E11 THE USE OF MOLECULAR MODELS

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1 E11 THE USE F MLECULAR MDELS THE TASK To use molecular models to develop a better understanding of molecular size, shape and bonding. THE SKILLS By the end of the experiment you should be able to: use molecular modelling kits and reasoning to study some of the properties of molecules, draw Lewis structures and to use these and the valence shell electron pair repulsion model to predict molecular shapes. THER UTCMES You will develop generic scientific skills including the use of models in scientific theories and an appreciation of their limitations. INTRDUCTIN Models versus Reality In this experiment you will use molecular models to explore a number of aspects of molecular structure and an important molecular property, polarity. You will also look at some cases in which the models prove too simple to represent the reality at the molecular level successfully. In order to be able to make proper use of models it is important to understand the relationship between a model and the reality it tries to represent. Models are imperfect things, often very useful when applied to the problems for which they were designed, frequently misleading when extended beyond their range of applicability. Let s consider the model kit used in this experiment. It is designed to represent some basic geometric features of bonding, specifically, bond angles and lengths. These models are at their most valuable in showing how the structure of a molecule arises as a consequence of the geometry of bonds about the individual atoms. Pauling and Corey s determination of the structure of simple proteins and Watson and Crick s discovery of the base pairing in DNA were some early triumphs of the use of simple model kits like the one you will use in this experiment. The model kits do not convey information about the energetics of bonding, i.e., why a particular bond forms between two atoms. The kits do not allow for the distortions of the geometry away from the perfect tetrahedral, octahedral, etc. The kits do not really attempt to depict the distribution of electrons around the atoms and in the bonds with any accuracy. You need to be aware of these limitations when you draw conclusions from your model building. (You will see what we mean in this experiment.) These limitations, however, are also the reason why the models are as simple to use (and, hence, as useful) as they are. E11-1

2 E11-2 LAB-WRK 20 (1) The Framework Molecular Model Kit (2) Bonding and Valence Shell Electrons (3) Building Bonds with the Model Kit (4) Measuring the Distance between Electron Pairs (5) Building Models of Molecules with Nitrogen and Hydrogen (6) Modelling Double and Triple Bonds (7) Molecular Shape and Polarity (8) A Puzzle: Which Isomer of ClF 3 is Correct? (9) N 3 : Resonance and the Limitations of Models (1) The Framework Molecular Model Kit Acknowledgment is made to Prentice-Hall Inc. for permission to reprint sections of the instruction manual that accompanies their Framework Molecular Models kits. The FMM kit is composed of two basic construction units, metal valence clusters and plastic tubing, by means of which atoms of all elements can be represented. Each type of atom is colour coded. For example, red tubing represents oxygen, white is hydrogen, blue is nitrogen, black is carbon, and so on. Valence clusters are coloured silver, brass and copper for easy identification. Each valence cluster has metal prongs joined at a common point representing the centre of the atom. These prongs, which fit snugly into the plastic tubing, point along the symmetry axes of the valence orbitals, and the angles formed determine bond angles in the assembled models. Models built from the FMM kit show to scale the mutual relations of atoms of a given structure. The scale is 1 inch = 1 angstrom (1 Å) = 10 1 nanometre = metre = 100 picometre. Note that one inch is approximately 25 millimetre on your metric ruler. The molecular framework is specified by * the internuclear distances, which measure the separation between the centres of two neighbouring atoms. * the bond angles which measure whether two atoms both bonded to a third are collinear or at an angle. W CVALENT RADIUS F CARBN H + X TETRAHEDRN FUR C H BNDS C X H W H X X H Figure E11-1 W W

