strongly they "stick" is governed by the kinds of intermolecular forces at work between them and the stationary phase. Next some kind of "solvent" or mobile phase is passed through the column. In gas chromatography the mobile phase is--surprise!--a gas such as helium or nitrogen, but any light gas will do if it does not react with the two phases. As the gas sweeps over the column packing it begins to dislodge some of the adsorbed sample molecules according to how tightly they are held by the stationary phase. Ones which are held more loosely move through the column more quickly and eventually arrive at some kind of detector. Molecules which are more strongly attracted to the stationary phase arrive later.

Thus a mixture can be separated on the column and even quantified. The time it takes a substance to emerge from the column is called the "retention time" and is characteristic of the compound under consistent experimental conditions.

In this experiment you will examine the behavior of three liquids (which vaporize readily in the stream of carrier gas) in a simple GC. The liquids are all hydrocarbons. Hydrocarbons are organic compounds (as distinguished from the inorganic compounds which you are more familiar with). Although only a relatively few elements are involved in the structure of hydrocarbons, there are many of them, principally because carbon is able to form up to four covalent bonds. Such carbon atoms are sp3 hybrids and so the bond angles around each carbon atom are 109.5o. The simplest case would be the compound methane (see right). This molecule is tetrahedral and non-polar due to symmetry. Additional similar compounds exist in which the carbon skeleton is simply longer. For example, C2H6 is ethane (see right). This molecule is also non-polar and the geometry around each carbon atom is tetrahedral. This compound and others similar to it are called alkanes. You will use several alkanes in this experiment. The compounds propane, butane, and octane are familiar examples of other alkanes. Note that these compounds are all used as fuels. Another reason for the large number of hydrocarbons is the existence of isomers. These are compounds with the same chemical formula but a different arrangement of atoms. Consider the two structures below, both of which are C4H10:

butane 2-methylpropane

These are but two examples of the many thousands of alkane isomers. Since their structures are different, you might expect their intermolecular forces to be different.

UNIT 6 12

The compounds used in this experiment are shown below:

hexane 3-methylpentane 2,2-dimethylbutane

These compounds are all isomers of hexane, C6H14. However, while their chemical formulas and molar

masses are the same, their structures are different enough to result in intermolecular forces of varying strengths. Thus they can be separated by a simple GC. A diagram of the GC you will use is shown below.

The column is made from glass tubing that has

been filled with Tide® laundry detergent. The detergent contains many different kinds of compounds but the actual detergent molecules are structured to have one end that is polar and one that is non-polar (see right).

Thus a variety of substances will interact to different degrees with the column packing in this simple GC.

But the long hydrocarbon chain of the soap will have strong dispersion interactions with the alkanes. One end of the column is connected via rubber tubing to a supply of carrier gas (the mobile phase) which is the natural gas used in the lab (mostly methane). The other end of the column terminates inside the plastic tube. It is attached to a glass tube drawn out as a jet. The carrier gas is lit there as it exits the column, producing a small flame. A cutaway diagram of the inside of the lower tube is shown below.

The CBL light probe is sensitive to both visible and infra-red radiation. The flame is initially adjusted to be small and nearly all blue so that mostly infra-red radiation is emitted. When one of the alkane isomers emerges at the end of the column it too burns, increasing both the luminosity and heat of the flame. The light probe converts this into a voltage which it sends to the CBL.

UNIT 6 13

Preparing to experiment You will be provided with the following materials: 1. samples of hexane, 3-methylpentane and 2,2-dimethylbutane 2. a mixture of two of the isomers 3. a simple gas chromatograph Design an experiment to determine the retention times of each pure substance in the column and the identity of the unknown mixture. The chemicals Hexane is one of the chief constituents of "petroleum ether", and is a colorless, very volatile liquid with a peculiar odor. It is insoluble in water. It is used for filling some "red-liquid" thermometers (as a substitute for mercury) and for determining the refractive index of minerals. Hexane is often sold as "hexanes" which is a mixture of common isomers, chiefly the actual hexane molecule but also including some of the isomers used in this experiment. Pure hexane is markedly more expensive due to the cost of purification. Analysis These questions should be answered in your laboratory notebook following your observations. 1. What are the retention times? 2. In light of the intermolecular forces possible for each compound and your prediction in the pre-lab take-home

quiz, explain the order in which the compounds emerged from the column. 3. What are the components of your unknown mixture? Explain your choices.

