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VSEPR: Molecular Shapes and Isomerism

Module by: Mary McHale

Lab 3:VSEPR: Molecular Shapes and Isomerism

Objective

  • To understand the 3-dimensional nature of molecules.
  • To learn about Molecular Symmetry.
  • To be able to identify the various isomers possible for one molecular formula.
  • To be able to identify enantiomers.

Grading

  • Pre-Lab (10%)
  • Lab Report Form (80%)
  • TA Points (10%)

Before Coming to Lab…

Look over the following to make sure you have a basic understanding of the topics presented. (Some of these topics are provided in the next several pages.) You can find information pertaining to most of these topics in Chapter 9 of your textbook.

  • Drawing Lewis Structures
  • Determining the Shapes of Molecules from their Lewis Structures
  • Some Basic Aspects of Bonding
  • Model Kits

Background Information

The shape of a molecule is extremely important in determining its physical properties and reactivity. A multitude of shapes are possible, and in today’s lab, you will be looking at some of them.

In Part I, you will be exploring the various symmetry elements that can be present in molecules. The symmetry elements you will be looking for are mirror planes, rotation axes, and inversion centers. Being able to determine which symmetry elements are present in a molecule helps in understanding its chemistry. If there is a plane present in the molecule that has the exact same arrangement of atoms on either side of the plane, then the molecule has a mirror plane (). It is important to note that a molecule can have more than one mirror plane. Rotation axes are represented as Cn (n = 1, 2, 3 . . .). The subscript indicates how many degrees of rotation (360o/n) are needed in order to return to the same orientation of atoms as you started with. So if there is a C2 axis, the rotation would be 180o. An example of a molecule having a C2 axis is H2O.

Figure 1
Figure 1 (Image3.png)

The third symmetry element is an inversion center (i). In such molecules, starting at any position and drawing a line through the center and an equal distance to the opposite side of the molecule, you will end up at a position with an identical environment to the one you started from.

Figure 2
Figure 2 (graphics1.png)
Part II of the lab introduces the concept of enantiomers. Enantiomers are molecules sharing the same molecular configuration, but they are non-superimposable images of each other. This concept should become clearer as you build the models for this part of the lab. Enantiomers share many of the same physical properties. The property which distinguishes them is the direction in which they rotate plane-polarized light. They will rotate the light in equal amounts but in different directions (plane-polarized light is just light in which all waves have been filtered out except for those in one plane). If both enantiomers are present in a 1:1 ratio, the effects of the rotation of light cancel and no net rotation is observed. Such a mixture of isomers is known as a racemic mixture or as a racemate. Because these isomers rotate plane-polarized light, they are also known as optical isomers. Compounds that form optical isomers are said to be chiral.

The chemistry of enantiomers is of great importance in the field of medicine. It has been discovered that with many drugs, one enantiomer will be biologically active while the other will be inactive or even produce undesired side effects. For this reason, it has become a challenge for pharmaceutical companies to try to synthesize the active enantiomer exclusively.

The next part of the lab deals with isomers. Isomers are molecules having the same molecular formula, but atoms are arranged in a different manner, while still obeying the rules of bonding. There are different classifications for isomers. For example, structural isomers differ from one another in the order in which the atoms are bound to each other (connectivity is different). On the other hand, geometrical isomers have the same order of atoms, but the spatial arrangement of atoms is different. A common example of geometrical isomers is the cis (same side) and trans (opposite side) forms of double bonds:

Figure 3
Figure 3 (diclethene.png)

Note: Remember that molecules having single carbon-carbon bonds cannot have cis/trans isomers because there is free rotation about single bonds.

By building the models of various molecules during this lab, you will be able to better understand molecular symmetry and isomers. Building models is not difficult; however, the chemical principles involved are very important and you may find some surprises in how atoms can be fit together.