3 E11-3 For example, take methane (see Figure E11-1). The points H and C are the centres of hydrogen and carbon atoms, and the distance between them is called the internuclear (or interatomic) distance. This distance can be conveniently thought of as composed of two parts: * the C to X or C X distance (0.77 Å), measuring the size of carbon in the direction of the bond, and known as the covalent radius of carbon. * The X H distance (0.30 Å), the covalent radius of hydrogen. Values of covalent radii are listed in Table E11-2 (page E11-4). The bond length is the distance between the two nuclei and is the sum of the covalent radii, in this case 1.07 Å. Use Table E11-2 to calculate the following bond lengths. BND LENGTH (Å) BND LENGTH (Å) C F C C1 C Br C I C N H S S In CH 4 the H C H bond angle is (109 28'), and is called the tetrahedral angle. In less symmetric molecules, such as CH 3 Cl or NH 3, the bond angles vary slightly from the tetrahedral angle. When molecules aggregate to form a crystal, or when they collide at moderate speed, their closest contact occurs at a distance corresponding to the van der Waals envelope. This envelope denotes how far the atoms within a molecule extend outwards. The sizes of van der Waals radii are given in Table E11-3. Note that such radii are not measurable precisely, but estimates, as given here, are nevertheless very useful. With the axis of the C H bond as example (see Figure E11-1), the distance C X denotes the covalent radius of carbon (black tubing), the distance X H denotes the covalent radius of hydrogen (0.30 Å), and the section H to W extending in the nonbonded direction corresponds to the van der Waals radius of hydrogen (1.20 Å). In Figure E11-1, one must imagine that the volume occupied by each hydrogen atom is a portion of a sphere with centre H and radius H W. Space-filling models, which show the outer surface of molecules, are available for comparison. With experience it is not difficult to visualise the volume of a molecule, given the framework of the molecule in an FMM model. Calculate the distance from the C nucleus to the farthest boundary of the CH 4 molecule.

4 E11-4 Calculate the distance from the C nucleus to the farthest boundary of each of the following molecules, which have the same shape as CH 4. MLECULE DISTANCE (Å) CF 4 CC1 4 CBr 4 For the preparation of bonds, coloured tubing is supplied, which can be used either by itself for a bond between like atoms, or joined by a linear fastener (see Figure E11-3 on page E11-8) to tubing of another colour to represent a bond joining different atoms. Tubing in various colours is provided (Table E11-1), and sections are cut on the chosen scale (1 inch = 1 angstrom), as specified by Tables E11-2 and E11-3. Tables E11-1, E11-2 and E11-3 give the data needed for construction of the structures dealt with in this exercise. Table E11-1 Colour Coding of Atoms H white red C black F light green N blue Cl dark green H 0.30 Table E11-2 Single Bond Covalent Radii (Å) Be 0.89 B 0.80 C 0.77 N F d transition metals 1.25 Si 1.17 P 1.10 S 1.04 Cl 1.00 Br 1.14 I 1.33 Xe 1.31 H 1.2 Table E11-3 Van der Waals Radii (Å) C 1.8 N F 1.4 P 1.9 S 1.8 Cl 1.8 Br 2.0 I 2.2 Xe 2.1

5 E11-5 (2) Bonding and Valence Shell Electrons When an atom forms a single covalent bond it in effect gains one electron into its valence shell. An isolated F atom has 7 valence electrons (F is in Periodic Group 17). An HF molecule can be represented by an electron dot structure as. o o o o oo H F o In the formation of the one covalent bond, both the H and the F atoms have gained one electron each and now have the stable electronic configuration of the next noble gas atom in the periodic table. An isolated atom has 6 valence electrons ( is in Periodic Group 16). The H 2 molecule can be represented as. o o o Ḣ oo H o In the formation of the two covalent bonds, the atom has gained two electrons and now has a stable octet of valence electrons. An isolated N atom has valence electrons (N is in Group 15). An NH 3 molecule can be represented as. In the formation of the covalent bonds, N has gained electrons and has a stable of valence electrons.

6 E11-6 An isolated C atom has valence electrons (C is in Group 14). A CH 4 molecule can be represented as. In the formation of the covalent bonds, C has gained electrons and has a stable of valence electrons. Demonstrator's Initials Where one of the electron pairs is a bonding pair, the coloured tubing representing it must be joined at its outer end to a length of the appropriate tubing representing the other atom. Where one of the electron pairs is a nonbonding pair, the coloured tubing (cut to the appropriate van der Waals radius length) represents the electron pair. A nonbonding electron pair is often called a lone pair. Complete the following table, showing the number of bonding pairs and nonbonding pairs in the valence shell of the atom indicated. Atom Number of bonding pairs Number of nonbonding pairs Total number of valence shell electron pairs F in HF in H 2 N in NH 3 B in BF 3 Cl in ClF 3 Demonstrator's Initials

