Lab #11- Molecular Geometry

Transcription

1 Objectives Chesapeake Campus Chemistry 111 Laboratory Lab #11- Molecular Geometry Determine the shape of a molecule using the VSEPR. Draw the Lewis structures of a molecule including bond angles and formal charges. Introduction The three dimensional shape of a molecule gives great insight into its physical and chemical properties. Indeed, modern pharmaceutical companies spend a great deal of time using computer models and manipulating the shapes of prospective drugs in cyberspace to mimic the shape of molecules with proven effectiveness. They can learn a lot about how a chemical will react, even before it is made by studying similarly shaped molecules. This approach potentially saves great deal of effort and expense; if a prospective drug does not function to specifications in a modeling program, the company will not pursue its use. In this lab, you will build models of several compounds and ions. First, draw Lewis dot structures for the molecules assigned, then construct the molecules out of the kits provided. It is recommended that you follow the general steps outlined below for each new structure. This is a surefire way to systematically obtain the correct geometry. Steps to Drawing Lewis Structures: 1. Determine the total number of valence electrons that must appear in the structure. All main group atoms have the same number of valence electrons as their group numbers (1-8). A positive charge on an ion removes an electron while if there is a negative charge, it is necessary to add an electron to the structure. 2. Determine the central element. The central element is usually the least electronegative element or the one which can form the most bonds. 3. Draw single bonds between all atoms and the central atom. 4. Fill in the octets on atoms that follow the octet rule by adding electrons in pairs. Most atoms want a total of 8 valence electrons. However, elements in groups 1-3 and metals in group 4 do not require 8 valence electrons, so it is not necessary to add an octet around these elements. 5. Count the valence electrons in the structure. Each bond counts as two electrons. a. If there are too few electrons, add lone pairs to the central atom until the correct number of electrons are present in the structure. Note that elements that are in groups 5-8 and which are in row 3 or lower may occasionally contain more than 8 electrons in their structures when following this rule. That is acceptable. b. If there are too many electrons, eliminate a lone pair of electrons from the central atom and a lone pair from a neighboring atom and draw a multiple bond between those atoms. 6. Count the formal charges. The formal charges of all atoms in the correct structure should be minimal (zero). If there must be a formal charge on an atom, a negative formal charge should be attached to the more electronegative atom. Formal charges should add up to be equal to the overall charge of the molecule or ion. The formula for formal charge is Formal Charge = # valence electrons ( 1 2 number of bonding electrons + number of lone pair electrons) Equation 1 Modified from Lumen Learning 2014 CC-BY Page 1

2 7. Use the VSEPR to determine the electronic and molecular geometries of the structure. An example of drawing a Lewis dot structure for NF 3 is as follows: 1. Determine the total number of valence electrons that must appear in the structure. Source (atom, ion or Number Valence Electrons Total charge) N F Total electrons that should appear in the structure: Determine the central element. Nitrogen is less electronegative than fluorine so it must be the central element. In addition, N only has 5 valence electrons (and has room for 3 bonds) while each F has 7 valence electrons (and will only make 1 bond). Therefore N must be the central element. 3. Draw single bonds between all atoms and the central atom. F N F 4. Fill in the octets on atoms that follow the octet rule by adding electrons in pairs. None of the elements in this structure are from groups 1-3. Neither are they a metal from group 4. Therefore these atoms all want at least 8 valence electrons. Note that : N already has 3 bonds or 6 valence electrons around it, so it only needs 1 lone F N F: pair to have a total of 8. Each F only has one bond and needs 3 lone pairs to : bring the total to 8. F: 5. Count the valence electrons in the structure. This structure has the 26 electrons originally calculated. Therefore we can move on to looking at formal charge. 6. Count the formal charges. The charge of the molecule is 0, so the formal charges should add up to be 0. The formal charge of N must take into account it is in group 5 (and should have 5 valence electrons) and has 6 bonding electrons and 1 lone pair of electrons: FC N = 5 ( 1 x 6 + 2) = 0 2 The formal charge of N is 0 which means it has exactly what it should have. The formal charge for F can be done for all three atoms. However, in this structure the F are all equivalent (1 bond, 3 lone pairs) and thus the calculation can be done one time. FC F = 7 ( 1 x 2 + 6) = 0 2 All the F also have a formal charge of 0. When the total formal charges for all atoms in the structure are tallied, it is clear the formal charges (0) add up to be the overall charge of the molecule (0). In addition, no atom contains a formal charge other than 0 making it unnecessary to evaluate other ways to draw the structure. 7. Use the VSEPR to determine the electronic and molecular geometries of the structure. The basis of the VSEPR is that like charges repel and try to get as far apart as possible. This structure has 4 regions of electron density around the N. Thus the electronic geometry of N (according to Table 1) is Tetrahedral. Since Modified from Lumen Learning 2014 CC-BY Page 2 F

