OBJECTIVES: EXPERIMENT #13 s and Draw Lewis structures of atoms, ions, and molecules Build models of linear, trigonal planar tetrahedral, trigonal bipyramidal, and octahedral arrangements of electron pairs Relate proper Lewis structure with the correct three dimensional structure of the molecule or ion Even before physicists and chemists understood the quantum nature of the electron and the chemical bond, G. N. Lewis observed that the valence electrons of stable molecules and ions are arranged in four pairs for a total of eight an octet of electrons. The Lewis structure of a molecule or ion emphasizes this observation. Since electrons are negatively charged and negative charges repel each other, we can assume these repulsions can be minimized by keeping the electrons in bonds as far from each other as possible. Sidgwick and Powell, and more recently Gillespie and Nyholm, have refined this assumption and have shown that well-defined geometric shapes are the result. These shapes maximize distances (minimize electron repulsions) in three-dimensional space. It is these shapes that a molecule or ion may take. Thus, a knowledge of the Lewis structure of a molecule allows us to make predictions about its shape. Once the shape of a molecule is known, properties dependent upon its shape, such as bond hybridization and dipole moment, can be predicted. The central atom in a molecule or ion may have two, three, four, five, or six electron domains (or groups) surrounding it. These electron domains may form bonds with other atoms or they may be non-bonding domains. A domain may consist of a single pair of electrons (single bond or non-bonding pair), two pairs (double bond) or three pairs (triple bond). For example, in the Lewis structure for ammonia, NH3, the central nitrogen atom is surrounded by four electron domains. Three of these domains are single bonds to hydrogen atoms, the remaining electron domain is the lone pair or nonbonding pair. The orientation of the bonds around the central atom, in other words the three dimensional shape, is determined by the number of electron domains surrounding the central atom. The chart below describes the electron domain geometry around the central atom. When using this chart, it is important to remember that multiple bonds (double and triple bonds) are considered as a single electron domain. H bonding pairs H N: H non-bonding or lone pairs The shape of the molecule or ion is related to the electron domain geometry. The name given to the shape of a molecule or ion is the shape defined by the number of bonding domains. This is summarized in the chart below. When using this chart, it is important to remember that multiple bonds (double and triple bonds) are considered as a single electron domain.) 133 P a g e
EXPERIMENT #13 LEWIS STRUCTURES TABLE I: ELECTRON DOMAIN GEOMETRY Number of Electron Around the Central Electron Domain Geometry Predicted Bond Angles Hybridization 2 linear 180 sp 3 trigonal planar 120 sp 2 4 tetrahedral 109.5 sp 3 5 trigonal bipyramidal 120 sp 3 d 90 6 octahedral 90 sp 3 d 2 TABLE II: MOLECULAR GEOMETRY Number of Electron Surrounding the Central Surrounding the Central Example 2 2 Linear CO2 3 3 trigonal planar BF3 2 angular, planar bent, V-shaped (120 bond angle) NO2-4 4 Tetrahedral CH4 3 Pyramidal NH3 2 angular, planar bent, V-shaped (109.5 bond angle) H2O 5 5 trigonal bipyramidal PCl5 4 Seesaw SF4 3 T-shaped ClF3 2 Linear XeF2 6 6 Octahedral SF6 5 square pyramidal BrF5 4 square planar XeF4 134 P a g e
EXPERIMENT #13 LEWIS STRUCTURES Linear Trigonal V-Shape, 120 Tetrahedral Pyramidal V-Shape, 109.5 Trigonal Bipyramid See-Saw T-Shape Linear (Distorted Tetrahedron) Octahedron Square Pyramid Square Planar FIGURE I: Molecular Shapes Derived from Solid Geometry 135 P a g e
EXPERIMENT #13 LEWIS STRUCTURES Rules for Drawing s 1. Count up the valence electrons for all atoms in formula. 2. Add an electron for each negative charge on an ion; subtract an electron for each positive charge on an ion. 3. Attach atoms in a way which makes chemical sense. Draw a line between each attached pair; this consumes two electrons from the total calculated in step 2. 4. Add pairs of electrons to fulfill the octet rule; duet rule for hydrogen. 5. If necessary make nonbonding pair(s) bonding pair(s), but do not alter the total number of electrons calculated in step 2. 6. If necessary, expand octet for atoms in third row or lower in the periodic table. PROCEDURE: (Review the rules for drawing Lewis structures before coming to lab.) 1. From the group of molecules/ions you are assigned, draw the Lewis structures using the rules you learned in class. This will allow you to determine the geometry around the central atom. 2. From the model kits find the spheres having 4, 5, and 6 holes drilled in them. These spheres will represent a central atom in a molecule or ion. The number of holes drilled into the ball will be usedto represent the number of electron domains around the central atom. 3. Using the sphere containing five holes, find the holes which are diametrically opposed. (These holes are 180 apart.) Place a wooden stick into each of these two holes. A molecule or ion containing a central atom surrounded by only two electron domains will have a linear arrangement of these electron domains. If these are bonding domains, the molecule will be linear; that is, it will have the shape of a straight line. 4. Using the sphere containing five holes, find the three holes in the same plane. (These holes are 120 apart.) Place a wooden stick into each of these three holes. A molecule or ion containing a central atom surrounded by three electron domains will have a trigonal planar arrangement of these electron domains. If these are bonding domains, the molecule will be trigonally planar; that is, it will have the shape of a triangle. (The triangle will be an equilateral triangle, if the three bonds are identical.) 5. Remove one of the wooden sticks. This is a model of a molecule or ion whose central atom is surrounded by three electron domains, but contains only two bonding domains. This molecule is V-shaped. It is a planar molecule with a bond angle of 120. 6. Using the sphere containing four holes, build a model of a tetrahedron by placing a wooden stick into each hole. The angle between each of the sticks is 109. It is important to be able to draw a three dimensional tetrahedron on a two dimensional piece of paper. Practice drawing a tetrahedron. Watch closely as your instructor demonstrates how to do this. [The solid line (FIGURE I) represents a bond or electron group in the plane of the paper. For a tetrahedron there are two such lines. 136 P a g e