UNIT 6 14

UNIT 6 15

Name:___________________________ Per.:____ Date:_______________

Plant Pigments and Paper Chromatography

Purpose: to determine the best solvent mixture ratio (2-propanone/petroleum ether) for the separation of the major pigments in common ivy by paper chromatography and attempt to relate the movement of the pigments to their structures Method: I will prepare 5 strips of chromatography paper by labeling them and marking pencil lines where the samples will be applied. I will use a quarter to roll over an ivy leaf pressed against the paper strips in order to transfer pigment. I will then prepare 3 mixtures of 2-propanone/petroleum ether and use them, along with the pure solvents, to attempt to separate the pigments. I will place the strips in test tubes containing the solvents, stopper the tubes and wait for the solvent to travel up the strips. Once the strips have been removed from the solvents I will mark the location of the pigments, identify them, and try to correlate their positions with the relative polarity of the solvent mixtures and structural features of the pigments. Data: (Tape paper strips below) 2-propanone : petroleum ether 20:0 6:14 4:16 2:18 0:20

UNIT 6 16

UNIT 6 17

LAB: Separation of Plant Pigments by Paper Chromatography

In a previous experiment you have seen how differences in the polarity of molecules can enable their separation through a process called chromatography. The name of the process indicates that originally it was a method having something to do with color. In gas or vapor phase chromatography this origin is obscured but in this experiment it is evident. Anyone who has ever played on grass or had an unplanned encounter with a bush knows that plants contain pigments which adhere to varying degrees to fabric (and skin!). The fact that a small industry has risen up in aid of attempts to remove stains like these and many others shows that not all of the molecules can be treated in the same way if their removal is desired. Part of the reason behind that is different fabrics and the varying polarity of their molecules. The constituent fibers of cotton, for example, are very different from those of silk or synthetics like rayon. The plant pigments themselves also have varying polarities and so there are a number of possible combinations to address in stain removal. The point of this experiment is not to develop a wash-day solution to all your grass stain problems but rather to look at the polarity of some of the common pigments in plant leaves (specifically common ivy) and how that polarity affects their interactions with the cellulose fibers in paper and a few solvents. In paper chromatography substances are applied to a piece of absorbent paper. A solvent is then allowed to move through the applied substance. Sometimes the motion is radial, sometimes descending, sometimes (as in this experiment) ascending. In all cases the "wicking" action of the paper disperses the solvent in a particular direction and depending on the solubility of the applied substance, may move it along as well. This kind of effect can easily be seen with an ordinary water-based marking pen and a piece of paper toweling. In many applications of paper chromatography the interaction of the paper, solvent and applied substance is very complex. The stationary phase would seem by simple logic to be the paper and the mobile phase which moves the applied substance along would seem to be the water. However, water can hydrogen bond with the cellulose in paper and in cases where the solvent is an aqueous solution, the water on the paper is often considered the stationary phase! The interpretation of such experiments in terms of the polarity of substances is often very difficult. In order to avoid such problems, we have chosen a non-aqueous solvent system for this experiment. One of the solvents is petroleum ether, a mixture of non-polar compounds including pentane, C5H12, hexane, C6H14, and 2-methylhexane, C7H16. Because of the inherent symmetry at each carbon of the molecules, the only significant intermolecular forces in these hydrocarbons are dispersion forces:

In contrast, 2-propanone (acetone), C3H6O, has a structure in which an oxygen atom on the center carbon of the three-carbon chain creates a definite molecular dipole (see right):

Unlike these small molecules, the cellulose in paper is a polymer (made up of many repeating unit structures connected as shown by the shaded oxygen atoms):

