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580306_Bonding Molecular Geometry_FINAL.pdf

Introduction to Molecules: A Molecular Bonding and Shapes Investigation

Carolina Distance Learning

Investigation Manual

Table of Contents

Overview ............................................................................................................................ 3

Objectives.......................................................................................................................... 3

Time Requirements ........................................................................................................... 3

Background ....................................................................................................................... 4

Materials............................................................................................................................. 6

Safety ................................................................................................................................. 6

Preparation ........................................................................................................................ 6

Activity 1: Lewis Dot Structures ........................................................................................ 7

Activity 2: Electron Domains.......................................................................................... 10

Activity 3: Molecular Geometry ................................................................................... 12

Activity 4: Bonding and Polarity .................................................................................... 14

Activity 5: Modeling Hydrocarbons.............................................................................. 19

Disposal and Cleanup ................................................................................................... 30

Data Tables and Extensions .......................................................................................... 31

Overview Learn about bonding and molecular geometry using molecular models. This

investigation includes classifying bonds, drawing Lewis structures, predicting molecular

geometry, constructing three-dimensional models, and determining polarity.

Resonance structures and isomers will also be identified as needed. The activities will

require approximately 2 hours to complete.

Objectives  Use electronegativity difference to classify bonds as ionic, polar covalent, or

nonpolar covalent.

 Apply valence bond theory to draw appropriate Lewis structures.

 Apply valence shell electron pair repulsion theory (VSEPR) to predict molecular

geometry.

 Construct three-dimensional molecular models.

 Sketch three-dimensional line structures using dashed lines and wedges.

 Determine whether the polarity of a molecule results in a dipole moment.

 Identify resonance structures and isomers as needed.

Time Requirements Preparation ................................................................... 10 minutes

Activity 1: Lewis Dot Structures ................................... 30 minutes

Activity 2: Electron Domain Geometries .................. 30 minutes

Activity 3: Molecular Geometry ................................ 30 minutes

Activity 4: Bonding and Polarity ................................. 20 minutes

Activity 5: Modeling Hydrocarbons........................... 40 minutes

Background

Few elements exist in nature as single atoms. Other than the noble gases, the atoms in

most elements are bonded to other atoms. This is because atoms are more stable if

their outer energy level is full. Electrons in the outer energy level are called valance

electrons. The halogens (Group 17 in Figure 1) have seven valance electrons in their

outer energy level.

Figure 1

Each element in group 18 has a full outer energy level. For most elements, a full outer energy

level has eight electrons, or an octet. Helium and hydrogen are exceptions in that they only

require two valence electrons.

Atoms can fill their outer energy level by transferring or sharing electrons to form either

ionic or covalent compounds (Figure 2.)

Ionic bond (electron is transferred to Y, giving it an octet)

Covalent bond (two electrons are shared, giving both atoms

a filled outer shell) Figure 2. Ionic and Covalent Bonds

The tendency for atoms to bond ionically or covalently is determined by the difference

in their electronegativity. Electronegativity is a dimensionless number that is a measure

of an atom’s ability to attract bonding valence electrons. Electronegativity values show

periodic trends on the periodic table, as shown in Figure 3.

Excluding the noble gases, the most electronegative element is fluorine, which is

assigned a value of 4.0. All the other elements have calculated values based on that of

fluorine. Across each period the electronegativity tends to increase, with the nonmetal

families of nitrogen, oxygen, and fluorine having the highest values. This is because

these atoms have smaller radii within which the positive nuclei exert a greater attraction

for the valence electrons. The alkali metals and alkaline earth metals (the groups on the

left side of the periodic table) have the lowest electronegativity because their atoms

have the largest radii, within which the positive nuclei exert a smaller attraction for

valence electrons. Cesium and francium, with the largest radii, have the lowest

electronegativity value of 0.7.

Figure 3

Materials

Included in the materials kit:

25 balloons

Needed from the modeling kit:

Molecular model set

Needed from the equipment kit:

Ruler

Needed, but not supplied:

Calculator

Digital camera or smart phone

Pencil

Reorder Information: A replacement kit for Bonding and Molecular Geometry item number

580306, can be ordered from Carolina Biological Supply Company. Call 1-800-334-5551 to order.