7 E11-7 A prediction of the geometry of a molecule can be made based on the simple idea that the electron pairs in the valence shell (non-bonding pairs as well as bonding pairs) seek to maximise their distance from one another in order to minimise the Coulombic repulsions. This idea is referred to as Valence Shell Electron Pair Repulsion (or VSEPR, for short). For example, consider an atom which, in a molecule, has 4 electron pairs in its valence shell. The VSEPR model would predict that those 4 pairs would arrange themselves tetrahedrally about the atom, as this is the arrangement which maximises the distance of all the pairs from each other. However, the shape of the molecule or ion is that figure defined by the atoms themselves and disregarding any lone pairs which may be present, even though those lone pairs have participated in determining the shape. Thus if the tetrahedral arrangement of valence shell electron pairs in the example above were all bonding pairs, then the shape of that molecule or ion would also be tetrahedral, but if for example two of those electron pairs were lone pairs, then the shape of the molecule or ion would be angular, the result of disregarding two of the corners of the tetrahedrally disposed valence shell pairs. Complete the following table. Central atom Electron dot structure Total number of valence electron pairs about central atom Geometry of electron pairs about central atom Number of bonding pairs Shape of molecule or ion in H 3 + B in BF 3 Cl in ClF 5 Demonstrator's Initials

8 E11-8 (3) Building Bonds with the Model Kit C C Bond A bond between a pair of like atoms is cut from tubing of one colour. The decimal-inch scale box can be used. Calculate the length of a C C single bond. C H Bond As shown in figure E11-2, a C H bond could be constructed by joining in sequence: a 0.77 inch length of black tubing representing the covalent radius of carbon. a 0.30 inch length of white tubing representing the covalent radius of hydrogen, and which has a permanently attached linear fastener at both ends. a 1.20 inch length of white tubing representing the van der Waals radius of hydrogen. CARBN NUCLEUS LINEAR FASTENER CVALENT RADIUS F CARBN (0.77 Å) CVALENT RADIUS F HYDRGEN (0.30 Å) HYDRGEN NUCLEUS VAN DER WAALS RADIUS F HYDRGEN (1.20 Å) Figure E11-2 The hydrogen atom occupies a sphere centred at the nucleus. You should mentally fill in the volume of the atom on this framework. C F Bond A 0.77 inch length of black tubing (covalent radius of carbon 0.77 Å) is attached to a 0.64 inch length of light-green tubing (covalent radius of fluorine 0.64 Å) by means of a linear fastener to form the carbon fluorine bond, as shown in Figure E11-3. LINEAR FASTENER CARBN NUCLEUS FLURINE NUCLEUS CVALENT RADIUS F CARBN (0.77 Å) CVALENT RADIUS F FLURINE (0.64 Å) ASSEMBLED C F BND (1.41 Å) Figure E11-3

9 E11-9 Single bonds have axial symmetry (like the tubing which represents them). Bonds with axial symmetry are termed σ bonds (sigma bonds). In this exercise, all tubing connecting two clusters, and all tubing connecting one cluster to an H nucleus, represents σ bonds. (4) Measuring the Distance between Electron Pairs In this exercise you will examine the properties of three of the five basic geometries adopted by valence shell electron pairs. (The two missing geometries are the linear and trigonal shapes.) From the model kit select metal clusters with 4, 5 and 6 prongs, respectively. Each cluster is associated with a solid figure in the following way. Take, for example, the 4 prong cluster and put it on the bench. The 3 points of contact with the bench make up the corners of a triangular face. The 4 prong cluster has 4 such equivalent triangular faces. The solid figure made up by these 4 faces is called a tetrahedron. By a similar sort of reasoning, the 5 prong cluster is described as trigonal bipyramidal (meaning two triangular pyramids joined by their bases) and the 6 prong cluster is described as octahedral. Diagrams of these three solids are shown in Figure E11-4 below. face prong edge corner tetrahedron trigonal bipyramid octahedron Figure E11-4 By examining these three clusters, complete Table E11-4. Table E11-4 Clusters and Their Associated Solid Figures SHAPES GEMETRIC PRPERTIES Tetrahedron Trigonal bipyramid ctahedron Number of prongs in associated model cluster Number of faces of this figure Number of corners of this figure Number of edges of this figure

10 E11-10 By inspection only, identify which of the clusters have all prongs equivalent? Sketch the cluster in which the prongs are not equivalent and identify the axial and equatorial prongs. Verify by measurement that the prongs on the tetrahedral cluster are farther apart than the neighbouring prongs in a cluster of four prongs in a square arrangement. Such an arrangement can be found in the octahedral cluster. Record here these distances for comparison. Tetrahedral: Square: Indicate on the diagram below all the different bond angles and give the size of each different angle. Demonstrator's Initials

