07Lab | Activity 03: Lewis Structure and VSEPR Theory
Objective: The purpose of this experiment is to 1) draw the structural formula for a molecule based on the chemical formula, 2) write the electron dot formula (Lewis Structure) corresponding to the structural formula and 3) construct a computer model of the molecules (VSEPR model).
Discussion: The tether between two atoms in a molecule is called a chemical bond. In a covalent bond, two nonmetallic elements bond to each other by sharing valence electrons. The valence electrons are the electrons furthest from the nucleus, and they occupy the highest s or p sublevels (for purpose of this class). The group number of an element indicates the number of valence electrons. For example, fluorine is in Group VIIA/17 and has seven valence electrons (7 e-).
Answer the questions below as your read through this activity, and then use the worksheet at the end of this activity to turn in to your instructor.
1. The valence electrons for an atom are obtained from the periodic table. Carbon is found in Group IVA and has 4 valence electrons. Refer to the periodic table and find the family (i.e., Group I) and the number of valence electrons for the following elements: Element N O P Ca Sb Ga
Family
Valence electrons
2 In this problem set, the Lewis structure is first constructed based on the chemical formula for the molecule. From the chemical formula, the atom connectivity for the structure is established.
Given a chemical formula, ABn, A is generally the central atom and the B atoms flank the A atom. i.e., NH3, NCl3, NO2. In these examples, N is central in the atom in the structure and H, Cl and O all flank the nitrogen atom in each molecule. Sometimes there are no central atoms, such as in the case of Cl2, and sometimes there is more than one central atom such in H-O-O-H. In this structure the atom connectivity is H-O-O-H. The two oxygen atoms are considered central. If the chemical formula contains hydrogen (H) or fluorine (F), these atoms are never central.
Determine which is the central atom and which are the flanking atoms. The flanking atoms are connected to the central atom. Chemical specie SO - CH PH HCN CH Cl 2 BrO3 4 3 3 Central Atom S
Flanking atoms Both O
Atomic Connectivity O S O
3 The Lewis structure is completed when all the valence electrons are shown in the structure and all atoms satisfy the octet rule. Note that there are exceptions to this rule that must be taken into account. A tutorial provides the methodology in which this is done. http://faculty.sdmiramar.edu/fgarces/ChemComon/Tutorial/Lewis/LewisTutorial/LewisTutorial.htm http://faculty.sdmiramar.edu/fgarces/ChemComon/Tutorial/Tutorial.htm The following example illustrates the structural formula and electron dot formula for molecular models having single, double and triple bonds.
Example 1 The model of a water molecule is shown below. The chemical formula for water is H2O. In box 1, the oxygen is central and the hydrogen atoms flank the oxygen. Box 2 shows a line between the oxygen and hydrogen atom, which represents the bond between the two atoms. A bond forms by two shared electrons, as shown in box 3. Box 4 takes into account all the valence electrons of water (8) and shows that all atoms satisfy the octet rule (hydrogen being satisfied with two). Finally, box 5 shows the molecular geometry of water. The molecular geometry is determined by VSEPR analysis.
1 2 3 4 5 Water, H2O
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To verify the above Lewis structure, the valence electrons are added from each atom in the molecule. Recall that hydrogen is in Group IA/1 and oxygen is in Group VIA/6.
2 H (2 * 1 e- ) = 2 e- 1 O (1 * 6 e- ) = 6 e- sum of valence electrons 8 e-
There are eight dots used to write the electron dot formula. Since this equals the number of valence electrons, the electron dot formula is correct.
Example 2 The Lewis structure analysis for chloroform is shown below. Chloroform possesses a total of 26 electrons. 1 H (1 * 1 e- ) = 1 e- 1 C (1 * 4 e- ) = 4 e- 3 Cl (3 * 7 e-) = 21 e- sum of valence electrons 26 e-
1 2 3 4 5
Chloroform, CHCl3
The procedure requires the establishment of the atomic connectivity (2), the addition of electrons to the central atom such that the octet rule is satisfied (3), and completion of the Lewis structure by adding the remaining 18 electrons such that the octet rule is satisfied for all atoms in the structure, excluding H which is satisfied with two electrons.