Safety Read all instructions for this laboratory activity before beginning. Follow the

instructions closely and observe established laboratory safety practices, including

use of appropriate personal protective equipment (PPE) as described in the Safety

and Procedure section.

Model pieces can be a choking hazard. Keep these and all laboratory materials away

from children.

Latex balloons are included in this kit and can cause an allergic reaction in some

individuals. Individuals sensitive to latex should not participate in any activities that may

result in exposure.

Preparation 1. Read the Student Guide.

2. Obtain all materials.

3. Separate model pieces into tray compartments based on shape, color, and number

of prongs.

Activities 1, 3, and 4 are inter-related. The compounds used in Activity 1 are also

used in Activities 3 and 4. Data in Table 1 will be used for these activities.

Activity 1

Lewis Dot Structures

A Lewis dot structure is one way to represent the arrangement of valence electrons in a

molecule. In Lewis structures, each element symbol is surrounded by a specific number

of dots representing valence electrons. Most atoms obey the octet rule* which states

that atoms tend to combine in such a way that they each have eight electrons in their

valence shells, giving them the same electronic configuration as a noble gas.

To fill their outer shells, elements form covalent bonds by sharing electrons. Lewis

structures are often used to show those bonds. Covalent compounds may have single,

double, or triple bonds between atoms. These bonds are represented in a Lewis

structure by dashes between the chemical symbols of the bonded elements. The

number of dashes corresponds to the number of bonds. Lewis structures also indicate

the lone pairs of electrons on different atoms. Guidelines for drawing a Lewis structure

for a given chemical formula are as follows:

*Exceptions: Hydrogen and helium need only two valence electrons to fill their outer

shell. Boron and beryllium form compounds with less than eight electrons and

elements in periods 3-6 may use more than eight.

Guidelines for Drawing Lewis Structures Example SiO2

1. Determine the number of valence electrons

present for each atom in the compound (consult

Figure 1 in the background section as needed).

Silicon (Si): 4 valence electrons

Oxygen (O): 6 valence electrons

2. Find the sum of valance electrons of all the atoms.

If there is a charge on a compound, add an

electron for each negative charge and subtract an

electron for each positive charge.

Silicon (Si): 4 valence electrons

Oxygen (O): 6 valence electrons

16 total electrons

3. Determine which atom is to be the central atom.

This is typically the least electronegative element

(that is not hydrogen) and/or the first element in the

chemical formula. Some compounds will have no

central atom, but the first element in the formula is

a good place to start.

4. Write the chemical symbol of the central atom,

followed by the chemical symbols for other atoms

around it.

5. Use dots to show the appropriate number of

valence electrons around each symbol in the

compound, based on their position in the periodic

table.

6. Determine the total number of electrons required

for each element. Allow for exceptions to the octet

rule as needed.

Silicon (Si) needs 8 valence electrons

Oxygen (O) needs 8 valence electrons

7. Create a skeleton structure by connecting

unpaired electrons from the central atom to

unpaired electrons on the atoms around it with a

solid line. Each solid line now represents a (two-

electron) covalent bond between the two atoms.

8. Redraw the structure as needed. If multiple bonds

exist between two atoms, reorient the chemical

symbols so that multiple bonds are with each other.

9. Sometimes there will be more than one way to add

the bonds. To determine which Lewis structure is the

best model, formal charge is often used. To

determine the formal charge of atoms:

a. Count the number of valence electrons that

each atom typically has when unbound.

b. In the Lewis structure being evaluated, count

the number of assigned electrons for an atom

by counting the number of electrons that exist

around that atom as lone pairs and then

counting the number of covalent bonds on that

atom. Add the number of lone pair electrons

and the number of bonds together.

c. Subtract the number of assigned electrons from

the number of valance electrons. The resulting

value is the formal charge for the atom in that

Lewis structure.