11 E11-11 (5) Building Models of Molecules with Nitrogen and Hydrogen In this exercise you will build the following molecules and ions: NH 3, NH 4 + and NH 2. These species all have the same number of electrons and, hence, are referred to as isoelectronic with one another. This exercise demonstrates the important role of the nonbonding electron pairs in determining molecular shape. Complete the following table. Do not forget to take into account the charge on the ions when determining the number of valence electrons. Electron dot Number of valence electron pairs Species diagram σ-bonding non-bonding Total NH 3 ammonia NH 4 + ammonium ion NH 2 amide ion Each molecule has valence electron pairs which are arranged about the nitrogen in a geometry. Using blue tubes for nitrogen and white tubes for hydrogen, build models of each of the three species, including the lone pairs. Indicate the shape of each species in the table below. Species NH 3 Geometrical arrangement of valence-shell electron pairs Shape of species NH 4 + NH 2 Demonstrator's Initials

12 E11-12 (6) Modelling Double and Triple Bonds Frequently there are molecules whose structure can only be explained by having more than one electron pair taking part in a bond. Consider, for example, the case of 2 which has a total number of valence electron pairs of 6. The only way these pairs can be distributed between the atoms such that each atom has a complete octet of electrons about it is with the following arrangement: The two shared electron pairs are said to make a double bond which is depicted as =. When three pairs are shared between two atoms we call this a triple bond and depict it as N N, to use N 2 as an example. For the sake of the VSEPR model, multiple bonds are assumed to act as a single electron pair when considering Coulombic repulsions. Draw electron dot diagrams, including all valence electron pairs, for the following molecules and indicate how many double and triple bonds there are. Molecule Electron dot diagram Total number of valence electron pairs (all atoms) Number and type of multiple bonds N 2 C 2 CH 2 In order to make models which include double and triple bonds, we need to know some more about how the electron pairs in multiple bonds are arranged in space. Let us start with the double bond. The two electron pairs in the double bond are distributed in quite different ways. The pair that occupies the space along the axis between the two atoms forms the σ-bond. The second pair occupies the space above and below the axis between the two atoms. This space is formed by the overlap of the two p-orbitals as shown in Figure E11-5. The resulting bond is called a π bond (pi bond). Figure E11-5.

13 E11-13 A triple bond consists of one σ-bond and two π-bonds, one π-bond above and below the σ-bond axis and the other π-bond in front and behind this axis, as shown in Figure E11-6. Two lobes of one π-bond Two lobes of one π-bond Axis of σ-bond Figure E11-6. To see how these two types of bond can be represented with the model kit, consider the case of ethylene H 2 C=CH 2. The lengths of tubing required are calculated from the appropriate π-bond covalent radii in Table E11-4 and the π-bond van der Waals radii in Table E11-5. As shown in Figure E11-7 the π bond is made from two bridges, each built out of three tubes and two right angle fasteners, one above and one below the plane of the molecule. (Pre-assembled π-bond bridges are included in your FMM kit. Do not attempt to pull them apart.) Note that the cluster used for each atom at the ends of the double bond, in this case carbon, is trigonal bipyramidal. Table E11-4 π-bond covalent radii Table E11-5 π-bond van der waals radii C N Period-2 atoms (C, N, ) 1.5 Double 0.67 double 0.62 Double 0.55 Period-2 atoms (Si, P, S) 1.9 Triple 0.60 triple 0.55 Triple 0.50 Figure E11-7. What kind of cluster from the model kit is needed to build models of the following molecules? Molecule Number and type of multiple bond Type of cluster required C 2 H 4 1 double C atom: N 2 N atom: C 2 C atom: atom: 2 Both atoms:

14 E11-14 Build models of N 2, C 2 and 2 including the nonbonding pairs and have them checked by your demonstrator. IMPRTANT: When constructing these models, make the σ framework first and then add the π-bonds. Demonstrator's Initials (7) Molecular Shape and Polarity The shape of a molecule determines many of its properties and chemical behaviour. Modern drug design, for example, makes extensive use of molecular models (typically built in a computer instead of by hand) in order to see how a drug will bind to large biological molecules like proteins or DNA. ne very important property, particularly for small molecules, is how the electrostatic charge is distributed about the molecule; specifically, does the molecule have a positive side and a negative side? A molecule which does have such a distribution is said to have a dipole moment and is called polar,... eg. H 2 δ H A molecule which has no such positive and negative sides has no dipole moment and is described as nonpolar, eg. N 2 N N The dipole moment of a molecule determines properties such as molecular solubility and whether the molecule will absorb microwave radiation. Examine the C 2 molecule built earlier. Would you expect this molecule to be polar or nonpolar? Explain your answer. H δ +