Example 3 The Lewis structure analysis for formaldehyde is shown below. Formaldehyde possesses a total of 12 electrons. 2 H (2 * 1 e- ) = 2 e- 1 C (1 * 4 e- ) = 4 e- 1 O (1 * 6 e- ) = 6 e- Sum of valence electrons 12 e-
1 2 3 4 5
Formaldehyde, H2CO
For formaldehyde, two bonds are placed between the carbon and the oxygen in order for the central atom, carbon, to posses an octet (2). Carbon shares a total of eight electrons (3) while the remaining four electrons are added on oxygen so that oxygen too satisfies the octet rule (4). The final Lewis structure shows all 12 valence electrons distributed so that all atoms (except H) satisfy the octet rule.
Example: 4 The Lewis structure analysis for hydrogen cyanide is shown below. Hydrogen cyanide possesses a total of 10 electrons. 1 H (1 * 1 e- ) = 1 e- 1 C (1 * 4 e- ) = 4 e- 1 N (1 * 5 e- ) = 5 e- Sum of valence electrons 10 e-
1 2 3 4 5 Hydrogen cyanide, HCN
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For hydrogen cyanide, three bonds are placed between the carbon and the nitrogen in order for the central atom, carbon, to posses an octet (2). Carbon shares a total of eight electrons (3), while an additional two electrons are added on to nitrogen so that nitrogen too satisfies the octet rule (4). The final Lewis structure shows all 10 valence electrons distributed such that all atoms (except H) satisfy the octet rule.
Complete the Lewis structure for the following. Use a separate sheet of paper if you need more room. 1 Compound, 2 3 4 Chemical formula Total Valence Connectivity Lewis Structure electron a) Ammonia, H
NH3 8 H N H
b) Fluoromethane, H
CH3F H C H F
d) Formic acid, O
HCOOH H C O H
e) Hydroxylamine, H
NH2OH H N O H
f) Nitrate ion, O
- O N O NO3
g) Acetate ion, H O
- H C C O CH3COO H
h) Sulfur dioxide
i) Dihydrogen dioxide H
(Hydrogen peroxide) O O
H2O2 H
4 The shape of a molecule has a tremendous impact on the physical properties and chemical reactivity of that molecule. The molecular shape of a molecule can be determined from its chemical formula by applying the Valence Shell Electron Pair Repulsion Theory (VSEPR) analysis. The core assumption of VSEPR is that the valence electrons around the central atom of the molecule are responsible for its shape, via their mutual repulsion; i.e., the shape of the molecule is determined by the arrangement of the central valence electrons which possess the lowest electronic repulsion and, hence, the lowest energy. VSEPR works well for main group covalent compounds and ions, less well for transition metal compounds, and not well at all for ionic compounds. In this class we will work with mostly main group covalent compounds.
A terminology, which must be clear, is the distinction made between electronic and molecular geometry. Electronic geometry refers to the distribution of the electron pairs around the central atom. Thus, if a molecule has 3 electron domains (regions of electron density) around its central atom, its electronic geometry is trigonal planar, regardless of how many of the pairs are bonding or are lone pair electrons. Molecular geometry refers to the arrangement of the atoms in the molecule. It is the molecular geometry that influences the physical property of the molecule.
The use of VSEPR clearly requires knowledge of how many electron domains (regions of electron density) are around the central atom, and also whether these pairs are bonding or are lone pairs. The simplest method for determining this is to draw the Lewis structure of the molecule and then count the number of electron domains surrounding the central atom.
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Table 7.1 The application of the VSEPR theory can be summarized by the following table. In this table, A represents the central atom, E represents electron domains, and B represents the bonded atom.