3 only 3 of these areas are bonds, while the other is a lone pair, it is possible to determine the molecular geometry of the molecule is Trigonal Pyrimidal. Note that according to Table 1, the areas of electron density will separate by ~ to maximize the distance between the electrons in 3 dimensional space. A further example of this is in comparing CH 4 to H 2O. Drawing the Lewis structure of each compound gives the following information: Note that both of these structure also contain four regions of electron density around the central atom (and therefore have the electronic geometry of Tetrahedral) while the number of bonds differs. This difference means that the molecular geometry of methane (which has 4 areas of electron density with all four of those areas bonds) is Tetrahedral. Water has only 2 bonds (the other two areas of electron density around the central oxygen are lone pairs) has the molecular geometry Bent. Table 1 contains a list of specific geometries and bond angles. Finally, it is necessary to note any polarity in the molecule. A covalent bond is a sharing of electrons. Sometimes, the partners in a bond don't share electrons equally. This unequal sharing of electrons is called a dipole. The electronegativity scale can be used to see if a dipole exists and which partner wins the greater share of the electrons. The greater the difference between values from the electronegativity scale, the greater the charge separation in the dipole. The most electronegative atom on the periodic table is fluorine. In general, electronegativity decreases as you move left and down away from fluorine. For our purposes, a bond between two different atoms will be considered polar. Dipoles are drawn as arrows along the bond pointing to the more electronegative atom. In water, the two bonds between O and H atoms are polar. We draw the arrows toward the more electronegative O. This gives an overall polar molecule (consider how vectors are added in mathematics-this is the same concept) with the more electronegative O pulling a partial negative charge (depicted δ - ) while the less electronegative H atoms each contain a δ +. However, it is also possible to have polar bonds and overall a nonpolar molecule. Consider the bonds in carbon dioxide. The bonds are polar but since they are equal and head in opposite direction, they cancel. O C O : : H O H However, in water, the polar bonds don't cancel out due to the molecular geometry and therefore water is polar. It may be helpful to remember that an asymmetrical arrangement of atoms will often result in a polar molecule. Modified from Lumen Learning 2014 CC-BY Page 3

4 Table 1: Electronic and Molecular Geometries Example Molecule Areas of e- Density Electronic Geometry # Bonding Areas # Non-Bonding Areas (Lone Pairs) Molecular Geometry Bond Angles Polar Bonds? Polar Molecule? CO 2 2 Linear 2 0 Linear 180 Yes No BCl Trigonal Planar 120 Yes No 3 Trigonal Planar SO Bent <120 Yes Yes CF Tetrahedral Yes No NH 3 4 Tetrahedral 3 1 Trigonal Pyrimidal <109.5 Yes Yes H 2O 2 2 Bent <109.5 Yes Yes PCl Trigonal Bipyrimidal 90, 120, 180 Yes No SCl See-Saw 90, <120, 180 Yes Yes Trigonal 5 Bipyrimidal IF T-Shaped 90 Yes Yes XeCl Linear 180 Yes No SF Octahedral 90 Yes No BrCl 5 6 Octahedral 5 1 Square Pyrimidal 90, 180 Yes Yes XeF Square Planar 90, 180 Yes No Modified from Lumen Learning 2014 CC-BY Page 4

5 Materials Molecular Model Kit Safety and Notes Return all materials to trays and drawers in a clean orderly condition for the next class. Experimental Procedure and Data 1. For each compound, a. Follow the directions for and write the correct (best) Lewis structure. b. Obtain your instructors approval, then build a molecular model from the kits provided. c. Answer the questions that describe the molecule. 2. Atoms are color coded within each kit. It may be beneficial to evaluate whether you would like to use an atom by the type listed or by areas of e density. In some instances (such as with carbon) the areas of e density do not vary by molecule so using the atom from the kit designated as carbon should work when making your molecular model. In other instances, such as with PCl 5, it may be more beneficial to look for an atom which can hold 5 areas of electron density for your central atom (there is most likely no atom set aside for phosphorus in the kits). 3. You might first want to do your work on a scratch piece of paper. However, if you choose to do this, transfer your answers to the data sheets provided. Modified from Lumen Learning 2014 CC-BY Page 5

6 Name Date Pre-lab Assignment/Questions N o t e this pre-lab must be finished before you come to lab. 1.Draw the dot structures for C,H,O, Cl, N, S, and P. 2. If covalent bonding occurs because an atom wants to achieve an octet and therefore fill empty spaces in its orbital, how many covalent bonds would you think are formed by each of the atoms in #1? 3. In some molecules the electron geometry and the molecular shape are the same, but in other molecules they are different. How do the Lewis structures of these two types of molecules differ? Look at Table 1 in the introduction, and consider for example CH4 and NH3. 4. List four elements that always obey the octet rule. List two elements which do not typically obey the octet rule. 5. From your knowledge of its electron structure, why doesn t hydrogen obey the octet rule? Modified from Lumen Learning 2014 CC-BY Page 6

7 Name Lab Partner Experimental Data and Results Molecule # Val. e- Lewis Structure Areas e Density Electronic Geometry # Bonding Areas # Non- Bonding Areas Date Bin # Molecular geometry Bond Angle(s) Polar Molecule? Yes or No NH2 - CH4 NO2 - CO CO2 H2CO Modified from Lumen Learning 2014 CC-BY Page 7

8 Name Lab Partner Date Bin # Molecule # Val. e- Lewis Structure Areas e Density Electronic Geometry # Bonds # Lone Pairs Molecular geometry Bond Angle(s) Polar Molecule? Yes or No SO2 XeF4 PCl3 NH4 + SCN - SF4 NO3 - SO4 2- SF3 + XeF3 + Modified from Lumen Learning 2014 CC-BY Page 8

9 Name Lab Partner Post-lab Questions Date Bin # 1. Without making a model, describe the electron geometry and molecular shape of carbon tetrabromide (CBr4). Would you expect the bonds in this molecule to be polar? Would you expect this molecule to be polar overall? Explain. 2. NH3 and H2CO each have three bonds about the central atom. However, their molecular geometries are not the same. Explain this difference. Modified from Lumen Learning 2014 CC-BY Page 9