Finally, in Part IV, you will be applying your knowledge of VSEPR (Valence Shell Electron Pair Repulsion) theory in order to determine the geometry of several different molecules. VSEPR theory is useful in helping to determine how atoms will orient themselves in molecules. Basically, the idea is that the arrangement adopted by a molecule will be the one in which the repulsions among the various electron domains are minimized. The two kinds of electron domains are bonding (electron pair shared by two atoms) and non-bonding (electron density centralized on one atom) pairs of electrons.

 Experimental Procedure

For Parts I & II: You and your lab partner are to work with one other lab group in preparing these models (no more than 4 students). Your TA will assign each group a certain set of molecules to make and answer questions pertaining to those molecules. Each group will then present their answers to the class. These models will need to be completed and answers determined within 30 minutes so that we can continue to Parts III & IV as soon as possible.

For Parts I-IV, the work should be divided among the group members. Be sure to discuss the questions and answers among yourselves, but put your own conclusions on the Report Form.

Part I. Symmetry Elements

Using the Molecular Framework models, make models of the following compounds:

  1. CH 4 CH 4 size 12{"CH" rSub { size 8{4} } } {}
  2. CH 3 Cl CH 3 Cl size 12{"CH" rSub { size 8{3} } "Cl"} {}
  3. CH 2 Cl 2 CH 2 Cl 2 size 12{"CH" rSub { size 8{2} } "Cl" rSub { size 8{2} } } {}
  4. CHCl 3 CHCl 3 size 12{"CHCl" rSub { size 8{3} } } {}
  5. CH 2 ClF CH 2 ClF size 12{"CH" rSub { size 8{2} } "ClF"} {}
  6. CHBrClF
  7. BF 3 BF 3 size 12{"BF" rSub { size 8{3} } } {}
  8. BF 2 Cl BF 2 Cl size 12{"BF" rSub { size 8{2} } "Cl"} {}
  9. PH 3 PH 3 size 12{"PH" rSub { size 8{3} } } {}
  10. PH 2 Cl PH 2 Cl size 12{"PH" rSub { size 8{2} } "Cl"} {}

Choose a color to represent each atom. The conventional scheme is: carbon (C) = black; hydrogen (H) = white; oxygen (O) = red, etc.

Once the models are created, look for symmetry elements that may be present. Ask yourselves the following questions:

  • Does the molecule contain a mirror plane ( σσ size 12{σ} {})? In other words, is there a plane within the molecule which results in one half being a mirror image of the other half?
  • Does the molecule contain a two-fold rotation axis ( C2C2 size 12{C rSub { size 8{2} } } {})? Remember from the Introduction that the subscript indicates the degrees of rotation necessary to reach a configuration that is indistinguishable from the original one. In this case, the rotation will be 180o.
  • Does the molecule contain any higher-order rotation axes?

C3C3 size 12{C rSub { size 8{3} } } {}– rotation by 120o

C4C4 size 12{C rSub { size 8{4} } } {}– rotation by 90o

C∞C∞ size 12{C rSub { size 8{ infinity } } } {} (infinity rotation axis) – rotation of any amount will result in an indistinguishable orientation

  • Does the molecule have an inversion center (i)?

Determine which of these symmetry elements are present in your assigned molecules. All of the columns of the table on the report form should be filled out. If you have any difficulty determining whether such symmetry elements are present in the molecules you are assigned, your TA can provide examples of each symmetry element.

Extra credit points can be earned by indicating in the table how many of each symmetry element is present for each molecule (i.e. How many mirror planes are present?).

Part II. Mirror Images

Using the model kits, build models which are the mirror images of the models you were assigned to build (a, b, c, d, e, f, g, h, i and j) in Part 1. With the two mirror images in hand, try to place the models on top of one another, atom for atom.

If you can do this, the model and its mirror image are superimposable mirror images of one another. If not, the molecule and its mirror image form nonsuperimposable mirror images. Nonsuperimposable mirror images are also known as enantiomers.

For each compound, decide whether the mirror image is superimposable or nonsuperimposable. Can you make a generalization about when to expect molecules to have nonsuperimposable mirror images?