A tutorial provides the methodology in which the molecular geometry is determined from VSEPR. http://faculty.sdmiramar.edu/fgarces/ChemComon/Tutorial/VSEPR/VSEPRTutorial/VSEPRTutorial.htm A summary on how to use VSEPR to determine molecular geometry is provided here.
A) Determine the Lewis Structure: i) Show valence electrons for each atom in the structure. ii) Determine the number of bonds in the molecule and identify the central atom.
B) Determine the electronic geometry (AEn system) from the Lewis structure: i) Count the electron domain (region) around the central atom. ii) Arrange electron domains around the central atom to minimize electron-electron repulsion. This occurs when electron pairs are as far apart as possible. iii) 2-domain yields a linear shape, 3-domains yield a trigonal shape, and 4-domains yield a tetrahedral shape
C) Determine the molecule geometry (ABmEn system) from its electronic geometry: i) The molecular geometry is based on the final position of the atoms. ii) The lone pair electrons are ignored when determining the molecular geometry.
1) AE2; Electronic geometry (linear) with this designation will yield a molecular geometry designation of (AB2) which is linear. This is accomplished by replacing the two-electron domains by 2 bonded atoms. Examples are BeH2 and CO2
2-a) AE3; Electronic geometry (trigonal) with this designation can yield a molecular geometry designation of (AB3) which is trigonal. This is accomplished by replacing the three-electron domains by 3 bonded atoms. Examples are BH3 and H2CO
2-b) AE3; Electronic geometry (trigonal) with this designation can yield a molecular geometry designation of (AB2E) which is bent. This is - accomplished by replacing two of the three electron domains by 2 bonded atoms. Examples are SO2 & NO2
3-a) AE4 Electronic geometry (tetrahedral) with this designation can yield a molecular geometry designation of (AB4) which is + tetrahedral. This is accomplished by replacing all four-electron domains by 4 bonded atoms. Examples are CH4 and NH4
3-b) AE4 Electronic geometry (tetrahedral) with this designation will yield a molecular geometry designation of (AB3E) which is pyramidal. This is accomplished by replacing three of the four electron domains by 3 bonded atoms. Example NH3
3-c) AE4 Electronic geometry (tetrahedral) with this designation will yield a molecular geometry designation of (AB2E2) which is bent. This is accomplished by replacing two of the four electron domains by 2 bonded atoms. Examples are H2O and SCl2
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The application of the VSEPR theory can be summarize by the following table. In this table, A represents the central atom, E represents electron domains, and B represents the bonded atom.
A tutorial provides the methodology in which the molecular geometry is determined from VSEPR. www.miramar.sdccd.net/faculty/fgarces/ChemComon/Tutorial/VSEPR/VSEPRTutorial/VSEPRTutorial.htm
A summary on how to use VSEPR to determine molecular geometry is provided here.