10. Lewis structures should show negative charge on

the most electronegative elements and positive

charges on the least electronegative elements.

is the preferred structure of SiO2

Use the Guidelines for Drawing Lewis Structures to complete the Data Table for Activity 1.

unbound electrons 6 4 6

assigned electrons -7 -4 -5

formal charge -1 0 +1

unbound electrons 6 4 6

assigned electrons -6 -4 -6

formal charge 0 0 0

Activity 2

Electron Domain Geometries

The three-dimensional shape of a molecule is predicted using the valence-shell

electron-pair repulsion (VSEPR) model. The VSEPR model is used to predict structures of

molecules or ions that contain only non-metals by minimizing the electrostatic repulsion

between the regions of high electron density. In this activity, balloons will model

electron domains. An electron domain is either a bond or a lone pair of electrons.

Double and triple bonds count as only one domain when using the VSEPR model. When

tied together, balloons naturally adopt the lowest energy arrangement predicted by

the VSEPR model.

Linear Geometry – Two Electron Domains

1. Inflate a balloon and tie the open end in a knot.

2. Repeat to prepare a second balloon of approximately the same size.

3. Tie the two balloons together in a knot.

4. Label the model and photograph it for future reference.

Trigonal Planar Geometry – Three Electron Domains

1. Inflate and tie three balloons.

2. Tie the three balloons together. Start by tying two balloons together and then use

the free ends to knot around the third.

3. Label the model and photograph for future reference.

Tetrahedral Geometry – Four Electron Domains

1. Inflate and tie four balloons.

2. Tie the balloons together to form two sets of two balloons.

3. Twist the two sets of two balloons together by crossing the knotted centers.

4. Label the model and photograph it for future reference.

Trigonal Bipyramidal Geometry – Five Electron Domains

1. Inflate and tie five balloons.

2. Tie four balloons together to form two sets of two balloons.

3. Tie a third balloon to one of the sets.

4. Twist the set of three balloons and the set of two balloons together by crossing

the knotted centers.

5. Label the model and photograph it for future reference.

Octahedral Geometry – Six Electron Domains

1. Inflate and tie six balloons.

2. Tie the balloons together to form three sets of two balloons.

3. Twist two sets of two balloons together by crossing the knotted centers. Twist a

third set of two balloons into the other two by crossing the knotted centers.

4. Label the model and photograph it for future reference.

Activity 3

Molecular Geometry

1. Construct a three-dimensional model of each molecule from Activity 1 using the

following molecular model pieces to represent atoms.

a. White one-pronged: Hydrogen

b. Light green one-pronged All other atoms

c. Dark brown three-pronged: Atoms with only three electron pairs and/or

bonds

d. Black, four-pronged: Carbon

e. Blue, four-pronged: Nitrogen

f. Red, four-pronged: Oxygen

g. Purple five-pronged: and/or bonds Atoms with five electron pairs

h. Silver six-pronged: Atoms with six electron pairs and/or bonds

Use white tubes to connect the pronged pieces and represent covalent bonds between

atoms in each molecule.

As needed, connect the white flat pieces for electron pairs on the central atom.

2. Use the white tubes to form bonds between the atoms. Each tube represents

one bond (two electrons). For a double bond, use two tubes and two prongs on

each atom.

3. Label each model and photograph it for future reference.

4. Sketch a perspective drawing of each compound. If necessary, use wedges and

dashes to show the three-dimensional appearance using the following table:

5. Build models for all the compounds in Activity 1.

Figure 4 Bonding Scale

Activity 4

Bonding

Chemical bonds are shared electrons between two atoms. If the atoms are different,

the attraction for these shared electrons will also be different. Chemical bonds are

classified as covalent or ionic. Electrons are equally shared in non-polar covalent

bonds, unequally shared in polar covalent bonds, and fully transferred to the more

electronegative atom in ionic bonds.

To determine the type of bond between two atoms, first calculate the difference

between their electronegativity values. Only the absolute difference is important.

Non-polar Covalent: If the difference in electronegativity is < 0.3, the electrons are

equally shared between the two atoms and the bond is a non-polar covalent bond.

Some examples of molecules with non-polar covalent bonds are Cl2, H2, CH4, and

CS2.

Polar Covalent: When the difference in electronegativity is between 0.4 and 1.7,

there is an unequal sharing of the electrons between the two atoms. Although all

electrons are constantly moving within the bonding orbital, in polar covalent bonds

the electrons are more attracted to the atom with higher electronegativity and will

spend more time closer to that atom. This unequal sharing creates poles of charges,

thus these bonds are termed polar covalent bonds. Some examples of polar

covalent molecules are NH3 and H2O.