Part III. Isomers

In this exercise you will build models of compounds which are structural and/or geometrical isomers of one another.

Make the following models:

A. Structural Isomers

  • Make a model(s) of C2H5ClC2H5Cl size 12{C rSub { size 8{2} } H rSub { size 8{5} } "Cl"} {}. How many different structural isomers are possible?
  • Make a model(s) of C3H7ClC3H7Cl size 12{C rSub { size 8{3} } H rSub { size 8{7} } "Cl"} {}. How many different structural isomers are possible?
  • Make a model(s) of C3H6Cl2C3H6Cl2 size 12{C rSub { size 8{3} } H rSub { size 8{6} } "Cl" rSub { size 8{2} } } {}. How many different structural isomers are possible?

B. Geometrical Isomers

  1. Make a model(s) of C2H3ClC2H3Cl size 12{C rSub { size 8{2} } H rSub { size 8{3} } "Cl"} {}. How many different structural and geometrical isomers are possible?
  2. Make a model(s) of C2H2Cl2C2H2Cl2 size 12{C rSub { size 8{2} } H rSub { size 8{2} } "Cl" rSub { size 8{2} } } {}. How many different structural and geometrical isomers are possible?
  3. Make a model(s) of cyclobutane ( C4H8C4H8 size 12{C rSub { size 8{4} } H rSub { size 8{8} } } {}). HINT: cyclo = ring of C atoms
  4. Now make dichlorocyclobutane ( C4H6Cl2C4H6Cl2 size 12{C rSub { size 8{4} } H rSub { size 8{6} } "Cl" rSub { size 8{2} } } {}) by replacing two H atoms on cyclopropane with Cl atoms. How many different structural and geometrical isomers are possible for dichlorocyclobutane? You may wish to make a couple of cyclobutane molecules so that you can compare the structures. Do any of the isomers have nonsuperimposable mirror images?

C. Aromatic Ring Compounds

  1. Make a model of benzene, C6H6C6H6 size 12{C rSub { size 8{6} } H rSub { size 8{6} } } {}. Even though your model will contain alternating double and single bonds, remember that in the real molecules of benzene all the C-C bonds are equivalent. What symmetry elements does benzene possess?
  2. Make a model(s) of chlorobenzene, C6H5ClC6H5Cl size 12{C rSub { size 8{6} } H rSub { size 8{5} } "Cl"} {}. How many different structural and geometrical isomers are possible?
  3. Make a model(s) of dichlorobenzene, C6H4Cl2C6H4Cl2 size 12{C rSub { size 8{6} } H rSub { size 8{4} } "Cl" rSub { size 8{2} } } {}. How many different structural and geometrical isomers are possible?
  4. Make a model(s) of trichlorobenzene, C6H3Cl3C6H3Cl3 size 12{C rSub { size 8{6} } H rSub { size 8{3} } "Cl" rSub { size 8{3} } } {}. How many different structural and geometrical isomers are possible?

Part IV. Hypervalent Structures

Hypervalent compounds are those that have more than an octet of electrons around them. Such compounds are formed commonly with the heavier main group elements such as Si, Ge, Sn, Pb, P, As, Sb, Bi, S, Se, Te, etc. but rarely with C, N or O. Refer to the rules for Electron Domain theory in order to assign Lewis structures to the following molecules. Describe possible isomeric forms and the bond angles between the atoms. How many lone pairs of electrons are present on the central atom of each molecule, if any? (** Your book will be very useful in aiding you with these structures **)

  1. PF 5 PF 5 size 12{"PF" rSub { size 8{5} } } {}
  2. PF 3 Cl 2 PF 3 Cl 2 size 12{"PF" rSub { size 8{3} } "Cl" rSub { size 8{2} } } {}
  3. SF 4 SF 4 size 12{"SF" rSub { size 8{4} } } {}
  4. XeF 2 XeF 2 size 12{"XeF" rSub { size 8{2} } } {}
  5. BrF 3 BrF 3 size 12{"BrF" rSub { size 8{3} } } {}