O

UNIT 6 18

In the diagram above each vertex is occupied by a carbon atom which is not explicitly shown. The most important characteristic of this huge structure (which extends out at both "ends" shown here) is the presence of the -OH groups that permit hydrogen bonding and other strongly polar interactions. There are many different pigments in plant parts. Most are large molecules with hydrocarbon skeletons like that in cellulose. With such large molecules it is often difficult to make judgments about polarity. So for this experiment we have selected four compounds with distinct structural features. These compounds are present in relatively large amounts in ivy leaves and they also have characteristic colors which are easily seen in a simple chromatogram.

chlorophyll a

chlorophyll b

-carotene

xanthophyll

The two forms of chlorophyll are identical except for the boxed part in chlorophyll b. Both compounds have a number of oxygen atoms on their hydrocarbon skeletons and that renders them somewhat polar. The additional structure on the b form is of interest in this experiment.

-carotene and xanthophyll are also identical except for the boxed structures on the xanthophyll. The -carotene molecule is essentially non-polar, having only a hydrocarbon skeleton like hexane (although much more complex). The additional structures on the xanthophyll molecule should give it different properties. In this experiment you will apply these plant pigments from ivy leaves to strips of chromatography paper. The strips will then be suspended in the two solvents already mentioned (petroleum ether and 2-propanone) as well as mixtures of the solvents. The pure solvents will readily distinguish between the polar and non-polar molecules.

UNIT 6 19

Petroleum ether should dissolve the non-polar compounds and move them along the paper.

2-propanone should dissolve the polar compounds and move them along the paper. Different degrees of polarity may be discerned by the gradual separation of compounds as they are carried along the paper. The more soluble (i.e., more polar) substances should move more closely with the solvent front, others will lag behind a little. Because cellulose itself is polar there may be some interaction with the polar pigment molecules but these should be considered minimal compared to the interaction with the 2-propanone. The mixed solvents may suggest an "optimum" mixture in which it should be possible to achieve a nearly complete separation of all four compounds as well as further illuminate the differing polarities of the pigments. Preparing to experiment You will be provided with the following materials: 1. five strips of chromatography paper 2. five test tubes w/rack and rubber stoppers 3. petroleum ether, 2-propanone 4. fresh ivy leaves 5. forceps 6. ruler Design an experiment to determine the mobility of plant pigments in various solvent mixtures using ascending paper chromatography. [See Technique section.] Technique 1. Applying the pigments You should make a pencil line about 1 cm from the end of each strip. This is where the pigments will be applied. The idea is to not have them sitting in the solvent when they are placed in the test tubes. To apply the pigments, place an ivy leaf, top side down, directly on a strip so you can just see where the pencil line is and roll a large coin (a quarter is excellent) firmly across the leaf, pressing pigment onto the paper. Choose a fresh surface of the leaf and repeat this for the other strips. Then return to the first strip and apply more pigment in the same place. Do this about 5-10 times on each strip, allowing some time for the lines of pigment to dry in between applications. The pigment stain must be TOTALLY dry before running the chromatogram. 2. Preparing solvent mixtures You only need a small amount of liquid in each test tube--enough to run up the paper strip but not so much as to have the applied pigment line sit in the liquid during the experiment. 20 drops total should be adequate. For the mixtures you can use drop ratios to approximate the following (listed as volume acetone : volume petroleum ether): 6:14, 4:16, 2:18. 3. Developing the chromatograms This is the easy part. Draw a second pencil line about 1 cm from the end the strip opposite to where the pigments have been applied,. Label each strip for the solvent mixture. After placing the strips in the appropriate test tubes, stopper the tubes and wait until the solvent has reached the upper pencil line. Remove the strips from the test tubes and allow them to dry. In the meantime empty your test tubes into the beakers in the fume hood. The spots have a tendency to fade when exposed to light so you might want to circle them (pencil) and maybe make a few notes before leaving the lab.