A) Determine the Lewis Structure. i) Valence electrons for each atom in the structure. ii) Determine the number of bonds in the molecule and identify the central atom
B) Determine electronic geometry (AEn system) from Lewis structure. i) Count the electron domain (region) around the central atom. ii) Arrange electron domain in order to minimize electron-electron repulsion. This occurs when electron pair are far apart as possible. iii) 2-domain yields a linear, 3-domains yield a trigonal, and 4-domains yield a tetrahedral
C) Determine the molecule geometry (ABmEn system) from its electronic geometry. i) The molecular geometry is based on the final position of the atoms. ii) The lone pair electrons are ignored when determining the molecular geometry. 1) AE2; Electronic geometry (linear) with this designation will yield a molecular geometry designation of (AB2) which is linear. This is accomplished by replacing all two-electron domains by 2 bonded atoms. Example BeH2 2-a) AE3; Electronic geometry (trigonal) with this designation will yield a molecular geometry designation of (AB3) which is trigonal. This is accomplished by replacing all three-electron domains by 3 bonded atoms. Example BH3 2-b) AE3; Electronic geometry (trigonal) with this designation will yield a molecular geometry designation of (AB2E) which is bent. This is accomplished by replacing two of the three electron domains by 2 bonded atoms. Example SO2 3-a) AE4 Electronic geometry (tetrahedral) with this designation will yield a molecular geometry designation of (AB4) which is tetrahedral. This is accomplished by replacing all four-electron domains by 4 bonded atoms. Example CH4 3-b) AE4 Electronic geometry (tetrahedral) with this designation will yield a molecular geometry designation of (AB3E) which is pyramidal. This is accomplished by replacing three of the four electron domains by 3 bonded atoms. Example NH3 3-c) AE4 Electronic geometry (tetrahedral) with this designation will yield a molecular geometry designation of (AB2E2) which is bent. This is accomplished by replacing two of the four electron domains by 2 bonded atoms. Example H2O
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07Lab | Activity 03: Molecular Models and VSEPR Theory Last Name ______First______/ _____ pts
Valence Electrons & Group Number Element N O P Ca Cl Ga
Group Number
# of Valence electrons
3 Complete the Lewis structure for the following. Place a square bracket around ions. # of # of Nonbonded Compound, Valence Electron pairs on Lewis Structure Chemical formula electrons all central atoms
Ammonia, NH3
Fluoromethane
CH3F
Carbon disulfide
O Formic acid H C O H HCOOH
H Hydroxylamine NH2OH H N O H
Nitrate ion, NO3-
Sulfur dioxide
Acetate Ion -1 CH3COO
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4. Shapes of Molecules VSEPR Theory Compound # of Lewis Structure # of e- # of NonBonded a.Electronic Geometry, Valence Groups pairs of e- b.Molecular Geometry, e- c.Bond angles, d. Hybridization AE4, Tetrahedral AB3E Pyramidal Tetrahedral 8 e- H-N-H angle Pyramidal NH3 < 109.5 ° < 109.5˚ sp3
CS2
HOCl
SiS2
- NO2
OF2
PI3
NO3-
CS2
HOCl
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5 Computer exercise (Your instructor will tell you if you are required to complete this as part of your assignment). Your instructor will need to provide you with more information for this activity. To complete this activity, you will need to go to the links below and find the three-dimensional model of the compounds listed.
Using Model 360 (first link), fill the table for each chemical.
Other links to 3D Molecular modeling software accessible through the web (Accessed July 2020): 1 http://www.chemeddl.org/resources/models360/models.php 2 http://molview.org/
3D Molecular Modeling 3D Structure: Chemed DL: Model 360 or Molview Bond length Bond angle Partial charge Molecular dipole Bond dipole a)
Carbon disulfide b)
HOCl c)
Silicon tetrahydride d)
Silicon Disulfide e)
Nitrite ion f)
Oxygen difluoride g)
Phosphorus triiodide h)
i)
Nitrate ion
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Resources for Lewis Structure and VSEPR (Accessed July 2020)
Chem Digital Library: http://www.chemeddl.org/
Model 360 from Chemed DL: http://www.chemeddl.org/resources/models360/index.php
PubChem Compounds: http://pubchem.ncbi.nlm.nih.gov/
Lewis structure made easy, Flash program: http://www.chemistry24.com/college_chemistry/lewis-structure.html
Lewis Structure Tutorial: http://www.ausetute.com.au/lewisstr.html
MolView Molecular Modeling Software: http://molview.org/
Google Links: http://www.google.com/search?hl=en&lr=&q=Lewis+Structures&btnG=Search
VSEPR help page: http://www.chem.purdue.edu/gchelp/vsepr/
VSEPR models: http://chemlabs.uoregon.edu/GeneralResources/models/vsepr.html
Google Links: http://www.google.com/search?hl=en&lr=&q=VSEPR+models&btnG=Search
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