Ionic: This type of bond occurs when there is complete transfer of the electrons

between the two atoms. This occurs when the difference in electronegativity is >1.8.

NaCl and MgCl2 are typical examples.

Sample Bond Type Calculations

Bonding Pair Electronegativity Difference Type of Bond

Li and F | 1.0 – 4.0 | = 3.0 Ionic

F and F | 4.0 – 4.0 | = 0 Non-polar covalent

C and F | 2.5 – 4.0 | = 1.5 Polar covalent

Bond Polarity

The term polar refers to the poles of partial charge. In a polar covalent bond, electrons

are more attracted to the atom with greater electronegativity. This difference results in

a partial negative charge on the more electronegative atom, and a partial positive

charge on the less electronegative atom. Consider the compound hydrochloric acid,

HCl. The electronegativity difference is 0.9 (|3.0 – 2.1| = 0.9). Chlorine is more

electronegative than hydrogen and therefore pulls the bonding pair closer, giving the

hydrogen a partial positive charge (delta plus, δ+) and the chlorine a partial negative

charge (delta minus, δ–), as shown in Figure 5.

Unequal sharing of electrons can also be represented by a vector indicating bond

polarity. To show bond polarity, draw a vector from the positively charged atom and

the negatively charged atom. Create a plus sign using a vertical line near the less

electronegative atom. The vector arrow points toward the more electronegative atom,

as in Figure 6.

Figure 6. Hydrochloric acid, HCl, showing bond polarity

Figure 5. Hydrochloric acid, HCl, showing partial charges

Molecular Polarity

Molecular polarity is similar to bond polarity. A polar molecule has regions of partial

positive and negative charge and will orient itself in an electric field. This orientation in

an electric field, called a dipole moment, is caused when charges are distributed

asymmetrically within the molecule. The molecular dipole moment is shown as a vector

to one side of the molecule. Once again, the vector arrow points toward the more

electronegative atom, as in Figure 7.

Figure 7. Hydrochloric acid, HCl, showing molecular dipole moment

Molecules that demonstrate a dipole moment are characterized as polar. Molecules of

hydrochloric acid are polar because the bond polarity is unbalanced.

Not all molecules containing polar covalent bonds are polar. If bond polarity is

balanced, the partial positive and partial negative charges are distributed

symmetrically. If the positive and negative charges are distributed symmetrically, the

molecule does not orient itself in an electric field and does not demonstrate a dipole

moment. Molecules that do not demonstrate a dipole moment are characterized as

non-polar.

Magnesium chloride (Cl – Mg – Cl) is an example of a non-polar molecule that contains

polar covalent bonds. Each Mg – Cl bond is polar covalent with an electronegativity

difference of 0.8 (|1.2 – 3.0| = 0.8). However, the linear molecular geometry of

magnesium chloride results in symmetrical bonds and a non-polar molecule.

Magnesium chloride does not have a dipole moment because the opposing bond

polarities cancel each other out (see Figure 8).

Figure 8. Magnesium chloride, MgCl2, showing bond polarity

It is also possible to arrange polar covalent bonds symmetrically in three-dimensional

space, as in carbon tetrachloride. Once again, each C – Cl bond is a polar covalent

bond; the electronegativity difference is 0.5 (|2.5 – 3.0| = 0.5). The molecular geometry

of carbon tetrachloride (tetrahedral) results in three-dimensional symmetry and a non-

polar molecule. Carbon tetrachloride does not have a dipole moment because the

opposing bond polarities cancel each other out (see Figure 9).

Polar molecules are always asymmetric; this can assist in identifying a molecule as

polar. Polar molecules are the result of unbalanced bond polarity and therefore all

polar molecules are asymmetrical in some way. If one chlorine atom in carbon

tetrachloride (CCl4) is replaced with a hydrogen atom, the molecule becomes

chloroform (CHCl3). The tetrahedral molecule is no longer symmetric; it is now both

asymmetric and polar. This molecule has a region of partial positive charge near the

hydrogen atom and a region of partial negative charge near the chlorine atoms. This

configuration results in a molecular dipole moment shown by the large vector on the far

right (see Figure 10).