UNIT 6 20

Analysis [In most chromatograms of this sort -carotene is an intense yellow while xanthophyll is a very pale yellow. The chlorophylls will be green, with b being somewhat “dirty” or olive green). It might also be helpful to note here that when molecules have similar polarities but very different sizes, the larger molecule will typically move more slowly in chromatography.] 1. Examine the strip which was developed with pure petroleum ether. Which pigment(s) appear(s) to have traveled

the farthest? Explain briefly. 2. Examine the strip which was developed with pure 2-propanone (acetone). Which pigment(s) appear(s) to have

traveled the farthest? Explain briefly. 3. Select the strip developed in the mixture which shows the best separation of the four pigments. What is the

solvent ratio? Considering the introductory remarks above and your answers to (1) and (2), explain briefly the order of the pigments.

4. Attach your five chromatograms to your lab report using scotch tape. DO NOT use staples!

UNIT 6 21

LAB: Classifying Solids

The behavior of substances in the solid state is governed mainly by the way in which the atoms, ions or molecules of the substance are arranged and the forces that hold them together. Although there are many different substances, the types of solids can be roughly divided into 4 categories: molecular network covalent (or macromolecular)

ionic metallic

Each of these is distinguished by the kinds of particles that make up the three-

dimensional structure of the solid. If the structure has a regular pattern, the solid is described as crystalline. A familiar example would be sodium chloride, NaCl, in which sodium and chlorine ions alternate in an extended structure known as a space lattice:

In this structure each ion occupies an imaginary point in space called a lattice point. This is an example of an ionic solid since ions are the structural unit. A more realistic representation of the arrangement of ions in the crystal would look more like the structure below:

Any crystalline solid would exhibit a similar orderly arrangement of particles. Some solids, however, do not have such organization on the molecular level. Substances such as glass, rubber and many plastics are called amorphous solids because they lack a typical space-lattice organization. In this experiment we are concerned only with crystalline solids.

Molecular crystals consist of either molecules or atoms in orderly arrangements. These arrangements are maintained by weak attractive forces such as hydrogen bonding, dipole forces and dispersion forces. Because of this, such solids generally have low to medium melting points (below about 400oC).

Since the individual units that make up the molecular crystal are neutral in charge and cannot move appreciably in the solid state, these solids are poor electrical conductors. Even when melted, electrical conductivity is negligible. And if the solids dissolve in water there is still very little, if any, conductivity due to lack of charges on the molecules or atoms. Some molecular crystals will dissolve in water (if the units of the crystal are polar) while others which are composed of non-polar molecules or atoms may dissolve in non-polar solvents such as hexane. (Recall the solubility of the non-polar halogens in hexane).

Most organic compounds (composed chiefly of C, H and O) form molecular crystals. While these generally melt at low temperatures (less than 250oC) they often decompose or char before melting.

Ionic crystals are, of course, composed of ions. These are held tightly in place in the space-lattice by strong electrostatic forces because of the opposite charges of the ions. These very strong forces contribute to high melting points, although some hydrated ionic compounds may appear to "melt" at lower temperatures (see below).

Although the ions are charged, they cannot move in the solid and so these crystals do not conduct electricity. However, when melted, most of the attractions have been overcome and molten ionic compounds conduct very well. Most ionic compounds also dissolve to at least some extent in water. The amount that dissolves is completely separated into ions and thus makes an electrolyte solution which is a good conductor.

Hydrated ionic compounds contain water molecules at some of the lattice points. These molecules are held in place by the attraction of the polar water molecules for the charged ions. However, heating will overcome these attractions and release the water (the crystal structure generally collapses or at least changes). Sometimes if the water is released slowly by gentle heating it can dissolve the ionic solid and make it appear as if it is melting at a low temperature.

Network covalent crystals (or macromolecular solids) are held together by a network of covalent bonds that extends throughout the solid sample. Diamond is a classic example in which each atom of carbon is attached to

UNIT 6 22

four other carbon atoms. This extensive framework of covalent bonds makes these solids very hard. Also, since each bond must be broken to melt the solid, melting points are typically very high (higher than 1000o). Network crystals do not conduct electricity because they lack movable charges, and their tight covalent structure makes them insoluble in essentially everything.