Some Simple Principles of Bonding

  1. Molecules are created when atoms form bonds to one another.
  2. Bonds between atoms are created when the atoms share electrons.
  3. An individual bond can contain no more than two electrons, although two atoms can have more than one bond between them. The common possibilities are single, double and triple bonds. It is also possible for metals to have quadruple bonds between them. (Pentuple bonds have been postulated but have not yet been observed.) The number of bonds between two atoms is known as the bond order. It is also possible for bonds to be fractional order so bond orders of 0.5, 1.5 and 2.5 are also possible. This occurs when two atom share 1, 3, or 5 electrons, respectively.
  4. Not all electrons possessed by an atom will be involved in bonding. The electrons that belong to filled shells are too strongly attached to the atom to be shared with other atoms. These electrons are known as core electrons.

Consider, for example, phosphorus (P, element #15) which has an electronic configuration of 1s22s22p63s23p31s22s22p63s23p3 size 12{"1s" rSup { size 8{2} } "2s" rSup { size 8{2} } "2p" rSup { size 8{6} } "3s" rSup { size 8{2} } "3p" rSup { size 8{3} } } {}. This means that in the phosphorus atom there are two electrons in the 1s orbital, two electrons in the 2s orbital, etc. The electrons in the 1s, 2s and 2p orbitals are in energy levels, which are completely filled. These electrons are so low in energy that they cannot be shared with other atoms. These electrons are the core electrons for P. The 3 level orbitals, however, are not completely filled. There are five electrons in the 3 level, two in the 3s and three in the 3p orbitals. These five electrons constitute the valence electrons. It is the valence electrons and only the valence electrons that become involved in bonding to other atoms.

The number of valence electrons can be determined readily from the periodic table. Find the element under consideration and, starting at the left, count the number of electrons (i.e., the elements) in that row up to and including the element under consideration. In the case of P, we would count Na, Mg, Al, Si and P which gives us a total of five, which is the number of valence electrons for P. In the case of an intervening group of d or f orbitals, there are a couple of possibilities. If the d or f subshell is not completely filled, then those electrons are included with all of the others in that row in the valence electron count. For example, iron (Fe, element #26) has eight valence electrons.

If a row or d- or f-orbitals is completely filled, then these electrons are not included in the valence count. For example, arsenic (As, element #33) has the 3d level filled and so those 10 electrons are not included. This makes As have the same number of valence electrons as P. As you will note, As and P lie in the same column in the periodic table. This is not a coincidence. All atoms in the same column of the periodic table will have the same number of valence electrons. Thus N, P, As, Sb and Bi all have five valence electrons; Fe, Ru and Os all have eight valence electrons, etc. The maximum positive oxidation state for an element cannot be greater than the number of its valence electrons.