Figure 10. Chloroform, CHCl3, a polar molecule Figure 9. Carbon tetrachloride, CCl4,

showing cancelation of polar bonds

The presence of lone pairs of electrons can also contribute to the polarity of a

molecule. When lone pairs are present, they are regions of partial negative charge.

However, not all molecules containing lone pairs are polar. The combination of

molecular geometry and multiple lone pairs may result in a symmetric, non-polar

molecule.

1. Complete Data Table 2 using the Periodic Table of Electronegativities and the

Bonding Scale to determine the type of bond that each set of atoms would exhibit if

they formed a bond. For polar covalent bonds only, draw a vector between atoms

showing bond polarity in the last column.

2. Determine the electronegativity difference and bond polarity for each bond in

compounds in Data Table 1. Record your calculations and bond polarity

determination under Activity 4 in Data Table 1.

3. Add appropriate bond polarity vectors beside each bond in the perspective

drawings under Activity 3 in Data Table 1.

4. Determine whether each molecule exhibits a dipole moment and write “yes” or

“no” in the space provided under Activity 4 in Data Table 1. If the drawing is not

clear, remake the model of the molecule to help visualize the shape and direction

of the charges.

5. Determine whether each molecule is polar and write “polar” or “non-polar” in the

space provided under Activity 4 in Data Table 1.

Activity 5

Modeling Hydrocarbons

Hydrocarbons are a large class of compounds that contain only hydrogen and carbon

atoms. They present a unique structural and Lewis dot challenge because the

compounds can be very large and do not have one central atom. Another challenge

is that each carbon atom can bond to four different atoms and a molecule such as

C5H12 can have three different structural formulas (see Figure 11). The different structural

formulas are called isomers.

pentane

(n-pentane)

2-methylbutane

(isopentane)

dimethylpropane

(neopentane)

Figure 11. Different structural arrangements of pentane, C5H12

Scientists have devised numerous methods to represent molecules that do not contain

a central atom. Each representation has certain advantages, as well as limitations.

Commonly used representations include the molecular formula, the structural formula,

the condensed structural formula, the skeletal model, the ball-and-stick model, and the

space-filling model. The model or formula used depends on the information that is to be

derived or conveyed. For example, to show how atoms are bonded together the

structural formula might be used, but if geometry is a concern a ball-and-stick model

might be preferable. The table below shows six different ways in which a pentane

(C5H12) molecule can be represented.

Model Example Advantages Limitations

Molecular

Formula C5H12

Lists the

number of

atoms

Does not

indicate

bonding

arrangements

(C5H12 can

represent any

one of three

different

isomers of this

molecule)

Structural

Formula

H H H H H

I I I I I

H — C — C — C — C — C — H

I I I I I

H H H H H

Shows the

number of

atoms and

how they are

bonded

Does not

show the

actual shape

of the

molecule

Condense

Structural

Formula

CH3-CH2-CH2-CH2-CH3

CH3(CH2)3CH3

Simpler than

the structural

formula;

shows the

number of

atoms and

which atoms

they are

bonded to

Does not

show the

actual order

bonding

order

Skeletal

Model

Simple model

that is useful

for

representing

large organic

molecules,

particularly

cyclic

compounds

Shows the

carbon

framework

but not the

molecule’s

shape; not all

of the atoms

and bonds

are

represented

Ball-and-

Stick

Model

Shows the

number of

atoms,

bonding

arrangements

, and shape

Does not

accurately

show how the

atoms bind

Space-

Filling

Model

Shows the

three-

dimensional

shape of the

molecule

The bonds

are not

clearly

indicated.

Individual

atoms are

difficult to see

in large

molecules.

Classes of Hydrocarbons

All hydrocarbons are composed entirely of carbon and hydrogen, yet their structures

can vary greatly from one to the next. Hydrocarbons are categorized into five groups:

alkanes, alkenes, alkynes, cyclic, and aromatic molecules.