Metallic crystals consist of metal cations (minus valence electrons) positioned at the lattice points of the crystal. The valence electrons are given up to a common "soup" or "sea" which flows more or less freely throughout the sample. The lack of directionality in metallic bonding makes most metals bendable and rather soft (certainly compared to a diamond!). Metals are excellent conductors because of all the mobile electrons. This conductivity exists in both the solid and liquid state. The melting points of metals vary widely from mercury at -38oC to tungsten at 3410oC. The effects that contribute to the wide-ranging melting points of metals are beyond the scope of this discussion.

Metals are essentially insoluble in common solvents, although they undergo vigorous chemical reactions with a variety of substances in which they become oxidized and move into aqueous solution as ions. And, of course, metals are recognizable by their optical properties which most other solids lack: lustre and shininess. Preparing to experiment You will be provided with the following materials:

1. four solid unknowns 2. heavy aluminum foil

3. 3 small test tubes 4. a 24-well microplate

5. conductivity device 6. hexane

Design an experiment to identify each solid unknown as either molecular (polar), molecular (non-polar), ionic, hydrated ionic, or network covalent. Technique 1. estimating melting point range The maximum temperature you can get from a bunsen burner is about 400oC. You can determine whether a solid melts above or below this temperature by placing a small amount (smaller than a match head) of the solid on a square of aluminum foil placed on the wire screen for your tripod. All of the solids can be tested simultaneously. Heat from below gently at first, then more strongly and observe the behavior of the solids carefully. Discard the foil square when it has cooled. 2. checking for water of hydration (ionic solids only) Solids suspected of being ionic can be checked for hydration by heating a small amount in a test tube and looking for signs of water condensation on the upper part of the test tube. Heat gently. 3. solubility in hexane (non-polar solvent) CAUTION!!!! Hexane is flammable. No flames should be used in the laboratory when hexane is in use. Hexane will also attack the 24-well plates and should only be used in glass test tubes. Use a minimal amount (1-2 mL) and a few crystals of solid. Discard used hexane in one of the containers found in the fume hoods. Analysis Prepare a table of the solids you have tested. Refer to each by its unknown number. Indicate the kind of solid you believe it to be and give a short summary of the experimental evidence that led you to your conclusion. Contradictory observations should be explained in more detail following the table.

UNIT 6 23

Name:___________________________ Per.:____ Date:_______________

Vapor Pressure and Structure

Purpose: to compare the relative vapor pressures of hexane, 3-pentanone and 1-pentanol over a range of temperatures and attempt to use the values to correlate vapor pressure with structure and intermolecular forces Method: After identifying which manometer is connected to which test tube I will gradually raise the temperature of the water bath containing the three samples by removing room-temperature water and replacing it with hot tap water. As the heights of liquid in the manometer arms stabilize at various temperature intervals I will record them. I will continue in this way until the sample with the apparent highest vapor pressure is near to ejecting the manometer fluid. With the relative vapor pressure values I will prepare a graph of the vapor pressures of the three liquids as a function of temperature. I will use the graph to compare the vapor pressures and relate them to structural differences among the compounds. Data:

Temp. (oC)

#1 ____________

(mm)

#2 ______________

(mm)

#3 _____________

(mm)

UNIT 6 24

Data (continued):

Temp. (oC)

#1 ____________

(mm)

#2 ______________

(mm)

#3 _____________

(mm)

UNIT 6 25

LAB: Vapor Pressure and Structure The physical changes in substances that we identify as phase changes may be generally viewed as results of changing the distance between the molecules. While this can be accomplished in a number of ways, surely the most common is to increase the kinetic energy of the molecules by heating. But even at temperatures between phase changes the molecules possess some kinetic energy and therefore some kind of motion occurs. This is a basic tenet of kinetic theory. Moreover, at the surface of a sample the forces that tend to keep a solid intact or a liquid together are not completely balanced. Some surface molecules are able to escape the sample. We recognize this in the evaporation of water in an open container which is at room temperature. Even some fairly common solids (such as moth balls) lose noticeable quantities of molecules from their surfaces below the melting or boiling point. If the container of water in the previous paragraph is closed, molecules still escape the surface on occasion but they will eventually return to the liquid state. Over time, a certain average proportion of molecules exist above the surface of the liquid where they collide with the walls and lid of the container and exert small forces. The sum of all these forces is a measure of the pressure exerted by the molecules in the vapor phase---what is commonly called vapor pressure. Why do some substances have higher vapor pressures than others? How can relative vapor pressure be measured? These are some of the questions you will investigate in this experiment using the compounds shown below:

hexane, C6H14 1-pentanol, C5H12O 3-pentanone, C5H10O Preparing to experiment You will be given the apparatus shown below:

Design an experiment to determine the vapor pressures of the various liquids at temperatures above room temperature.