  1. It is possible for all valence electrons to be involved in bonding to other elements. Some valence electrons, however, may not be involved in bonding. These electrons, usually found as pairs, are known as nonbonding electrons.
  2. A single bond between two atoms may be formed when the two atoms donate electrons equally; i.e. each atom contributes one electron to the bond. Such bonds are known as covalent bonds. Such a bond is found in the hydrogen molecule. It is possible that one atom supplies both the electrons in a bond between two atoms. In this case the bonds are called dative or donor bonds. Such a bond is formed between the N and B atoms in the complex H3N:BH3H3N:BH3 size 12{H rSub { size 8{3} } "N:BH" rSub { size 8{3} } } {}. Often an arrow is drawn from the atom that donates the electrons to the one that receives them. Thus H3N:BH3H3N:BH3 size 12{H rSub { size 8{3} } "N:BH" rSub { size 8{3} } } {} may also be written H3N→BH3H3N→BH3 size 12{H rSub { size 8{3} } N rightarrow "BH" rSub { size 8{3} } } {}.
  3. For the p-block elements (those elements with p-valence electrons which includes the area between B, He, Tl and Rn), there is a very strong preference for the atoms to have a total of 8 electrons around them. This is known as the octet rule. This occurs because these elements will have eight electrons around them when the valence orbitals (one s orbital, which accommodates 2 electrons each, and three p orbitals) are filled. This leads to a very stable electron configuration (like that of the noble gases). It is possible, however, especially for the heavier elements in the p-block to have more than eight electrons present, since they possessed unfilled d orbitals. Compounds of the p-block elements that have more than an octet of electrons are historically referred to as hypervalent compounds. This phenomenon is also referred to as valence shell expansion or octet expansion. The term hypervalent is not completely satisfactory for a rigorous theoretical bonding picture, but it is useful for simple descriptions and it is widely used. Examples include PF5PF5 size 12{"PF" rSub { size 8{5} } } {}, SF4SF4 size 12{"SF" rSub { size 8{4} } } {} and the [PF6]−[PF6]− size 12{ \[ "PF" rSub { size 8{6} } \] rSup { size 8{ - {}} } } {} ion. As you may note, hypervalent compounds very often have a heavier p-block element attached to highly electronegative elements such as the halogens or oxygen.
  4. All elements have (a) a tendency to attract electrons to themselves and (b) a resistance to having electron removed from them. The energy associated with adding an electron to an atom is known as its electron affinity (EA) measured for gas phase atoms. The energy required to remove an electron from an atom is known as its ionization potential (IP), also measured for gas phase atoms. A third quantity that is often encountered is the electronegativity of an element. This quantity is the average of the element’s EA and IP (using one scheme — there are other methods of calculating electronegativity). Elements that are very electronegative have a strong attractive force for electrons. The most electronegative element is fluorine. The least electronegative or most electropositive stable element is cesium (Cs) Electronegativity tends to increase as one moves from the bottom of the periodic table to the top and as one moves across the periodic table from the left to the right (except for the noble gases).
  5. Bond polarity: We noted above that when two atoms each donate one electron to a bond; this bond is a covalent bond. If the two atoms have the same electronegativity, then the electrons are rigorously shared equally between the two atoms, as in the case of H2H2 size 12{H rSub { size 8{2} } } {}. Such a bond is known as a nonpolar bond. If the atoms are different, however, they will have different electronegativities. This means that the two atoms are competing for the electrons and the one with the stronger electronegativity will win. When this happens, the bonds become polarized so that the electrons reside more often on the more electronegative element. This is the case in diatomic molecules such as HF. The H-F bond is a polar bond. Sometimes the Greek symbol δδ size 12{δ} {}is used to represent a partial charge that builds up as a result of the electrons being unequally shared. Thus we could write HF as Hδ+−Fδ−Hδ+−Fδ− size 12{H rSup { size 8{δ+{}} rSub { size 8{-{}} } } F rSup { size 8{δ-{}} } } {} to emphasize the polar nature of the bond and to indicate that the electrons will be pulled away from the H atom and towards the F atom. This shifting of electron density within a molecule creates what is known as a dipole moment. Molecules which are nonpolar tend to have lower boiling points than those which are polar because the dipole moments created in the individual molecules tend to attract the dipole moments present in other molecules: Hδ+−Fδ−⋯Hδ+−Fδ−Hδ+−Fδ−⋯Hδ+−Fδ− size 12{H rSup { size 8{δ+{}} rSub { size 6{ - {}} } } F rSup { size 6{δ - {}} } dotsaxis H rSup { size 6{δ+{}} rSub { size 6{ - {}} } } F rSup { size 8{δ - {}} } } {}. Here we can see that the slightly negative charge created at the F of one HF molecule will be slightly attracted to the slightly positively charged H of a different HF molecule. The boiling point of H2H2 size 12{H rSub { size 8{2} } } {} (non-polar) is 20.6 K while that of HF (polar) is 292.7 K. The dipole moment of a molecule with a number of atoms and more than one bond is simply the vector sum of all the dipole moments of the individual bonds that make up the molecule. Water, H2OH2O size 12{H rSub { size 8{2} } O} {}, for example, has a dipole moment as shown in Figure 1.
Figure 4
Figure 4 (graphics2.png)

Figure 1. Individual bond dipoles (which are treated as vector quantities) add to give an overall molecular dipole moment.