15 E11-15 (8) A Puzzle: Which isomer of ClF 3 is the most stable? In this exercise we will see how a combination of modelling and experiment can be used to predict a molecular structure. The molecule in question is ClF 3. The central Cl atom has valence electron pairs around it which, according to the VSEPR model adopt a geometry. There are 3 different ways of arranging the three fluorine atoms about the chlorine resulting in 3 different structural isomers. Build a model of any one of the isomers of ClF 3. Make sure to include the non-bonding pairs on the chlorine atom. What is the approximate angle between two equatorially-oriented electron pairs? What is the approximate angle between an equatorially-oriented electron pair and an axially-oriented electron pair? In which of these two types of interactions will the repulsion between the two electron pairs be greater? As non-bonding electron pairs have, effectively, a stronger Coulombic repulsion than do bonding pairs, it follows that the lowest energy structure will be that in which the interactions involving the non-bonding pairs and all the other electron pairs have been minimised. That is, the most stable isomer is likely to be the one in which the number of 90 interactions between a non-bonding pair and the other electron pairs is minimised. Consider SF 4 as an example. It is known to have the see-saw shape with the non-bonding pair occupying the equatorial position. F F F F F S S F F F 3 90 interactions 2 90 interactions The problem is to predict which of the three possible isomers of ClF 3 is the most stable. Sketch each of the three isomers of ClF 3 in the spaces provided in the following table and, for each isomer, indicate the number of axial and equatorial non-bonding pairs on the central chlorine atom. Record the total number of 90 interactions between the non-bonding pairs and any other electron pair. (Change your model to represent each of the other isomers, if necessary.)

16 E11-16 I II III Number of axial non-bonding pairs Number of equatorial non-bonding pairs Total number of 90 interactions between the Cl non-bonding pairs and any other electron pair Arrange the three isomers above in order of increasing energy. Use the labels I, II and III to indicate the 3 isomers. lowest energy highest energy It has been determined experimentally that ClF 3 is a polar molecule. Does this additional information eliminate any of the isomers? If so, which one(s)? By selecting the lowest energy isomer, sketch your prediction for the structure of ClF 3 in the space below. Indicate on your diagram the approximate values of the bond angles. Demonstrator's Initials

17 E11-17 (9) N 3 : Resonance and the Limitations of Simple Models In this exercise we will look at one of the limitations of simple geometrical models. Consider the nitrate ion, N 3. This molecular ion has a total of electron pairs. valence Draw an electron dot diagram of N 3 and indicate how these electron pairs are distributed. Based on this electron dot diagram, build a model of N 3 including any multiple bonds and nonbonding electron pairs. Before you start, consider the following: are all the N bonds equivalent? Give the cluster type that should be used for each of the atoms. N atom: cluster Double bonded atom: cluster Both single bonded atoms: clusters Measurements of the rotation of the nitrate ion (probed using microwaves) indicate that all three N bonds are equivalent in length. This is (or should be) in contradiction of your model. The shortcoming of the model (and the electron dot representation) is that the valence electrons do not have to be always located either between two atoms (in the case of a bonding pair) or on a single atom (in the case of a nonbonding pair). Electrons can be distributed between 3 or more atoms. In the case of N 3 we can represent this as shown in Figure E11-8. _ N Figure E11-8 As usual each solid line represents a bonding electron pair. The dotted lines represent three electron pairs distributed between all 4 atoms. Note that each N bond is now equivalent, in agreement with experiment. N 3 is an example of the limitation of the bonding model we have used. Specifically, it shows up the weakness of the simplifying assumption that bonding electron pairs must be located only between a specific pair of atoms.

18 E11-18 In order to get around this problem while retaining the simplicity of this model, the idea of resonance structures was introduced. Examine your model of the nitrate ion. There were three choices for which N bond would be the double bond and, hence, three possible structures (related to each other by a rotation through 120 ) as shown in Figure E11-9. In the resonance picture, the structure of certain species (such as N 3 ) cannot be depicted accurately by a single electron dot formula. Rather, the true structure of N 3 is an average (or resonance hybrid) of all the contributing or resonance forms shown in Figure E11-9. N - N - N - Figure E11-9 The formula of formic acid is usually written HCH, whereas the formula of the formate ion is usually given as HC 2. Draw the electron dot formulas of these two species and explain why the two atoms in formic acid are different, but the two atoms in the formate ion are equivalent. Demonstrator's Initials