EXPERIMENT #13 LEWIS STRUCTURES The wedge indicates a bond or electron domain coming out of the plane of the page toward the reader, while the dashed line indicates a bond or electron domain going behind the plane of the paper, away from the reader.] 7. Remove one stick from the tetrahedron. The resulting three dimensional shape is a pyramid. The angle between each electron domain of the pyramid is approximately 109. If you can draw a tetrahedron, you can also draw a pyramid. 8. Remove a second stick from the tetrahedron. The resulting three dimensional shape is a planar, bent shape. It is also called V-shaped, but the angle between the two bonds is approximately 109 (see above). 9. Go back to the sphere containing five holes. Place a wooden stick into each of the five holes. A molecule or ion containing central atom surrounded by five electron domains will have a trigonal bipyramidal arrangement of these electron domains. If these are all bonding domains, the molecule will be a trigonal bipyramid. Notice that there are two types of electron domains, depending on their orientation in space. The three electron domains 120 from each other are in the same plane. They form a triangle and are said to occupy the equatorial positions of a trigonal bipyramid. The two electron domains 180 from each other are said to occupy the axial positions of a trigonal bipyramid. [Watch closely as your instructor demonstrates how to draw a trigonal bipyramid. The solid (FIGURE I) line represents a bond or electron group in the plane of the paper. Note that there are three such lines. The wedge indicates a bond or electron domain coming out of the plane of the page, while the dashed line indicates a bond or electron domain going behind the plane of the paper.] 10. The shape of a molecule or ion having a central atom surrounded by five electron domains, but containing only four bonds, is made from the trigonal bipyramid by removing one of the wooden sticks. But which stick do we remove, the axial or the equatorial? It turns out that removal of one of the equatorial sticks (or electron domains) gives the shape having the minimum amount of electron domain repulsions. (You may want to convince yourself of this by discussions with your classmates or your instructor.) The shape of a molecules whose central atom is surrounded by five electron groups, but contains only four bonds is a distorted tetrahedron or see-saw. Build such a model by removing an equatorial bond from your model of a trigonal bipyramid. The bond angles are 180, 120, and 90. 11. If we remove one more equatorial bond, we see the shape of a molecule whose central atom is surrounded by five electron domains, but contains only three bonds. Such a molecule is T-shaped. The bond angles are 180 and 90. 12. Using the sphere containing six holes, build a model of an octahedron by placing a wooden s tick into each hole. The angle between each of the sticks is 90. It is important to be able to draw a three dimensional octahedron on a two dimensional piece of paper. Practice drawing an octahedron. [Watch closely as your instructor demonstrates how to do this. The solid line (FIGURE I) represents a bond or electron domain in the plane of the paper. For an octahedron 137 P a g e
EXPERIMENT #13 LEWIS STRUCTURES there are two such lines. The wedge indicates a bond or electron domain coming out of the plane of the page; there are two such lines. The dashed line indicates a bond or electron domain going behind the plane of the paper; there are two such lines.] If all the atoms bonded to the central atom in the octahedron are the same, then each of the bonds in the octahedron is identical. Build such a model to show that this statement is true. When drawn as shown in FIGURE I, an octahedron appears to have two types of bonds. The four bonds forming the square are equatorial bonds, and the bond above the square and the bond below the square are axial bonds. 13. The shape of a molecule or ion having a central atom surrounded by six electron domains, but containing only five bonds, is made from the octahedron by removing one of the wooden sticks. But which stick do we remove, the axial or the equatorial? It turns out that removal of one of the axial sticks (or bonds) gives the shape having the minimum amount of electronic repulsions. (You may want to convince yourself of this by discussions with your classmates or your instructor.) The shape of a molecule whose central atom is surrounded by six electron domains, but contains only five bonds is a square pyramid. The bond angles are 90. Build such a model by removing an axial bond from your model of an octahedron. 14. If we remove one more axial bond, we see the shape of a molecule whose central atom is surrounded by six electron domains, but contains only four bonds. Such a molecules is a square plane. The bond angles are 90. 138 P a g e
NAME Section Date s and In the boxes provided below draw the Lewis structure of each ion or molecule from the groups you are assigned. Also, include in the space indicated, the formula, the number of valence electrons, electron domain geometry, number of bonds, molecular geometry, and the hybridization of the central atom. Group I: CO 2, BCl 3, SO 3, CHCl 3, SO 4 2, H 2S, NI 3, SO 2, PCl 5, SF 4, ClF 3, I 3, C 2H 2, SF 6, BrF 5, XeF 4 Group II: CS 2, BF 3, NO 2, CH 2Cl 2, ClO 4, H 2O, AsH 3, O 3, AsF 5, XeO 2F 2, BrF 3, XeF 2, C 2H 4, IOF 5, TeF 5, ICl 4 Group V H 2CCCH 2, CH 2CH 2, C 2H 2Cl 2, H 2CO, CH 3C(=O)OH, C 6H 6 (all carbons sp 2 ) 139 P a g e
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NAME Section Date 141 P a g e
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NAME Section Date 143 P a g e
H 2 CCCH 2 144 P a g e
NAME Section Date CH 2 CH 2 C 2 H 2 Cl 2 H 2 CO 145 P a g e
CH 3 C(=O)OH C 6 H 6 (all carbons sp 2 ) 146 P a g e