O

O

UNIT 6 26

Pre-lab take home quiz 1. What kinds of intermolecular forces are most significant for each of the three compounds on the facing page? 2. In which of the liquids would you expect the intermolecular attractive forces to be strongest and WHY? 3. Would you expect molar mass to be a significant factor in the results of this experiment with these compounds?

WHY? Technique 1. Setting up the apparatus Most of the set-up is already done for you when you come into the lab. You simply need to equalize the pressure inside the test tubes with air pressure. To equalize the pressure, turn the handle on the little valve so that it points away from the rubber tubing. This opens the test tube to the air temporarily. The manometer fluid should more or less self-level. Then turn the valve handle to point toward the open stem (i.e., the air). This closes the system to the air. Finally adjust the sliding rulers near the manometers so that each starts at 0.0 mm. 2. Measuring relative pressure As you will have read, heard in class, or soon will find out, pressure is measured in a number of ways. Historically the units of choice have been mm of Mercury (mmHg). You may be more familiar with units of pounds per square inch (psi) which are commonly used in this country for the air pressure in automobile tires. In this experiment you will measure in mm, but of water. Since the intent of the experiment is only to compare vapor pressures, no attempt is made to convert these units into others. In fact, the absolute vapor pressures of the liquids cannot be measured in such an apparatus used as suggested below since a fixed sample of air is also trapped in the test tubes and expands as the temperature rises. This adds to the "apparent" pressure indicated by the manometer; however, since the air volume in each test tube is roughly the same, we can assume any differences in pressure are due to the different liquids. Also, the liquids have finite vapor pressures at room temperature. When the apparatus is assembled, you "zero" out any of that. Consequently, you are truly only comparing the changes in vapor pressure as each liquid is heated. 3. Temperature regulation Finding the right rate to heat the samples is a little tricky. One simple method is to start with a water bath at room temperature. You can use a turkey baster to remove the room temperature water and replace it with hot water (the hottest water from the faucet--it can be stored in the styrofoam ice buckets). By using the magnetic stirrer you can help keep the water mixed well. When the new temperature levels off, you can record the readings on the manometers. Then withdraw another sample of water and replace it with more hot water. In this fashion, you can slowly increase the temperature of the samples without heating so much that the manometer fluid is expelled! Analysis 1. At a given temperature, which liquid shows the greatest vapor pressure? Why do you think this is so? [hint:

consider the intermolecular forces in these liquids] 2. Which liquid would you predict to have the highest boiling point? [hint: during boiling all of the molecules are

escaping from the liquid] 3. What factors can you think of that will influence the relative vapor pressures of liquids or solids? 4. Graph the manometer readings vs. temperature for all three liquids on the same graph. Describe the change in

vapor pressure with temperature (i.e., is it directly proportional, etc.)

UNIT 6 27

So you want to use EXCEL to plot the vapor pressure curves.......

The procedure is similar to what you were shown in class and given on the handout. The temperature data goes into the first column (A) and the "vapor pressures" (the mm readings from the rulers) go into columns B, C, and D. When you select the data to prepare to plot, select all four columns.

This data is not linear so linear regression is not appropriate. Instead you want a smooth curve through the data points. When you try to add a trendline, Excel suggests options it “thinks” are appropriate for your data. The actual function for vapor pressure and temperature is not one of those suggested but you can approximate the correct curve with the “polynomial fit”. You will have to do this for each data series (i.e., three times). [remember, you can't do anything to the graph unless it is "selected"--double-click anywhere on the graph to "get in"]

If you want a legend (a little box that says what each line is for), you can go to the Insert menu (when the graph is selected) and select Legend. The little box will appear with uninspiring labels like "Series 1". You can change these (to something like the chemical name?) by clicking once on any data point in a series (so all the points are selected) and then going to the Format menu and clicking on the first selection (format selected series or something like that). A box opens up and you select the Name and Values tab. There will be a little blank box with Name: next to it. Type what you want in the legend for that data series and click on OK. Follow the same procedure for the other series.