Dipole moments when hydrogen atoms are attached to very electronegative atoms can be comparatively large. This makes it very easy for dipole-dipole interactions in molecules such as water, ammonia and hydrogen fluoride. When hydrogen is involved in strong dipole-dipole interactions, it is known as hydrogen bonding. Hydrogen bonding is extremely important in determining the chemistry and reactivity of biological systems. Hydrogen bonding is generally only significant for H attached to O, F or N. The other elements are usually not electronegative enough to give rise to hydrogen bonding.

The greater the difference in the electronegativities of the two atoms involved in a bond, the greater the polarity. Consider C-F and Na-F bonds. The electronegativity of C is 2.55, that of Na is 0.93 and that of F is 3.98. Since the difference between the electronegativities of C and F is less than that between Na and F, the C-F bond will be less polar than the Na-F bond.

In the extreme picture where the electrons are completely transferred from one atom to another, the bonding becomes ionic rather than covalent. It is important to remember that ionic and covalent bonding are simply extremes of one bonding continuum. Most bonds have some covalent and some ionic character.

Drawing Lewis Structures (Read before coming to lab)

  1. Sum up the total number of valence electrons for all atoms present in the structure.
  2. First assume that all the atoms of the structure obey the octet rule except hydrogen, H, which will have only two electrons.
  3. Subtract the number determined in part 1 from the number obtained in part 2. This number represents how many electrons too few we have if we were to give each nonhydrogen atom its own set of 8 electrons and each H atom 2 electrons. This number is then the number of shared electrons that we must have, which is twice the number of bonds that must be present in the molecule.
  4. Determine (or guess) the connectivity of atoms in the molecule. Place electron pairs between as many atoms that are connected to each other as possible in order to generate single bonds. Compare the number of single bonds that you drew to the number you obtained in 3 above. If you need more bonds, then add as many multiple bonds as is necessary to get to the number of bonds determined in 3.
  5. Fill in the rest of the electrons remaining as nonbonding electron pairs on the nonhydrogen atoms.
  6. In some cases, there may be more than one possibility for the way that the atoms are connected or the way the electrons are arranged around the atoms. In these cases, there are some extra numbers that we can calculate to help us determine whether the structure is reasonable or not. These numbers are the formal charges on the atoms in the structure.

 FC = VE — NBE — 1/2 BE

Where:

FC = formal charge

VE = number of valence electrons

NBE = number of nonbonding electrons

BE = number of bonding electrons

If the formal charges on individual atoms are too high (usually anything greater than ±1±1 size 12{ +- 1} {} is too high), then the Lewis structure drawn is higher in energy than a similar structure with lower charges.

The sum of the formal charges on all atoms in the molecule or ion must equal the charge on the molecule (0) or ion. For example, the formal charges on C and O in carbon monoxide are -1 on C and +1 on O. The sum is 0 which is correct for the neutral molecule CO. Likewise the formal charges for NO+NO+ size 12{"NO" rSup { size 8{+{}} } } {} are 0 on N and +1 on O giving a total of +1 which is the overall charge on NO+NO+ size 12{"NO" rSup { size 8{+{}} } } {}.

Caution

Do not confuse formal charge with oxidation state. Oxidation state assignments are based on a completely different set of rules and are not directly related to formal charge.

VSEPR:Molecular Shapes and Isomerism

Pre-Lab: (Total 10 Points)

Pre-Lab 3: VSEPR Hopefully here for the Pre-Lab

Name(Print then sign): ___________________________________________ Lab Day: ___________________Section: ________TA__________________________

This assignment must be completed individually and turned in to your TA at the beginning of lab. You will not be allowed to begin the lab until you have completed this assignment.