Alkanes

Alkanes are saturated hydrocarbons because all carbon atoms are connected by

single bonds. They have the general formula CnH2n+2. Alkanes can form straight

chains or branched chains. All carbon atoms in an alkane adopt a tetrahedral

geometry and can rotate freely in three-dimensional space. The figures below show

different structural arrangements for butane, an alkane with the formula C4H10.

Figure 12. Different structural arrangements of butane, C4H10

The first four-carbon, straight-chain molecule is normal butane, or n-butane. The

second four-carbon, branched-chain molecule is isobutane (iso- meaning same or

an isomer), or 2-methylpropane. Isobutane and n-butane are structural isomers:

molecules that possess the same type and number of atoms but different bonding

arrangements. Because of their structural differences, these molecules have

different physical and chemical properties.

Alkenes

Alkenes are unsaturated hydrocarbons because they contain at least one double

bond. Monounsaturated alkenes are characterized by the general formula CnH2n.

The presence of double bonds results in carbon atoms that adopt a rigid trigonal

planar geometry and cannot rotate freely. Because of these rigid bonds, alkenes

can form stereoisomers in addition to structural isomers. Stereoisomers have the

same bonding sequence, but the atoms are oriented differently in space. The

diagram below shows three isomers of butene.

Figure 13. Different structural arrangements of butene, C4H10

Because the order of its bonding arrangements is different, 1-butene is a structural

isomer of cis 2-butene and trans 2-butene. However, cis 2-butene and trans 2-

butene are stereoisomers, not structural isomers, because the atoms have the same

bonding sequence but are oriented differently in space. When an alkene has two

groups bonded to a carbon double bond on the same side, the molecule is in the

cis position. When an alkene has two groups on opposite sides of the C=C bond, the

molecule is in the trans position. The bonds forming the C=C bond are rigid,

Therefore the CH3 and H groups attached to the carbon cannot rotate to match

the other carbon bond. Not all alkenes exhibit cis-trans isomerization. If one carbon

of the double bond holds two identical groups, the molecule does not exhibit cis-

trans isomerization.

Geometric cis-trans isomers are only one type of stereoisomer; hydrocarbons may

also exhibit E/Z isomerization, chirality, and optical isomerization.

Figure 14. Molecule not exhibiting cis-trans isomerization

Alkynes

Alkynes are unsaturated hydrocarbons that contain at least one triple bond. The

general formula for alkynes with one triple bond is CnH2n–2. Ethyne*, or acetylene*

(CHCH), is the simplest of the alkynes and is shown in Figure 15. Acetylene is linear

in shape, with rigid bonds that do not rotate freely. Alkynes adopt a linear shape

and thus do not form geometric isomers but can form structural isomers.

Figure 15. Ethyne (acetylene), an alkyne

Ethyne is a systematic name using the suffix –yne to indicate the presence of a triple bond. In

biology, ethyne is often referred to by its common name, acetylene.

Cyclic and Aromatic Hydrocarbons

In a cyclic hydrocarbon, the carbon atoms are arranged in a ring (a closed loop).

The general formula for cyclic hydrocarbons is CnH2(n+2-g), where g is the number of

rings. Cyclohexane, in Figure 16, is one example of a cyclic hydrocarbon and has

the formula C6H12. Alkenes can also form cyclic compounds. Cyclohexene is an

example of a cyclic alkene.

Aromatic hydrocarbons are a special class of cyclic hydrocarbons. A cyclic

hydrocarbon is aromatic if the carbon atoms alternate between single and double

bonds. Aromatic hydrocarbons are very stable and form the backbone of many

hydrocarbons. Benzene (C6H6) is the simplest aromatic compound, taking the form

of a planar ring consisting of three double bonds and three single bonds. The figure

below shows four different ways in which a benzene molecule can be represented.

Figure 17. Four representations of a benzene molecule

Figure 16. Cyclohexane, a cyclic hydrocarbon

Basic Hydrocarbons

1. Form covalent bonds between a carbon and a hydrogen atom by attaching a

`black four-prong piece to one end of a white tube and a white one-prong piece to

the other end. Throughout this activity the black pieces will represent carbon atoms

and the white pieces will represent hydrogen atoms.

2. Bond hydrogen atoms to the three remaining prongs on the carbon atom.

3. This structure is a “ball-and-stick” model of the simplest hydrocarbon found in nature,

a methane molecule. There are several other ways to represent this molecule.