How do you know if something is supposed to be linear or curved?! Actually, you can do some rudimentary analyis for at least your data with Excel. If you choose the linear trendline and ask Excel to display the R2 value on the chart (not the equation—you are not going to use it and it just clutters up the graph), you will see a number that is probably slightly less than 1.0. The closer it is to 1.0, the better “fit” you data is to a linear function. Now, delete that trendline (right click on it and select “Clear”) and try again but this time choose “polynomial” and remember to check the “R2” box.

Now compare the new value of R2 to the old (linear) one. Whichever is closer to 1.0 is a better approximation of the actual function. In this particular experiment you have three data sets and some may “look” more linear than others because the change in vapor pressure is smaller with temperature. To decide whether the function is linear or curvved, then, you should always select the most extreme change (hexane?). There is no reason for the same property to be linear in one case and curved in another, so if you find a curve in one case, they must all be curves.

Finally, what is a “direct proportion”? In your math classes at one point or another you should have learned that if you multiply x by some constant to get y then y is directly proportional to x. Another way to put that is y = mx. This, of course, is the equation of a line with intercept = 0. Just a reminder!

UNIT 6 28

UNIT 6 29

Reminder……..Instructions for Graphing with Microsoft Excel 1. Double-click the Microsoft Excel icon. A new spreadsheet will appear with vertical columns,

horizontal rows, and a tool bar along the top. In the first column, type your X (horizontal) axis data into each cell. In the second column, type in your Y (vertical) axis data.

2. Position the mouse pointer (it looks like a cross) on the first cell in the first column and press

the left mouse button. Hold this button down and drag the pointer diagonally to the bottom Y-value in the second column. The section will now be highlighted (white on black). The first cell is also selected but will not look like it.

3. From the tool bar, select Insert. From the Insert menu, select the Scatter plot option and

among the types of Scatter plot, select Scatter with only Markers as your plot of choice. Format your graph using the following steps:

a. Delete the legend (Series 1 box) by clicking on it and hitting delete on your keyboard.

b. Under the Chart Tools toolbox menu (only visible when you have clicked on the graph),

select the Layout tab. This will allow you to label your axes and title your graph.

To title your chart, select Chart Title under the Layout tab. Select the Above Chart option. Click and type directly in the chart title text box to re-title your chart. Recall: Graphs should always be titled in a Y vs. X format!

To title each axis, select Axis Titles under the Layout tab. For the x-axis, select Primary Horizontal Axis Title and then Title Below Axis. For the y-axis, select Primary Vertical Axis Title and then Rotated Title. Then, click and type directly into each axis title text box to re-title each x-axis.

c. To zoom in on your data points to present a nice, clear graph, right-click on any of the number values on the x-axis. Select Format Axis… From the pop-up menu, change your minimum and maximum axis options from Auto to Fixed. Then, choose minimum and maximum values that will nicely frame your x values. When done, hit the close button on the pop-up menu. Repeat this process with the y-axis.

d. To add a trendline, right-click on any of the data points on your graph and select Add Trendline… From the trendline options menu in the pop-up menu, select Linear. Also, check the boxes for Display Equation on chart and Display R-squared* value on chart. When done, hit the close button on the pop-up menu and reposition your equation so that it does not cover up any of your data points.

e. To move your graph to full page size, right-click on any white space on the chart outside of the graph and select Move Chart… From the pop-up menu, select New Sheet and hit OK. Your full-page graph will now have its own tab along the bottom of the worksheet.