__________ 1. In HCN the bond between C and N is

(a) a single bond (b) a double bond (c) a triple bond

__________ 2. In CO2CO2 size 12{"CO" rSub { size 8{2} } } {}, the two CO bonds are

(a) both single bonds (b) both double bonds (c) 1 single and 1 double bond

 __________ 3. The H-N-H bond angle in NH3NH3 size 12{"NH" rSub { size 8{3} } } {} is expected to be

(a) about 90° (b) about 109° (c) about 120° (d) about 180°

 __________ 4. The bond order between the two carbon atoms in C2H2C2H2 size 12{C rSub { size 8{2} } H rSub { size 8{2} } } {} is

(a) a single bond (b) a double bond (c) a triple bond

 __________ 5. The O-C-O bond angle in CO 32−CO 32− size 12{"CO" {rSub { size 8{3} }/rSub} {rSup { size 8{2 - } } /rSup} {} is expected to be

(a) about 90° (b) about 109° (c) about 120° (d) about 180°

 __________ 6. BF3BF3 size 12{"BF" rSub { size 8{3} } } {} is expected to be

(a) planar (b) trigonal pyramidal (c) T-shaped

 __________ 7. CH4CH4 size 12{"CH" rSub { size 8{4} } } {} is expected to be

(a) square planar (b) tetrahedral (c) see-saw shaped

 __________ 8. The O-S-O bond angle in SO2SO2 size 12{"SO" rSub { size 8{2} } } {} is expected to be

(a) about 90° (b) about 109° (c) about 120° (d) about 180°

 __________ 9. The property of enantiomers which differs is

  1. melting point (b) density (c) rotation of light

_________ 10. In a molecule with 5 electron-domains (i.e. trigonal bipyramidal), a non-bonding electron pair will experience more repulsion in the

(a) axial position (b) equatorial position (c) neither – both the same

VSEPR: Molecular Shapes and Isomerism

Report (80 points)

Report 3: VSEPR - Molecular Shapes and Isomerism Hopefully here for the Report Form Note: In preparing this report you are free to use references and consult with others. However, you may not copy from other students’ work (including your laboratory partner) or misrepresent your own data (see honor code).

Name(Print then sign): ___________________________________________________

Lab Day: ___________________Section: ________TA__________________________

Part I. Symmetry Elements

Compound
Symmetry Elements
C2C2 size 12{C rSub { size 8{2} } } {} axis C3C3 size 12{C rSub { size 8{3} } } {} axis C4C4 size 12{C rSub { size 8{4} } } {} axis Mirror plane Inversion Center
  CH4CH4 size 12{ ital "CH" rSub { size 8{4} } } {}  _______  _______  _______  ____________  _______________
  CH3ClCH3Cl size 12{ ital "CH" rSub { size 8{3} } ital "Cl"} {}  _______  _______  _______  ____________  _______________
  CH2Cl2CH2Cl2 size 12{ ital "CH" rSub { size 8{2} } ital "Cl" rSub { size 8{2} } } {}  _______  _______  _______  ____________  _______________
  CH2ClFCH2ClF size 12{ ital "CH" rSub { size 8{2} } ital "ClF"} {}  _______  _______  _______  ____________  _______________
 CHBrClF  _______  _______  _______  ____________  _______________
  BF3BF3 size 12{ ital "BF" rSub { size 8{3} } } {}  _______  _______  _______  ____________  _______________
  BF2ClBF2Cl size 12{ ital "BF" rSub { size 8{2} } ital "Cl"} {}  _______  _______  _______  ____________  _______________
  PH3PH3 size 12{ ital "PH" rSub { size 8{3} } } {}  _______  _______  _______  ____________  _______________
  PH2ClPH2Cl size 12{ ital "PH" rSub { size 8{2} } ital "Cl"} {}  _______  _______  _______  ____________  _______________

Part II. Mirror Images

Compound  Is the mirror image superimposable?
CH 3 Cl CH 3 Cl size 12{ ital "CH" rSub { size 8{3} } ital "Cl"} {}  ________________________________________________
CH 2 Cl 2