Chemical

formula

Condensed

structural

formula

Expanded

structural

formula

Perspective

drawing

Ball-and-stick

model

Space

filling

model

CH4 CH4

H — C — H

4. Label the model as methane and photograph for future reference.

Carbon can also form single bonds with other carbon atoms to form long hydrocarbon

chains. These molecules are saturated hydrocarbons because all carbon atoms are

connected by single bonds. The structural formula of propane is

5. Build propane (C3H8) by bonding three carbon atoms together.

6. Attach a covalent bond and a hydrogen atom to all the open prongs.

7. Label the model as propane and photograph it for future reference.

8. Compare the ball and stick model that was constructed with the expanded

structural formula. Describe the similarities and differences between the two

representations in the Data Table.

9. Disassemble the molecular model.

Straight-Chained Hydrocarbons

1. Construct molecular models of each straight chain hydrocarbon in the table below.

For double and triple bonds, bend the white tubes until the required number of

connections between carbon atoms has been made.

Class

(bonding) Example General

Formula

Molecular

Formula

Structural

Formula

Alkanes

(all single

C–C bonds)

ethane CnH2n+2 C2H6

Alkenes

(at least

one

C=C bond)

ethene CnH2n C2H4

Alkynes

(at least

one

CC bond)

ethyne CnH2n–2 C2H2

2. Lay the ethane, ethene, and ethyne models side-by-side for comparison. In the

Data Table, describe the following differences between the molecules:

a. Three-dimensional structure

b. Shape

c. Rotation of the bonds (i.e., can the carbon–carbon bonds rotate freely, or

are they rigid?)

d. Bond strength (Which molecule has the strongest bonds?)

3. Label each model and photograph them for future reference.

4. Disassemble the molecular models.

Isomers and Branched-Chain Hydrocarbons

Isomers are molecules that have the same type and number of atoms but different

bonding arrangements (structural isomers) or different orientations in space (geometric

isomers).

Alkanes

1. Construct molecular models of both n-butane and isobutene using the structural

models below as a guide.

2. Compare the ball-and-stick models of butane. In the Data Table, describe the

similarities and differences between the models.

3. Label each model and photograph it for future reference.

4. Disassemble the molecular models.

Alkenes

1. Construct molecular models of each of the three isomers of the alkene C4H8 using

the structural models below as a guide.

2. Draw the structural formula and skeletal model for each compound on the Data

Table.

3. Answer the Observation Questions on the Data Sheet.

4. Label the models as methane and photograph them for future reference.

5. Disassemble the molecular models.

Alkynes

1. In the space provided on the Data Sheet, draw the structural models of all of the

isomers of the alkyne C4H6. Remember that alkynes contain at least one CC bond.

2. Determine which structures are the same, and then eliminate the duplicates. Using

the model set, build all the isomers of C4H6 using the structural models as reference.

3. In the space provided for observations on the Data Sheet, describe any similarities

between these models and the modeled alkanes or alkenes.

4. Label each model and photograph them for future reference.

5. Disassemble the molecular models.

Cyclic and Acyclic Compounds

1. Construct a molecular model of cyclohexane and hexane using the structural

models below as a guide.

Cyclohexane Hexane

2. Draw the skeletal model of cyclohexane and hexane on the Data Sheet.

3. Compare the structure of cyclohexane with the structure of hexane. Which

arrangement is more rigid?

4. Record observations on the Data Sheet.

5. Label the models and photograph them for future reference.

6. Disassemble the molecular models.

Aromatic Compounds - Benzene

1. Using the model set, build two benzene molecules using the structural models as a

guide.

2. Label each model as benzene and photograph them for future reference.

3. On the Data Sheet, draw the skeletal model for both of these models of benzene.

4. Compare the two benzene models. Are these molecules isomers? Rotate both

molecules and determine whether they are the same.

5. Use the two benzene models for the next activity.

Aromatic Compounds - Dichlorobenzene

1. Using the model sets from the previous activity, create the following three

dichlorobenzene isomers: ortho, meta, and para.

2. One additional benzene ring will have to be made.

3. On each benzene ring, replace the hydrogen atoms with a light green one-

pronged atom to represent a chlorine atom.