4. Print your finalized graph by selecting the File tab in the top toolbar, followed by Print.

UNIT 6 30

UNIT 6 31

Unit 6 Sample Test The test will be similar in format to previous tests with five multiple choice questions, two required problems, a choice of one out of three smaller problems, a choice of two out of four chemical reactions to write and one essay question. Remember that on this test the first five solubility rules will be missing. An electronegativity table will be included. The following are representative of typical multiple choice questions but do not necessarily indicate topics to be addressed on your actual test. Use these choices for the following TWO questions: a. BCl3 b. H2O c. NH3 d. PCl3 e. all _____1. Which of the molecules is non-polar? _____2. For which are dispersion forces the only intermolecular forces that would exist? _____3. Select the geometry that best describes NH4

+ a. linear b. bent (v-shaped) c. trigonal planar d. trigonal pyramid e. tetrahedron _____4. The molecule CH4 is tetrahedral in shape. The most likely hybridization of the central

atom is: a. sp b. sp2 c. sp3 d. sp3d e. sp3d2 _____ 5. Assuming roughly equal masses of the particles occupying the lattice points, which

type of solid listed below is likely to have the lowest melting point? a. ionic b. covalent (network) c. molecular d. metallic

UNIT 6 32

The next section consists of typical problems like those that might be found on the test. 6. For each of the following:

a. Draw a correct Lewis structure b. Draw the molecule or ion showing the correct geometry using 3-D conventions c. Give the correct VSEPR classification (AxByEz) d. Tell what kind of hybrid orbitals would be used e. Indicate whether the neutral species are polar or non-polar a. SO2

2- b. IF5 c. NF3 7. Consider the compounds whose structures are represented below. Give the strongest type of

intermolecular force expected in each case and arrange them in order of lowest to highest vapor pressure, based on the nature of the intermolecular forces expected.

KEY:

C H O Cl

C5H12 C4H9OH C3H7Cl C5H12

Cl O

Cl O

UNIT 6 33

8. Consider the reaction AlCl3 + Cl - AlCl4 -

Describe the change (if any) in hybridization of the Al atom and the change in geometry (if any) of the aluminum/chlorine species. Support your answer with orbital diagrams.

9. If you were given a sample of a white solid and asked to determine whether the compound

probably was an ionic solid, a polar molecular solid or a non-polar molecular solid, explain how you would make your decision based on the information given below in sequence (i.e., consider each piece of evidence in addition to whatever has been given before):

a. the solid melts at 110oC b. the solid dissolves in water c. the water solution of the solid does not conduct electricity 10. In each of the following groups of substances, pick the one that has the given property.

Justify each answer. a. highest boiling point: NaCl, N2, H2O b. weakest surface tension: H2O, CH3OH, C5H12 (pentane) c. lowest freezing point: H2, CH4, I2 d. smallest vapor pressure at 25oC: SiO2 (network), CO2, H2O e. greatest viscosity: CH3CH2CH2CH3, CH3CH2OH, HOCH2CH2OH f. strongest hydrogen bonding: NH3, PH3, SbH3 g. greatest Hvap : HF, HCl, HBr, HI h. smallest Hfus : H2, CO2, MgO

UNIT 6 34

The next section consists of representative reactions to complete and write balanced net-ionic equations for. Note that some reactions do not occur in aqueous solution and thus molecular equations are all that would be needed. Each student is expected to choose two from this section. 11. For each of the following, complete the word equation, write a balanced net-ionic reaction,

and tell what type of reaction it is. (All reactions occur in aqueous solution.) a. potassium chloride + silver nitrate __________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ type:______________________ b. iron + hydrochloric acid (a compound containing iron(III) is among the products)____ ___________________________________________________________________________ ___________________________________________________________________________ type:______________________ c. bromine + sodium chloride _______________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ type:______________________ d. iodate ions (IO3

-) and sulfide ions are mixed in a basic solution; iodide ions and sulfur are among the products

___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ The final section of the test will consist of one essay question selected from the following topics: ----concepts behind VSEPR and preferred shapes for molecules and ions ----concepts behind existence of expanded octets ----polar bonds vs. polar molecules ----correlation of structure with liquid properties

bonding, molecular geometry, intermolecular forces, classification of · pdf...

Documents