Orthodichlorobenzene Metadichlorobenzene Paradichlorobenzene

Three isomers of dichlorobenzene

4. Label each model with the correct name and photograph them for future

reference.

5. On the Data Sheet, draw the skeletal model for both of these models of

dichlorobenzene.

6. Compare the three dichlorobenzene models. Are these molecules isomers? Rotate

both molecules and determine whether they are the same.

7. Can any other isomers be made with a benzene ring and two chlorine atoms?

8. Disassemble the molecular models.

Disposal and Cleanup

1. Return molecular model parts to the resealable bag for storage.

Bonding and Molecular Geometry

Choose three compounds in Group 1, three compounds in Group 2, and three

compounds in Group 3 to use in Activities 1, 3, and 4. Copy each chemical formula into

the spaces provided in the first column of Data Table 1.

Group 1

1. BeCl2

2. BF3

3. CH4

4. PCl5

5. SF6

Group 2

6. CO2

7. NH3

8. H2O

9. SF4

10. ClF3

11. XeF2

12. IF5

13. XeF4

Group 3

14. O2

15. HCN

16. H2CO

17. CH3Br

18. SnCl2

Data Table 1 (Make Copies of This Table)

Activity 1

Lewis Dot Structures

Activity 3

Molecular Geometry

Activity 4

Bonding and Polarity

Chemical Formula:

Dot Diagrams:

Electrons:

Central Atom:

Lewis Structure:

Domains:

Lone Pairs:

Geometry:

Bond Angle:

Line Drawing: Electronegativity:

Bond Polarity:

Dipole Moment:

Polarity:

Chemical Formula:

Dot Diagrams:

Electrons:

Central Atom:

Lewis Structure:

Domains:

Lone Pairs:

Geometry:

Bond Angle:

Line Drawing: Electronegativity:

Bond Polarity:

Dipole Moment:

Polarity:

Figure 4 Bonding Scale

Activity 4: Bonding and Polarity

Data Table 2

Bonding Pair Electronegativity

Difference Type of Bond Bond Polarity

H and H

B and F

Ge and F

C and H

N and H

H and O

P and Cl

S and F

Xe and F

O and O

C and O

C and F

Sn and Cl

Al and O

Na and Cl

Mg and O

Mg and Cl

Activity 5: Modeling Hydrocarbons

Basic Hydrocarbons

Observation Questions:

1. Does the ball-and-stick model look like the structural formula?

2. What differences are there between the ball-and-stick model of methane and the

molecular and structural formulas above?

3. List the information that the ball-and-stick model provides that the structural and

molecular formulas do not.

Straight-Chained Hydrocarbons

Class Type of Bonding General Formula

Alkanes

Alkenes

Alkynes

Example Structural

Formula

3-Dimensional

Structure Shape

Rotation of

the Bonds

Bond

Strength

Ethane

C2H6

Ethene

C2H4

Ethyne

C2H2

Modeling Isomers and Branched-Chain Hydrocarbons

Alkanes

Structural Isomers of C4H10

Alkenes

Isomers of C4H8

Molecule 1 Molecule 2 Molecule 3

Structural Formula

Condensed

Structural Formula

Skeletal Model*

*Skeletal models show the carbon-to-carbon bonds. Carbon atoms exist at the ends of

the line segments. All hydrogen atoms and carbon-to-hydrogen bonds are understood.

Observation Questions:

1. Can Molecule 1 be made to look exactly like Molecule 2? Why or why not?

2. Describe the differences in structures between the three alkenes.

3. Are Molecule 1, Molecule 2, and Molecule 3 chemically identical?

Alkynes

Isomers of C4H6

Observations

Cyclic and Acyclic Compounds

Cyclohexane Hexane

Skeletal

Model

Which arrangement is more stable, cyclohexane or hexane? Why?

Aromatic Compounds

Skeletal Models of Benzene

Are these molecules isomers or are they the same molecule? Explain.

Aromatic Compounds

Skeletal Models of Dichlorobenzene

Are these molecules isomers or are they the same molecule? Explain.

lab

580306_Bonding Molecular Geometry_FINAL.pdf

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